Unconditional uniqueness and non-uniqueness for Hardy–Hénon parabolic equations

Abstract

We study the problems of uniqueness for Hardy–Hénon parabolic equations, which are semilinear heat equations with the singular potential (Hardy type) or the increasing potential (Hénon type) in the nonlinear term. To deal with the Hardy–Hénon type nonlinearities, we employ weighted Lorentz spaces as solution spaces. We prove unconditional uniqueness and non-uniqueness, and we establish uniqueness criterion for Hardy–Hénon parabolic equations in the weighted Lorentz spaces. The results extend the previous works on the Fujita equation and Hardy equations in Lebesgue spaces.

1 Introduction and main results

1.1 Introduction and our setting

We consider the Cauchy problem of the Hardy-Hénon parabolic equation

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u - \Delta u = |x|^{\gamma } |u|^{\alpha -1} u, &{}(t,x)\in (0,T)\times {\mathbb {R}}^d, \\ u(0) = u_0 \in L^{q,r}_{s}({\mathbb {R}}^d), \end{array}\right. } \end{aligned}$$

(1)

where $T>0,$ $d\in {\mathbb {N}}$, $\gamma \in {\mathbb {R}}$, $\alpha >1$, $q\in [1,\infty ]$, $r \in (0,\infty ]$ and $s\in {\mathbb {R}}$. Here, $\partial _t:=\frac{\partial }{\partial t}$ is the time derivative, $\Delta :=\sum _{j=1}^d\frac{\partial ^2}{\partial x_j^2}$ is the Laplace operator on ${\mathbb {R}}^d$, $u=u(t,x)$ is an unknown complex-valued function on $(0,T)\times {\mathbb {R}}^d$, $u_0=u_0(x)$ is a prescribed complex-valued function on ${\mathbb {R}}^d$, and $L^{q,r}_s({\mathbb {R}}^d)$ is the weighted Lorentz space (see Definition 2.2), which includes the Lebesgue space $L^q({\mathbb {R}}^d)= L^{q,q}_0({\mathbb {R}}^d)$ as a special case $r=q$ and $s=0$. The equation (1) in the case $\gamma = 0$ is the Fujita equation, which has been extensively studied in various directions. The equation (1) with $\gamma <0$ is known as a Hardy parabolic equation, while that with $\gamma >0$ is known as a Hénon parabolic equation. The corresponding stationary problem to (1), that is,

$$\begin{aligned} -\Delta U=|x|^{\gamma }|U|^{\alpha -1}U, \end{aligned}$$

was proposed by Hénon as a model to study the rotating stellar systems (see [25]), and has also been extensively studied in the mathematical context, especially in the fields of nonlinear analysis and variational methods (see [18] for example).

In this paper we study the problem on unconditional uniqueness and non-uniqueness for (1) in weighted Lorentz spaces $L^{q,r}_s({\mathbb {R}}^d)$. Here, unconditional uniqueness means uniqueness of the solution to (1) for any initial data $u_0 \in L^{q,r}_s({\mathbb {R}}^d)$ in the sense of the integral form

$$\begin{aligned} u(t) = e^{t\Delta }u_0 + \int _{0}^t e^{(t-\tau )\Delta }(|\cdot |^{\gamma } |u(\tau )|^{\alpha -1} u(\tau ))\, d\tau \end{aligned}$$

(2)

in $L^\infty (0,T; L^{q,r}_s({\mathbb {R}}^d))$ or $C([0,T]; L^{q,r}_s({\mathbb {R}}^d))$, where $T>0$ and $\{e^{t\Delta }\}_{t>0}$ is the heat semigroup. We say that non-uniqueness holds for (1) if unconditional uniqueness fails. In contrast, we say that conditional uniqueness holds if uniqueness of the solution to (1) holds in the entire space with some auxiliary function spaces. In addition, we also study uniqueness criterion which is a necessary and sufficient condition on the Duhamel term (i.e. the second term in the right-hand side of (2)) for uniqueness to hold.

Let us here state previous works on uniqueness for (1). For (1) with $\gamma \le 0$, the problem on uniqueness has been well studied (see [3,4,5, 10,11,12, 20, 24, 36, 39, 47, 48, 50, 54, 55] for example). In the study of unconditional uniqueness for (1) in Lebesgue spaces $L^q({\mathbb {R}}^d)$ or Lorentz spaces $L^{q,r}({\mathbb {R}}^d)$, the following two critical exponents are known to be important. The first one is the so-called scale-critical exponent $q_c$ given by

$$\begin{aligned} q_c = q_c (d,\gamma ,\alpha ):= \frac{d(\alpha -1)}{2+\gamma }, \end{aligned}$$

(3)

and we say that the problem (1) is scale-critical if $q=q_c$, scale-subcritical if $q>q_c$, and scale-supercritical if $q<q_c$. The second one is the critical exponent $Q_c$ given by

$$\begin{aligned} Q_c =Q_c(d,\gamma ,\alpha ):= \frac{d\alpha }{d+\gamma }, \end{aligned}$$

(4)

which is related to well-definedness of the Duhamel term in (2) in $L^{q,r}({\mathbb {R}}^d)$. In fact, the nonlinear term $|x|^\gamma |u|^{\alpha -1}u \in L^1_{\textrm{loc}}({\mathbb {R}}^d)$ for any $u \in L^{q,r}({\mathbb {R}}^d)$ if and only if “$q>Q_c$” or “$q=Q_c$ and $r \le \alpha$”. In the case $\gamma =0$, unconditional uniqueness for (1) in $C([0,T]; L^q({\mathbb {R}}^d))$ was proved in the double subcritical case $q > \max \{q_c, Q_c\}$ by Weissler [54] and in the single critical cases $q = Q_c > q_c$ and $q = q_c>Q_c$ by Brezis and Cazenave [10]. In the double critical case $q = q_c = Q_c$, non-uniqueness was proved for some initial data $u_0 \in L^q({\mathbb {R}}^d)$ by Terraneo [50], and then, for any initial data $u_0 \in L^q({\mathbb {R}}^d)$ by Matos and Terraneo [36]. In [50], uniqueness criterion was also obtained in the double critical case. Recently, Takahashi [47] proved the existence of an uncountably infinite number of solutions to (1) with moving singularities for some initial data in the double critical case. In the scale-supercritical case $q<q_c$, non-uniqueness for (1) was proved for initial data $u_0=0$ by Haraux and Weissler [24]. Uniqueness and non-uniqueness have also been studied for heat equations with exponential nonlinearities (see [27, 29] and references therein). In the Hardy case $-\min \{2,d\}< \gamma < 0$, similar results were obtained by [5, 48], where the Lorentz spaces $L^{q,r}({\mathbb {R}}^d)$ is used to study unconditional uniqueness in the critical case $q=Q_c$ in [48]. The above previous works are summarized in Fig. 1. In contrast, the Hénon case $\gamma >0$ has not been well studied. This is due to the difficulty of treating the increasing potential $|x|^\gamma$ in the nonlinear term at infinity. To overcome this difficulty, the weighted spaces are effective, and recently, conditional uniqueness was obtained in $L^{q}_s({\mathbb {R}}^d)=L^{q,q}_s({\mathbb {R}}^d)$ in [13]; however, unconditional uniqueness and non-uniqueness are completely open. The main purpose of this paper is to prove unconditional uniqueness, non-uniqueness and uniqueness criterion for (1) with all $\gamma > -\min \{2,d\}$, including the Hénon case, in $L^{q,r}_s({\mathbb {R}}^d)$.

Fig. 1

The figure shows the domain of $(\alpha ,q)$ for $d\ge 3$ and $\gamma \le 0$, where $\alpha _0:= 1 + \frac{\gamma }{d}$, $\alpha _F:= 1 + \frac{2+\gamma }{d}$ is the Fujita exponent, $\alpha ^*:= \frac{d+\gamma }{d-2}$ is the Serrin exponent and $\alpha _{HS}:= \frac{d+2+2\gamma }{d-2}$ is the Hardy-Sobolev exponent. Table 1 and Table 2 summarize the previous results on uniqueness for (1) with $\gamma \le 0$

1.2 Statement of the results

To describe our results, let us give some definitions and notation. For $T \in (0,\infty ]$ and a quasi-normed space X, we denote by $L^\infty (0,T;X)$ the space of functions $u: (0,T) \rightarrow X$ such that

$$\begin{aligned} \Vert u\Vert _{L^\infty (0,T;X)}:= \underset{t \in (0,T)}{\mathrm {ess\, sup}}\, \Vert u(t)\Vert _{X} < \infty , \end{aligned}$$

and by C([0, T]; X) the space of continuous functions $u: [0,T] \rightarrow X$ with respect to the quasi-norm of X. The space ${\mathcal {L}}^{q,r}_s ({\mathbb {R}}^d)$ is defined as the completion of $L^{q,r}_s ({\mathbb {R}}^d) \cap L^\infty _0({\mathbb {R}}^d)$ with respect to $\Vert \cdot \Vert _{L^{q,r}_s}$, where $L^\infty _0({\mathbb {R}}^d)$ denotes the set of all functions in $L^\infty ({\mathbb {R}}^d)$ with compact support in ${\mathbb {R}}^d$ (see Definition 2.2).

Definition 1.1

Let $T>0$ and $X = L^{q,r}_s ({\mathbb {R}}^d)$ or ${\mathcal {L}}^{q,r}_s ({\mathbb {R}}^d)$. We say that a function $u=u(t,x)$ on $(0,T) \times {\mathbb {R}}^d$ is a mild solution to (1) with initial data $u_0 \in X$ in C([0, T]; X) ($L^\infty (0,T; X)$ resp.) if u belongs to C([0, T]; X) ($L^\infty (0,T; X)$ resp.) and satisfies the integral equation (2) for almost everywhere $(t,x) \in (0,T)\times {\mathbb {R}}^d$.

Note that the Duhamel term in (2) converges in $L^{q,r}_s({\mathbb {R}}^d)$ under conditions on functions $u=u(t,x)$ and parameters q, r, s given in Lemma 4.1 or Lemma 4.2.

We define two critical cases in the framework of $L^{q,r}_s({\mathbb {R}}^d)$ in a similar manner to $q_c$ and $Q_c$, respectively. The equation (1) is invariant under the following scale transformation:

$$\begin{aligned} u_\lambda (t,x):= \lambda ^\frac{2+\gamma }{\alpha -1}u(\lambda ^2 t, \lambda x),\quad \lambda >0. \end{aligned}$$

More precisely, if u is a solution to (1), then so is $u_\lambda$ with the rescaled initial data $\lambda ^\frac{2+\gamma }{\alpha -1}u_0(\lambda x)$. Moreover, we calculate

$$\begin{aligned} \Vert u_\lambda (0)\Vert _{L^{q,r}_s} = \lambda ^{-s + \frac{2+\gamma }{\alpha -1}- \frac{d}{q}} \Vert u_0\Vert _{L^{q,r}_s}= \lambda ^{-d (\frac{s}{d} + \frac{1}{q}- \frac{1}{q_c})} \Vert u_0\Vert _{L^{q,r}_s},\quad \lambda >0. \end{aligned}$$

Hence, if q and s satisfy

$$\begin{aligned} \frac{s}{d} + \frac{1}{q} = \frac{1}{q_c}, \end{aligned}$$

then $\Vert u_\lambda (0)\Vert _{L^{q,r}_{s}} = \Vert u_0\Vert _{L^{q,r}_{s}}$ for any $\lambda >0$, i.e., the norm $\Vert u_\lambda (0)\Vert _{L^{q,r}_{s}}$ is invariant with respect to $\lambda$. Therefore, we say that the problem (1) is scale-critical if $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c}$, scale-subcritical if $\frac{s}{d} + \frac{1}{q} < \frac{1}{q_c}$, and scale-supercritical if $\frac{s}{d} + \frac{1}{q} > \frac{1}{q_c}$. Another critical case is when the following holds:

$$\begin{aligned} \frac{s}{d} + \frac{1}{q} = \frac{1}{Q_c}. \end{aligned}$$

This is related to local integrability of the nonlinear term $|x|^\gamma |u|^{\alpha -1}u$. In fact, $|x|^\gamma |u|^{\alpha -1}u \in L^1_{\textrm{loc}}({\mathbb {R}}^d)$ for any $u \in L^{q,r}_s({\mathbb {R}}^d)$ if and only if

$$\begin{aligned} \frac{s}{d} + \frac{1}{q} < \frac{1}{Q_c} \quad \text {or} \quad \frac{s}{d} + \frac{1}{q} = \frac{1}{Q_c} \text { and } r\le \alpha . \end{aligned}$$

(5)

Then, it is ensured for the Duhamel term in (2) to be well-defined in $L^{q,r}_s({\mathbb {R}}^d)$.

In terms of the two critical cases, we divide the problem into the following four cases: Double subcritical case ($\frac{s}{d} + \frac{1}{q} < \min \{\frac{1}{q_c}, \frac{1}{Q_c}\}$), single critical case I ($\frac{s}{d} + \frac{1}{q} = \frac{1}{Q_c} < \frac{1}{q_c}$), single critical case II ($\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} <\frac{1}{Q_c}$), and double critical case ($\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} = \frac{1}{Q_c}$). Moreover, we define the exponent $\alpha ^*$ by

$$\begin{aligned} \alpha ^* =\alpha ^* (d,\gamma ):= \left\{ \begin{array}{ll} \frac{d+\gamma }{d-2}\quad &{}\text {if }d\ge 3,\\ \infty &{}\text {if }d=1,2,\\ \end{array} \right. \end{aligned}$$

which is often referred to the Serrin exponent (see [45, 46] and also [19]). The exponents $\alpha ^*$, $q_c$ and $Q_c$ are related as follows:

$$\begin{aligned} \alpha \lesseqgtr \alpha ^* \quad \text {if and only if }\quad q_c \lesseqgtr Q_c. \end{aligned}$$

In our results on unconditional uniqueness below, we assume that

$$\begin{aligned} \left\{ \begin{array}{ll} &{}d\in {\mathbb {N}}, \quad \gamma> -\min \{2,d\}, \quad \displaystyle \alpha > \max \left\{ 1,1 + \frac{\gamma }{d}\right\} ,\\ &{} \alpha \le q \le \infty ,\quad \frac{\gamma }{\alpha -1} \le s< d,\quad 0<r \le \infty . \end{array} \right. \end{aligned}$$

(6)

Our results on unconditional uniqueness are the following:

Theorem 1.2

(Scale-subcritical case) Let $T>0$, and let $d,\gamma ,\alpha ,q,r,s$ be as in (6). Assume either $(\textrm{i})$ or $(\textrm{ii})$:

• i. (Double subcritical case) $r \le \alpha$ if $q=\alpha$, and $0< \frac{s}{d} + \frac{1}{q} < \min \{\frac{1}{q_c}, \frac{1}{Q_c}\}$.

• ii. (Single critical case I) $\alpha < \alpha ^*$, $q\not =\infty$, $r\le \alpha$ and $\frac{s}{d} + \frac{1}{q} = \frac{1}{Q_c} < \frac{1}{q_c}$.

Then unconditional uniqueness holds for (1) in $L^\infty (0,T; L^{q,r}_s({\mathbb {R}}^d))$.

Theorem 1.3

(Scale-critical case) Let $T>0$, and let $d,\gamma ,\alpha ,q,r,s$ be as in (6). Assume $d\ge 3$, $q\not =\infty$, and either $(\textrm{i})$ or $(\textrm{ii})$:

• i. (Single critical case II) $\alpha > \alpha ^*$ and $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} < \frac{1}{Q_c}$ (replace $L^{q,\infty }_{s}({\mathbb {R}}^d)$ by ${\mathcal {L}}^{q,\infty }_{s}({\mathbb {R}}^d)$ if $r=\infty$).

• ii. (Double critical case) $\alpha = \alpha ^*$, $r\le \alpha ^* -1$ and $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} = \frac{1}{Q_c}$.

Then unconditional uniqueness holds for (1) in $C([0,T]; L^{q,r}_s({\mathbb {R}}^d))$.

Remark 1

In Theorem 1.2 (i), the condition “$r \le \alpha$ if $q=\alpha$” comes from the restriction on parameters in linear estimates. More precisely, the condition is due to the restriction $r_1=1$ for linear estimates with $q_1=1$ in Proposition 3.1 (see (12) and also Lemma 4.1 (ii)).

Next, we consider the following two cases where the unconditional uniqueness is not obtained in the above theorems: $r>\alpha$ in the single critical case I; $r>\alpha ^*-1$ in the double critical case.

In the single critical case I, the condition $r \le \alpha$ naturally appears from the viewpoint of well-definedness of mild solutions to (1) as seen in (5). On the other hand, when $r>\alpha$, we can define mild solutions to (1) with the auxiliary condition and we know that conditional uniqueness holds (see [13, Theorem 1.13]). We are interested in the questions whether the conditional uniqueness can be improved. In fact, we can give the following sufficient condition for uniqueness to hold which improves the conditional uniqueness [13, Theorem 1.13].

Proposition 1.4

Let $T>0$, and let $d,\gamma ,\alpha ,q,r,s$ be as in (6). Assume that $\alpha < \alpha ^*$, $q\not =\infty$, $\alpha < r \le \infty$, and $\frac{s}{d} + \frac{1}{q} = \frac{1}{Q_c} < \frac{1}{q_c}$. Let $u_0\in L^{q,r}_{s}({\mathbb {R}}^d)$. Then, if $u_1, u_2 \in L^\infty (0,T; L^{q, r}_{s}({\mathbb {R}}^d))$ are mild solutions to (1) with $u_1(0)=u_2(0)=u_0$ such that

$$\begin{aligned} u_i(t) - e^{t\Delta } u_0 \in L^\infty (0,T; L^{q, r'(\alpha -1)}_{s}({\mathbb {R}}^d)) \quad \text {for i=1,2}, \end{aligned}$$

then $u_1 = u_2$ on [0, T]. Here, $r'$ is the Hölder conjugate of r, i.e., $1=\frac{1}{r}+\frac{1}{r'}$.

In the double critical case, we prove the result on non-uniqueness for (1) if $\alpha ^*-1< r \le \infty$. More precisely, we have the following:

Theorem 1.5

(Double critical case) Let $d\ge 3$, $\gamma >-2$, $\alpha = \alpha ^*$, $\alpha ^* \le q < \infty$, $\alpha ^*-1< r \le \infty$, and $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} = \frac{1}{Q_c}$. Then, for any initial data $u_0 \in L^{q,r}_{s}({\mathbb {R}}^d)$, there exists $T=T(u_0)>0$ such that the problem (1) has at least two different solutions in $C([0,T]; L^{q,r}_{s}({\mathbb {R}}^d))$ (replace $L^{q,r}_{s}({\mathbb {R}}^d)$ by ${\mathcal {L}}^{q,\infty }_{s}({\mathbb {R}}^d)$ if $r=\infty$).

By Theorem 1.3 (ii) and Theorem 1.5, we clarify that the exponent $r=\alpha ^*-1$ is a threshold of dividing unconditional uniqueness and non-uniqueness for (1) in the double critical case. The importance of $r=\alpha ^*-1$ was pointed out by [50, Theorem 0.10 and Proposition 5.4] in the Fujita case $\gamma =0$ (see [48, Theorem 1.4 and Proposition 8.2] for the Hardy case $\gamma <0$). The idea of proof of Theorem 1.5 is based on the method by [36, 50], i.e., we construct two different solutions which are regular and singular at $x=0$ to (1) for any initial data $u_0$. The regular solution can be found in a similar way to [13] and the singular solution can be constructed from the singular stationary solution to

$$\begin{aligned} \Delta U + |x|^{\gamma } U^{\frac{d+\gamma }{d-2}} = 0\quad \text {in } B\setminus \{0\}, \quad U>0, \end{aligned}$$

where $B:= \{x \in {\mathbb {R}}^d ;\, |x| < 1\}$. The threshold $r=\alpha ^*-1$ comes essentially from the logarithmic rate of the singularity at $x=0$ of the singular stationary solution (see (67) and (68) in Sect. 5.2). The existence and behavior near the origin of singular stationary solutions have been studied in [1, 2, 8, 9, 15, 19, 21, 23, 45, 46] for instance. See Sect. 5.2 for the details.

In addition, we give the following uniqueness criterion.

Theorem 1.6

Let $T>0$, and let $d,\gamma ,\alpha ,q,r,s$ be as in (6). Assume that $d\ge 3$, $\gamma >-2$, $\alpha = \alpha ^*$, $\alpha ^* \le q < \infty$, $\alpha ^*-1< r \le \infty$, and $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} = \frac{1}{Q_c}$. Let $u_0\in L^{q, r}_{s}({\mathbb {R}}^d)$. Then, if $u_1,u_2 \in C([0,T]; L^{q, r}_{s}({\mathbb {R}}^d))$ are mild solutions to (1) with $u_1(0)=u_2(0)=u_0$ such that

$$\begin{aligned} u_i(t) - e^{t\Delta } u_0 \in C([0,T]; L^{q, \alpha ^*-1}_{s}({\mathbb {R}}^d)) \quad \text {for i=1,2}, \end{aligned}$$

(7)

then $u_1 = u_2$ on [0, T] (replace $L^{q, r}_{s}({\mathbb {R}}^d)$ by ${\mathcal {L}}^{q, \infty }_{s}({\mathbb {R}}^d)$ if $r=\infty$).

Remark 2

The exponent $r = \alpha ^*-1$ in (7) of Theorem 1.6 is optimal for the same reason as above (see Theorem 5.3).

In the scale-supercritical case, we have the following result on non-uniqueness for (1). Here, we define the exponents $\alpha _F$ and $\alpha _{HS}$ by

$$\begin{aligned} \alpha _F = \alpha _F(d,\gamma ):= 1+ \frac{2+\gamma }{d} \quad \text {and}\quad \alpha _{HS} = \alpha _{HS}(d,\gamma ):= \frac{d+2+2\gamma }{d-2}, \end{aligned}$$

which are often referred to as the Fujita exponent (see [42, 43]) and the critical Hardy–Sobolev exponent (see [33]).

Proposition 1.7

(Scale-supercritical case) Let $d\ge 3$, $\gamma >-2$, $\alpha > 1$, $1<q\le \infty$, $1\le r \le \infty$ and $s\in {\mathbb {R}}$ be such that

$$\begin{aligned} \gamma \le {\left\{ \begin{array}{ll} \sqrt{3}-1 &{}\text {if }d=3,\\ 0 &{}\text {if }d\ge 4, \end{array}\right. } \quad \alpha _F< \alpha<\alpha _{HS} \quad \text {and}\quad \frac{1}{q_c}< \frac{s}{d} + \frac{1}{q} < 1. \end{aligned}$$

Then Eq. (1) has a global positive solution in $C([0,\infty ); L^{q,r}_s({\mathbb {R}}^d))$ with initial data 0.

To visually understand our above results, we give Fig. 2 for the case $\gamma <0$ and $\min \{{1\over q_c},{1\over Q_c}\}<\max \{{1\over q_c},{1\over Q_c}\}$.

Fig. 2

The figure shows the domain of $({s\over d},{1\over q})$ in the case $\gamma <0$ and $\min \{{1\over q_c},{1\over Q_c}\}<\max \{{1\over q_c},{1\over Q_c}\}$. (U.U.) and (N.U.) mean unconditional uniqueness and non-uniqueness, respectively. The cases $\gamma =0$ and $\gamma >0$ are deduced by moving the line ${s\over d}={\gamma \over d(\alpha -1)}$ to the right

Fig. 3

The figure shows the domain of $(\alpha ,s)$ for $d\ge 3$ and $q >1$. Here, $\alpha _0:= \min \{1,1 + \frac{\gamma }{d}\}$, $\alpha _F, \alpha ^*, \alpha _{HS}$ are given in Fig. 1, $s_c, S_c$ are given in (8), and $s^*:= d-2-\frac{d}{q}$. Table 3 and Table 4 summarize our results on uniqueness for (1)

Herein, we compare our results with previous ones. Our results generalize the previous works [5, 10, 24, 36, 48, 50, 54], since s can be taken as $s=0$ if $\gamma \le 0$ in our results. More precisely, our results on unconditional uniqueness (Theorem 1.2 and Theorem 1.3 (i)) include the results in [54, Theorem 4] and [10, Theorem 4] ($\gamma =0$ and $s=0$) and [5, Theorem 1.1] and [48, Theorem 1.1] ($\gamma <0$ and $s=0$), and our result on non-uniqueness (Theorem 1.5) generalizes the previous works [36, Theorem 1] ($\gamma =0$ and $s=0$) and [48, Theorem 1.3] ($\gamma <0$ and $s=0$). Moreover, our results on the double critical case (Theorem 1.3 (ii) and Theorem 1.5) also clarify the threshold $r=\alpha ^*-1$ of dividing unconditional uniqueness and non-uniqueness. Regarding the uniqueness criterion, Theorem 1.6 generalizes the previous works [50, Theorem 0.10] ($\gamma =0$ and $s=0$) and [48, Theorem 1.4] ($\gamma <0$ and $s=0$), and Proposition 1.4 has not been mentioned in the previous works. In the scale-supercritical case, Proposition 1.7 corresponds to [24, Theorem 1] ($\gamma =0$ and $s=0$) and [48, Proposition B.1] ($\gamma <0$ and $s=0$).

To easily compare our results with the previous work [13] which includes the Hénon case $\gamma >0$, we can rewrite our results by using the two critical exponents on s:

$$\begin{aligned} s_c=s_c(d,\gamma ,\alpha , q):= \frac{2+\gamma }{\alpha -1} - \frac{d}{q} \quad \text {and}\quad S_c=S_c(d,\gamma ,\alpha ,q):= \frac{d+\gamma }{\alpha } - \frac{d}{q}. \end{aligned}$$

(8)

The exponents $s_c$ and $S_c$ correspond to $q_c$ and $Q_c$ in the case without weights, respectively. In fact, we can see that

$$\begin{aligned} s_c=0\text { if and only if }q=q_c \quad \text {and}\quad S_c=0\text { if and only if }q=Q_c. \end{aligned}$$

Hence, we can also say that the problem (1) is scale-critical if $s=s_c$, scale-subcritical if $s<s_c,$ and scale-supercritical if $s>s_c$. Moreover, the four cases can be rewritten as follows: Double subcritical case ($s< \min \{s_c, S_c \}$), single critical case I ($s = S_c < s_c$), single critical case II ($s = s_c < S_c$), and double critical case ($s = s_c = S_c$). The results in [13] show local well-posedness, including the conditional uniqueness, for (1) if $s\le s_c$ and non-existence of positive mild solution to (1) for some initial data $u_0 \ge 0$ if $s>s_c$. However, unconditional uniqueness and non-uniqueness are not mentioned in [13]. Our results are summarized in Fig. 3.

