On the Continuity of Pseudo-Differential Operators on Multiplier Spaces Associated to Herz-type Triebel–Lizorkin Spaces

Abstract

In this paper, for a certain range of parameters, we prove that there exist symbols in the Hörmander class $S_{1,0}^{0}$ which do not define bounded operators on $M\big ({\dot{K}}_{q}^{\alpha ,p}{F_{\beta }^{s}}\big )$. To do these, we need the characterization of Herz–Besov spaces by ball means of differences and some properties of pointwise multipliers for Herz–Triebel–Lizorkin spaces.

Appendix

Here, we present the proof of Theorem 2.9 and Proposition 4.4.

Proof of Theorem 2.9

We split it into three steps.

Step 1. Let $f\in {\dot{K}}_{q}^{\alpha ,p}B_{\beta }^{s}$. Since $s>0$, we see that

$$\begin{aligned} \big \Vert f|{\dot{K}}_{q}^{\alpha ,p}\big \Vert \leqslant \sum _{j=0}^{\infty }\big \Vert \Lambda _{j}f|{\dot{K}}_{q}^{\alpha ,p}\big \Vert \leqslant c\big \Vert f|{\dot{K}}_{q}^{\alpha ,p}B_{\beta }^{s}\big \Vert . \end{aligned}$$

Step 2. We put

$$\begin{aligned} H=\Bigg (\int _{0}^{1}t^{-s\beta }\big \Vert d_{t}^{M}f|{\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{\beta }\frac{\mathrm{d}t}{t}\Bigg )^{\frac{1}{\beta }} \end{aligned}$$

After a change of variable $t=2^{-y}$, we get

$$\begin{aligned} H=(\log 2)^{\frac{1}{\beta }}\Bigg (\int \limits _{1}^{+\infty }2^{ys\beta }\big \Vert d_{2^{-y}}^{M}f|{\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{\beta }\mathrm{d}y\Bigg )^{\frac{1}{ \beta }}. \end{aligned}$$

Then

$$\begin{aligned} H\leqslant c\Bigg (\sum \limits _{k=0}^{\infty }2^{ks\beta }\big \Vert d_{2^{-k}}^{M}f| {\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{\beta }\Bigg )^{\frac{1}{\beta }}. \end{aligned}$$

Let $\left\{ \varphi _{j}\right\} _{j}$ be a smooth dyadic resolution of unity. Obviously we need to estimate

$$\begin{aligned} \Bigg \{2^{ks}\sum \limits _{j=0}^{k}\big \Vert d_{2^{-k}}^{M}(\Lambda _{j}f)|{\dot{K}} _{q}^{\alpha ,p}\big \Vert \Bigg \}_{k} \end{aligned}$$

(5.1)

and

$$\begin{aligned} \Bigg \{2^{ks}\sum \limits _{j=k+1}^{\infty }\big \Vert d_{2^{-k}}^{M}(\Lambda _{j}f)| {\dot{K}}_{q}^{\alpha ,p}\big \Vert \Bigg \}_{k}, \end{aligned}$$

(5.2)

in $\ell ^{\beta }$-norm. Using the same arguments of [32, 2.5.10] we get

$$\begin{aligned} \left| \Delta _{h}^{M}\Lambda _{j}f\left( x\right) \right| \leqslant c 2^{\left( j-k\right) M}\Lambda _{j}^{*,a}f\left( x\right) , \end{aligned}$$

if $0\leqslant j\leqslant k$, $x\in \mathbb {R}^{n}$ and $\left| h\right| \leqslant 2^{-k}$, with $c>0$ independent of j, k, h and x, which yields that

$$\begin{aligned} d_{2^{-k}}^{M}(\Lambda _{j}f)(x)\leqslant c2^{\left( j-k\right) M}\Lambda _{j}^{*,a}f\left( x\right) , \end{aligned}$$

if $0\leqslant j\leqslant k,k\in \mathbb {N}_{0}$ and $x\in \mathbb {R}^{n}$. Since $s<M$ , (5.1) in $\ell ^{\beta }$-norm does not exceed

$$\begin{aligned} \Bigg (\sum \limits _{j=0}^{\infty }2^{js\beta }\big \Vert \Lambda _{j}^{*,a}f| {\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{\beta }\Bigg )^{\frac{1}{\beta }}\lesssim \big \Vert f|{\dot{K}}_{q}^{\alpha ,p}B_{\beta }^{s}\big \Vert , \end{aligned}$$

by Theorem 2.5. Let $j>k$. Recalling the definition of $d_{2^{-k}}^{M}(\Lambda _{j}f)$ and observe that

$$\begin{aligned} \int _{B}\left| \Lambda _{j}f(x+(M-i)2^{-k}h)\right| dh\leqslant c{\mathcal {M}}(\Lambda _{j}f)(x), \end{aligned}$$

if $j,k\in \mathbb {N}_{0},i\in \{0,\ldots ,M\},\left| h\right| \leqslant 2^{-k}$ and $x\in \mathbb {R}^{n}$. Therefore,