This paper is organized as follows. In Sect. 2, we summarize the definitions and fundamental lemmas on Lorentz spaces and weighted Lorentz spaces. In Sect. 3, we establish the two kinds of weighted linear estimates. In Sect. 3.1, we extend the usual $L^{q_1}$-$L^{q_2}$ estimates to the weighted Lorentz spaces, which are fundamental tools in this paper. In Sect. 3.2, we prove a certain space-time estimate in the weighted Lorentz spaces. We call it the weighted Meyer inequality. This inequality corresponds to a certain endpoint case of the weighted Strichartz estimates, and it is an important tool in studying the scale-critical case. In Sect. 4, we prove our results on unconditional uniqueness and uniqueness criterion (Theorem 1.2, Theorem 1.3, Proposition 1.4 and Theorem 1.6), based on the weighted linear estimates. In Sect. 5, we prove our result on non-uniqueness (Theorem 1.5). In Sect. 6, we discuss the non-uniqueness in the scale-supercritical case and prove Proposition 1.7. In Sect. 7, we give a remark on the number of solutions in the double critical case, and additional results on the critical singular case $\gamma = -\min \{2,d\}$ and the exterior problem on domains not containing the origin.

1.3 Notation

Throughout this paper, we use the notation C for a positive constant which may change from line to line for convenience. We use the symbols $a \lesssim b$ and $b \gtrsim a$ for $a, b \ge 0$ which mean that there exists a constant $C>0$ such that $a \le C b$. The symbol $a \sim b$ means that $a \lesssim b$ and $b \lesssim a$ happen simultaneously. We denote by ${\overline{\Omega }}$ the closure of a domain $\Omega$ in ${\mathbb {R}}^d$. For $a \in {\mathbb {R}}$ and a sequence $\{a_n\}_{n\in {\mathbb {N}}} \subset {\mathbb {R}}$, the symbol $a_n \nearrow a$ as $n\rightarrow \infty$ means that $a_n \le a_{n+1}$ for any $n \in {\mathbb {N}}$ and $a_n \rightarrow a$ as $n\rightarrow \infty$. For functions f and g, the symbol $f *g$ denotes the convolution of f and g:

$$\begin{aligned} (f *g)(x):= \int _{{\mathbb {R}}^d} f(x-y) g(y)\, dy,\quad x \in {\mathbb {R}}^d. \end{aligned}$$

For quasi-normed spaces X and Y, the notation $\Vert \cdot \Vert _{X \rightarrow Y}$ denotes the operator norm from X to Y, i.e.,

$$\begin{aligned} \Vert T\Vert _{X\rightarrow Y}:= \sup _{\Vert f\Vert _{X} = 1} \Vert Tf\Vert _{Y} \end{aligned}$$

for an operator T from X into Y, and the notation $X \hookrightarrow Y$ denotes that X is continuously embedded in Y, i.e., X is a subset of Y and there exists a constant $C>0$ such that

$$\begin{aligned} \Vert f\Vert _{Y} \le C\Vert f\Vert _{X} \quad \text {for any }f\in X. \end{aligned}$$

For a domain $\Omega$ in ${\mathbb {R}}^d$, we denote by $C_0^\infty (\Omega )$ the set of all $C^\infty$-functions having compact support in $\Omega$, by $L^0(\Omega )$ the set of all Lebesgue measurable functions on $\Omega$, by $L^\infty _0(\Omega )$ the set of all functions in $L^\infty (\Omega )$ with compact support in $\Omega$, and by ${\mathcal {S}}'({\mathbb {R}}^d)$ the space of tempered distributions on ${\mathbb {R}}^d$.

2 Weighted Lorentz spaces

We define the distribution function $d_f$ of a function f by

$$\begin{aligned} d_f (\lambda ):= \left| \left\{ x \in \Omega ;\, |f(x)| > \lambda \right\} \right| , \end{aligned}$$

where |A| denotes the Lebesgue measure of a set A.

Definition 2.1

For $0<q,r \le \infty$, the Lorentz space $L^{q,r}(\Omega )$ is defined by

$$\begin{aligned} L^{q,r}(\Omega ):=\left\{ f \in L^0(\Omega ) ;\, \Vert f\Vert _{L^{q,r}(\Omega )} < \infty \right\} \end{aligned}$$

endowed with a quasi-norm

$$\begin{aligned} \Vert f\Vert _{L^{q,r}(\Omega )}:= {\left\{ \begin{array}{ll} \displaystyle \left( \int _0^\infty ( t^{\frac{1}{q}} f^*(t) )^r\frac{dt}{t} \right) ^{\frac{1}{r}} \quad &{} \text {if } r<\infty ,\\ \displaystyle \sup _{t>0} t^\frac{1}{q} f^*(t) &{} \text {if }r=\infty , \end{array}\right. } \end{aligned}$$

where $f^*$ is the decreasing rearrangement of f given by

$$\begin{aligned} f^*(t):= \inf \{ \lambda > 0 ; \, d_f(\lambda ) \le t \}. \end{aligned}$$

We refer to [22] for the properties of the distribution function, the decreasing rearrangement and the Lorentz space.

Remark 3

For $0<q,r<\infty$, the quasi-norm of $L^{q,r}(\Omega )$ is equivalent to

$$\begin{aligned} \Vert f\Vert _{L^{q,r}(\Omega )} = q^{\frac{1}{r}} \left( \int _0^{\infty } (d_f (\lambda )^{\frac{1}{q}} \lambda )^r \frac{d\lambda }{\lambda } \right) ^{\frac{1}{r}}. \end{aligned}$$

For $0<q<\infty$ and $r=\infty$,

$$\begin{aligned} \begin{aligned} \Vert f\Vert _{L^{q,\infty }(\Omega )}&= \sup \left\{ \lambda d_f(\lambda )^\frac{1}{q} ;\, \lambda>0\right\} \\&=\inf \left\{ C>0 ;\, \lambda d_f(\lambda )^{\frac{1}{q}} \le C \quad \text {for all }\lambda >0 \right\} . \end{aligned} \end{aligned}$$

Definition 2.2

Let $0<q,r \le \infty$ and $s \in {\mathbb {R}}$.

• (i) The weighted Lebesgue space $L^{q}_s(\Omega )$ is defined by

  $$\begin{aligned} L^{q}_s(\Omega ):= \left\{ f \in L^0(\Omega ) \,;\, \Vert f\Vert _{L^{q}_s} <\infty \right\} \end{aligned}$$

  endowed with a quasi-norm

  $$\begin{aligned} \Vert f\Vert _{L^{q}_s(\Omega )}:= {\left\{ \begin{array}{ll} \displaystyle \left( \int _{\Omega } (|x|^s|f(x)|)^q\, dx \right) ^\frac{1}{q}\quad &{}\text {if }q<\infty ,\\ \underset{x \in \Omega }{\mathrm {ess\,sup}}\, |x|^s|f(x)| &{} \text {if }q=\infty . \end{array}\right. } \end{aligned}$$

  The space ${\mathcal {L}}^{q}_s (\Omega )$ is defined as the completion of $L^{q}_s (\Omega ) \cap L^\infty _0(\Omega )$ with respect to $\Vert \cdot \Vert _{L^{q}_s(\Omega )}$.

• (ii) The weighted Lorentz space $L^{q,r}_s(\Omega )$ is defined by

  $$\begin{aligned} L^{q,r}_s(\Omega ):= \left\{ f \in L^0(\Omega ) \,;\, \Vert f\Vert _{L^{q,r}_s(\Omega )} <\infty \right\} \end{aligned}$$

  endowed with a quasi-norm

  $$\begin{aligned} \Vert f\Vert _{L^{q,r}_s(\Omega )}:= \Vert |\cdot |^s f\Vert _{L^{q,r}(\Omega )}. \end{aligned}$$

  The space ${\mathcal {L}}^{q,r}_s (\Omega )$ is defined as the completion of $L^{q,r}_s (\Omega ) \cap L^\infty _0(\Omega )$ with respect to $\Vert \cdot \Vert _{L^{q,r}_s(\Omega )}$.

Only when $\Omega = {\mathbb {R}}^d$, we omit $\Omega$ and we write $\Vert \cdot \Vert _{L^{q,r}_s} = \Vert \cdot \Vert _{L^{q,r}_s({\mathbb {R}}^d)}$ for simplicity.

Remark 4

There are several ways to define weighted Lorentz spaces. For example, the definitions in [14, 17, 31] are different from ours.

Remark 5

We give several properties and remarks on $L^{q,r}_s(\Omega )$. Let $0<q,r \le \infty$ and $s\in {\mathbb {R}}$.

• (a) $L^{q,q}_s(\Omega )=L^{q}_s(\Omega )$ and ${\mathcal {L}}^{q,q}_s(\Omega )={\mathcal {L}}^{q}_s(\Omega )$.

• (b) $L^{\infty ,r}_s(\Omega ) = \{0\}$ for any $r<\infty$. Hence, in this paper, we always take $r=\infty$ when $q=\infty$ in $L^{q,r}_s(\Omega )$ even if it is not explicitly stated.

• (c) $L^{q,r}_s(\Omega )$ is a quasi-Banach space (see Remark 15 below), and it is normable if $1<q<\infty$ and $1\le r \le \infty$.

• (d) $L^{q,r}_s(\Omega ) \cap L_0^\infty (\Omega )$ is dense in $L^{q,r}_s(\Omega )$ if $q<\infty$ and $r < \infty$, which implies that ${\mathcal {L}}^{q,r}_s (\Omega ) = L^{q,r}_s (\Omega )$ (see Lemma A.2 below). On the other hand, ${\mathcal {L}}^{q,r}_s (\Omega ) \subsetneq L^{q,r}_s (\Omega )$ if $q=\infty$ or $r=\infty$.

• (e) $L^{q,r}_s (\Omega )$ has the following embedding:

  $$\begin{aligned} L^{q,r_1}_s (\Omega ) \hookrightarrow L^{q,r_2}_s (\Omega ) \end{aligned}$$

  for $0<r_1\le r_2 \le \infty$ (see e.g. [22, Proposition 1.4.10]).

• (f) Let $0\in {\overline{\Omega }}$. Then $L^{q,r}_s(\Omega ) \subset L^1_{\textrm{loc}}(\Omega )$ if and only if either of (f-1)–(f-3) holds:

  • (f-1) $q>1$ and $\frac{s}{d} + \frac{1}{q} < 1$;

  • (f-2) $q>1$, $\frac{s}{d} + \frac{1}{q} = 1$ and $r\le 1$;

  • (f-3) $q=1$, $\frac{s}{d} + \frac{1}{q} \le 1$ and $r\le 1$.

• (g) Let $a, b \in {\mathbb {R}}$. Then

  $$\begin{aligned} |x|^{-a} \left| \log |x|\right| ^{-b} \in L^{q,r}_s(\{|x| \le e^{-1}\}) \end{aligned}$$

  if and only if either of (g-1)–(g-3) holds:

  • (g-1) $a < s + \frac{d}{q}$;

  • (g-2) $a = s + \frac{d}{q}$, $b > \frac{1}{r}$ and $r<\infty$;

  • (g-3) $a = s + \frac{d}{q}$, $b \ge 0$ and $r=\infty$.

• (h) Let $a, b \in {\mathbb {R}}$. Then

  $$\begin{aligned} |x|^{-a} \left| \log |x|\right| ^{-b} \in L^{q,r}_s(\{|x| \ge e\}) \end{aligned}$$

  if and only if either of (h-1)–(h-3) holds:

  • (h-1) $a > s + \frac{d}{q}$;

  • (h-2) $a = s + \frac{d}{q}$, $b > \frac{1}{r}$ and $r<\infty$;

  • (h-3) $a = s + \frac{d}{q}$, $b \ge 0$ and $r=\infty$.

For (g) and (h), see e.g. [22, Exercise 1.4.8] and also the calculations of proof of [48, Proposition 8.4].

We have the Hölder and Young inequalities in Lorentz spaces.

Lemma 2.3

(Generalized Hölder’s inequality) Let $0< q, q_1,q_2 < \infty$ and $0<r,r_1,r_2 \le \infty$. Then the following assertions hold:

• (i) If

  $$\begin{aligned} \frac{1}{q} = \frac{1}{q_1} + \frac{1}{q_2}\quad \text {and}\quad \frac{1}{r} \le \frac{1}{r_1} + \frac{1}{r_2}, \end{aligned}$$

  then there exists a constant $C>0$ such that

  $$\begin{aligned} \Vert f g\Vert _{L^{q,r}} \le C \Vert f\Vert _{L^{q_1,r_1}}\Vert g\Vert _{L^{q_2,r_2}} \end{aligned}$$

  for any $f \in L^{q_1,r_1}({\mathbb {R}}^d)$ and $g \in L^{q_2,r_2}({\mathbb {R}}^d)$.

• (ii) There exists a constant $C>0$ such that

  $$\begin{aligned} \Vert f g\Vert _{L^{q,r}} \le C \Vert f\Vert _{L^{q,r}}\Vert g\Vert _{L^{\infty }} \end{aligned}$$

  for any $f \in L^{q,r}({\mathbb {R}}^d)$ and $g \in L^{\infty }({\mathbb {R}}^d)$.

Lemma 2.4

(Generalized Young’s inequality) Let $1< q,q_1,q_2<\infty$ and $0 < r,r_1,r_2\le \infty$. Then the following assertions hold:

• (i) If

  $$\begin{aligned} \frac{1}{q} = \frac{1}{q_1} + \frac{1}{q_2} -1 \quad \text {and}\quad \frac{1}{r} \le \frac{1}{r_1} + \frac{1}{r_2}, \end{aligned}$$

  then there exists a constant $C>0$ such that

  $$\begin{aligned} \Vert f * g\Vert _{L^{q,r}} \le C \Vert f\Vert _{L^{q_1,r_1}}\Vert g\Vert _{L^{q_2,r_2}} \end{aligned}$$

  for any $f \in L^{q_1,r_1}({\mathbb {R}}^d)$ and $g \in L^{q_2,r_2}({\mathbb {R}}^d)$.

• (ii) If

  $$\begin{aligned} 1= \frac{1}{q_1} + \frac{1}{q_2} \quad \text {and}\quad 1 \le \frac{1}{r_1} + \frac{1}{r_2}, \end{aligned}$$

  then there exists a constant $C>0$ such that

  $$\begin{aligned} \Vert f * g\Vert _{L^{\infty }} \le C \Vert f\Vert _{L^{q_1,r_1}}\Vert g\Vert _{L^{q_2,r_2}} \end{aligned}$$

  for any $f \in L^{q_1,r_1}({\mathbb {R}}^d)$ and $g \in L^{q_2,r_2}({\mathbb {R}}^d)$.

• (iii) There exists a constant $C>0$ such that

  $$\begin{aligned} \Vert f * g\Vert _{L^{q,r}} \le C \Vert f\Vert _{L^{q,r}}\Vert g\Vert _{L^{1}} \end{aligned}$$

  for any $f \in L^{q,r}({\mathbb {R}}^d)$ and $g \in L^{1}({\mathbb {R}}^d)$.

Lemmas 2.3 and 2.4 are originally proved by O’Neil [41] (see also Yap [56] for Lorentz spaces with second exponents less than one). Lemma 2.4 (iii) is known in the abstract setting (cf. Lemarié-Rieusset [32, Chapter 4, Proposition 4.1]). It is also recently proved by Wang, Wei and Ye [53, Lemma 2.2].

We also have the interpolation inequality in Lorentz spaces (see e.g. [53, (2.4) on page 8]).

Lemma 2.5

Let $0<q_1<q<q_2\le \infty$, $0<r\le \infty$ and $0<\theta <1$ satisfy

$$\begin{aligned} \frac{1}{q} = \frac{\theta }{q_1}+\frac{1-\theta }{q_2}. \end{aligned}$$

Then

$$\begin{aligned} \Vert f\Vert _{L^{q,r}} \le \left( \frac{(q_2-q_1)q^2}{(q_2-q)(q-q_1) r} \right) ^{\frac{1}{r}} \Vert f\Vert _{L^{q_1,\infty }}^\theta \Vert f\Vert _{L^{q_2,\infty }}^{1-\theta } \end{aligned}$$

for any $f \in L^{q_1,\infty }(\mathbb R^d) \cap L^{q_2,\infty }(\mathbb R^d)$.

3 Linear estimates

In this section, we summarize linear estimates for the heat semigroup in the weighted Lorentz spaces.

3.1 Smoothing and time decay estimates in weighted spaces

Let $\{e^{t\Delta }\}_{t>0}$ be the heat semigroup whose element is defined by

$$\begin{aligned} e^{t\Delta } f:= G_t *f,\quad f\in {\mathcal {S}}'({\mathbb {R}}^d) \end{aligned}$$

with the Gaussian kernel

$$\begin{aligned} G_t(x):= (4\pi t)^{-\frac{d}{2}} e^{-\frac{|x|^2}{4t}},\quad t>0,\ x \in {\mathbb {R}}^d. \end{aligned}$$

In this subsection, we prove the following:

Proposition 3.1

Let $d \in {\mathbb {N}}$, $1\le q_1\le \infty$, $1<q_2\le \infty$, $0< r_1,r_2 \le \infty$ and $s_1,s_2\in {\mathbb {R}}$. Then there exists a constant $C>0$ such that

$$\begin{aligned} \Vert e^{t\Delta }\Vert _{L^{q_1,r_1}_{s_1} \rightarrow L^{q_2,r_2}_{s_2}} = C t ^{-\frac{d}{2} (\frac{1}{q_1} - \frac{1}{q_2}) - \frac{s_1 - s_2}{2}} \end{aligned}$$

(9)

for any $t>0$ if and only if $q_1,q_2,r_1,r_2,s_1,s_2$ satisfy

and

Remark 6

The estimate (9) can be also obtained for $0<q_2\le 1$. More precisely, let $d \in {\mathbb {N}}$, $1\le q_1\le \infty$, $0<q_2\le 1$, $0< r_1,r_2 \le \infty$ and $s_1,s_2\in {\mathbb {R}}$, and assume (10)–(15) with the additional condition

$$\begin{aligned} r_2\ge 1 \quad \text {if }\frac{s_2}{d} + \frac{1}{q_2} = \frac{s_1}{d} + \frac{1}{q_1} = 1. \end{aligned}$$

(16)

Then we have (9) for any $t>0$. The additional condition (16) is required due to use of the embedding $L^1({\mathbb {R}}^d) \hookrightarrow L^{1,r_2}({\mathbb {R}}^d)$ for $r_2\ge 1$ and Young’s inequality $\Vert f * g\Vert _{L^1} \le \Vert f\Vert _{L^1}\Vert g\Vert _{L^1}$ in the case $\frac{s_2}{d} + \frac{1}{q_2} = \frac{s_1}{d} + \frac{1}{q_1} = 1$. On the other hand, we can also prove the necessity of (10)–(15), but we do not know if (16) is necessary. The proof is similar to that of Proposition 3.1, and we omit it. In the proofs of the nonlinear estimates (Lemmas 4.1, 4.2, 5.2 and 5.7), we do not use the case $0<q_2 \le 1$.

Remark 7

The estimates (9) are known in some particular cases, for example, the case $s_2=0$ in Lebesgue spaces in [5], the case $s_2 \ge 0$ in Lorentz spaces in [49], and the case $q_1 \le q_2$ in Lebesgue spaces in [40, 51] (see also [13]). Similar estimates are proved in Herz spaces and weak Herz spaces in [40, 51].

Remark 8

Proposition 3.1 gives a precision of [48, Proposition 3.3] in the endpoint case (14) with $s_2=0$ and $s_1>0$. This implies that [48, Remark 3.4, (2)] does not hold. However, this does not change the results in [48] as this endpoint case is not used in [48].

To reduce (9) for $e^{t\Delta }$ into that for $e^\Delta$, we give the following lemma.

Lemma 3.2

Let $d \in {\mathbb {N}}$, $1\le q_1,q_2 \le \infty$, $0<r_1,r_2 \le \infty$ and $s_1,s_2\in {\mathbb {R}}$. Then $e^{\Delta }$ is bounded from $L^{q_1,r_1}_{s_1}({\mathbb {R}}^d)$ into $L^{q_2,r_2}_{s_2}({\mathbb {R}}^d)$ if and only if $e^{t\Delta }$ is bounded from $L^{q_1,r_1}_{s_1}({\mathbb {R}}^d)$ into $L^{q_2,r_2}_{s_2}({\mathbb {R}}^d)$ with

$$\begin{aligned} \Vert e^{t\Delta }\Vert _{L^{q_1,r_1}_{s_1} \rightarrow L^{q_2,r_2}_{s_2}} = t ^{-\frac{d}{2} (\frac{1}{q_1} - \frac{1}{q_2}) - \frac{s_1 - s_2}{2}} \Vert e^{\Delta }\Vert _{L^{q_1,r_1}_{s_1} \rightarrow L^{q_2,r_2}_{s_2}} \end{aligned}$$

(17)

for any $t>0$.

Proof

It is enough to show (17) if $e^{\Delta }$ is bounded from $L^{q_1,r_1}_{s_1}({\mathbb {R}}^d)$ into $L^{q_2,r_2}_{s_2}({\mathbb {R}}^d)$, since the converse is trivial. The proof is based on the scaling argument. Let $f \in L^{q_1,r_1}_{s_1}({\mathbb {R}}^d)$. Since

$$\begin{aligned} (e^{t\Delta } f)(x)= & {} \left( e^{\Delta } (f(t^{\frac{1}{2}} \cdot ))\right) (t^{-\frac{1}{2}} x), \\ (e^{\Delta } f)(x)= & {} \left( e^{t\Delta } (f(t^{-\frac{1}{2}} \cdot ))\right) (t^{\frac{1}{2}} x), \end{aligned}$$

for $t>0$ and $x\in {\mathbb {R}}^d$, we have

$$\begin{aligned} \Vert e^{t\Delta } f\Vert _{L^{q_2,r_2}_{s_2}}\le & {} t ^{-\frac{d}{2} (\frac{1}{q_1} - \frac{1}{q_2}) - \frac{s_1 - s_2}{2}} \Vert e^{\Delta }\Vert _{L^{q_1,r_1}_{s_1} \rightarrow L^{q_2,r_2}_{s_2}} \Vert f\Vert _{L^{q_1,r_1}_{s_1}}, \\ \Vert e^{\Delta } f\Vert _{L^{q_2,r_2}_{s_2}}\le & {} t ^{\frac{d}{2} (\frac{1}{q_1} - \frac{1}{q_2}) + \frac{s_1 - s_2}{2}} \Vert e^{t\Delta }\Vert _{L^{q_1,r_1}_{s_1} \rightarrow L^{q_2,r_2}_{s_2}} \Vert f\Vert _{L^{q_1,r_1}_{s_1}}. \end{aligned}$$

Hence, (17) is proved. $\square$

Proof of the necessity part of Proposition 3.1

For the condition (15), see Remark 5 (b).

Step 1: Conditions $\frac{s_1}{d} + \frac{1}{q_1}\le 1$ in (10) and (12). If either of these fails, then $L^{q_1, r_1}_{s_1} ({\mathbb {R}}^d)$ is not included in $L^1_{\textrm{loc}}({\mathbb {R}}^d)$ (see Remark 5 (f)), which implies that $e^{t\Delta }: L^{q_1, r_1}_{s_1} ({\mathbb {R}}^d) \rightarrow L^{q_2,r_2}_{s_2}({\mathbb {R}}^d)$ is not well-defined.

Step 2: Conditions $\frac{s_2}{d}+\frac{1}{q_2}\ge 0$ in (10) and (13). Suppose either of these fails, i.e.,

$$\begin{aligned} \frac{s_2}{d}+\frac{1}{q_2}< 0\quad \text {or}\quad \frac{s_2}{d}+\frac{1}{q_2}=0 \text { and } r_2 < \infty . \end{aligned}$$

We consider the case $\frac{s_2}{d}+\frac{1}{q_2}=0$ and $r_2 < \infty$. By Lemma A.3, if $f \in L^{q_2, r_2}_{s_2}({\mathbb {R}}^d)$, then

$$\begin{aligned} \liminf _{|x|\rightarrow 0} |f(x)| \le \liminf _{|x| \rightarrow 0} |x|^{s_2+\frac{d}{q_2}} |\log |x||^{\frac{1}{r_2}} |f(x)| = 0. \end{aligned}$$

However, there exists an $f_0 \in L^{q_1,r_1}_{s_1}({\mathbb {R}}^d)$ such that

$$\begin{aligned} \liminf _{|x|\rightarrow 0} |e^{\Delta } f_0(x)| \not = 0, \end{aligned}$$

which implies $e^{\Delta } f_0 \not \in L^{q_2, r_2}_{s_2}({\mathbb {R}}^d)$. Hence, it is impossible to obtain (9). The case $\frac{s_2}{d} + \frac{1}{q_2} < 1$ is similarly proved.

Step 3: Condition $\frac{s_1}{d}+\frac{1}{q_1}\le \frac{s_2}{d}+\frac{1}{q_2}$ in (10). Suppose that $\frac{s_1}{d}+\frac{1}{q_1}\le \frac{s_2}{d}+\frac{1}{q_2}$ does not hold. Let $f \in C^\infty _0({\mathbb {R}}^d)$ with $f\not =0$. Then we have

$$\begin{aligned} \Vert e^{t\Delta } f\Vert _{L^{q_2,r_2}_{s_2}} \le C t ^{\delta } \Vert f \Vert _{L^{q_1, r_1}_{s_1}},\quad t>0, \end{aligned}$$

where

$$\begin{aligned} \delta := -\frac{d}{2} \left( \frac{1}{q_1} - \frac{1}{q_2}\right) - \frac{s_1 - s_2}{2}>0. \end{aligned}$$

Hence, $e^{t\Delta } f \rightarrow 0$ in ${\mathcal {S}}'({\mathbb {R}}^d)$ as $t\rightarrow 0$. Combining this with the continuity $e^{t\Delta } f \rightarrow f$ in ${\mathcal {S}}'({\mathbb {R}}^d)$ as $t\rightarrow 0$, we have $f=0$ by uniqueness of the limit. However, this is a contradiction to $f\not =0$. Thus, $\frac{s_1}{d}+\frac{1}{q_1}\le \frac{s_2}{d}+\frac{1}{q_2}$ is necessary.

Step 4: Condition (11). The proof is based on the translation argument as in [14, 49]. In fact, take a non-negative function $f \in C^\infty _0({\mathbb {R}}^d)$ with $\text {supp}\,f \subset \{x = (x_1, x') \in {\mathbb {R}} \times {\mathbb {R}}^{d-1} \,;\, x_1 \ge 0\}$, and let $x_0 =(1,0,\ldots ,0) \in {\mathbb {R}}^d$ and $\tau >0$. Since $(e^{\Delta } f(\cdot - \tau x_0))(x) = (e^{\Delta } f)(x-\tau x_0)$, it follows from (9) that

$$\begin{aligned} \left\| |\cdot |^{s_2}(e^{\Delta } f)(\cdot -\tau x_0)\right\| _{L^{q_2,r_2}} \le C \left\| |\cdot |^{s_1}f(\cdot -\tau x_0)\right\| _{L^{q_1,r_1}}. \end{aligned}$$

By making the changes of variables, we have

$$\begin{aligned} \tau ^{-(s_1-s_2)}\left\| \left| {\cdot \over \tau }+ x_0\right| ^{s_2}e^{\Delta } f\right\| _{L^{q_2,r_2}}\le C \left\| \left| {\cdot \over \tau }+ x_0\right| ^{s_1}f\right\| _{L^{q_1,r_1}}. \end{aligned}$$

(18)

The weight $|{\cdot \over \tau }+ x_0|^{s_2}$ has the uniform lower bounds with respect to sufficient large $\tau$:

$$\begin{aligned} \left| {x\over \tau }+ x_0\right| ^{s_2} \ge {\left\{ \begin{array}{ll} \displaystyle \left( 1- {|x| \over \tau } \right) ^{s_2} \ge 2^{-s_2} \quad &{} \text {for }|x|\le 1,\ \tau \ge 2 \quad \text {if }s_2\ge 0,\\ \displaystyle \left( {|x|\over \tau }+ 1\right) ^{s_2} \ge (|x|+1)^{s_2} &{}\text {for }\tau \ge 1 \quad \text {if }s_2\le 0. \end{array}\right. } \end{aligned}$$

Hence, once

$$\begin{aligned} \limsup _{\tau \rightarrow \infty } \left\| \left| {\cdot \over \tau }+ x_0\right| ^{s_1}f\right\| _{L^{q_1,r_1}} < \infty \end{aligned}$$

(19)

is obtained, we deduce $s_2 \le s_1$ from (18), (19) and positivity of $e^{\Delta }f$. Therefore, it is enough to show (19). In the case $s_1 \ge 0$, we have the uniform upper bound

$$\begin{aligned} \left| {x\over \tau }+ x_0\right| ^{s_1} \le \left( {|x|\over \tau }+ 1\right) ^{s_1} \le (|x|+1)^{s_1} \quad \text {for }\tau \ge 1, \end{aligned}$$

which implies (19). In the other case $s_1 < 0$, the weight

$$\begin{aligned} \left| {x\over \tau }+ x_0 \right| ^{s_1} = \left[ \left( \frac{x_1}{\tau } + 1\right) ^2 + \frac{|x'|^2}{\tau ^2} \right] ^\frac{s_1}{2} \end{aligned}$$

has a singularity only at $x=x^*(\tau )=(-\tau ,0,\ldots ,0)$, and is increasing with respect to $\tau$ for each $x \in \{x_1\ge 0\}$. Here, we note that $|{\cdot \over \tau }+ x_0|^{s_1}f \in L^{q_1,r_1}$ for any $\tau >0$, since the singular points $x^*(\tau )$ are not included in $\text {supp}\,f$ for any $\tau >0$. Hence,

$$\begin{aligned} \left| {\cdot \over \tau }+ x_0\right| ^{s_1}f \nearrow f\quad \text {a.e.}\, x \in \{x_1\ge 0\} \quad \text {as }\tau \rightarrow \infty , \end{aligned}$$

and we can use Lemma A.1 to obtain

$$\begin{aligned} \lim _{\tau \rightarrow \infty } \left\| \left| {\cdot \over \tau }+ x_0\right| ^{s_1}f\right\| _{L^{q_1,r_1}} =\Vert f\Vert _{L^{q_1,r_1}}< \infty . \end{aligned}$$

This implies (19). Thus, the necessity of $s_2 \le s_1$ is proved.

Final step: Condition (14). Let

$$\begin{aligned} f (x) = (1+|x|)^{-\frac{d}{q_1}-s_1} ( \log (e+|x|))^{-b} \end{aligned}$$

where $b > \frac{1}{r_1}$ if $r_1 < \infty$ and $b=0$ if $r_1=\infty$. Then $f\in L^{q_1,r_1}_{s_1}({\mathbb {R}}^d)$ (see Remark (5) (h)). Since f is a positive, radially symmetric and decreasing function, we have

$$\begin{aligned} e^{\Delta }f(x) \ge \int _{|y|\le 1} G_1(y) f(x-y)\, dy \ge Cf(x) \end{aligned}$$

(20)

for $|x| \ge 1$ sufficiently large. Now, if (14) fails, i.e., $r_1 > r_2$ and $\frac{s_1}{d} + \frac{1}{q_1}=\frac{s_2}{d} + \frac{1}{q_2}$, then $f \not \in L^{q_2,r_2}_{s_2}({\mathbb {R}}^d)$ since b can be taken as $\frac{1}{r_1}< b < \frac{1}{r_2}$ if $r_1<\infty$ and $0= b < \frac{1}{r_2}$ if $r_1=\infty$. Hence, by (20), we also have $e^{\Delta }f \not \in L^{q_2,r_2}_{s_2}({\mathbb {R}}^d)$, which means $\Vert e^{\Delta }\Vert _{L^{q_1,r_1}_{s_1} \rightarrow L^{q_2,r_2}_{s_2}} = \infty$. Thus, the necessity of (14) is shown by contraposition. The proof of the necessity part is finished. $\square$

Proof of the sufficiency part of Proposition 3.1

By Lemma 3.2, it is enough to prove (9) with $t=1$:

$$\begin{aligned} \Vert e^{\Delta } f\Vert _{L^{q_2,r_2}_{s_2}} \le C\Vert f\Vert _{L^{q_1,r_1}_{s_1}}. \end{aligned}$$

(21)

We start the proof with the case $1< q_1, q_2<\infty$. We first prove (21) with the non-endpoint case:

$$\begin{aligned} 0< \frac{s_2}{d} + \frac{1}{q_2}< \frac{s_1}{d} + \frac{1}{q_1} < 1 \quad \text {and}\quad s_2\le s_1. \end{aligned}$$

(22)

From Lemma 3.2 and the embedding $L^{q_1,r_1}({\mathbb {R}}^d) \hookrightarrow L^{q_1,\infty }({\mathbb {R}}^d)$ for any $0< r_1 \le \infty$, it is sufficient to show that $e^{\Delta }$ is bounded from $L^{q_1,\infty }_{s_1}({\mathbb {R}}^d)$ into $L^{q_2,r_2}_{s_2}({\mathbb {R}}^d)$. We divide the proof into three cases:

$$\begin{aligned} s_2 \ge 0, \quad s_2< 0 \le s_1 \quad \text {and}\quad s_1 < 0. \end{aligned}$$

In the case $s_2\ge 0$, we use the inequality $|x|^{s_2} \le C( |x-y|^{s_2} + |y|^{s_2})$ to obtain

$$\begin{aligned} \begin{aligned} ||x|^{s_2}e^{\Delta } f (x)|&= |x|^{s_2} \left| (G_1 * f) (x)\right| \\&\le C \left\{ (G_1 * (|\cdot |^{s_2} |f|))(x) + ((|\cdot |^{s_2} G_1) * |f|))(x) \right\} . \end{aligned} \end{aligned}$$

Then we use Lemma 2.3 (i) and Lemma 2.4 (i) to estimate

$$\begin{aligned} \Vert G_1 * (|\cdot |^{s_2} |f|)\Vert _{L^{q_2,r_2}}\le & {} C\Vert G_1\Vert _{L^{p_1,r_2}} \Vert |\cdot |^{s_2} |f|\Vert _{L^{p_2,\infty }} \nonumber \\\le & {} C\Vert G_1\Vert _{L^{p_1,r_2}} \Vert |\cdot |^{s_2-s_1}\Vert _{L^{\frac{d}{s_1-s_2}, \infty }} \Vert |\cdot |^{s_1}f\Vert _{L^{q_1,\infty }} \nonumber \\\le & {} C\Vert f\Vert _{L^{q_1,\infty }_{s_1}}, \end{aligned}$$

(23)

where $p_1$ and $p_2$ satisfy $1<p_1<(\frac{s_2}{d} + \frac{1}{q_2})^{-1}$, $(1-\frac{s_2}{d})^{-1}<p_2 <q_2$, $\frac{1}{q_2} = \frac{1}{p_1} + \frac{1}{p_2} -1$ and $\frac{1}{p_2} = \frac{s_1-s_2}{d} + \frac{1}{q_1}$, and

$$\begin{aligned} \Vert (|\cdot |^{s_2} G_1) * |f|\Vert _{L^{q_2,r_2}}\le & {} C \Vert |\cdot |^{s_2} G_1\Vert _{L^{p_3,r_2}} \Vert f\Vert _{L^{p_4,\infty }} \nonumber \\\le & {} C \Vert |\cdot |^{s_2} G_1\Vert _{L^{p_3,r_2}}\Vert |\cdot |^{-s_1}\Vert _{L^{\frac{d}{s_1},\infty }} \Vert |\cdot |^{s_1}f\Vert _{L^{q_1,\infty }} \nonumber \\\le & {} C\Vert f\Vert _{L^{q_1,\infty }_{s_1}}, \end{aligned}$$

(24)

where $p_3$ and $p_4$ satisfy $(1-\frac{s_2}{d})^{-1}< p_3 < q_2$, $1<p_4<(\frac{s_2}{d} + \frac{1}{q_2})^{-1}$, $\frac{1}{q_2} = \frac{1}{p_3} + \frac{1}{p_4} -1$ and $\frac{1}{p_4} = \frac{s_1}{d} + \frac{1}{q_1}$. Here, we note that such $p_1$, $p_2$, $p_3$ and $p_4$ exist if (22) and $s_2\ge 0$ hold. Hence, (21) is proved in this case.

In the case $s_2 < 0 \le s_1$, we use Lemma 2.3 (i) to obtain

$$\begin{aligned} \Vert e^{\Delta } f \Vert _{L^{q_2,r_2}_{s_2}} \le C \Vert |\cdot |^{s_2} \Vert _{L^{-\frac{d}{s_2},\infty }} \Vert e^{\Delta } f \Vert _{L^{p_5,r_2}}, \end{aligned}$$

(25)

where $p_5$ satisfies $(\frac{s_1}{d} + \frac{1}{q_1})^{-1}<p_5<\infty$ and $\frac{1}{q_2} = -\frac{s_2}{d} + \frac{1}{p_5}$, and such a $p_5$ exists under the conditions (22) and $s_2 < 0 \le s_1$. Now, noting $p_5$ satisfies $0< \frac{1}{p_5}< \frac{s_1}{d} + \frac{1}{q_1} < 1$ and $0\le s_1$, we can apply the estimate shown in the previous case with $s_2=0$ to obtain

$$\begin{aligned} \Vert e^{\Delta } f \Vert _{L^{p_5,r_2}} \le C \Vert f\Vert _{L^{q_1,\infty }_{s_1}}. \end{aligned}$$

Thus, the case $s_2 < 0 \le s_1$ is also proved.

In the case $s_1< 0$, setting $g:= |x|^{s_1}|f|$, and using the inequality $|y|^{-s_1} \le C( |x-y|^{-s_1} + |x|^{-s_1})$, we have

$$\begin{aligned} \begin{aligned} \left| |x|^{s_2}e^{\Delta } f (x)\right|&\le |x|^{s_2} \int _{{\mathbb {R}}^d} G_1 (x-y) |y|^{-s_1}g(y)\, dy\\&\le C\left( |x|^{s_2-s_1} e^{\Delta } g(x) + |x|^{s_2}\left( (|\cdot |^{-s_1} G_1) * g\right) (x)\right) . \end{aligned} \end{aligned}$$

(26)

Then we use Lemma 2.3 (i) and Lemma 2.4 (i) to estimate

$$\begin{aligned} \begin{aligned} \Vert |\cdot |^{s_2-s_1} e^{\Delta } g \Vert _{L^{q_2,r_2}}&\le C \Vert |\cdot |^{s_2-s_1} \Vert _{L^{\frac{d}{s_1-s_2},\infty }} \Vert e^{\Delta } g\Vert _{L^{p_6,r_2}}\\&\le C \Vert |\cdot |^{s_2-s_1} \Vert _{L^{\frac{d}{s_1-s_2},\infty }} \Vert G_1\Vert _{L^{p_7,r_2}} \Vert g\Vert _{L^{{q_1, \infty }}}\\&\le C\Vert f\Vert _{L^{q_1,\infty }_{s_1}}, \end{aligned}\nonumber \\ \end{aligned}$$

(27)

where $p_6$ and $p_7$ satisfy $q_1<p_6 < -\frac{d}{s_1}$, $1< p_7 < (\frac{s_2}{d} + \frac{1}{q_2})^{-1}$, $\frac{1}{q_2} = \frac{s_1-s_2}{d} + \frac{1}{p_6}$ and $\frac{1}{p_6} = \frac{1}{p_7} + \frac{1}{q_1}-1$, and

$$\begin{aligned} \begin{aligned} \left\| |\cdot |^{s_2}\left( (|\cdot |^{-s_1} G_1) * g\right) \right\| _{L^{q_2,r_2}}&\le C \Vert |\cdot |^{s_2}\Vert _{L^{-\frac{d}{s_2},\infty }} \Vert (|\cdot |^{-s_1} G_1) * g\Vert _{L^{p_8, r_2}}\\&\le C \Vert |\cdot |^{s_2}\Vert _{L^{-\frac{d}{s_2},\infty }} \Vert |\cdot |^{-s_1} G_1\Vert _{L^{p_9, r_2}} \Vert g\Vert _{L^{q_1,\infty }}\\&\le C\Vert f\Vert _{L^{q_1,\infty }_{s_1}}, \end{aligned}\nonumber \\ \end{aligned}$$

(28)

where $p_8$ and $p_9$ satisfy $(\frac{s_1}{d} + \frac{1}{q_1})^{-1}< p_8< \infty$, $(\frac{s_1}{d}+1)^{-1}< p_9 < (1-\frac{1}{q_1})^{-1}$, $\frac{1}{q_2} = - \frac{s_2}{d} + \frac{1}{p_8}$ and $\frac{1}{p_8} = \frac{1}{p_9} + \frac{1}{q_1}-1$. Here, we note that such $p_6$, $p_7$, $p_8$ and $p_9$ exist if (22) and $s_1<0$ hold. Thus, the case $s_1<0$ is also proved.

Next, we consider the endpoint cases (12), (13) or (14) with $1<q_1,q_2<\infty$. Here, we give only sketch of proofs of single endpoint cases. If two or more endpoints overlap, simply combine them.

As to the case (12), i.e., $\frac{s_1}{d} + \frac{1}{q_1}=1$ and $r_1 \le 1$, we note that $s_1\ge 0$, and the proof is almost the same as the non-endpoint case (22) with $s_1\ge 0$. In fact, we can take $p_1=(\frac{s_2}{d} + \frac{1}{q_2})^{-1}$, $p_2=(1-\frac{s_2}{d})^{-1}$, $p_3=q_2$ and $p_4=1$, and use Lemma 2.4 (iii) (instead of Lemma 2.4 (i)) in (24), where $\Vert f\Vert _{L^{p_4,\infty }}$ is replaced by $\Vert f\Vert _{L^1}$ and the restriction $r_1 \le 1$ appears.

As to the case (13), i.e., $\frac{s_2}{d} + \frac{1}{q_2}=0$ and $r_2 = \infty$, we note that $s_2< 0$, and the proof is similar to the non-endpoint case (22) with $s_2 < 0 \le s_1$ or $s_2 \le s_1 <0$. For $s_2 < 0 \le s_1$, we use Lemma 2.3 (ii) to obtain

$$\begin{aligned} \Vert e^{\Delta } f \Vert _{L^{q_2,\infty }_{s_2}} \le C \Vert |\cdot |^{s_2} \Vert _{L^{-\frac{d}{s_2},\infty }} \Vert e^{\Delta } f \Vert _{L^{\infty }} \le C \Vert e^{\Delta } f \Vert _{L^{\infty }} \end{aligned}$$

(this corresponds to taking $p_5 = \infty$ in (25)). The estimate $\Vert e^{\Delta } f \Vert _{L^{\infty }} \le C \Vert f\Vert _{L^{q_1,r_1}_{s_1}}$ will be given later (see the proof of the case $q_2=\infty$ below). For $s_2 \le s_1 <0$, we also have

$$\begin{aligned} \Vert e^{\Delta } f \Vert _{L^{q_2,\infty }_{s_2}}&\le \Vert |\cdot |^{s_2-s_1} e^{\Delta } g \Vert _{L^{q_2,\infty }} + \left\| |\cdot |^{s_2}\left( (|\cdot |^{-s_1} G_1) * g\right) \right\| _{L^{q_2,\infty }}\\&\le C \Big (\Vert |\cdot |^{s_2-s_1} \Vert _{L^{\frac{d}{s_1-s_2},\infty }} \Vert e^{\Delta } g\Vert _{L^{-\frac{d}{s_1},\infty }}\\&\quad + \Vert |\cdot |^{s_2}\Vert _{L^{-\frac{d}{s_2},\infty }} \Vert (|\cdot |^{-s_1} G_1) * g\Vert _{L^{\infty }} \Big ) \\&\le C \Big (\Vert |\cdot |^{s_2-s_1} \Vert _{L^{\frac{d}{s_1-s_2},\infty }} \Vert G_1\Vert _{L^{p_7,\infty }} \Vert g\Vert _{L^{{q_1, \infty }}}\\&\quad + \Vert |\cdot |^{s_2}\Vert _{L^{-\frac{d}{s_2},\infty }} \Vert |\cdot |^{-s_1} G_1\Vert _{L^{p_9,1}} \Vert g\Vert _{L^{q_1,\infty }} \Big )\\&\le C\Vert f\Vert _{L^{q_1,\infty }_{s_1}}, \end{aligned}$$

where we take $p_6 = -\frac{d}{s_1}$, $p_7 = [1- (\frac{s_1}{d} + \frac{1}{q_1})]^{-1}$, $p_8 = \infty$ and $p_9 = (1-\frac{1}{q_1})^{-1}$.

As to the case (14), i.e., $\frac{s_1}{d} + \frac{1}{q_1}=\frac{s_2}{d} + \frac{1}{q_2}$ and $r_1 \le r_2$, we can use Lemma 2.4 (iii) to make a similar argument to the non-endpoint case. In fact, when $s_2\ge 0$, this case corresponds to taking $p_1=1$, $p_2=q_2$, $p_3=(1-\frac{s_2}{d})^{-1}$ and $p_4 = (\frac{s_2}{d} + \frac{1}{q_2})^{-1}$ in (23) and (24). In particular, in (23), Lemma 2.4 (iii) is used and the restriction $r_1 \le r_2$ is required:

$$\begin{aligned} \begin{aligned} \Vert G_1 * (|\cdot |^{s_2} |f|)\Vert _{L^{q_2,r_2}}&\le C\Vert G_1\Vert _{L^1} \Vert |\cdot |^{s_2} |f|\Vert _{L^{q_2,r_2}}\\&\le C\Vert G_1\Vert _{L^1} \Vert |\cdot |^{s_2-s_1}\Vert _{L^{\frac{d}{s_1-s_2}, \infty }} \Vert |\cdot |^{s_1}f\Vert _{L^{q_1,r_2}}\\&\le C\Vert f\Vert _{L^{q_1,r_2}_{s_1}}\le C\Vert f\Vert _{L^{q_1,r_1}_{s_1}}. \end{aligned} \end{aligned}$$

The case $s_2<0$ is similar, and we may omit it.

In the rest of the proof, we consider the cases $q_1=1$, $q_1=\infty$ or $q_2=\infty$. The case $q_1=1$ and $q_2 = \infty$ is just $L^1$-$L^\infty$ estimate. The case $q_1=q_2=\infty$ has been already proved (see, e.g., [13, Lemma 2.1]).

The case $1<q_1<\infty$ and $q_2=\infty$ is the estimate (21) with

$$\begin{aligned} 0\le s_2\le s_1,\quad \frac{s_1}{d} + \frac{1}{q_1} \le 1 \quad \text {and}\quad r_1\le 1\text { if }\frac{s_1}{d} + \frac{1}{q_1} = 1. \end{aligned}$$

Since $s_2\ge 0$, this case is proved in a similar way to (23) and (24). In fact, we deduce from Lemma 2.4 (ii) and Lemma 2.3 (i) that

$$\begin{aligned} \begin{aligned} \Vert G_1 * (|\cdot |^{s_2} |f|)\Vert _{L^{\infty }}&\le C\Vert G_1\Vert _{L^{p_{10},1}} \Vert |\cdot |^{s_2} |f|\Vert _{L^{p_{11},\infty }}\\&\le C\Vert |\cdot |^{s_2-s_1}\Vert _{L^{\frac{d}{s_1-s_2}, \infty }} \Vert |\cdot |^{s_1}f\Vert _{L^{q_1,\infty }}\\&\le C\Vert f\Vert _{L^{q_1,\infty }_{s_1}}, \end{aligned} \nonumber \\ \end{aligned}$$

(29)

where $p_{10}$ and $p_{11}$ satisfy $1\le p_{10}<\frac{d}{s_2}$, $\frac{d}{d-s_2}<p_{11}\le \infty$, $1 = \frac{1}{p_{10}} + \frac{1}{p_{11}}$ and $\frac{1}{p_{11}} = \frac{s_1-s_2}{d} + \frac{1}{q_1}$, and

$$\begin{aligned} \begin{aligned} \Vert (|\cdot |^{s_2} G_1) * |f|\Vert _{L^{\infty }}&\le C \Vert |\cdot |^{s_2} G_1\Vert _{L^{p_{12},1}} \Vert f\Vert _{L^{p_{13},\infty }}\\&\le C\Vert |\cdot |^{-s_1}\Vert _{L^{\frac{d}{s_1},\infty }} \Vert |\cdot |^{s_1}f\Vert _{L^{q_1,\infty }}\\&\le C\Vert f\Vert _{L^{q_1,\infty }_{s_1}}, \end{aligned} \end{aligned}$$

where $p_{12}$ and $p_{13}$ satisfy $\frac{d}{d-s_2} \le p_{12} < \infty$, $1<p_{13}\le \frac{d}{s_2}$, $1 = \frac{1}{p_{12}} + \frac{1}{p_{13}}$ and $\frac{1}{p_{12}} = \frac{s_1}{d} + \frac{1}{q_1}$. Here, we note that such $p_{10}$, $p_{11}$, $p_{12}$ and $p_{13}$ exist if $0\le s_2\le s_1$ and $\frac{s_1}{d} + \frac{1}{q_1}<1$. For the case $\frac{s_1}{d} + \frac{1}{q_1}=1$, the first term can be estimated in the same way as (23) (where we take $p_8=\frac{d}{s_2}$ and $p_9=\frac{d}{d-s_2}$). For the second term, we take $p_{10}=\infty$ and $p_{11}=1$ and we use Lemma 2.4 (ii) to obtain

$$\begin{aligned} \begin{aligned} \Vert (|\cdot |^{s_2} G_1) * |f|\Vert _{L^{\infty }}&\le C \Vert |\cdot |^{s_2} G_1\Vert _{L^{\infty }} \Vert f\Vert _{L^{1}}\\&\le C\Vert |\cdot |^{-s_1}\Vert _{L^{\frac{d}{s_1},\infty }} \Vert |\cdot |^{s_1}f\Vert _{L^{q_1,1}}\\&\le C\Vert f\Vert _{L^{q_1,1}_{s_1}}. \end{aligned} \end{aligned}$$

Thus, the estimate (21) is proved in the case $q_2=\infty$.

The case $q_1=1$ and $1<q_2 <\infty$ is the estimate (21) with

$$\begin{aligned} s_2\le s_1\le 0,\quad 0\le \frac{s_2}{d} + \frac{1}{q_2} < \frac{s_1}{d} + 1, \quad r_1\le 1 \quad \text {and}\quad r_2 = \infty \text { if } \frac{s_2}{d} + \frac{1}{q_2}=0. \end{aligned}$$

The proof is similar to (27) and (28). Let $\frac{s_2}{d} + \frac{1}{q_2}>0$. As to the first term, it follows from Lemma 2.3 (i) and Lemma 2.4 (ii) that

$$\begin{aligned} \begin{aligned} \Vert |\cdot |^{s_2-s_1} e^{\Delta } g \Vert _{L^{q_2,r_2}}&\le C \Vert |\cdot |^{s_2-s_1} \Vert _{L^{\frac{d}{s_1-s_2},\infty }} \Vert e^{\Delta } g\Vert _{L^{p_{14},r_2}}\\&\le C \Vert |\cdot |^{s_2-s_1} \Vert _{L^{\frac{d}{s_1-s_2},\infty }} \Vert G_1\Vert _{L^{p_{14},r_2}} \Vert g\Vert _{L^1}\\&\le C\Vert f\Vert _{L^1_{s_1}}, \end{aligned} \nonumber \\ \end{aligned}$$

(30)

where $p_{14}$ satisfies $1<p_{14} < -\frac{d}{s_1}$ and $\frac{1}{q_2} = \frac{s_1-s_2}{d}+ \frac{1}{p_{14}}$. The second term can be estimated as

$$\begin{aligned} \begin{aligned} \left\| |\cdot |^{s_2}\left( (|\cdot |^{-s_1} G_1) * g\right) \right\| _{L^{q_2,r_2}}&\le C \Vert |\cdot |^{s_2}\Vert _{L^{-\frac{d}{s_2},\infty }} \Vert (|\cdot |^{-s_1} G_1) * g\Vert _{L^{p_{15}, r_2}}\\&\le C \Vert |\cdot |^{s_2}\Vert _{L^{-\frac{d}{s_2},\infty }} \Vert |\cdot |^{-s_1} G_1\Vert _{L^{p_{15}, r_2}} \Vert g\Vert _{L^{1}}\\&\le C\Vert f\Vert _{L^1_{s_1}}, \end{aligned} \end{aligned}$$

where $p_{15}$ satisfies $(\frac{s_1}{d} +1)^{-1}< p_{15}< \infty$ and $\frac{1}{q_2} = - \frac{s_2}{d} + \frac{1}{p_{15}}$. Here, we note that such $p_{14}$ and $p_{15}$ exist if $s_2\le s_1\le 0$ and $0< \frac{s_2}{d} + \frac{1}{q_2} < \frac{s_1}{d} + 1$. For the case $\frac{s_2}{d} + \frac{1}{q_2}=0$, the first term can be estimated in the same way as (30) (where we take $p_{14}=-\frac{d}{s_1}$). For the second term, we have only to take $p_{15}=\infty$ and $r_2=\infty$ and use Young’s inequality $\Vert f * g\Vert _{L^\infty }\le \Vert f\Vert _{L^1}\Vert g\Vert _{L^\infty }$. Thus, the estimate (21) is proved in the case $q_1=1$ and $1<q_2 <\infty$. The proof of Proposition 3.1 is finished. $\square$

3.2 Weighted Meyer inequality

In this subsection, we shall prove the following proposition, which is a key tool to study unconditional uniqueness and uniqueness criterion in the scale-critical case and the construction of a singular solution in the double critical case.

Proposition 3.3

Let $T\in (0,\infty ]$, and let $d\ge 3$, $1 \le q_1 \le \infty$, $1<q_2<\infty$, $0<r_1 \le \infty$ and $s_1,s_2 \in {\mathbb {R}}$ satisfy

and

Then there exists a constant $C>0$ such that

$$\begin{aligned} \left\| \int _0^{t} e^{(t-\tau )\Delta } f(\tau ) \,d\tau \right\| _{L^{q_2,\infty }_{s_2}} \le C \sup _{0<\tau <t} \Vert f(\tau )\Vert _{L^{q_1,r_1}_{s_1}} \end{aligned}$$

(36)

for any $t\in (0,T)$ and $f \in L^\infty (0,T; L^{q_1,r_1}_{s_1}({\mathbb {R}}^d))$.

The case $s_1=s_2=0$ is known as Meyer’s inequality and is proved by Meyer [37] (see also [50]).

Proof

We shall prove only the case $q_1>1$ and $\frac{s_1}{d} + \frac{1}{q_1}<1$, since the proofs of the other cases are similar. By the argument in [37], it suffices to prove that

$$\begin{aligned} \Vert g\Vert _{L^{q_2,\infty }_{s_2}} \le C, \end{aligned}$$

(37)

where we define

$$\begin{aligned} g(x):= \int _0^{\infty } e^{t\Delta } f(t,x) \,dt \end{aligned}$$

and we may assume that

$$\begin{aligned} \sup _{t\ge 0} \Vert f(t,\cdot ) \Vert _{L^{q_1,r_1}_{s_1}} \le 1 \end{aligned}$$

without loss of generality. Let $\lambda \in (0,\infty )$ be arbitrarily fixed. For $\tau \in (0,\infty ),$ which is to be determined later, we divide g into two parts:

$$\begin{aligned} g(x) = \int _{0}^{\tau } e^{t\Delta } f(t,x) \,dt +\int _{\tau }^{\infty } e^{t\Delta } f(t,x) \,dt =: h(x) + \ell (x). \end{aligned}$$

Let $p_0$ and $p_1$ be such that

$$\begin{aligned} 1< p_1< q_2 < p_0 \le \infty \quad \text {and}\quad 0 \le \frac{s_2}{d} + \frac{1}{p_i} \le \frac{s_1}{d} + \frac{1}{q_1}\quad \text {for }i=0,1. \end{aligned}$$

Then, by Proposition 3.1, we have

$$\begin{aligned} \begin{aligned} \Vert \ell \Vert _{L^{p_0,\infty }_{s_2}}&\le \int _{\tau }^{\infty } \Vert e^{t\Delta } f(t) \Vert _{L^{p_0,\infty }_{s_2}} \,dt \\&\le C \int _{\tau }^{\infty } t^{-\frac{d}{2}(\frac{1}{q_1} - \frac{1}{p_0}) - \frac{s_1-s_2}{2} } \Vert f(t) \Vert _{L^{q_1,\infty }_{s_1}} \,dt\\&\le C \tau ^{ - \frac{d}{2} (\frac{1}{q_2}-\frac{1}{p_0})} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Vert h\Vert _{L^{p_1,\infty }_{s_2}}&\le \int _{0}^{\tau } \left\| e^{t\Delta } f(t) \right\| _{L^{p_1,\infty }_{s_2}} \,dt \\&\le \int _{0}^{\tau } t^{-\frac{d}{2} (\frac{1}{q_1} - \frac{1}{p_1}) - \frac{s_1 -s_2}{2}} \left\| f(t) \right\| _{L^{q_1,\infty }_{s_1}} \,dt \\&\le C \tau ^{ \frac{d}{2} (\frac{1}{p_1}-\frac{1}{q_2})}. \end{aligned} \end{aligned}$$

Now, the definition of the Lorentz norms yields

$$\begin{aligned} d_{|\cdot |^{s_2} \ell }\left( \frac{\lambda }{2}\right) \le \left( \frac{ \Vert \ell \Vert _{L^{p_0,\infty }_{s_2}} }{ \lambda /2 } \right) ^{p_0} \le \left( \frac{ C \tau ^{ - \frac{d}{2} (\frac{1}{q_2}-\frac{1}{p_0})} }{ \lambda } \right) ^{p_0} \end{aligned}$$

and similarly,

$$\begin{aligned} d_{|\cdot |^{s_2} h}\left( \frac{\lambda }{2}\right) \le \left( \frac{ \Vert h \Vert _{L^{p_1,\infty }_{s_2}} }{ \lambda /2 } \right) ^{p_1} \le \left( \frac{ C \tau ^{ \frac{d}{2} (\frac{1}{p_1}-\frac{1}{q_2})} }{ \lambda } \right) ^{p_1}. \end{aligned}$$

Thus, choosing $\tau$ such that $\tau = \lambda ^{-\frac{2q_2}{d}}$, we deduce

$$\begin{aligned} d_{|\cdot |^{s_2} g}(\lambda ) \le d_{|\cdot |^{s_2} h}\left( \frac{\lambda }{2}\right) + d_{|\cdot |^{s_2} \ell }\left( \frac{\lambda }{2}\right) \le \frac{C}{\lambda ^{q_2} }, \end{aligned}$$

which implies (37). Thus, we conclude Proposition 3.3. $\square$

4 Unconditional uniqueness and uniqueness criterion

In this section, we prove Theorems 1.2, 1.3, Proposition 1.4 and Theorem 1.6.

4.1 Nonlinear estimates

We define the Duhamel term N(u) by

$$\begin{aligned} N(u) (t):= \int _{0}^t e^{(t-\tau )\Delta }(|\cdot |^{\gamma } |u(\tau )|^{\alpha -1} u(\tau ))\, d\tau . \end{aligned}$$

Then we have the following nonlinear estimates, which are used to prove unconditional uniqueness in the double subcritical case and in the single critical case I.

Lemma 4.1

Let $d,\gamma ,\alpha ,q,s$ be as in (6). Let $T \in (0,\infty ]$ and $\delta$ be given by

$$\begin{aligned} \delta := \frac{d(\alpha -1)}{2}\left[ \frac{1}{q_c} - \left( \frac{s}{d}+\frac{1}{q} \right) \right] . \end{aligned}$$

(38)

Then the following assertions hold:

• (i) If $0<\frac{s}{d} + \frac{1}{q} < \min \{\frac{1}{q_c}, \frac{1}{Q_c}\}$ and $q>\alpha$, then there exists a constant $C>0$ such that

  $$\begin{aligned} \begin{aligned}&\Vert N(u_1)(t) - N(u_2)(t)\Vert _{L^{q,\infty }_{s}} \\&\quad \le C t^{\delta } \max _{i=1,2} \Vert u_i\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})} \end{aligned} \end{aligned}$$

  for any $t\in (0,T)$ and $u_1, u_2 \in L^\infty (0,T; L^{q,\infty }_{s}({\mathbb {R}}^d))$.

• (ii) If either “$0<\frac{s}{d} + \frac{1}{q} < \min \{\frac{1}{q_c}, \frac{1}{Q_c}\}$ and $q=\alpha$” or “$\frac{s}{d} + \frac{1}{q} = \frac{1}{Q_c} < \frac{1}{q_c}$”, then there exists a constant $C>0$ such that

  $$\begin{aligned} \begin{aligned}&\Vert N(u_1)(t) - N(u_2)(t)\Vert _{L^{q,\alpha }_{s}} \\&\quad \le C t^{\delta } \max _{i=1,2} \Vert u_i\Vert _{L^\infty (0,t; L^{q,\alpha }_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\alpha }_{s})} \end{aligned} \end{aligned}$$

  for any $t\in (0,T)$ and $u_1, u_2 \in L^\infty (0,T; L^{q,\alpha }_{s}({\mathbb {R}}^d))$, provided that $q\not = \infty$.

Remark 9

In (ii), the space of $u_1, u_2$ is restricted to $L^\infty (0,T; L^{q,\alpha }_{s}({\mathbb {R}}^d))$. Here, note that $L^\infty (0,T; L^{q,\alpha }_{s}({\mathbb {R}}^d)) \subsetneq L^\infty (0,T; L^{q,\infty }_{s}({\mathbb {R}}^d))$ (see Remark 5 (e)). This restriction is due to the condition (12) in Proposition 3.1.

Proof

We define $\sigma := \alpha s - \gamma$. First, we prove the assertion (i). Let $T \in (0,\infty ]$ and $u_1, u_2 \in L^\infty (0,T; L^{q,\infty }_{s}({\mathbb {R}}^d))$. We assume (6) and $0<\frac{s}{d} + \frac{1}{q} < \min \{\frac{1}{q_c}, \frac{1}{Q_c}\}$. Then the parameters $q,s,\sigma$ satisfy

$$\begin{aligned}{} & {} 1\le \frac{q}{\alpha }, q \le \infty ,\quad 0< \frac{s}{d} + \frac{1}{q}< \frac{\sigma }{d} + \frac{\alpha }{q}< 1,\quad s \\{} & {} \quad \le \sigma \quad \text {and}\quad d \left( \frac{\alpha }{q} - \frac{1}{q}\right) +\sigma - s < 2. \end{aligned}$$

Hence, we use Proposition 3.1 with $(q_1,r_1,s_1) = (\frac{q}{\alpha }, \infty , \sigma )$ and $(q_2,r_2,s_2) = (q, \infty ,s)$, and then, Lemma 2.3 with $(q,r)=(\frac{q}{\alpha },\infty )$, $(q_1,r_1) = (\frac{q}{\alpha -1},\infty )$ and $(q_2,r_2)=(q,\infty )$ to obtain

$$\begin{aligned} \begin{aligned}&\Vert N(u_1)(t) - N(u_2)(t)\Vert _{L^{q,\infty }_{s}} \\&\quad \le C \int _0^t (t-\tau ) ^{-\frac{d}{2} (\frac{\alpha }{q} - \frac{1}{q}) - \frac{\sigma - s}{2}}\\&\qquad \times \Vert |\cdot |^{\gamma } (|u_1(\tau )|^{\alpha -1}u_1(\tau ) - |u_2(\tau )|^{\alpha -1}u_2(\tau )) \Vert _{L^{\frac{q}{\alpha },\infty }_{\sigma }} \, d\tau \\&\quad \le C \int _0^t (t-\tau ) ^{-\frac{d}{2} (\frac{\alpha }{q} - \frac{1}{q}) - \frac{\sigma - s}{2}}\\&\quad \quad \times \Vert |\cdot |^\gamma (|u_1(\tau )|^{\alpha -1} + |u_2(\tau )|^{\alpha -1})|u_1(\tau )-u_2(\tau )\Vert _{L^{\frac{q}{\alpha },\infty }_{\sigma }}\, d\tau \\&\quad \le C \int _0^t (t-\tau ) ^{-\frac{d}{2} (\frac{\alpha }{q} - \frac{1}{q}) - \frac{\sigma - s}{2}} \, d\tau \\&\quad \quad \times \max _{i=1,2} \Vert u_i\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}\\&\quad \le C t^{\delta } \max _{i=1,2} \Vert u_i\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}. \end{aligned} \nonumber \\ \end{aligned}$$

(39)

Therefore, the assertion (i) is proved.

The assertion (ii) is also proved in the same way. In fact, when $\frac{s}{d} + \frac{1}{q} = \frac{1}{Q_c} < \frac{1}{q_c}$, we use Proposition 3.1 with the endpoint case (12) to obtain

$$\begin{aligned} \begin{aligned}&\Vert N(u_1) (t)- N(u_2)(t) \Vert _{L^{q,\alpha }_{s}}\\&\le C \int _0^t (t-\tau ) ^{-\frac{d}{2} (\frac{\alpha }{q} - \frac{1}{q}) - \frac{\sigma - s}{2}}\\&\quad \times \Vert |\cdot |^\gamma (|u_1(\tau )|^{\alpha -1} + |u_2(\tau )|^{\alpha -1})|u_1(\tau )-u_2(\tau )\Vert _{L^{\frac{q}{\alpha },1}_{\sigma }}\, d\tau \\&\le C t^\delta \max _{i=1,2} \Vert u_i\Vert _{L^\infty (0,t; L^{q,\alpha }_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\alpha }_{s})}. \end{aligned} \end{aligned}$$

Note that this case corresponds to taking the endpoint $\frac{\sigma }{d} + \frac{\alpha }{q} = 1$ in (39), which causes the restriction $r\le \alpha$. Here, the exponent $q=\infty$ is excluded (see Remark 5 (b)). The proof in the case $0<\frac{s}{d} + \frac{1}{q} < \min \{\frac{1}{q_c}, \frac{1}{Q_c}\}$ and $q=\alpha$ is similar and also uses (12). Thus, the proof of Lemma 4.1 is finished. $\square$

In addition, we prepare the nonlinear estimates of the following type. These estimates are used to prove uniqueness criterion in the single critical case I, and unconditional uniqueness and uniqueness criterion in the scale-critical case.

Lemma 4.2

Let $d,\gamma ,\alpha ,q,r,s$ be as in (6). Assume that ${{\tilde{q}}} \in (q, \infty )$ satisfies

$$\begin{aligned} \frac{s}{d} + \frac{1}{q} - \frac{2}{d(\alpha -1)}< \frac{s}{d} + \frac{1}{{{\tilde{q}}}} < \min \left\{ \frac{1}{q_c}, \frac{1}{Q_c} - \frac{1}{\alpha -1} \left[ \frac{1}{Q_c} - \left( \frac{s}{d} + \frac{1}{q}\right) \right] \right\} .\nonumber \\ \end{aligned}$$

(40)

Let $T \in (0,\infty ]$ and $\beta$ be defined by

$$\begin{aligned} \beta = \beta (d, q,{{\tilde{q}}}):= \frac{d}{2}\left( \frac{1}{q} - \frac{1}{{{\tilde{q}}}} \right) . \end{aligned}$$

(41)

Then the following assertions hold:

• (i) If $\frac{s}{d} + \frac{1}{q} = \frac{1}{Q_c} < \frac{1}{q_c}$ and $r>\alpha$, then there exists a constant $C>0$ such that

  $$\begin{aligned} \begin{aligned}&\Vert N(u_1)(t) - N(u_2)(t)\Vert _{L^{q,r}_{s}} \\&\quad \le C t^{\delta } \Big ( \max _{i=1,2}\Vert u_i-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,r'(\alpha -1)}_{s})} \\&\qquad + \sup _{0<\tau <t}\tau ^{ \beta }\Vert e^{\tau \Delta } u_0\Vert _{L_{s}^{{\tilde{q}},\infty }} \Big )^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,r}_{s})} \end{aligned} \end{aligned}$$

  (42)

  for any $t\in (0,T)$ and $u_1, u_2 \in L^\infty (0,T; L^{q,r}_{s}({\mathbb {R}}^d))$ satisfying $u_i-e^{\tau \Delta } u_0 \in L^\infty (0,T; L^{q,r'(\alpha -1)}_{s}({\mathbb {R}}^d))$ for $i=1,2$, where $\delta$ is given by (38).

• (ii) If $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} < \frac{1}{Q_c}$, then there exists a constant $C>0$ such that

  $$\begin{aligned} \begin{aligned}&\Vert N(u_1)(t) - N(u_2)(t)\Vert _{L^{q,\infty }_{s}} \\&\quad \le C \Big ( \max _{i=1,2}\Vert u_i-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,\infty }_{s})} \\&\qquad + \sup _{0<\tau <t}\tau ^{ \beta }\Vert e^{\tau \Delta } u_0\Vert _{L_{s}^{{\tilde{q}},\infty }} \Big )^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})} \end{aligned} \end{aligned}$$

  for any $t\in (0,T)$ and $u_1, u_2 \in L^\infty (0,T; L^{q,\infty }_{s}({\mathbb {R}}^d))$.

• (iii) If $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} = \frac{1}{Q_c}$, then there exists a constant $C>0$ such that

  $$\begin{aligned}{} & {} \Vert N(u_1)(t) - N(u_2)(t)\Vert _{L^{q,\infty }_{s}}\\{} & {} \quad \le C \Big ( \max _{i=1,2}\Vert u_i-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,\alpha ^*-1}_{s})} + \sup _{0<\tau <t}\tau ^{ \beta }\Vert e^{\tau \Delta } u_0\Vert _{L_{s}^{{\tilde{q}},\infty }} \Big )^{\alpha ^*-1}\\{} & {} \qquad \times \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})} \end{aligned}$$

  for any $t\in (0,T)$ and $u_1, u_2 \in L^\infty (0,T; L^{q,\infty }_{s}({\mathbb {R}}^d))$ satisfying $u_i-e^{\tau \Delta } u_0 \in L^\infty (0,T; L^{q,\alpha ^*-1}_{s}({\mathbb {R}}^d))$ for $i=1,2$.

Remark 10

In (ii) and (iii), the restriction on the second exponent $r=\infty$ in the left-hand side is due to use of the weighted Meyer inequality (36) in Proposition 3.3. In (iii), the reason why the space of $u_i-e^{\tau \Delta } u_0$ is restricted to be $L^\infty (0,T; L^{q,\alpha ^*-1}_{s}({\mathbb {R}}^d))$ is the endpoint condition (34) in Proposition 3.3.

Proof

Let $T \in (0,\infty ]$. For two functions $u_1$ and $u_2$ on $(0,T)\times {\mathbb {R}}^d$, we estimate

$$\begin{aligned} \begin{aligned}&|N(u_1)(t) - N(u_2)(t)| \\&\quad \le C\int _0^t e^{(t-\tau )\Delta } \left[ |\cdot |^{\gamma }\left( |u_1(\tau )|^{\alpha -1}+|u_2(\tau )|^{\alpha -1}\right) |u_1(\tau ) - u_2(\tau )|\right] \, d\tau \\&\quad \le C\int _0^t e^{(t-\tau )\Delta } \left[ |\cdot |^{\gamma }|u_1(\tau )-e^{\tau \Delta } u_0|^{\alpha -1}|u_1(\tau ) - u_2(\tau )|\right] d\tau \\&\qquad + C\int _0^t e^{(t-\tau )\Delta } \left[ |\cdot |^{\gamma }|u_2(\tau )-e^{\tau \Delta } u_0|^{\alpha -1}|u_1(\tau ) - u_2(\tau )|\right] d\tau \\&\qquad + C\int _0^t e^{(t-\tau )\Delta } \left[ |\cdot |^{\gamma }|e^{\tau \Delta } u_0|^{\alpha -1}|u_1(\tau ) - u_2(\tau )|\right] d\tau \\&\quad =: I(t) + II(t) + III(t). \end{aligned} \end{aligned}$$

(43)

First, we prove the assertion (i). Set $\sigma = \alpha s - \gamma$. In a similar way to the proof of Lemma 4.1 (ii), we estimate

$$\begin{aligned} \begin{aligned} \Vert I(t)\Vert _{L^{q,r}_{s}}&\le \int _0^t \Vert e^{(t-\tau )\Delta } (|\cdot |^\gamma |u_1(\tau )-e^{\tau \Delta } u_0|^{\alpha -1}|u_1(\tau ) - u_2(\tau )|)|\Vert _{L^{q,r}_{s}} \, d\tau \\&\le C \int _0^t (t-\tau ) ^{-\frac{d}{2} (\frac{\alpha }{q} - \frac{1}{q}) - \frac{\sigma - s}{2}}\, d\tau \\&\qquad \times \sup _{0<\tau < t} \Vert |\cdot |^\gamma |u_1(\tau )-e^{\tau \Delta } u_0|^{\alpha -1} |u_1(\tau ) - u_2(\tau )|\Vert _{L^{\frac{q}{\alpha },1}_{\sigma }}\\&\le C t ^\delta \ \Vert u_1-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,r'(\alpha -1)}_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,r}_{s})} \end{aligned} \end{aligned}$$

(44)

for any $t \in (0,T)$, where $1=\frac{1}{r}+\frac{1}{r'}$ and $\delta >0$ is given in (38). Similarly, we have

$$\begin{aligned} \Vert II(t)\Vert _{L^{q,r}_{s}} \le C t ^\delta \ \Vert u_2(\tau )-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,r'(\alpha -1)}_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,r}_{s})} \end{aligned}$$

(45)

for any $t \in (0,T)$. To estimate III(t), we take auxiliary parameters p, ${{\tilde{q}}}$ and $\sigma$ satisfying

$$\begin{aligned}{} & {} 1<p<\infty ,\quad q< {{\tilde{q}}}< \infty ,\quad 0< \frac{s}{d} + \frac{1}{q}< \frac{\sigma }{d} + \frac{1}{p} < 1, \quad s\le \sigma , \end{aligned}$$

(46)

$$\begin{aligned}{} & {} \frac{1}{p} = \frac{\alpha -1}{{{\tilde{q}}}} + \frac{1}{q}, \end{aligned}$$

(47)

$$\begin{aligned}{} & {} -\frac{d}{2} \left( \frac{1}{p} - \frac{1}{q}\right) - \frac{\sigma - s}{2}>-1, \quad -(\alpha -1)\beta >-1. \end{aligned}$$

(48)

Here, the above p, ${{\tilde{q}}}$ and $\sigma$ exist if (6) and (40) hold. We use Proposition 3.1 with $(q_1,r_1,s_1) = (p, \infty , \sigma )$ and $(q_2,r_2,s_2) = (q,r,s)$ to obtain

$$\begin{aligned} \Vert III(t)\Vert _{L^{q,r}_{s}} \le C \int _0^t (t-\tau ) ^{-\frac{d}{2} (\frac{1}{p} - \frac{1}{q}) - \frac{\sigma - s}{2}} \Vert |\cdot |^\gamma |e^{\tau \Delta } u_0|^{\alpha -1} |u_1(\tau ) - u_2(\tau )|\Vert _{L^{p,\infty }_{\sigma }}\, d\tau , \end{aligned}$$

where (46) is required. Moreover, it follows from Lemma 2.3 with $(q,r) = (p,\infty )$, $(q_1,r_1)=(\frac{{{\tilde{q}}}}{\alpha -1}, \infty )$ and $(q_2,r_2)=(q,\infty )$ that

$$\begin{aligned} \Vert |\cdot |^\gamma |e^{\tau \Delta } u_0|^{\alpha -1} |u_1(\tau ) - u_2(\tau )|\Vert _{L^{p,\infty }_{ \sigma }} \le C \Vert e^{\tau \Delta } u_0\Vert _{L^{{\tilde{q}},\infty }_{s}}^{\alpha -1}\Vert u_1(\tau ) - u_2(\tau )\Vert _{L_{s}^{{q},\infty }}, \end{aligned}$$

where (47) is required. Combining the above two estimates, and using the equality

$$\begin{aligned} -\frac{d}{2} \left( \frac{1}{p} - \frac{1}{q}\right) - \frac{\sigma - s}{2} -(\alpha -1)\beta + 1 = \delta \end{aligned}$$

which is a combination of (38), (41) and (47), we have

$$\begin{aligned} \begin{aligned} \Vert III(t)\Vert _{L^{q,r}_{s}}&\le Ct^{-\frac{d}{2} (\frac{1}{p} - \frac{1}{q}) - \frac{\sigma - s}{2} -(\alpha -1)\beta + 1} \left( \int _0^1 (1-\tau )^{-\frac{d}{2} (\frac{1}{p} - \frac{1}{q}) - \frac{\sigma - s}{2}} \tau ^{-(\alpha -1)\beta }d\tau \right) \\&\quad \times \left( \sup _{0<\tau<t}\tau ^{ \beta } \Vert e^{\tau \Delta } u_0\Vert _{L_{{{\tilde{s}}}}^{{\tilde{q}},\infty }}^{\alpha -1}\right) \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,r}_{s})}\\&\le C t^{\delta } \left( \sup _{0<\tau <t}\tau ^{ \beta }\Vert e^{\tau \Delta } u_0\Vert _{L_{s}^{{\tilde{q}},\infty }}^{\alpha -1}\right) \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,r}_{s})}, \end{aligned} \end{aligned}$$

(49)

where (48) is required. Hence, summarizing (43)–(45) and (49), we obtain (42). Therefore, the assertion (i) is proved.

Next, we prove the assertion (ii). Let $T \in (0,\infty ]$ and $u_1, u_2 \in L^\infty (0,T; L^{q,\infty }_{s}({\mathbb {R}}^d))$. In this case, the parameters $q,s,\sigma$ satisfy

$$\begin{aligned} 0< \frac{s}{d} + \frac{1}{q}< \frac{\sigma }{d} + \frac{\alpha }{q} < 1, \quad s \le \sigma \quad \text {and}\quad \quad \frac{d}{2} \left( \frac{\alpha }{q} - \frac{1}{q} \right) + \frac{\sigma -s}{2} = 1. \end{aligned}$$

(50)

We use Proposition 3.3 with the non-endpoint case as $(q_1,r_1,s_1) = (\frac{q}{\alpha }, \infty , \sigma )$ and $(q_2,s_2) = (q,s)$ to obtain

$$\begin{aligned} \begin{aligned} \Vert I(t)\Vert _{L^{q,\infty }_{s}}&\le C \sup _{0<\tau <t} \Vert |\cdot |^\gamma |u_1(\tau )-e^{\tau \Delta } u_0|^{\alpha -1}|u_1(\tau ) - u_2(\tau )| \Vert _{L^{\frac{q}{\alpha },\infty }_{\sigma }} \, d\tau \\&\le C \Vert u_1-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}. \end{aligned} \end{aligned}$$

Similarly, we have

$$\begin{aligned} \Vert II(t)\Vert _{L^{q,\infty }_{s}} \le C \Vert u_2-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}. \end{aligned}$$

For the term III(t), we can proceed as (49) to obtain

$$\begin{aligned} \Vert III(t)\Vert _{L^{q,\infty }_{s}} \le C\left( \sup _{0<\tau <t}\tau ^{ \beta }\Vert e^{\tau \Delta } u_0\Vert _{L_{s}^{{\tilde{q}},\infty }}^{\alpha -1}\right) \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})} \end{aligned}$$

under the conditions (46)–(48). Therefore, the assertion (ii) is proved.

For the assertion (iii), the proof can be done in the same way as the above (ii), but it corresponds to the endpoint case $\frac{\sigma }{d} + \frac{\alpha ^*}{q} = 1$ in (50). For this, we use Proposition 3.3 with the endpoint case (34), which requires the stronger restriction on r, to obtain

$$\begin{aligned} \begin{aligned}&\Vert I(t)\Vert _{L^{q,\infty }_{s}} + \Vert II(t)\Vert _{L^{q,\infty }_{s}} \\&\quad \le C \max _{i=1,2}\Vert u_i-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,\alpha ^*-1}_{s})}^{\alpha ^*-1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}, \end{aligned} \end{aligned}$$

where the condition $r = \alpha ^*-1$ is required in the first norm of the right-hand side. The estimate for III(t) is the same as in (ii). Thus, (iii) is also proved. $\square$

4.2 Proofs of Theorems 1.2, 1.3, 1.6 and Proposition 1.4

To begin with, we prepare the following lemma.

Lemma 4.3

Let $d\in {\mathbb {N}}$, $1\le q, {{\tilde{q}}}\le \infty$, $0<r\le \infty$ and $s\in {\mathbb {R}}$, and let $\beta$ be given by (41). Then, given a compact set ${\mathcal {K}}$ of $L^{q,r}_{s}({\mathbb {R}}^d)$, there exists a function $\mu : (0,1)\rightarrow (0,\infty )$ such that

$$\begin{aligned} \lim _{t\rightarrow 0}\mu (t)=0 \end{aligned}$$

and

$$\begin{aligned} t^{\beta }\Vert e^{t\Delta }f\Vert _{L_{s}^{{{\tilde{q}}},\infty }}\le \mu (t) \end{aligned}$$

for any $t\in (0,1)$ and any $f\in {\mathcal {K}}$ (replace $L^{q, r}_{s}({\mathbb {R}}^d)$ by ${\mathcal {L}}^{q, r}_{s}({\mathbb {R}}^d)$ if $q=\infty$ or $r=\infty$).

The proof of this lemma can be done as in [10, Lemma 8, page 283] (see also [36]) and uses the density of $L^{q, r}_{s}({\mathbb {R}}^d) \cap L^\infty _0({\mathbb {R}}^d)$ in $L^{q, r}_{s}({\mathbb {R}}^d)$ or ${\mathcal {L}}^{q, r}_{s}({\mathbb {R}}^d)$ (see Remark 5 (d) and Lemma A.2 in Appendix A).

We are now in a position to prove the theorems.

Proof of Theorem 1.2

We give the proof only for the case (ii), since the proof of the case (i) is similar. Let $T>0$ and $u_1,u_2 \in L^\infty (0,T; L^{q,\alpha }_{s}({\mathbb {R}}^d))$ be mild solutions to (1) with initial data $u_1(0) = u_2(0)$. By Lemma 4.1 (ii), we have

$$\begin{aligned} \Vert u_1(t) - u_2(t)\Vert _{L^{q,\alpha }_{s}} \le C_0 t^{\delta } \max _{i=1,2} \Vert u_i\Vert _{L^\infty (0,t; L^{q,\alpha }_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\alpha }_s)} \end{aligned}$$

for any $t \in (0,T)$, where $\delta >0$ is given in (38). If we choose $t_0 \in (0,T)$ such that

$$\begin{aligned} C_0 t_0^{\delta } \max _{i=1,2} \Vert u_i\Vert _{L^\infty (0,T; L^{q,\alpha }_{s})}^{\alpha -1} < 1, \end{aligned}$$

then we can derive that $u_1=u_2$ on $[0,t_0]$. We can repeat this argument until we reach $t=T$, and hence, we arrive at $u_1=u_2$ on [0, T]. Thus, we conclude Theorem 1.2. $\square$

The proof of Proposition 1.4 is similar to that of Theorem 1.2, and we have only to use Lemma 4.2 (i) instead of Lemma 4.1 (ii).

Proof of Theorem 1.3

We give the proof only for the case (ii), since the proof of the case (i) is similar. Let $T>0$ and $u_1, u_2 \in C([0,T]; L^{q,\alpha ^*-1}_{s}({\mathbb {R}}^d))$ be mild solutions to (1) with initial data $u_1(0) = u_2(0)=u_0 \in L^{q,\alpha ^*-1}_{s}({\mathbb {R}}^d)$. By Lemma 4.2 (iii), we have

$$\begin{aligned} \begin{aligned} \Vert u_1(t) - u_2(t)\Vert _{L^{q,\infty }_{s}}&\le C \Big ( \max _{i=1,2}\Vert u_i-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,\alpha ^*-1}_{s})} \\&\quad + \sup _{0<\tau <t}\tau ^{ \beta }\Vert e^{\tau \Delta } u_0\Vert _{L_{s}^{{\tilde{q}},\infty }} \Big )^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})} \end{aligned} \end{aligned}$$

(51)

for any $t\in (0,T)$, where ${{\tilde{q}}} \in (q,\infty )$. Since $u_0 \in L^{q,\alpha ^*-1}_{s}({\mathbb {R}}^d)$, we see that

$$\begin{aligned} \begin{aligned} \Vert u_i-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,\alpha ^*-1}_{s})} \le \Vert u_i-u_0\Vert _{L^\infty (0,t; L^{q,\alpha ^*-1}_{s})} + \Vert u_0-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,\alpha ^*-1}_{s})} \end{aligned} \end{aligned}$$

for $i=1,2$. Since $u_1, u_2, e^{\tau \Delta } u_0 \in C([0,T]; L^{q,\alpha ^*-1}_{s}({\mathbb {R}}^d))$, the right-hand side converges to zero as $t\rightarrow 0$, and hence,

$$\begin{aligned} \lim _{t\rightarrow 0} \max _{i=1,2}\Vert u_i-e^{\tau \Delta } u_0\Vert _{L^\infty (0,t; L^{q,\alpha ^*-1}_{s})}=0. \end{aligned}$$

(52)

On the other hand, we deduce from Lemma 4.3 that

$$\begin{aligned} \lim _{t\rightarrow 0}\sup _{0<\tau <t}\tau ^{ \beta }\Vert e^{\tau \Delta } u_0\Vert _{L_{s}^{{\tilde{q}},\infty }}=0. \end{aligned}$$

(53)

Hence, by (51), (52) and (53), there exists $t_0 \in (0,T]$ such that $u_1=u_2$ on $[0,t_0]$. The extension of uniqueness to the whole interval [0, T] can be done by the continuity argument as in [48, Proof of Theorem 1.4]. Thus, we conclude Theorem 1.3. $\square$

Theorem 1.6 is similarly proved to Theorem 1.3, and so we omit the proof.

5 Non-uniqueness

In this section, we prove Theorem 1.5, i.e., non-uniqueness for (1) in the double critical case $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} = \frac{1}{Q_c}$ (i.e. $\alpha =\alpha ^*$). For this purpose, we shall show the existence of two kind of mild solutions (regular and singular) to (1) for arbitrary initial data $u_0 \in L^{q,r}_{s}({\mathbb {R}}^d)$. For convenience, we define

$$\begin{aligned} q^*(\gamma ):=\frac{d(\alpha ^*-1)}{2} = \frac{d(2+\gamma )}{2(d-2)}. \end{aligned}$$

Then we note that

$$\begin{aligned} q_c=Q_c = \frac{d}{d-2}=q^*(0). \end{aligned}$$

5.1 Existence of the regular solution

In this subsection, we prove the local in time existence of a mild solution u to (1) in $C([0,T]; L^{q,r}_{s}({\mathbb {R}}^d))$ with the auxiliary condition

$$\begin{aligned} \Vert u\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}:= \sup _{0<t<T} t^{\beta } \Vert u(t)\Vert _{L^{{{\tilde{q}}},\infty }_{s}} < \infty \end{aligned}$$

(54)

for ${{\tilde{q}}} > q$, where $\beta$ is given in (41). The goal of this subsection is to prove the following:

Proposition 5.1

Let $d\ge 3$, $\gamma >-2$, $\alpha = \alpha ^*$, $\alpha ^* \le q < \infty$, $0 < r \le \infty$, and $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} = \frac{1}{Q_c}$. Assume that ${{\tilde{q}}}$ satisfies

$$\begin{aligned} \max \left\{ 0, \frac{1}{q}-\frac{1}{q^*(0)}, \frac{1}{q}-\frac{2}{d\alpha ^*} \right\}< \frac{1}{{{\tilde{q}}}} < \frac{1}{q}. \end{aligned}$$

(55)

Then, for any $u_0 \in L^{q,r}_{s}({\mathbb {R}}^d)$, there exist a time $T=T(u_0)>0$ and a unique mild solution $u \in C([0,T]; L^{q,r}_{s}({\mathbb {R}}^d))$ to (1) with $u(0)=u_0$ satisfying (54) (replace $L^{q,r}_{s}({\mathbb {R}}^d)$ by ${\mathcal {L}}^{q,\infty }_{s}({\mathbb {R}}^d)$ if $r=\infty$).

The proof is based on the standard fixed point argument as in [13, Subsection 3.1]. We prepare the following estimates on the Duhamel term N(u).

Lemma 5.2

Let $T>0$, and let $d\ge 3$, $\gamma >-2$, $\alpha = \alpha ^*$, $\alpha ^* \le q < \infty$, $0 < r \le \infty$ and $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} = \frac{1}{Q_c}$.

• (i) Assume that ${{\tilde{q}}}$ satisfies (55). Then there exists a constant $C>0$ such that

  $$\begin{aligned} \Vert N(u_1) - N(u_2)\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} \le C \max _{i=1,2} \Vert u_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}^{\alpha -1} \Vert u_1-u_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} \end{aligned}$$

  for any $t\in (0,T)$ and any functions $u_1,u_2$ satisfying

  $$\begin{aligned} \Vert u_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}<\infty , \quad i=1,2. \end{aligned}$$

  (56)

• (ii) Assume that

  $$\begin{aligned} \max \left\{ 0, \frac{1}{q}-\frac{2}{\alpha ^*} \right\}< \frac{1}{{{\tilde{q}}}} < \frac{1}{q}. \end{aligned}$$

  (57)

  Then there exists a constant $C>0$ such that

  $$\begin{aligned} \Vert N(u_1) - N(u_2)\Vert _{L^\infty (0,T; L^{q,r}_s)} \le C \max _{i=1,2} \Vert u_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}^{\alpha -1} \Vert u_1-u_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} \end{aligned}$$

  for any $t\in (0,T)$ and any functions $u_1,u_2$ satisfying (56).

Remark 11

Note that (55) implies (57).

Proof

We first prove the assertion (i). We set $\sigma := \alpha s - \gamma$ and take

$$\begin{aligned}{} & {} 1< {{\tilde{q}}}, \frac{{{\tilde{q}}}}{\alpha }< \infty , \quad 0< \frac{s}{d} + \frac{1}{{{\tilde{q}}}}< \frac{\sigma }{d} + \frac{\alpha }{{{\tilde{q}}}} < 1, \quad s \le \sigma , \end{aligned}$$

(58)

$$\begin{aligned}{} & {} -\frac{d}{2} \left( \frac{\alpha }{{{\tilde{q}}}} - \frac{1}{{{\tilde{q}}}} \right) - \frac{\sigma -s}{2}>-1,\quad -\beta \alpha >-1. \end{aligned}$$

(59)

Here, there exists a ${{\tilde{q}}}$ as above if (55) holds. In a similar way to (39), we estimate

$$\begin{aligned} \begin{aligned}&\Vert N(u_1)(t) - N(u_2)(t)\Vert _{L^{{{\tilde{q}}},\infty }_s}\\&\quad \le C \left( \int _0^t (t-\tau )^{-\frac{d}{2} \left( \frac{\alpha }{{{\tilde{q}}}} - \frac{1}{{{\tilde{q}}}} \right) - \frac{\sigma -s}{2}} \tau ^{-\beta \alpha }\, d\tau \right) \max _{i=1,2} \Vert u_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}^{\alpha -1} \Vert u_1-u_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}\\&\quad \le C t^{-\beta } \max _{i=1,2} \Vert u_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}^{\alpha -1} \Vert u_1-u_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}, \end{aligned} \end{aligned}$$

where (58) and (59) are required in the first and second steps, respectively.

Similarly, we can prove the assertion (ii). In fact, taking

$$\begin{aligned}{} & {} 1< q, \frac{{{\tilde{q}}}}{\alpha }< \infty , \quad 0< \frac{s}{d} + \frac{1}{q}< \frac{\sigma }{d} + \frac{\alpha }{{{\tilde{q}}}} < 1, \quad s \le \sigma , \end{aligned}$$

(60)

$$\begin{aligned}{} & {} -\frac{d}{2} \left( \frac{\alpha }{{{\tilde{q}}}} - \frac{1}{q} \right) - \frac{\sigma -s}{2}>-1,\quad -\beta \alpha >-1, \end{aligned}$$

(61)

we estimate

$$\begin{aligned} \begin{aligned}&\Vert N(u_1)(t) - N(u_2)(t)\Vert _{L^{q,r}_s}\\&\quad \le C \left( \int _0^t (t-\tau )^{-\frac{d}{2} \left( \frac{\alpha }{{{\tilde{q}}}} - \frac{1}{q} \right) - \frac{\sigma -s}{2}} \tau ^{-\beta \alpha }\, d\tau \right) \max _{i=1,2} \Vert u_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}^{\alpha -1} \Vert u_1-u_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}\\&\quad \le C \max _{i=1,2} \Vert u_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}^{\alpha -1} \Vert u_1-u_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}, \end{aligned} \end{aligned}$$

where (60) and (61) are required in the first and second steps, respectively. Here, there exists a ${{\tilde{q}}}$ as above if (57) holds. Thus, the proof is finished. $\square$

Proof of Proposition 5.1

We give only a sketch of proof, as the proof is almost the same as in [13, Subsection 3.1]. Let $u_0 \in L^{q,r}_{s}({\mathbb {R}}^d)$, and let $\rho$ and M be positive constants such that

$$\begin{aligned} \rho + C_0M^{\alpha ^*} \le M \quad \text {and} \quad C_1 M^{\alpha ^*-1} <\frac{1}{2}, \end{aligned}$$

where $C_0$ and $C_1$ are positive constants given in (62) and (63) below. In addition, we take $T>0$ as

$$\begin{aligned} \Vert e^{t\Delta }u_0\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} \le \rho . \end{aligned}$$

Now, we define a nonempty complete metric space $X_M$ by

$$\begin{aligned} X_{M}:= \{ u \in {\mathcal {K}}^{{{\tilde{q}}}}(T)\,; \Vert u\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} \le M \} \end{aligned}$$

with a metric $d(u_1,u_2):= \Vert u_1 -u_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}$. Define a mapping $\Phi$ by

$$\begin{aligned} \Phi (u)(t):= e^{t\Delta }u_0 + N(u)(t) \end{aligned}$$

for $u \in X_M$. Then it follows from Lemma 5.2 (i) that

$$\begin{aligned} \begin{aligned}&\Vert \Phi (u)\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}\\&\le \Vert e^{t\Delta }u_0\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} + \Vert N(u)\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} \\&\le \Vert e^{t\Delta }u_0\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} + C_0\Vert u\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}^{\alpha ^*} \\&\le \rho + C_0M^{\alpha ^*} \le M, \end{aligned} \end{aligned}$$

(62)

and

$$\begin{aligned} \begin{aligned} d(u_1,u_2)&= \Vert N(u_1-u_2)\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} \\&\le C_1 \max _{i=1,2} \Vert u_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}^{\alpha ^*-1} \Vert u_1-u_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}\\&\le C_1 M^{\alpha ^*-1} d(u_1,u_2) \le \frac{1}{2} d(u_1,u_2). \end{aligned} \end{aligned}$$

(63)

for $u, u_1, u_2 \in X_M$. Hence, $\Phi$ is a contraction mapping from $X_M$ into itself. Thus, Banach’s fixed point theorem ensures the existence of a unique fixed point $u \in X_M$ of $\Phi$. Finally, $u \in C([0,T]; L^{q,r}_{s}({\mathbb {R}}^d))$ follows from Lemma 5.2 (ii), Lemma A.4 and the well-known argument as in [40, 51] for instance. The proof of Proposition 5.1 is finished. $\square$

5.2 Existence of singular solution

The mild solution u obtained in Sect. 5.1 is a bounded solution (see [5, Remark 1.1 and Proposition 3.2] and also [52, the remark after Definition 2.1]). In this subsection, we find a singular mild solution v to (1) for any initial data $u_0 \in L^{q,r}_s({\mathbb {R}}^d)$. Here, the singular mild solution means that $v(t) \not \in L^{{{\tilde{q}}},\infty }_s({\mathbb {R}}^d)$ for any $t \in [0,T]$ and for any ${{\tilde{q}}}$ satisfying (55) (in particular, this solution has a singularity at $x=0$). The goal of this subsection is to prove the following:

Theorem 5.3

Let $d\ge 3$, $\gamma >-2$, $\alpha = \alpha ^*$, $\alpha ^*\le q < \infty$, $\alpha ^*-1< r \le \infty$, and $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} = \frac{1}{Q_c}$. Then, for any $u_0 \in L^{q,r}_{s}({\mathbb {R}}^d)$, there exist $T=T(u_0)>0$ and a mild solution $v \in C([0,T]; L^{q, r}_{s}({\mathbb {R}}^d))$ to (1) with $v(0)=u_0$ such that $v \not \in L^{{{\tilde{q}}},\infty }_s({\mathbb {R}}^d)$ for any ${{\tilde{q}}}$ satisfying (55) and

$$\begin{aligned} v(t) - e^{t\Delta } u_0 \in L^{q, r}_{s}({\mathbb {R}}^d) \setminus L^{q, \alpha ^*-1}_{s}({\mathbb {R}}^d) \text { for any }r>\alpha ^*-1 \end{aligned}$$

(64)

for any $t\in (0,T]$ (replace $L^{q,r}_{s}({\mathbb {R}}^d)$ by ${\mathcal {L}}^{q,\infty }_{s}({\mathbb {R}}^d)$ if $r=\infty$).

The proof is based on the argument in [36, 50]. In order to construct the singular solution v, we use a positive, radially symmetric and singular stationary solution of

$$\begin{aligned} \Delta U + |x|^{\gamma } U^{\frac{d+\gamma }{d-2}} = 0\quad \text {in } B\setminus \{0\}, \quad U>0, \end{aligned}$$

(65)

where $d\ge 3$, $\gamma >-2$ and $B:= \{x \in {\mathbb {R}}^d \,;\, |x| < 1\}$. We have the results on the existence of the singular stationary solution and the sharp bound of its behavior at $x=0$.

Theorem 5.4

Let $d\ge 3$ and $\gamma >-2$. The the following assertions hold:

• (i) The equation (65) has a positive, radial, and singular solution at $x=0$, where the singular solution means that it diverges at $x=0$.

• (ii) Let $U \in C^2 (B{\setminus } \{0\})$ be a positive radial solution to (65). Then, U has either a removable singularity at $|x|=0$ or a singularity at $|x|=0$ as

  $$\begin{aligned} \lim _{x\rightarrow 0} |x|^{d-2} |\log |x||^{\frac{d-2}{\gamma +2}}U(x) = \left( \frac{(d-2)^2}{2+\gamma } \right) ^{\frac{d-2}{2+\gamma }}. \end{aligned}$$

  (66)

Remark 12

The constant (66) appears in [9, Theorem 2.1] for $-2<\gamma <2$, and it gives the precise value to that in [2, Theorem A] and hence to that in [48, Remark 6.2].

The proofs of (i) and (ii) can be found in [1, Example 1] and [15, Theorem 1.1 (ii)], respectively. For completeness, we give the proof of (ii) in Appendix B. Therefore, we denote by $U_0$ the singular stationary solution with

$$\begin{aligned} U_0(x) \sim |x|^{-(d-2)} |\log |x||^{-\frac{d-2}{\gamma +2}} =|x|^{-(d-2)} |\log |x||^{-\frac{1}{\alpha ^* -1}} \end{aligned}$$

(67)

near $x=0$. Then we note from Remark 5 (g) that

$$\begin{aligned} U_0 \in L^{q,r}_{s}({\mathbb {R}}^d) \setminus L^{q,\alpha ^*-1}_{s}(B) \end{aligned}$$

(68)

for any $r > \alpha ^*-1$.

We extend $U_0$ to a function $V_0$ on ${\mathbb {R}}^d$ as follows.

Proposition 5.5

Let $d,\gamma ,\alpha ,q,s$ be as in Theorem 5.3. Then there exists a function $V_0 \ge 0$ on ${\mathbb {R}}^d{\setminus } \{0\}$ with compact support such that

$$\begin{aligned} V_0(x) \sim |x|^{-(d-2)} |\log |x||^{-\frac{1}{\alpha ^* -1}} \end{aligned}$$

in a neighborhood of $x=0$, and

$$\begin{aligned} R:= \Delta V_0 + |x|^{\gamma } V_0^{\alpha ^*} \text { is of }C^1\text { with compact support}. \end{aligned}$$

(69)

Moreover,

The proof of Proposition 5.5 is the same as in [50, Theorem 0.7] (see also [48, Proposition 6.1]).

To prove Theorem 5.3, we find a singular mild solution v to (1) of the form

$$\begin{aligned} v(t) = w(t) + V_0. \end{aligned}$$

(72)

Here, $w=w(t)$ is a (regular) solution to the perturbed problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle w(t) = e^{t\Delta }w_0 + {\mathcal {N}}(w)(t) + \int _0^t e^{(t-\tau )\Delta } R\, d\tau ,\\ w(0) =w_0:= u_0 - V_0, \end{array}\right. } \end{aligned}$$

(73)

where

$$\begin{aligned} {\mathcal {N}}(w)(t):= \int _0^t e^{(t-\tau )\Delta } \left( |x|^\gamma |w(\tau ) + V_0|^{\alpha ^*-1} (w(\tau ) + V_0) - |x|^\gamma V_0^{\alpha ^*} \right) \, d\tau . \end{aligned}$$

More precisely, we have the following:

Lemma 5.6

Let $d\ge 3$, $\gamma >-2$, $\alpha = \alpha ^*$, $\alpha ^* \le q < \infty$, $0 < r \le \infty$, and $\frac{s}{d} + \frac{1}{q} = \frac{1}{q_c} = \frac{1}{Q_c}$. Then, for any $w_0 \in L^{q,r}_{s}({\mathbb {R}}^d)$, there exist $T>0$ and a unique solution $w \in C([0,T]; L^{q,r}_{s}({\mathbb {R}}^d))$ to (73) with $w(0)=w_0$ such that it satisfies (54) for any ${{\tilde{q}}}$ satisfying (55) (replace $L^{q,r}_{s}({\mathbb {R}}^d)$ by ${\mathcal {L}}^{q,\infty }_{s}({\mathbb {R}}^d)$ if $r=\infty$).

The proof of this lemma is based on the fixed point argument as in [36]. Hence, we need to show some estimates for the term ${\mathcal {N}}(w)$. To prove the estimates, we use the following decomposition of $V_0$. By the property (71) of $V_0$ and Lemma A.2 (i), for any $\varepsilon >0$, there exist functions $h\in L^{q,\infty }_{s}({\mathbb {R}}^d) \cap L^\infty _0({\mathbb {R}}^d)$ and ${\overline{V}}_0 \in L^{q,\infty }_{s}({\mathbb {R}}^d)$ such that

$$\begin{aligned} V_0=h+{\overline{V}}_0,\quad \Vert {\overline{V}}_0\Vert _{L^{q,\infty }_{s}}<\varepsilon . \end{aligned}$$

(74)

Then we have the following estimates for ${\mathcal {N}}(w)$.

Lemma 5.7

Let $d,\gamma ,\alpha ,q,r,s$ be as in Lemma 5.6, $\gamma _+:= \max \{0,\gamma \}$ and $\gamma _-:= -\min \{0,\gamma \}$. Assume ${{\tilde{q}}}$ satisfies (55). Then there exists a constant $C>0$ such that

$$\begin{aligned} \begin{aligned}&\Vert {\mathcal {N}}(w_1) - {\mathcal {N}}(w_2)\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)}\\&\quad \le C \left( \max _{i=1,2} \Vert w_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)}^{\alpha ^*-1} + \Vert {\overline{V}}_0\Vert _{L^{q,\infty }_{s}}^{\alpha ^*-1} + t^{1-\frac{\gamma _-}{2}} \Vert |\cdot |^{\gamma _+}|h|^{\alpha ^*-1}\Vert _{L^\infty } \right) \\&\qquad \times \Vert w_1-w_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)} \end{aligned} \end{aligned}$$

(75)

and

$$\begin{aligned} \begin{aligned}&\Vert {\mathcal {N}}(w_1) - {\mathcal {N}}(w_2)\Vert _{L^\infty (0,t; L^{q,r}_s)}\\&\quad \le C \left( \max _{i=1,2} \Vert w_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)}^{\alpha ^*-1} + \Vert {\overline{V}}_0\Vert _{L^{q,\infty }_{s}}^{\alpha ^*-1} + \Vert h\Vert _{L^{{{\tilde{q}}},\infty }_s}^{\alpha ^*-1} \right) \Vert w_1-w_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)} \end{aligned} \end{aligned}$$

(76)

for any two functions $w_1,w_2$ satisfying (54) and for any $t>0$.

Proof

We write

$$\begin{aligned} \begin{aligned} {\mathcal {N}}(w_1)(t)-{\mathcal {N}}(w_2)(t)&= \int _0^t e^{(t-\tau )\Delta }\big [|\cdot |^{\gamma }\left| w_1(\tau )+V_0\right| ^{\alpha ^*-1}\left( w_1(\tau )+V_0\right) \\&\quad -|\cdot |^{\gamma }\left| w_2(\tau )+V_0\right| ^{\alpha ^*-1}\left( w_2(\tau )+V_0\right) \big ]d\tau . \end{aligned} \end{aligned}$$

By the decomposition (74) together with the inequality

$$\begin{aligned} \begin{aligned}&\left| |x+y|^{\alpha ^*-1}(x+y) - |x'+y|^{\alpha ^*-1}(x'+y)\right| \\&\quad \le C|x-x'|\left( |x|^{\alpha ^*-1}+|x'|^{\alpha ^*-1}+|y|^{\alpha ^*-1}\right) \end{aligned} \end{aligned}$$

for $x,x',y \in {\mathbb {R}}$, we have

$$\begin{aligned} \begin{aligned} |{\mathcal {N}}(w_1)(t)-{\mathcal {N}}(w_2)(t)|&\le C\int _0^te^{(t-\tau )\Delta }[|\cdot |^{\gamma }|w_1(\tau )|^{\alpha ^*-1}|w_1(\tau )-w_2(\tau )|]\, d\tau \\&\quad +C\int _0^te^{(t-\tau )\Delta }[|\cdot |^{\gamma }|w_2(\tau )|^{\alpha ^*-1}|w_1(\tau )-w_2(\tau )|]\, d\tau \\&\quad +C\int _0^te^{(t-\tau )\Delta }[|\cdot |^{\gamma }|h|^{\alpha ^*-1}|w_1(\tau )-w_2(\tau )|]\, d\tau \\&\quad +C\int _0^te^{(t-\tau )\Delta }[|\cdot |^{\gamma }|{\overline{V}}_0|^{\alpha ^*-1}|w_1(\tau )-w_2(\tau )|]\, d\tau \\&=: I(t) + II(t) + III(t) + IV(t). \end{aligned} \end{aligned}$$

(77)

First, we prove the estimate (75). In the same way as in the proof of Lemma 4.2, the norms of the terms I(t) and II(t) can be estimated as

$$\begin{aligned} \Vert I\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)} + \Vert II\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)} \le C \max _{i=1,2} \Vert w_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)}^{\alpha ^*-1} \Vert w_1-w_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)}. \end{aligned}$$

(78)

As to the term III(t), we use Proposition 3.1 with $(q_1,r_1,s_1) = ({{\tilde{q}}}, \infty , s + \gamma _-)$ and $(q_2,r_2,s_2) = ({{\tilde{q}}},\infty ,s)$ to obtain

$$\begin{aligned} \begin{aligned} \Vert III(t)\Vert _{L^{{{\tilde{q}}},\infty }_{s}}&\le C \int _0^t (t-\tau )^{- \frac{\gamma _-}{2}} \Vert |\cdot |^{\gamma }|h|^{\alpha ^*-1}|w_1(\tau )-w_2(\tau )| \Vert _{L^{{{\tilde{q}}}}_{s + \gamma _-}}\, d\tau \\&= C \int _0^t (t-\tau )^{- \frac{\gamma _-}{2}} \Vert |\cdot |^{\gamma _+}|h|^{\alpha ^*-1}|w_1(\tau )-w_2(\tau )| \Vert _{L^{{{\tilde{q}}},\infty }_{s}}\, d\tau \\&\le C \Vert |\cdot |^{\gamma _+}|h|^{\alpha ^*-1}\Vert _{L^\infty } \int _0^t (t-\tau )^{- \frac{\gamma _-}{2}} \Vert w_1(\tau )-w_2(\tau ) \Vert _{L^{{{\tilde{q}}},\infty }_{s}}\, d\tau \\&\le C \Vert |\cdot |^{\gamma _+}|h|^{\alpha ^*-1}\Vert _{L^\infty } \left( \int _0^t (t-\tau )^{- \frac{\gamma _-}{2}} \tau ^{-\beta }\, d\tau \right) \Vert w_1-w_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)}\\&\le C t^{1-\frac{\gamma _-}{2}-\beta } \Vert |\cdot |^{\gamma _+}|h|^{\alpha ^*-1}\Vert _{L^\infty } \Vert w_1-w_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)}, \end{aligned}\nonumber \\ \end{aligned}$$

(79)

where we required that

$$\begin{aligned} 0< \frac{s}{d} + \frac{1}{{{\tilde{q}}}} \le \frac{s+\gamma _-}{d} + \frac{1}{{{\tilde{q}}}} <1\quad \text {and}\quad s\le s+\gamma _-. \end{aligned}$$

Here, thanks to (55) and $\gamma _-\in [0,2)$, the above conditions are satisfied.

As to the term IV(t), thanks to (55), we can take $\sigma := \alpha ^* s - \gamma$ and

$$\begin{aligned} 0< \frac{s}{d} + \frac{1}{{{\tilde{q}}}} \le \frac{\sigma }{d} + \frac{1}{p} < 1, \quad s\le \sigma ,\quad \frac{1}{p} = \frac{\alpha ^*-1}{q}+\frac{1}{{{\tilde{q}}}}. \end{aligned}$$

Then we use Proposition 3.3 with $(q_1,r_1,s_1) = (p, \infty , \sigma )$ and $(q_2,r_2,s_2) = ({{\tilde{q}}},\infty ,s)$ to obtain

$$\begin{aligned} \begin{aligned} \Vert IV(t)\Vert _{L^{{{\tilde{q}}},\infty }_{s}}&\le C \sup _{0<\tau<t} \Vert |\cdot |^{\gamma }|{\overline{V}}_0|^{\alpha ^*-1}|w_1(\tau )-w_2(\tau )|\Vert _{L^{p,\infty }_{\sigma }}\\&\le C \sup _{0<\tau <t} \Vert (|\cdot |^{s}|{\overline{V}}_0|)^{\alpha ^*-1}|\cdot |^{s}||w_1(\tau )-w_2(\tau )|\Vert _{L^{p,\infty }}\\&\le C t^{-\beta } \Vert {\overline{V}}_0\Vert _{L^{q,\infty }_{s}}^{\alpha ^*-1} \Vert w_1-w_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(t)}. \end{aligned} \nonumber \\ \end{aligned}$$

(80)

By combining (77), (78), (79) and (80), we obtain (75).

For the estimate (76), it is enough to use the interpolation argument with Lemma  5.2 such as the proof of Lemma 5.7 (ii) to deal with all $r \in (0,\infty ]$. The terms I(t), II(t) and IV(t) can be be estimated in a similar way to (75). The estimate for III(t) is proved in a similar way to Lemma 5.2 (ii). The proof of Lemma 5.7 is finished. $\square$

Proof of Lemma 5.6

Let $w_0 \in L^{q,r}_{s}({\mathbb {R}}^d)$, and let $\rho , \delta , \varepsilon$ and M be positive constants such that

$$\begin{aligned} \rho + C_1 (M^{\alpha ^*-1} + \varepsilon ^{\alpha ^*-1} + \delta ) M \le M \quad \text {and} \quad C_2 \left( M^{\alpha ^*-1} + \varepsilon ^{\alpha ^*-1} + \delta \right) <\frac{1}{2}, \end{aligned}$$

where $C_1$ and $C_2$ are positive constants given in (81) and (82) below. In addition, we take $T>0$ as

$$\begin{aligned} \Vert e^{t\Delta }w_0\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} + C_0 T \Vert R\Vert _{L^{q,\infty }_{s}} \le \rho \end{aligned}$$

and

$$\begin{aligned} T^{1-\frac{\gamma _-}{2}} \Vert |\cdot |^{\gamma _+}|h|^{\alpha ^*-1}\Vert _{L^\infty } \le \delta , \end{aligned}$$

where $C_0$ is a positive constant given in (81) below. Now, we define a nonempty complete metric space $X_M$ by

$$\begin{aligned} X_{M}:= \{ w \in {\mathcal {K}}^{{{\tilde{q}}}}(T)\,; \Vert w\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} \le M \} \end{aligned}$$

with a metric $d(u_1,u_2):= \Vert u_1 -u_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}$. Define a mapping $\Phi$ by

$$\begin{aligned} \Phi (w)(t):= e^{t\Delta }w_0 +{\mathcal {N}}(w)(t) + \int _0^t e^{(t-\tau )\Delta } R\, d\tau \end{aligned}$$

for $w \in X_M$. By (75) in Lemma 5.7 and (74), it follows that

$$\begin{aligned} \begin{aligned}&\Vert \Phi (w)\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}\\&\le \Vert e^{t\Delta }w_0\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)} + C_0 T \Vert R\Vert _{L^{q,\infty }_{s}}\\&\quad + C_1 \left( \Vert w\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}^{\alpha ^*-1} + \Vert {\overline{V}}_0\Vert _{L^{q,\infty }_{s}}^{\alpha ^*-1} + T^{1-\frac{\gamma _-}{2}} \Vert |\cdot |^{\gamma _+}|h|^{\alpha ^*-1}\Vert _{L^\infty } \right) \Vert w\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}\\&\le \rho + C_1 (M^{\alpha ^*-1} + \varepsilon ^{\alpha ^*-1} + \delta ) M \le M \end{aligned} \end{aligned}$$

(81)

and

$$\begin{aligned} \begin{aligned}&d(\Phi (w_1), \Phi (w_2)) \le \Vert {\mathcal {N}}(w_1)- {\mathcal {N}}(w_2)\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}\\&\quad \le C_2 \left( \max _{i=1,2} \Vert w_i\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}^{\alpha ^*-1} + \Vert {\overline{V}}_0\Vert _{L^{q,\infty }_{s}}^{\alpha ^*-1} + T^{1-\frac{\gamma _-}{2}} \Vert |\cdot |^{\gamma _+}|h|^{\alpha ^*-1}\Vert _{L^\infty } \right) \\&\qquad \times \Vert w_1-w_2\Vert _{{\mathcal {K}}^{{{\tilde{q}}}}(T)}\\&\quad \le C_2 \left( M^{\alpha ^*-1} + \varepsilon ^{\alpha ^*-1} + \delta \right) d(w_1,w_2) \le \frac{1}{2}d(w_1,w_2) \end{aligned} \end{aligned}$$

(82)

for $w,w_1,w_2 \in X_M$. Hence, $\Phi$ is a contraction mapping from $X_M$ into itself. Thus, Banach’s fixed point theorem ensures the existence of a unique fixed point $w \in X_M$ of $\Phi$. Finally, $w \in C([0,T]; L^{q,r}_{s}({\mathbb {R}}^d))$ follows from (76) in Lemma 5.7, Lemma A.4 and the well-known argument as in [40, 51] for instance. The proof of Lemma 5.6 is finished. $\square$

Proof of Theorem 5.3

The existence part of Theorem 5.3 immediately follows from a combination of Lemma 5.6 with Proposition 5.5 and (72). The remaining part, i.e., the properties (64) of v, can be proved in a similar way to the proof of [48, Proposition 8.2]. In fact, we decompose the Duhamel term $v(t) - e^{t\Delta }u_0$ into the following three terms:

$$\begin{aligned} v(t) - e^{t\Delta }u_0 = (w(t) - e^{t\Delta }w_0) - e^{t\Delta }V_0 + V_0. \end{aligned}$$

The first term $w(t) - e^{t\Delta }w_0$ can be rewritten as

$$\begin{aligned} w(t) - e^{t\Delta }w_0 = {\mathcal {N}}(w)(t) + \int _0^t e^{(t-\tau )\Delta } R\, d\tau . \end{aligned}$$

We see from Lemma 5.7 and the property (69) of R that both terms in the right-hand side belong to $L_{s}^{q,{{\tilde{r}}}}({\mathbb {R}}^d)$ for any ${{\tilde{r}}}>0$ and any $t \in (0,T]$, and hence, $w(t) - e^{t\Delta }w_0$ also belongs to $L_{s}^{q,{{\tilde{r}}}}({\mathbb {R}}^d)$. As to the second term $e^{t\Delta } V_0$, we estimate

$$\begin{aligned} \Vert e^{t\Delta } V_0\Vert _{L_{s}^{q,{{\tilde{r}}}}} \le {\left\{ \begin{array}{ll} \displaystyle C t^{-\frac{d}{2}(1- \frac{1}{q}) + \frac{s}{2}} \Vert V_0\Vert _{L^1} = C t^{-1} \Vert V_0\Vert _{L^1} \quad &{}\text {if }s\le 0,\\ \displaystyle C t^{-\frac{d}{2}(1- \frac{1}{q})} \Vert V_0\Vert _{L^1_{s}} = C_{V_0} t^{-\frac{d}{2}(1- \frac{1}{q})} \Vert V_0\Vert _{L^1} \quad &{}\text {if }s > 0, \end{array}\right. } \end{aligned}$$

where we used Propositions 3.1 and $V_0 \in L^1({\mathbb {R}}^d)$ with compact support in Proposition 5.5. Hence, $e^{t\Delta }V_0 \in L_{s}^{q,{{\tilde{r}}}}({\mathbb {R}}^d)$ is also shown for any ${{\tilde{r}}}>0$ and $t \in (0,T]$. In contrast, the third term $V_0$ satisfies $V_0 \not \in L_{s}^{q,\alpha ^*-1}({\mathbb {R}}^d)$ and $V_0 \in L_{s}^{q,r}({\mathbb {R}}^d)$ for any $r>\alpha ^*-1$ by Proposition 5.5. Therefore, (64) is proved for any $t \in (0,T]$. Thus, Theorem 5.3 is proved. $\square$

Proof of Theorem 1.5

The proof is a combination of Proposition 5.1 and Theorem 5.3. In fact, by these results, there exist a regular mild solution u and singular mild solution v to (1) with the same initial data $u_0$. When $r=\infty$, the above arguments are also valid if $L^{q,r}_{s}({\mathbb {R}}^d)$ is replaced by ${\mathcal {L}}^{q,\infty }_{s}({\mathbb {R}}^d)$. $\square$

6 Scale-supercritical case

In this section we discuss the scale-supercritical case. We use the self-similar solution of (1) to show the existence of a non-trivial mild solution of (1) with initial data 0. More precisely, we have the following:

Proposition 6.1

Let $d\ge 3$, $\gamma >-2$, $\alpha >\alpha _F$, $1< q \le \infty$, $1\le r\le \infty$ and $s\in {\mathbb {R}}$ be such that

$$\begin{aligned} \frac{1}{q_c}< \frac{s}{d} + \frac{1}{q} < 1. \end{aligned}$$

Assume that there exists a solution W of

$$\begin{aligned} \Delta W+{1\over 2}x\cdot \nabla W+{2+\gamma \over 2(\alpha -1)}W+|x|^{\gamma }|W|^{\alpha -1} W=0,\quad x \in {\mathbb {R}}^d\setminus \{0\} \end{aligned}$$

(83)

such that

• (i) $W>0$ and $W\in C({\mathbb {R}}^d)\cap C^2({\mathbb {R}}^d{\setminus }\{0\}),$

• (ii) $\displaystyle \lim _{|x|\rightarrow 0}|x||\nabla W|=0,$

• (iii) $\displaystyle \lim _{|x|\rightarrow \infty }|x|^mW(x)=0$ and $\displaystyle \lim _{|x|\rightarrow \infty }|x|^m|\nabla W(x)|=0$ for all $m>0$.

Let $\Psi (t,x)=t^{-{2+\gamma \over 2(\alpha -1)}}W(x/\sqrt{t})$ be the positive self-similar solution of (1). Then $\Psi \in C([0,\infty ); L^{q,r}_s({\mathbb {R}}^d))$ satisfies the equation

$$\begin{aligned} \Psi (t)= \int _0^t e^{(t-\tau )\Delta }\left( |\cdot |^{\gamma }|\Psi (\tau )|^\alpha \Psi (\tau )\right) d\tau \end{aligned}$$

for any $t \in (0,\infty )$. In particular, $\Psi$ is a non-trivial mild solution to (1) with initial data 0 in $C([0,\infty ); L^{q,r}_s({\mathbb {R}}^d))$.

Proof

By the assumptions (i)–(iii) on W,  it follows that

$$\begin{aligned} {1\over 2}x\cdot \nabla W+{2+\gamma \over 2(\alpha -1)}W+|x|^{\gamma }|W|^{\alpha -1} W\in L^1({\mathbb {R}}^d). \end{aligned}$$

Then W satisfies the Eq. (83) in ${\mathcal {D}}'({\mathbb {R}}^d)$ and $\Psi (t,x)=t^{-{2+\gamma \over 2(\alpha -1)}}W(x/\sqrt{t})$ satisfies the equation (1) in ${\mathcal {D}}'((0,\infty )\times {\mathbb {R}}^d)$. Here, ${\mathcal {D}}'(X)$ is the space of distributions on an open set X. Hence,

$$\begin{aligned} \Psi (t)=e^{(t-\varepsilon )\Delta }\Psi (\varepsilon )+\displaystyle \int _\varepsilon ^te^{(t-\tau )\Delta }\left( |\cdot |^{\gamma }|\Psi (\tau )|^{\alpha -1} \Psi (\tau )\right) d\tau \end{aligned}$$

for $0<\varepsilon <t$ in the sense of distributions. It is clear that

$$\begin{aligned} \Vert \Psi (t)\Vert _{L^{q,r}_{s}}=t^{\frac{d}{2} ((\frac{s}{d} + \frac{1}{q}) - \frac{1}{q_c})}\Vert W\Vert _{L^{q,r}_{ s}} <\infty ,\quad t>0, \end{aligned}$$

where $0<\frac{s}{d}+\frac{1}{q} <1$. Then

$$\begin{aligned} \lim _{t\rightarrow 0}\Vert \Psi (t)\Vert _{L^{q,r}_s}=0 \quad \text {for }\frac{1}{q_c}< \frac{s}{d} + \frac{1}{q} < 1. \end{aligned}$$

(84)

Finally, we prove that the integral

$$\begin{aligned} \int _0^t e^{(t-\tau )\Delta }(|\cdot |^{\gamma }|\Psi (\tau )|^\alpha \Psi (\tau ))\, d\tau \end{aligned}$$

(85)

converges absolutely in $L^{q,r}_s({\mathbb {R}}^d)$. By Proposition 3.1, we have

$$\begin{aligned}&\left\| e^{(t-\tau )\Delta }\left( |\cdot |^{\gamma }|\Psi (\tau )|^{\alpha -1} \Psi (\tau )\right) \right\| _{L^{q,r}_s}\\&\quad \le C(t-\tau )^{- {d\over 2}({\alpha \over {\tilde{q}}}-{1\over q})-{\alpha {\tilde{s}}-\gamma -s\over 2}}\Vert \Psi (\tau )\Vert _{L^{{\tilde{q}},r}_{{\tilde{s}}}}^{\alpha }\\&\quad = C(t-\tau )^{- {d\over 2}({\alpha \over {\tilde{q}}}-{1\over q})-{\alpha {\tilde{s}}-\gamma -s\over 2}}\tau ^{\alpha ({{d\over 2{\tilde{q}}}+{{\tilde{s}}\over 2}-{2+\gamma \over 2(\alpha -1)}})}\Vert W\Vert _{L^{{\tilde{q}},r}_{{\tilde{s}}}}^{\alpha }, \end{aligned}$$

where we require that

$$\begin{aligned} \alpha< {{\tilde{q}}}<\infty ,\quad 0< \frac{s}{d}+\frac{1}{q}< \frac{\alpha {{\tilde{s}}}-\gamma }{d} + \frac{\alpha }{{{\tilde{q}}}} < 1, \quad s \le \alpha {{\tilde{s}}}-\gamma . \end{aligned}$$

(86)

If $\alpha$, q, s, ${{\tilde{q}}}$ and ${{\tilde{s}}}$ satisfy

$$\begin{aligned} {d\over 2}\left( {\alpha \over {\tilde{q}}}-{1\over q}\right) +{\alpha {\tilde{s}}-\gamma -s\over 2}<1,\quad \alpha \left( {2+\gamma \over 2(\alpha -1)}-{d\over 2}\left( {{1\over {\tilde{q}}}+{{\tilde{s}}\over d}}\right) \right) <1, \end{aligned}$$

(87)

then (85) converges absolutely in $L^{q,r}_s({\mathbb {R}}^d)$. To check these conditions, let us choose ${\tilde{q}}$ and ${\tilde{s}}$ such that

$$\begin{aligned} {s+\gamma \over \alpha }\le & {} {\tilde{s}},\quad 0<{\alpha \over {\tilde{q}}}<1,\quad {\alpha \over {\tilde{q}}}+{\alpha {\tilde{s}}\over d}<{\gamma +d\over d}, \end{aligned}$$

(88)

$$\begin{aligned} {1\over q}+{\gamma +s\over d}< & {} {\alpha \over {\tilde{q}}}+{\alpha {\tilde{s}}\over d}<{2\over d}+{1\over q}+{\gamma +s\over d}. \end{aligned}$$

(89)

It is obvious that under the assumptions in Proposition 6.1, it is possible to take ${\tilde{q}},$ ${\tilde{s}}$ satisfying (88) and (89). We now show that (86) and (87) hold if (84), (88) and (89) are satisfied. Indeed, (86) is already in (88) and the first inequality in (89). For (87), we have

$$\begin{aligned} \begin{aligned} {d\over 2}\left( {\alpha \over {\tilde{q}}}-{1\over q}\right) +{\alpha {\tilde{s}}-\gamma -s\over 2}&= {d\alpha \over 2{\tilde{q}}}-{d\over 2q}+{\alpha {\tilde{s}}-\gamma -s\over 2}\\&= {d\alpha \over 2{\tilde{q}}}+{\alpha {\tilde{s}}\over 2}-{d\over 2q}-{\gamma +s\over 2}\\&<1+ {d\over 2q}+{\gamma +s\over 2}-{d\over 2q}-{\gamma +s\over 2}\\&= 1, \\ \alpha \left( {2+\gamma \over 2(\alpha -1)}-{d\over 2}\left( {{1\over {\tilde{q}}}+{{\tilde{s}}\over d}}\right) \right)&= {\alpha (2+\gamma )\over 2(\alpha -1)}-{d\over 2}\left( {\alpha \over {\tilde{q}}}+{\alpha {\tilde{s}}\over d}\right) \\&< {\alpha (2+\gamma )\over 2(\alpha -1)}-{d\over 2}\left( {1\over q}+{\gamma +s\over d}\right) \\&=1+ {\gamma \over 2}+{2+\gamma \over 2(\alpha -1)}-{d\over 2q}-{\gamma +s\over 2}\\&< 1. \end{aligned} \end{aligned}$$

Thus, we conclude Proposition 6.1. $\square$

The existence of positive self-similar solutions $\Psi$ of (1) with (i)–(iii) in Proposition 6.1 is proved for any $\alpha$ satisfying

$$\begin{aligned} \alpha _F< \alpha <\alpha _{HS} \end{aligned}$$

(90)

with $\gamma =0$ by [24, Propositions 3.1, 3.4 and 3.5] and with $\gamma$ satisfying

$$\begin{aligned} -2 < \gamma \le {\left\{ \begin{array}{ll} \sqrt{3}-1\quad &{}\text {if }d=3,\\ 0 &{}\text {if }d\ge 4 \end{array}\right. } \end{aligned}$$

(91)

by Hirose [26, Theorem 1.2 (ii)]. Furthermore,

$$\begin{aligned} W(x)=C|x|^{\frac{2+\gamma }{\alpha -1}-d}e^{-{|x|^2\over 4}}\left( 1+O\left( |x|^{-2}\right) \right) \quad \text {as }|x|\rightarrow \infty . \end{aligned}$$

From Proposition 6.1 and this result, it immediately follows that the equation (1) has three different solutions 0 and $\pm \Psi$ with initial data 0 in $C([0,\infty ); L^{q,r}_s({\mathbb {R}}^d))$ under the assumptions (90) and (91) for $d,\gamma ,\alpha ,q,r,s$ as in Proposition 6.1. Thus, Proposition 1.7 is proved.

Remark 13

When $\gamma$ does not satisfy (91), the existence of self-similar solutions with (i)–(iii) in Proposition 6.1 under the condition (90) is an open problem.

The situation of the case $\alpha > \alpha _{HS}$ is different from the case (90). In this case, the nonexistence of positive self-similar solution $\Psi$ satisfying (i)–(iii) in Proposition 6.1 is proved by the following result on uniqueness in the Sobolev space $H^1({\mathbb {R}}^d)$:

Lemma 6.2

Let $T>0$ and $u=u(t,x)$ be a mild solution to (1) satisfying

$$\begin{aligned} u \in C^1((0,T); L^2({\mathbb {R}}^d)) \cap C^1((0,T); L^{\alpha +1}_{\frac{\gamma }{\alpha +1}}({\mathbb {R}}^d)) \cap C((0,T); H^2({\mathbb {R}}^d)). \end{aligned}$$

Assume that $u(t) \rightarrow 0$ in $H^1({\mathbb {R}}^d)$ as $t\rightarrow 0$. Then $u\equiv 0$ on [0, T].

The proof of Lemma 6.2 is almost the same as that of [24, Theorem 2], and so we omit the proof. If $\alpha > \alpha _{HS}$ and there exists a positive self-similar solution $\Psi$ satisfying (i)–(iii) in Proposition 6.1, then $\Psi$ satisfies all assumptions in Lemma 6.2, and hence, $\Psi \equiv 0$. This contradicts $\Psi >0$. Thus, we see the nonexistence of such a $\Psi$.

7 Additional results and remarks

7.1 Double critical case

We give a remark on the number of solutions in the double critical case. Theorem 1.5 shows that the problem (1) has two different solutions, where one is regular and the other is singular (see Sect. 5). In fact, however, (1) has an uncountably infinite number of different mild solutions in $C([0,T]; L^{q,r}_{s}({\mathbb {R}}^d))$ for any initial data $u_0 \in L^{q,r}_{s}({\mathbb {R}}^d)$. This can be confirmed by constructing the family $\{u_{t_0}\}_{t_0\in (0,T)}$ of solutions to (1) such that $u_{t_0}$ is a singular solution for $0<t \le t_0$ and a regular solution for $t_0<t<T$.

7.2 Case $\gamma = -\min \{2,d\}$

The problem on well-posedness for (1) in the critical singular case $\gamma = -\min \{2,d\}$ has not been studied. Establishing the weighted linear estimates (9) with the double endpoint $\frac{s_1}{d}+\frac{1}{q_1}=1$ and $\frac{s_2}{d}+\frac{1}{q_2}=0$, we can present the following results on uniqueness for the case $d=1$ and $\gamma =-1$.

Theorem 7.1

Let $T>0$, and let $d=1$, $\gamma =-1$, $\alpha >1$, $\alpha \le q < \infty$, and $\frac{s}{d} + \frac{1}{q} = 0$. Then the following assertions hold:

• (i) Let $0<r \le \alpha -1$. Then unconditional uniqueness holds for (1) in $L^\infty (0,T; L^{q,r}_s({\mathbb {R}}))$.

• (ii) Let $r> \alpha -1$ and $u_0\in L^{q,r}_{s}({\mathbb {R}})$. Then, if $u_1,u_2 \in L^\infty (0,T; L^{q, r}_{s}({\mathbb {R}}))$ are mild solutions to (1) with $u_1(0)=u_2(0)=u_0$ such that

  $$\begin{aligned} u_i(t) - e^{t\Delta } u_0 \in L^\infty (0,T; L^{q, \alpha -1}_{s}({\mathbb {R}})) \quad \text {for }i=1,2, \end{aligned}$$

  then $u_1 = u_2$ on [0, T].

Proof

The proofs of (i) and (ii) are similar to those of Theorem 1.2 (ii) and Proposition 1.4, respectively. The only difference is use of Proposition 3.1 with the double endpoint case (12) and (13), where the restriction on r is required. $\square$

In the case $d=1$ and $\gamma =-1$, the existence of a solution has not been proved, but Theorem 7.1 implies that only one solution exists at most. It remains open whether unconditional uniqueness holds in the critical singular case $d\ge 2$ and $\gamma =-2$. Once the weighted Meyer inequality (36) with the endpoint case $\frac{s_2}{d}+\frac{1}{q_2}=0$ is proved, this problem is solved, but we do not know if the endpoint inequality holds.

7.3 Case of the exterior problem

It is also interesting to analyze in more detail the influence of the potential $|x|^\gamma$ at the origin or at infinity. For this, we discuss unconditional uniqueness for the initial-boundary value problem of the Hardy-Hénon parabolic equation on the exterior domain $\Omega :=\{x \in {\mathbb {R}}^d;\, |x| > 1\}$.

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u - \Delta u = |x|^{\gamma } |u|^{\alpha -1} u, &{}(t,x)\in (0,T)\times \Omega , \\ u=0, &{}(t,x)\in (0,T)\times \partial \Omega ,\\ u(0) = u_0 \in L^{q,r}_{s}(\Omega ), \end{array}\right. } \end{aligned}$$

(92)

where $T>0,$ $d\in {\mathbb {N}}$, $\gamma \in {\mathbb {R}}$, $\alpha >1$, $q\in [1,\infty ]$, $r \in (0,\infty ]$ and $s\in {\mathbb {R}}$. Here, $\partial \Omega$ denotes the boundary of $\Omega$. In conclusion, the critical exponents (3) and (4) with $\gamma =0$ (i.e. $q_c(0)=\frac{d(\alpha -1)}{2}$ and $Q_c(0) = \alpha$) appear in the results on unconditional uniqueness for (92), since the effect near the origin $x=0$ has been eliminated. The results of this subsection can be extended to more general situations such as the initial-boundary value problem on general domains $\Omega$ not containing the origin with the Robin boundary condition (cf. [28, Section 5]).

In the following, we shall prove the result on unconditional uniqueness.

Proposition 7.2

Let $d\in {\mathbb {N}}$, $\gamma \in {\mathbb {R}}$, $\alpha >1$, $q \in [1,\infty ]$ and $s\in {\mathbb {R}}$ be such that

$$\begin{aligned} \alpha \le q \le \infty ,\quad -\frac{d}{q}< s < d\left( 1-\frac{\alpha }{q}\right) \quad \text {and}\quad \frac{\gamma }{\alpha -1} \le s. \end{aligned}$$

(93)

Then the following assertions hold:

• (i) Assume either

  $$\begin{aligned} q> \min \left\{ q_c(0), Q_c(0)\right\} \quad \text {and}\quad r=\infty \end{aligned}$$

  (94)

  or

  $$\begin{aligned} q = Q_c(0) > q_c(0) \quad \text {and}\quad r=\alpha . \end{aligned}$$

  Then unconditional uniqueness holds for (92) in $L^\infty (0,T; L^{q,r}_s(\Omega ))$.

• (ii) Assume either

  $$\begin{aligned} q = q_c(0) > Q_c(0) \quad \text {and}\quad r=\infty . \end{aligned}$$

  or

  $$\begin{aligned} q = q_c(0) = Q_c(0) \quad \text {and}\quad r=\alpha -1. \end{aligned}$$

  Then unconditional uniqueness holds for (92) in $C([0,T]; L^{q,r}_s(\Omega ))$.

Remark 14

Since $L^{q,r}_{s_1}(\Omega ) \subset L^{q,r}_{s_2}(\Omega )$ if $s_2\le s_1$, the exponent s should be taken as close to $\max \{-\frac{d}{q}, \frac{\gamma }{\alpha -1}\}$ as possible in the above proposition from the point of view of unconditional uniqueness.

We denote by $-\Delta _D$ the Laplace operator with the homogeneous Dirichlet boundary condition on $\Omega$ and by $\{e^{t\Delta _D}\}_{t>0}$ the semigroup generated by $-\Delta _D$. The integral kernel $G_D(t,x,y)$ of $e^{t\Delta _D}$ satisfies the Gaussian upper bound

$$\begin{aligned} 0\le G_{D}(t, x,y) \le G_t(x-y) \end{aligned}$$

(95)

for any $t >0$ and almost everywhere $x,y \in \Omega$. Then, we have the following linear estimates.

Lemma 7.3

Let $d \in {\mathbb {N}}$, $1 \le q_1\le \infty$, $1<q_2\le \infty$, $0< r_1,r_2 \le \infty$ and $s_1,s_2\in {\mathbb {R}}$.

• (i) Assume (10)–(15). Then there exists a constant $C>0$ such that

  $$\begin{aligned} \Vert e^{t\Delta _D} f\Vert _{L^{q_2,r_2}_{s_2}(\Omega )} \le C t ^{-\frac{d}{2} (\frac{1}{q_1} - \frac{1}{q_2}) - \frac{s_1 - s_2}{2}} \Vert f\Vert _{L^{q_1,r_1}_{s_1} (\Omega )} \end{aligned}$$

  for any $t >0$ and $f \in L^{q_1,r_1}_{s_1} (\Omega )$.

• (ii) Assume (15) and

  $$\begin{aligned} \left\{ \begin{array}{l} -\frac{d}{q_2} \le s_2 \le \min \left\{ s_1, d\left( 1-\frac{1}{q_1}\right) \right\} ,\\ q_1 \le q_2, \\ r_1\le 1 \quad \text {if } s_2 = d\left( 1 -\frac{1}{q_1}\right) \text { or } q_1=1,\\ r_2=\infty \quad \text {if } s_2 = -\frac{d}{q_2},\\ r_1 \le r_2 \quad \text {if } q_1=q_2. \end{array} \right. \end{aligned}$$

  Then there exists a constant $C>0$ such that

  $$\begin{aligned} \Vert e^{t\Delta _D} f\Vert _{L^{q_2,r_2}_{s_2}(\Omega )} \le C t ^{-\frac{d}{2} (\frac{1}{q_1} - \frac{1}{q_2})} \Vert f\Vert _{L^{q_1,r_1}_{s_1} (\Omega )} \end{aligned}$$

  for any $t >0$ and $f \in L^{q_1,r_1}_{s_1} (\Omega )$.

Proof

The assertion (i) is obtained by combining the upper bound (95) with the argument of proof of Propositions 3.1. The assertion (ii) is proved by combining the assertion (i) with $s_1=s_2$ and the inclusion $L^{q,r}_{s_1}(\Omega ) \subset L^{q,r}_{s_2}(\Omega )$ if $s_2\le s_1$. $\square$

Similarly, we also have the following:

Lemma 7.4

Let $T\in (0,\infty ]$, and let $d\in {\mathbb {N}}$, $q_1 \in [1,\infty ]$, $q_2\in (1,\infty )$, $r_1 \in (0,\infty ]$ and $s_1,s_2 \in {\mathbb {R}}$.

• (i) Assume (31)–(35). Then there exists a constant $C>0$ such that

  $$\begin{aligned} \left\| \int _0^{t} e^{(t-\tau )\Delta _D} f(\tau ) \,d\tau \right\| _{L^{q_2,\infty }_{s_2}(\Omega )} \le C \sup _{0<\tau <t} \Vert f(\tau )\Vert _{L^{q_1,r_1}_{s_1}(\Omega )} \end{aligned}$$

  (96)

  for any $t\in (0,T)$ and $f \in L^\infty (0,T; L^{q_1,r_1}_{s_1}(\Omega ))$.

• (ii) Assume that

  $$\begin{aligned} \left\{ \begin{array}{l} - \frac{d}{q_2} < s_2 \le \min \left\{ s_1, d\left( 1-\frac{1}{q_1}\right) \right\} ,\\ \frac{d}{2} \left( \frac{1}{q_1} - \frac{1}{q_2} \right) = 1,\\ r_1 \le 1 \quad \text {if }s_2 = d\left( 1-\frac{1}{q_1} \right) \text { or } q_1=1. \end{array} \right. \end{aligned}$$

  Then the estimate (96) holds.

Proof of Proposition 7.2

The proof is simply the same argument as the proofs of Theorems 1.2, 1.3 and 1.6 with Propositions 3.1 and 3.3 replaced by Lemmas 7.3 and 7.4. In fact, in the double subcritical case (94), we set $\sigma := \alpha s - \gamma$ and use Lemma 7.3 (ii) with $(q_1,r_1,s_1) = (\frac{q}{\alpha }, \infty , \sigma )$ and $(q_2,r_2,s_2) = (q,\infty ,s)$ and Lemma 2.3 with $(q,r)=(\frac{q}{\alpha },\infty )$, $(q_1,r_1) = (\frac{q}{\alpha -1},\infty )$, $(q_2,r_2)=(q,\infty )$ to obtain

$$\begin{aligned} \begin{aligned}&\Vert u_1(t) - u_2(t)\Vert _{L^{q,\infty }_{s}} \\&\quad \le C \int _0^t (t-\tau ) ^{-\frac{d}{2} (\frac{\alpha }{q} - \frac{1}{q})} \Vert |\cdot |^{\gamma } (|u_1(\tau )|^{\alpha -1}u_1(\tau ) - |u_2(\tau )|^{\alpha -1}u_2(\tau )) \Vert _{L^{\frac{q}{\alpha },\infty }_{\sigma }} \, d\tau \\&\quad \le C \int _0^t (t-\tau ) ^{-\frac{d}{2} (\frac{\alpha }{q} - \frac{1}{q})} \, d\tau \times \max _{i=1,2} \Vert u_i\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}\\&\quad \le C t^{\delta } \max _{i=1,2} \Vert u_i\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}^{\alpha -1} \Vert u_1 - u_2\Vert _{L^\infty (0,t; L^{q,\infty }_{s})}. \end{aligned} \end{aligned}$$

Here, the conditions

$$\begin{aligned} 1< \frac{q}{\alpha } \le q \le \infty ,\quad -\frac{d}{q}< s< d\left( 1-\frac{\alpha }{q}\right) , \quad s \le \sigma , \quad \frac{d}{2} \left( \frac{\alpha }{q} - \frac{1}{q}\right) < 1 \end{aligned}$$

are required in order to use Proposition 3.1 and for the above integral in $\tau$ to be finite. Note that the conditions amount to (93) and (94). Similarly to the proof of Theorem 1.2, we conclude the assertion (i) in the case (94). The other cases can be also proved in a similar way, and so we may omit the details. The proof of Proposition 7.2 is finished. $\square$

Appendices

Some lemmas on weighted Lorentz spaces

In this appendix we provide several lemmas on weighted Lorentz spaces. First, we will show the Fatou property of $L^{q,r}_s(\Omega )$. A quasi-normed space $X\subset L^0(\Omega )$ is said to satisfy the Fatou property if the following holds: Suppose that $f_n \in X$, $f_n\ge 0$ ($n\in {\mathbb {N}}$) and $f_n\nearrow f$ a.e. as $n\rightarrow \infty$. If $f \in X$, then $\Vert f_n\Vert _{L^{q,r}_{s}} \nearrow \Vert f\Vert _{L^{q,r}_s}$ as $n\rightarrow \infty$, whereas if $f \not \in X$, then $\Vert f_n\Vert _{L^{q,r}_s} \nearrow \infty$ as $n\rightarrow \infty$. Then we have the following lemma.

Lemma A.1

Let $s\in {\mathbb {R}}$, $0<q,r\le \infty$ and $r=\infty$ if $q=\infty$. Then $L^{q,r}_s(\Omega )$ satisfies the Fatou property.

Remark 15

From Lemma A.1, we can immediately see that $L^{q,r}_s(\Omega )$ is a quasi-Banach space by using the fact that a quasi-normed space $X \subset L^0(\Omega )$ is complete if it satisfies the Fatou property (see e.g. [35, Remark 2.1 (ii) and Proposition 2.2]).

Proof

Suppose that $f_n \in X$, $f_n\ge 0$ ($n\in {\mathbb {N}}$) and $f_n\nearrow f$ a.e. as $n\rightarrow \infty$. Then $0 \le (|x|^s f_n)^* \nearrow (|x|^s f)^*$ as $n\rightarrow \infty$ (see [22, Proposition 1.4.5 (8)]). If $q,r < \infty$, then the monotone convergence theorem (over $(0,\infty )$ with Lebesgue measure) yields

$$\begin{aligned} \Vert f_n\Vert _{L^{q,r}_s} = \left( \int _0^\infty ( t^{\frac{1}{q}} (|x|^s f_n)^*(t) )^r\frac{dt}{t} \right) ^{\frac{1}{r}} \nearrow \left( \int _0^\infty ( t^{\frac{1}{q}} (|x|^s f)^*(t) )^r\frac{dt}{t} \right) ^{\frac{1}{r}} = \Vert f\Vert _{L^{q,r}_s} \end{aligned}$$

as $n\rightarrow \infty$. If $q<\infty$ and $r=\infty$, then we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert f_n\Vert _{L^{q,\infty }_s} = \sup _{n \in {\mathbb {N}}}\sup _{t>0} t^\frac{1}{q} (|x|^sf_n)^*(t) = \sup _{t>0} \sup _{n \in {\mathbb {N}}} t^\frac{1}{q} (|x|^sf_n)^*(t) =\Vert f\Vert _{L^{q,\infty }_s}, \end{aligned}$$

since the limit as $n\rightarrow \infty$ is the supremum over $n\in {\mathbb {N}}$ by the monotone increase of $\{\Vert f_n\Vert _{L^{q,\infty }_s}\}_n$. The case $q=r=\infty$ is similarly proved as above. The proof of Lemma A.1 is finished. $\square$

Next, we will prove the following density result, which is used in Remark 5 (d) and Lemma 4.3.

Lemma A.2

Let $\Omega$ be a domain in ${\mathbb {R}}^d$, and let $s \in {\mathbb {R}}$ and $0<q,r<\infty$. Then the following statements hold:

• (i) $L^{q,r}_s(\Omega ) \cap L^\infty _0(\Omega )$ is dense in $L^{q,r}_s(\Omega )$.

• (ii) If $1<q<\infty$ and $1 \le r <\infty$, then $C^\infty _0(\Omega )$ is dense in $L^{q,r}_s(\Omega )$.

Remark 16

Let us give some remarks on Lemma A.2.

• (a) There are many results on density for weighted spaces such as weighted Lebesgue space and weighted Banach function spaces (see e.g. [38, Theorems 1.1 and 1.2], [30, Lemmas 2.4, 2.10 and 2.12] and [16, Proposition 3.1 and Remark 3.2]). However, applying the previous results to our weighted Lorentz spaces requires restrictions on the parameters s, q, r in the density result. In particular, the condition $0<\frac{s}{d}+\frac{1}{q}<1$ is imposed on s. On the other hand, we note that Lemma A.2 requires no additional conditions on s, q, r.

• (b) The case $r=q$ of Lemma A.2 (i) can be found in [34, Lemma 2.12].

• (c) A similar density result to Lemma A.2 (i) is proved in the first paragraph of the proof of [35, Proposition 3.13].

• (d) For $0<q \le 1$ or $0<r<1$, the density of $C^\infty _0(\Omega )$ in $L^{q,r}_s(\Omega )$ is not proved. In fact, the approximate functions $f_n$ in (99) (which approximate the target function f for $0<q \le 1$) belong to $L^{q,r}_s(\Omega ) \cap L^\infty _0(\Omega )$, but not to $C^\infty _0(\Omega )$.

Proof

We divide the proof into three steps.

Step 1: In this step we show that $L^{q,r}_s(\Omega )$ has an absolutely continuous quasi-norm, i.e., for any function $f\in L^{q,r}_s(\Omega )$,

$$\begin{aligned} \Vert f \chi _{E_n} \Vert _{L^{q,r}_s(\Omega )} \rightarrow 0 \quad \text {as }n\rightarrow \infty \end{aligned}$$

(97)

holds for any sequence $\{E_n\}_n$ of measurable subsets of $\Omega$ such that $\chi _{E_n} \rightarrow 0$ a.e. as $n\rightarrow \infty$. See e.g. [6, Chapter 1], [16, Section 2] and [30, Section 2] for the details of absolutely continuous (quasi-)norm.

If (97) is shown for any non-negative function in $L^{q,r}_s(\Omega )$, then (97) is also shown for all functions $f\in L^{q,r}_s(\Omega )$ by the decomposition $f = f_+ - f_-$ with the positive part $f_+\ge 0$ and negative part $f_- \ge 0$ of f. Hence, we may assume that $f\in L^{q,r}_s(\Omega )$ is non-negative on $\Omega$ without loss of generality. Let $g_n:= |x|^s f \chi _{E_n}$. Then

$$\begin{aligned} \Vert f \chi _{E_n} \Vert _{L^{q,r}_s(\Omega )} = \Vert g_n\Vert _{L^{q,r}(\Omega )} = q^{\frac{1}{r}} \left( \int _0^\infty \left( d_{g_n} (\lambda )^\frac{1}{q} \lambda \right) ^r \frac{d\lambda }{\lambda } \right) ^\frac{1}{r}, \end{aligned}$$

where we recall

$$\begin{aligned} d_{g_n} (\lambda ) = \left| \{ x \in \Omega \,;\, |g_n(x)|>\lambda \}\right| = \int _\Omega \chi _{\{|g_n| > \lambda \}}(x) \, dx. \end{aligned}$$

Since $\chi _{E_n} \rightarrow 0$ a.e. as $n\rightarrow \infty$, we see that $\chi _{\{|g_n| > \lambda \}} \rightarrow 0$ a.e. as $n\rightarrow \infty$. There is a dominating function of $\chi _{\{|g_n| > \lambda \}}$, i.e., $\chi _{\{|g_n|> \lambda \}} \le \chi _{\{|x|^s f > \lambda \}}$ and $\chi _{\{|x|^s f > \lambda \}} \in L^1(\Omega )$ for any $n \in {\mathbb {N}}$. Hence, it follows from Lebesgue’s dominated convergence theorem that $d_{g_n} (\lambda ) \rightarrow 0$ for any $\lambda >0$. Furthermore, since $d_{g_n} (\lambda ) \le d_{|x|^s f}(\lambda )$ for any $\lambda >0$, we can again apply Lebesgue’s dominated convergence theorem to obtain

$$\begin{aligned} \Vert g_n\Vert _{L^{q,r}(\Omega )} = q^{\frac{1}{r}} \left( \int _0^\infty \left( d_{g_n} (\lambda )^\frac{1}{q} \lambda \right) ^r \frac{d\lambda }{\lambda } \right) ^\frac{1}{r} \rightarrow 0\quad \text {as }n\rightarrow \infty , \end{aligned}$$

which implies (97).

Step 2: In this step we prove the case $s=0$ of (i) and (ii).

If $1<q<\infty$ and $1\le r < \infty$, then $L^{q,r}(\Omega )$ is a Banach function space,¹ and moreover, it is already shown in Step 1 that $L^{q,r}(\Omega )$ has an absolutely continuous norm. Hence, it follows from [16, Proposition 3.1 and Remark 3.2] that $C_0^\infty (\Omega )$ is dense in $L^{q,r}(\Omega )$. Thus, the case $s=0$ of (ii) is proved.

The cases $0<q\le 1$ or $0<r<1$ can be obtained by the argument of proof of [34, Lemma 2.12]. In fact, we take a constant $m \in {\mathbb {N}}$ such that $2^m q >1$ and $2^m r \ge 1$ and define $g:= |f|^{2^{-m}}$. Then $g \in L^{2^mq,2^mr}(\Omega )$, since

$$\begin{aligned} \Vert g\Vert _{L^{2^mq,2^mr}(\Omega )}^{2^m} = \left\| |g|^{2^m}\right\| _{L^{q,r}(\Omega )} = \Vert f\Vert _{L^{q,r}(\Omega )} < \infty \end{aligned}$$

(see [22, Remark 1.4.7] for the first equality). Hence, there exists a sequence $\{g_n\}_n \subset C^\infty _0(\Omega )$ such that

$$\begin{aligned} \Vert g - g_n \Vert _{L^{2^mq,2^mr}(\Omega )} \rightarrow 0 \quad \text {as }n\rightarrow \infty . \end{aligned}$$

(98)

Define

$$\begin{aligned} f_n:= g_n^{2^m} \textrm{sgn}\, (f) \end{aligned}$$

(99)

for $n \in {\mathbb {N}}$, where $\textrm{sgn}\, (f)$ denotes the sign function of f. Then $f_n \in L^{q,r}(\Omega ) \cap L^\infty _0(\Omega )$ and

$$\begin{aligned} |f - f_n| = \left| g^{2^m} - g_n^{2^m} \right| = | g - g_n| \prod _{\ell =0}^{m-1} \left| g^{2^\ell } + g_n^{2^\ell } \right| . \end{aligned}$$

By Lemma 2.3 (i), we estimate

$$\begin{aligned} \begin{aligned} \Vert f - f_n\Vert _{L^{q,r}(\Omega )}&= \left\| | g - g_n| \prod _{\ell =0}^{m-1} \left| g^{2^\ell } + g_n^{2^\ell } \right| \right\| _{L^{q,r}(\Omega )} \\&\le C \Vert g -g_n\Vert _{L^{2^mq,2^mr}(\Omega )} \prod _{\ell =0}^{m-1} \Vert g^{2^\ell } + g_n^{2^\ell }\Vert _{L^{2^{m-\ell }q,2^{m-\ell }r}(\Omega )}. \end{aligned} \end{aligned}$$

(100)

Here, $\prod _{\ell =0}^{m-1} \Vert g^{2^\ell } + g_n^{2^\ell }\Vert _{L^{2^{m-\ell }q,2^{m-\ell }r}(\Omega )}$ is bounded in sufficiently large n by combining the convergence (98) and the inequality

$$\begin{aligned} \begin{aligned} \Vert g^{2^\ell } + g_n^{2^\ell }\Vert _{L^{2^{m-\ell }q,2^{m-\ell }r}(\Omega )}&\le \Vert g^{2^\ell } \Vert _{L^{2^{m-\ell }q,2^{m-\ell }r}(\Omega )} + \Vert g^{2^\ell }_n \Vert _{L^{2^{m-\ell }q,2^{m-\ell }r}(\Omega )}\\&= \Vert g \Vert _{L^{2^{m}q,2^{m}r}(\Omega )} + \Vert g_n \Vert _{L^{2^{m}q,2^{m}r}(\Omega )}. \end{aligned} \end{aligned}$$

Therefore, it follows from (98) and (100) that $\Vert f - f_n\Vert _{L^{q,r}(\Omega )} \rightarrow 0$ as $n\rightarrow \infty$. Thus, the density in the case $0<q\le 1$ is proved, and hence, the case $s=0$ of (i) is also proved.

Final step: Finally, we prove the case $s\not =0$ of (i) and (ii). Let $f \in L^{q,r}_s(\Omega )$. We decompose f into

$$\begin{aligned} f = f \chi _{\{|x|\le \delta \}} + f \chi _{\{\delta<|x| < R\}} + f \chi _{\{|x|\ge R\}} \end{aligned}$$

for $0< \delta < R$. Then $\chi _{\{|x|\le \delta \}}$ and $\chi _{\{|x|\ge R\}}$ converge to 0 almost everywhere in $\Omega$ as $\delta \rightarrow 0$ and $R\rightarrow \infty$, respectively. Hence it follows from (97) that, for any $\varepsilon >0$, there exist $\delta , R>0$ such that

$$\begin{aligned} \Vert f \chi _{\{|x|\le \delta \}}\Vert _{L^{q,r}_s(\Omega )} + \Vert f \chi _{\{|x|\ge R\}}\Vert _{L^{q,r}_s(\Omega )} < \varepsilon . \end{aligned}$$

(101)

Furthermore, since $L^{q,r}_s(\{\delta<|x|< R\}) = L^{q,r}(\{\delta<|x| < R\})$, it follows from the density of $L^{q,r}_s(\{\delta<|x|< R\}) \cap L^\infty _0(\{\delta<|x| < R\})$ in $L^{q,r}(\{\delta<|x| < R\})$ which is shown in Step 2, that there exists a function $g = g_{\varepsilon , \delta , R} \in L^{q,r}_s(\{\delta<|x|< R\}) \cap L^\infty _0(\{\delta<|x| < R\})$ satisfying

$$\begin{aligned} \Vert f - g\Vert _{L^{q,r}(\{\delta<|x|< R\})} < \frac{\varepsilon }{\max \{\delta ^s, R^s\}}. \end{aligned}$$

This implies that

$$\begin{aligned} \Vert f - g\Vert _{L^{q,r}_s(\{\delta<|x|< R\})} \le \max \{\delta ^s, R^s\} \Vert f - g\Vert _{L^{q,r}(\{\delta<|x|< R\})} < \varepsilon . \end{aligned}$$

(102)

Let ${\tilde{g}}$ be the zero extension of g to $\Omega$. Then ${\tilde{g}} \in L^{q,r}_s(\Omega ) \cap L^\infty _0(\Omega )$ and

$$\begin{aligned} \begin{aligned} \Vert f - {\tilde{g}}\Vert _{L^{q,r}_s(\Omega )}&\le \Vert f \Vert _{L^{q,r}_s(\{|x| \le \delta \})} + \Vert f - g\Vert _{L^{q,r}_s(\{\delta<|x|< R\})} + \Vert f \Vert _{L^{q,r}_s(\{|x| \ge R\})}\\&< 2\varepsilon \end{aligned} \end{aligned}$$

by (101) and (102). Therefore, the case $s\not =0$ of (i) is proved. For (ii), we also prove the density in the same way. Only difference is to be able to take an approximate function $g \in C^\infty _0(\{\delta<|x| < R\})$ and the zero extension ${\tilde{g}} \in C^\infty _0(\Omega )$. Thus the proof of Lemma A.2 is finished. $\square$

Next, we will show the following lemma on behavior of $f \in L^{q,r}_s({\mathbb {R}}^d)$ as $|x|\rightarrow 0$ or $|x|\rightarrow \infty$, which is used in Step 2 of the proof of the necessity part of Proposition 3.1.

Lemma A.3

Let $0<q,r<\infty$ and $s\in {\mathbb {R}}$. If $f \in L^{q,r}_s({\mathbb {R}}^d)$, then

$$\begin{aligned} \liminf _{|x| \rightarrow 0} |x|^{s+\frac{d}{q}} |\log |x||^{\frac{1}{r}} |f(x)| = \liminf _{|x| \rightarrow \infty } |x|^{s+\frac{d}{q}} |\log |x||^{\frac{1}{r}} |f(x)|=0. \end{aligned}$$

Remark 17

The case of weighted Lebesgue space $L^q_s({\mathbb {R}}^d)$ can be found in [13, Corollary A.4] for instance.

Proof

For simplicity, we give the proof only for the case $d=1$ and $s=0$. Suppose that

$$\begin{aligned} \liminf _{|x| \rightarrow 0} |x|^{\frac{1}{q}} |\log |x||^{\frac{1}{r}} |f(x)|=c>0. \end{aligned}$$

Then there exists a positive constant $\delta$ such that $c/2 \le |x|^{\frac{1}{q}} |\log |x||^{\frac{1}{r}} |f(x)|$ for $|x|\le \delta$. Using [22, Proposition 1.4.5 (4) and (5)], we have

$$\begin{aligned} (f \chi _{|x|\le \delta })^*(t) \ge \frac{c}{2} (|x|^{-\frac{1}{q}} |\log |x||^{-\frac{1}{r}} \chi _{\{|x|\le \delta \}})^*(t) = \frac{c}{2} t^{-\frac{1}{q}} |\log t|^{-\frac{1}{r}} \chi _{\{t\le \delta '\}} \end{aligned}$$

for some $\delta '>0$. Hence,

$$\begin{aligned} \begin{aligned} \Vert f \chi _{|x|\le \delta } \Vert _{L^{q,r}}&= \left( \int _0^\infty ( t^{\frac{1}{q}} (f \chi _{|x|\le \delta })^*(t) )^r\frac{dt}{t} \right) ^{\frac{1}{r}}\\&\ge \frac{c}{2} \left( \int _0^{\delta '} |\log t|^{-1} \frac{dt}{t} \right) ^{\frac{1}{r}} = +\infty , \end{aligned} \end{aligned}$$

which implies $f \not \in L^{q,r}({\mathbb {R}})$. The second equality is similarly proved. $\square$

Finally, we will prove the continuity of heat semigroup at $t=0$, which is used to prove the continuity of mild solutions at $t=0$ in Proposition 5.1 and Lemma 5.6.

Lemma A.4

Let $s\in {\mathbb {R}}$, $1<q \le \infty$ and $0<r \le \infty$ satisfy

$$\begin{aligned} \left\{ \begin{array}{l} 0 \le \frac{s}{d} + \frac{1}{q} \le 1, \\ r \le 1\quad \text {if }\frac{s}{d} + \frac{1}{q}=1,\\ r = \infty \quad \text {if } \frac{s}{d} + \frac{1}{q}=0 \text { or } q=\infty . \end{array} \right. \end{aligned}$$

Then

$$\begin{aligned} \lim _{t\rightarrow 0}\Vert e^{t\Delta } f - f\Vert _{L^{q,r}_s} =0 \end{aligned}$$

holds for any $f \in L^{q,r}_s({\mathbb {R}}^d)$ (replace $L^{q, r}_{s}({\mathbb {R}}^d)$ by ${\mathcal {L}}^{q, r}_{s}({\mathbb {R}}^d)$ if $q=\infty$ or $r=\infty$).

Proof

The case $r=q$ follows from the standard argument (see e.g. [44, Theorem 5.5 on page 198]). The case $r\not =q$ can be proved by a real interpolation argument. In fact, it is known that $L^{q,r}({\mathbb {R}}^d)$ coincides with the real interpolation space $(L^{q_0}({\mathbb {R}}^d), L^{q_1}({\mathbb {R}}^d))_{\theta ,r}$, where $0<\theta <1$ and $1<q_0<q<q_1<\infty$ (see e.g. [7, 5.3.1 Theorem on page 113]). This implies that for any $g \in L^{q,r}({\mathbb {R}}^d)$, there exist $g_i \in L^{q_i}({\mathbb {R}}^d)$ ($i=0,1$) satisfying $g=g_0+g_1$ and

$$\begin{aligned} \Vert g\Vert _{L^{q,r}}\le \left( \int _0^\infty \left( \lambda ^\theta (\Vert g_0\Vert _{L^{q_0}} + \lambda ^{-1}\Vert g_1\Vert _{L^{q_1}}) \right) ^r\frac{d\lambda }{\lambda } \right) ^\frac{1}{r} \le 2 \Vert g\Vert _{L^{q,r}}. \end{aligned}$$

Let $f \in L^{q,r}_s$. Then, taking $g = |x|^s f$, and applying the above interpolation result to this g, we have two functions $g_i$ ($i=0,1$) as above. Defining $f_i$ by $f_i:= |x|^{-s} g_i$ for $i=0,1$, we see that $f_i \in L^{q_i}_s({\mathbb {R}}^d)$ for $i=0,1$, the decomposition $f = f_0 + f_1$ and the inequalities

$$\begin{aligned} \Vert f\Vert _{L^{q,r}_s}\le \left( \int _0^\infty \left( \lambda ^\theta (\Vert f_0\Vert _{L^{q_0}_s} + \lambda ^{-1}\Vert f_1\Vert _{L^{q_1}_s}) \right) ^r\frac{d\lambda }{\lambda } \right) ^\frac{1}{r} \le 2 \Vert f\Vert _{L^{q,r}_s}. \end{aligned}$$

(103)

Now, we have

$$\begin{aligned} \Vert e^{t\Delta } f - f\Vert _{L^{q,r}_s} \le \left( \int _0^\infty \left( \lambda ^\theta (\Vert e^{t\Delta } f_0 - f_0\Vert _{L^{q_0}_s} + \lambda ^{-1}\Vert e^{t\Delta } f_1 - f_1\Vert _{L^{q_1}_s}) \right) ^r\frac{d\lambda }{\lambda } \right) ^\frac{1}{r}. \end{aligned}$$

The integrand in the right-hand side converges to 0 almost everywhere in $\lambda \in (0,\infty )$ as $t\rightarrow 0$ by Lemma A.4 with $r=q$ (which is already proved). Moreover, we see that the integrand has a dominating function by a combination of (103) and the inequality

$$\begin{aligned} \Vert e^{t\Delta } f_0 - f_0\Vert _{L^{q_0}_s} + \lambda ^{-1}\Vert e^{t\Delta } f_1 - f_1\Vert _{L^{q_1}_s} \le C ( \Vert f_0\Vert _{L^{q_0}_s} + \lambda ^{-1}\Vert f_1\Vert _{L^{q_1}_s}), \end{aligned}$$

where we used the triangle inequality and boundedness of $e^{t\Delta }$ on $L^{q,r}_s({\mathbb {R}}^d)$ (Proposition 3.1). Therefore, we can use Lebesgue’s dominated convergence theorem to obtain

$$\begin{aligned} \lim _{t\rightarrow 0}\Vert e^{t\Delta } f - f\Vert _{L^{q,r}_s} =0. \end{aligned}$$

Thus, the proof of Lemma A.4 is finished. $\square$

Proof of Theorem 5.4 (ii)

In this appendix, we give a proof of Theorem 5.4 (ii) for completeness. The proof is based on the argument of the proof of [23, Theorem 4.1]. For simplicity, we write $U = U(r)$, where $r=|x|$. Then U satisfies the problem

$$\begin{aligned} -(r^{d-1} U')' = r^{d-1+\gamma }U^{\frac{d + \gamma }{d-2}},\quad r \in (0,1). \end{aligned}$$

Then the upper bound of U near $x=0$ was already obtained.

Theorem B.1

(Theorem 1.1 (iv) in [8]) There exists a constant $C>0$ such that

$$\begin{aligned} U(r) \le C r^{-(d-2)} |\log r|^{-\frac{d-2}{\gamma +2}},\quad r \in (0,1). \end{aligned}$$

We make the change of variable

$$\begin{aligned} U(r) = r^{-(d-2)} {\mathfrak {u}}(t),\quad t = -\log r. \end{aligned}$$

(104)

The properties of ${\mathfrak {u}}$ are as follows.

Lemma B.2

The function ${\mathfrak {u}}$ is of $C^2$ on $(0,\infty )$ and is a positive and strictly decreasing solution of the nonlinear ordinary differential equation

$$\begin{aligned} \frac{d}{dt}\left( \frac{d{\mathfrak {u}}}{dt}(t)+ (d-2) {\mathfrak {u}}(t) \right) + {\mathfrak {u}}(t)^{\frac{d + \gamma }{d-2}} = 0, \quad t \in (0,\infty ) \end{aligned}$$

(105)

with

$$\begin{aligned} {\mathfrak {u}}(0) = \lim _{r\rightarrow 1}U(r)\quad \text {and}\quad \frac{d{\mathfrak {u}}}{dt}(0) = - \lim _{r\rightarrow 1}\left( \frac{dU}{dr}(r) -(d-2) U(r)\right) \end{aligned}$$

(and hence, ${\mathfrak {u}}$ is a $C^1$-diffeomorphism from $(0,\infty )$ to $(0,{\mathfrak {u}}(0))$). Moreover, ${\mathfrak {u}}^{\frac{d + \gamma }{d-2}} \in L^1((0, \infty ))$ and

$$\begin{aligned} \frac{d{\mathfrak {u}}}{dt}(t)+ (d-2) {\mathfrak {u}}(t) > 0, \quad t \in (0,\infty ). \end{aligned}$$

(106)

Proof

It is obvious that ${\mathfrak {u}}$ is positive and of $C^2$, and a straightforward calculation gives that ${\mathfrak {u}}$ satisfies the nonlinear ordinary differential equation (105). It is shown by Theorem B.1 that ${\mathfrak {u}}^{\frac{d + \gamma }{d-2}} \in L^1((0, \infty ))$.

We shall prove that ${\mathfrak {u}}$ is strictly decreasing on $(0,\infty )$ by contradiction. Suppose that ${\mathfrak {u}}$ is not strictly decreasing on $(0,\infty )$. Then there exist $t_0, t_1$ such that $0<t_0<t_1$ and

$$\begin{aligned} {\mathfrak {u}}_t(t_0)={\mathfrak {u}}_t(t_1) = 0\quad \text {and}\quad {\mathfrak {u}}_t \ge 0 \text { on } (t_0,t_1). \end{aligned}$$

(107)

Since ${\mathfrak {u}}$ is positive, we find from (107) that

$$\begin{aligned} \begin{aligned} (d-2) \left\{ {\mathfrak {u}}(t_1) - {\mathfrak {u}}(t_0)\right\}&= \left[ {\mathfrak {u}}_t(\tau )+ (d-2) {\mathfrak {u}}(\tau )\right] _{\tau =t_0}^{\tau = t_1}\\&= - \int _{t_0}^{t_1} {\mathfrak {u}}(\tau )^{\frac{d + \gamma }{d-2}}\, d\tau < 0, \end{aligned} \end{aligned}$$

which implies that ${\mathfrak {u}}(t_0) >{\mathfrak {u}}(t_1)$. This is a contradiction to ${\mathfrak {u}}_t \ge 0$ on $(t_0,t_1)$. Therefore, ${\mathfrak {u}}$ is strictly decreasing on $(0,\infty )$. In addition, it is also shown by the inverse function theorem that ${\mathfrak {u}}$ is a $C^1$-diffeomorphism from $(0,\infty )$ to $(0,{\mathfrak {u}}(0))$.

Lastly, since ${\mathfrak {u}} \in C^2((0,\infty ))$, the fundamental theorem of calculus gives

$$\begin{aligned} {\mathfrak {u}}(t') - {\mathfrak {u}}(t) = \int _t^{t'} {\mathfrak {u}}_t(\tau )\, d\tau \end{aligned}$$

for $t'\ge t > 0$, and as $t' \rightarrow \infty$,

$$\begin{aligned} {\mathfrak {u}}(t) = - \int _t^{\infty } {\mathfrak {u}}_t(\tau )\, d\tau \end{aligned}$$

for $t>0$. Since ${\mathfrak {u}}_t<0$, the convergence ${\mathfrak {u}}_t(t) \rightarrow 0$ as $t\rightarrow \infty$ must hold. Noting ${\mathfrak {u}}(t), {\mathfrak {u}}_t(t) \rightarrow 0$ as $t\rightarrow \infty$, and integrating (105) over $[t,\infty )$, we have

$$\begin{aligned} {\mathfrak {u}}_t(t)+ (d-2) {\mathfrak {u}}(t) = \int _{t}^{\infty } {\mathfrak {u}}(\tau )^{\frac{d + \gamma }{d-2}}\, d\tau \end{aligned}$$

(108)

for any $t>0$, which implies (106). The proof of Lemma B.2 is finished. $\square$

Lemma B.3

Let $d\ge 3$ and $\gamma >-2$. Assume that

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{{\mathfrak {u}}_t(t)}{{\mathfrak {u}}(t)} = 0 \text { or }-(d-2). \end{aligned}$$

(109)

Then the assertion (ii) in Theorem 5.4 holds.

Proof

In the case where

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{{\mathfrak {u}}_t(t)}{{\mathfrak {u}}(t)} = -(d-2) \quad \left( \text {i.e. }\lim _{t\rightarrow \infty } (\log {\mathfrak {u}}(t))_t =-(d-2) \right) , \end{aligned}$$

then for any $\varepsilon \in (0,d-2)$, there exists $T=T(\varepsilon )>0$ such that

$$\begin{aligned} -(d-2)-\varepsilon<(\log {\mathfrak {u}}(t))_t < -(d-2)+\varepsilon \end{aligned}$$

(110)

for any $t\ge T$. By integrating (110) over [T, t], we estimate

$$\begin{aligned} {\mathfrak {u}}(t)< {\mathfrak {u}}(T)e^{(-(d-2)+\varepsilon )(t-T)} < {\mathfrak {u}}(0)e^{(-(d-2)+\varepsilon )(t-T)}, \end{aligned}$$

and by recalling (104), we find that

$$\begin{aligned} U(r) \le C e^{((d-2)- \varepsilon )T} r^{-\varepsilon },\quad r\in (0,1). \end{aligned}$$

Hence, U can be extended as a $C^1$ function on B (see [46, Theorem 1] and also [15, Lemma 2.1 and Section 3]).

Next, we consider the other case:

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{{\mathfrak {u}}_t(t)}{{\mathfrak {u}}(t)} = 0. \end{aligned}$$

(111)

Set

$$\begin{aligned} \psi (t):= \int _t^\infty {\mathfrak {u}}(\tau )^{\frac{d + \gamma }{d-2}} d\,\tau . \end{aligned}$$

Then the following hold:

$$\begin{aligned} \lim _{t\rightarrow \infty } \frac{\psi (t)^{\frac{d + \gamma }{d-2}}}{\psi _t(t)} = - (d-2)^{\frac{d + \gamma }{d-2}}, \end{aligned}$$

(112)

and

$$\begin{aligned} \lim _{t\rightarrow \infty } t^{\frac{d-2}{2+\gamma }} \psi (t) = (2+\gamma )^{-\frac{d-2}{2+\gamma }} (d-2)^{\frac{2d + \gamma -2}{2+\gamma }}. \end{aligned}$$

(113)

In fact, noting that $\lim _{t\rightarrow \infty } {\mathfrak {u}} (t) =\lim _{t\rightarrow \infty } {\mathfrak {u}}_t (t) =0$, we see from (105) that

$$\begin{aligned} \psi (t) = {\mathfrak {u}}_t(t)+ (d-2) {\mathfrak {u}}(t) \quad \text {and}\quad \psi _t(t) = -{\mathfrak {u}}(t)^{\frac{d+\gamma }{d-2}}. \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\psi (t)^{\frac{d + \gamma }{d-2}}}{\psi _t(t)} = \frac{({\mathfrak {u}}_t(t)+ (d-2) {\mathfrak {u}}(t))^{\frac{d + \gamma }{d-2}}}{-{\mathfrak {u}}(t)^{\frac{d+\gamma }{d-2}}} = -\left( \frac{{\mathfrak {u}}_t(t)}{{\mathfrak {u}}(t)}+ (d-2) \right) ^{\frac{d + \gamma }{d-2}}. \end{aligned}$$

This and (111) imply (112). Moreover, we see from (112) that

$$\begin{aligned} \lim _{t\rightarrow \infty }(\psi ^{-\frac{2+\gamma }{d-2}}(t))_t = -\frac{2+\gamma }{d-2} \lim _{t\rightarrow \infty }\frac{\psi _t(t)}{\psi (t)^{\frac{d + \gamma }{d-2}}} = (2+\gamma )(d-2)^{-\frac{d + \gamma }{d-2}-1}. \end{aligned}$$

Integrating the above yields

$$\begin{aligned} \lim _{t\rightarrow \infty } t^{-1}\psi ^{-\frac{2+\gamma }{d-2}}(t) = (2+\gamma )(d-2)^{-\frac{d + \gamma }{d-2}-1}, \end{aligned}$$

which implies (113). By using (112) and (113) and noting that ${\mathfrak {u}}(t) = (-\psi _t(t))^{\frac{d-2}{d+\gamma }}$, we obtain

$$\begin{aligned} \begin{aligned} \lim _{t\rightarrow \infty } t^{\frac{d-2}{2+\gamma }}{\mathfrak {u}}(t) = \lim _{t\rightarrow \infty } t^{\frac{d-2}{2+\gamma }}(-\psi _t(t))^{\frac{d-2}{d+\gamma }} = \lim _{t\rightarrow \infty } t^{\frac{d-2}{2+\gamma }} \left( \frac{\psi (t)}{d-2} \right) = \left( \frac{(d-2)^2}{2+\gamma } \right) ^{\frac{d-2}{2+\gamma }}. \end{aligned} \end{aligned}$$

Thus, we conclude Lemma B.3. $\square$

Finally, we conclude the proof of (ii) of Theorem 5.4 by showing the following.

Lemma B.4

Let $d\ge 3$ and $\gamma >-2$. Then (109) holds.

Proof

Since ${\mathfrak {u}}$ is a $C^1$-diffeomorphism from $(0,\infty )$ to $(0,{\mathfrak {u}}(0))$ by Lemma B.2, we can define

$$\begin{aligned} \rho = {\mathfrak {u}}(t)\quad \text {and}\quad {\mathfrak {v}}(\rho ) = {\mathfrak {u}}_t (t) \quad \left( \text {i.e. }{\mathfrak {v}}(\rho ) = {\mathfrak {u}}_t({\mathfrak {u}}^{-1}(\rho ))\right) . \end{aligned}$$

For convenience, we set

$$\begin{aligned} {\mathfrak {w}}(\rho ):= \frac{{\mathfrak {v}}(\rho )}{\rho }. \end{aligned}$$

Then our goal is to prove that

$$\begin{aligned} \lim _{\rho \rightarrow +0}{\mathfrak {w}}(\rho ) = 0 \text { or } - (d-2). \end{aligned}$$

(114)

First, we will show there exists a limit of ${\mathfrak {w}}$ as $\rho \rightarrow +0$ such that

$$\begin{aligned} \lim _{\rho \rightarrow +0} {\mathfrak {w}}(\rho ) = m \in [-(d-2),0]. \end{aligned}$$

(115)

Since ${\mathfrak {w}}$ is continuous and $-(d-2)< {\mathfrak {w}} < 0$, (115) is obvious if ${\mathfrak {w}}$ is monotone in $(0,{\mathfrak {u}}(0))$. Moreover, even if it is not, we can prove that

$$\begin{aligned} {\mathfrak {w}}_{\rho \rho }(a) >0 \quad \text {if there is }a \in (0,{\mathfrak {u}}(0)) \text { such that }{\mathfrak {w}}_\rho (a)=0. \end{aligned}$$

(116)

In fact, by Lemma B.2, ${\mathfrak {w}}$ satisfies $-(d-2)< {\mathfrak {w}} < 0$ and

$$\begin{aligned} \begin{aligned} 0&= \rho {\mathfrak {w}}(\rho )\frac{d}{d\rho }\left( \rho {\mathfrak {w}}(\rho )+ (d-2) \rho \right) + \rho ^{\frac{d + \gamma }{d-2}}\\&=\rho {\mathfrak {w}}(\rho ) \left\{ \left( {\mathfrak {w}}(\rho )+ (d-2) \right) + \rho {\mathfrak {w}}_\rho (\rho )\right\} + \rho ^{\frac{d + \gamma }{d-2}}, \end{aligned} \end{aligned}$$

that is,

$$\begin{aligned} {\mathfrak {w}}_\rho = - \frac{1}{\rho } \left\{ \frac{\rho ^{\frac{\gamma +2}{d-2}}}{{\mathfrak {w}}} + ( {\mathfrak {w}}+ (d-2) )\right\} = - \frac{{\mathfrak {w}}^2+(d-2) {\mathfrak {w}} +\rho ^{\frac{\gamma +2}{d-2}}}{\rho {\mathfrak {w}}} =: F(\rho , {\mathfrak {w}}). \end{aligned}$$

Then, ${\mathfrak {w}}_\rho (a)=0$ implies that

$$\begin{aligned} {\mathfrak {w}}_{\rho \rho }(a) = F_\rho (a, {\mathfrak {w}}(a)) = - \frac{\gamma +2}{d-2} \frac{a^\frac{\gamma +2}{d-2}}{a^2 {\mathfrak {w}}(a)}>0. \end{aligned}$$

Hence, (116) is proved. Since (116) implies that the sign of ${\mathfrak {w}}_\rho$ is constant near $\rho =+0$, ${\mathfrak {w}}$ is monotone near $\rho =+0$. Hence, since $-(d-2)< {\mathfrak {w}} < 0$, there exists a limit of ${\mathfrak {w}}$ as $\rho \rightarrow +0$ satisfying (115).

Next, we will show that $m=0$ or $-(d-2)$. Suppose that $-(d-2)<m<0$ for contradiction. We calculate

$$\begin{aligned} m= \lim _{\rho \rightarrow +0} {\mathfrak {w}}(\rho ) = \lim _{\rho \rightarrow +0} \frac{{\mathfrak {v}}(\rho )}{\rho } = \lim _{t\rightarrow +\infty } \frac{{\mathfrak {u}}_t(t)}{{\mathfrak {u}}(t)} = \lim _{t\rightarrow +\infty } (\log {\mathfrak {u}}(t))_t. \end{aligned}$$

Then, for any $\varepsilon \in (0,-m)$, there exists $T>0$ such that

$$\begin{aligned} (m-\varepsilon ) {\mathfrak {u}}(t)< {\mathfrak {u}}_t(t) < (m+\varepsilon ) {\mathfrak {u}}(t) \end{aligned}$$

(117)

and

$$\begin{aligned} m-\varepsilon<(\log {\mathfrak {u}}(t))_t < m+\varepsilon \end{aligned}$$

(118)

for any $t> T$. Integrating (118) over [t, T] gives

$$\begin{aligned} {\mathfrak {u}}(T)e^{(m-\varepsilon )(t-T)}< {\mathfrak {u}}(t)< {\mathfrak {u}}(T)e^{(m+\varepsilon )(t-T)} < {\mathfrak {u}}(0)e^{(m+\varepsilon )(t-T)} \end{aligned}$$

(119)

for any $t> T$, and hence,

$$\begin{aligned} {\mathfrak {u}}(t)^{\frac{d + \gamma }{d-2}} < {\mathfrak {u}}(0)^{\frac{d + \gamma }{d-2}} e^{\frac{(m+\varepsilon )(d + \gamma )}{d-2}(t-T)} \end{aligned}$$

(120)

for any $t> T$. By (117) and (119), we also have

$$\begin{aligned} \begin{aligned} {\mathfrak {u}}_t(t) + (d-2){\mathfrak {u}}(t)&> \{ (d-2) + m - \varepsilon \} {\mathfrak {u}}(t)\\&>\{ (d-2) + m - \varepsilon \} {\mathfrak {u}}(T) e^{(m-\varepsilon )(t-T)}. \end{aligned} \end{aligned}$$

(121)

By combining (108), (120) and (121), we have

$$\begin{aligned} \begin{aligned} \{ (d-2) + m - \varepsilon \} {\mathfrak {u}}(T) e^{(m-\varepsilon )(t-T)}&< {\mathfrak {u}}_t(t)+ (d-2) {\mathfrak {u}}(t)\\&= \int _t^\infty {\mathfrak {u}}(\tau )^{\frac{d + \gamma }{d-2}} d\,\tau \\&< \int _t^\infty {\mathfrak {u}}(0)^{\frac{d + \gamma }{d-2}} e^{\frac{(m+\varepsilon )(d + \gamma )}{d-2}(\tau -T)}d\,\tau \\&= C e^{\frac{(m+\varepsilon )(d + \gamma )}{d-2}(t-T)} \end{aligned} \end{aligned}$$

for any $t> T$. Then, if we further assume $(d-2) + m - \varepsilon >0$, we have

$$\begin{aligned} {\mathfrak {u}}(T) < C e^{\{\frac{(m+\varepsilon )(d + \gamma )}{d-2} - (m-\varepsilon )\}(t-T)} \end{aligned}$$

for any $t> T$. However, as $t \rightarrow \infty$, this contradicts that

$$\begin{aligned} {\mathfrak {u}}(T) \ge C_\varepsilon \quad \text {for some constant }C_\varepsilon >0, \end{aligned}$$

if we fix $\varepsilon$ sufficiently small so that

$$\begin{aligned} \frac{(m+\varepsilon )(d + \gamma )}{d-2} - (m-\varepsilon )< 0\quad \text {i.e.}\quad 0< \varepsilon < \frac{(2+ \gamma )|m|}{2d + \gamma -2}. \end{aligned}$$

Therefore, m must be 0 or $-(d-2)$, which means (114). Thus, we prove Lemma B.4. $\square$