$$\begin{aligned} d_{2^{-k}}^{M}(\Lambda _{j}f)(x)\leqslant c{\mathcal {M}}(\Lambda _{j}f)(x) \end{aligned}$$

for any $j>k$ and $x\in \mathbb {R}^{n}$. Hence

$$\begin{aligned} 2^{ks}\sum \limits _{j=k+1}^{\infty }\big \Vert d_{2^{-k}}^{M}(\Lambda _{j}f)|{\dot{K}}_{q}^{\alpha ,p}\big \Vert \leqslant c2^{ks}\sum \limits _{j=k+1}^{\infty }\big \Vert {\mathcal {M}}(\Lambda _{j}f)|{\dot{K}}_{q}^{\alpha ,p}\big \Vert . \end{aligned}$$

Using Lemma 3.2, we obtain that (5.2) in $\ell ^{\beta }$-norm can be estimated from above by

$$\begin{aligned} \Bigg (\sum \limits _{j=0}^{\infty }2^{js\beta }\big \Vert {\mathcal {M}}(\Lambda _{j}f)| {\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{\beta }\Bigg )^{\frac{1}{\beta }}\lesssim \big \Vert f|{\dot{K}}_{q}^{\alpha ,p}B_{\beta }^{s}\big \Vert , \end{aligned}$$

where we used Lemma 3.4.

Step 3. Let $\Psi$ be a function in ${\mathcal {S}}(\mathbb {R}^{n})$ satisfying $\Psi (x)=1$ for $|x|\leqslant 1$ and $\Psi (x)=0$ for $|x|\geqslant \frac{3}{2}$, in addition radialsymmetric. We make use of an observation made by Nikol’skij [26] (see also [29]). We put

$$\begin{aligned} \psi \left( x\right) :=\left( -1\right) ^{M+1}\sum \limits _{i=0}^{M-1}\left( -1\right) ^{i}\left( \begin{array}{c} M \\ i \end{array} \right) \Psi \left( x\left( M-i\right) \right) . \end{aligned}$$

The function $\psi$ satisfies $\psi \left( x\right) =1$ for $\left| x\right| \leqslant 1/M$ and $\psi \left( x\right) =0$ for $\left| x\right| \geqslant 3/2$. Then, taking $\varphi _{0}\left( x\right) =\psi \left( x\right) ,\varphi _{1}\left( x\right) =\psi \left( x/2\right) -\psi \left( x\right)$ and $\varphi _{j}\left( x\right) =\varphi _{1}\left( 2^{-j+1}x\right)$ for $j=2,3,\ldots$, we obtain that $\left\{ \varphi _{j}\right\}$ is a smooth dyadic resolution of unity. This yields that

$$\begin{aligned} \Bigg (\sum _{j=0}^{\infty }2^{js\beta }\big \Vert {\mathcal {F}}^{-1}\varphi _{j}*f|{\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{\beta }\Bigg )^{1/\beta } \end{aligned}$$

is an norm equivalent in ${\dot{K}}_{q}^{\alpha ,p}{B_{\beta }^{s}}$. Let us prove that the last expression is bounded by

$$\begin{aligned} C\big \Vert f|{\dot{K}}_{q}^{\alpha ,p}{B_{\beta }^{s}}\big \Vert _{M}. \end{aligned}$$

(5.3)

Obviously from (2.2)

$$\begin{aligned} \big \Vert {\mathcal {F}}^{-1}\varphi _{0}*f|{\dot{K}}_{q}^{\alpha ,p}\big \Vert \lesssim \big \Vert f|{\dot{K}}_{q}^{\alpha ,p}\big \Vert \lesssim \big \Vert f| {\dot{K}}_{q}^{\alpha ,p}{B_{\beta }^{s}}\big \Vert _{M}. \end{aligned}$$

(5.4)

Moreover, it holds for $x\in \mathbb {R}^{n}$ and $j=1,2,\ldots$

$$\begin{aligned} {\mathcal {F}}^{-1}\varphi _{j}*f\left( x\right) :=\left( -1\right) ^{M+1}\int _{\mathbb {R}^{n}}\Delta _{2^{-j}y}^{M}f\left( x\right) {\widetilde{\Psi }}\left( y\right) \mathrm {d}y, \end{aligned}$$

with ${\widetilde{\Psi }}\left( \cdot \right) ={\mathcal {F}}^{-1}\Psi \left( \cdot \right) -2^{-n}{\mathcal {F}}^{-1}\Psi \left( \cdot /2\right)$. Now, for $j\in \mathbb {N}$ we write

$$\begin{aligned} \int _{\mathbb {R}^{n}}|\Delta _{2^{-j}y}^{M}f\left( x\right) ||{\widetilde{\Psi }}\left( y\right) |\mathrm {d}y= & {} \int _{\left| y\right| \leqslant 1}|\Delta _{2^{-j}y}^{M}f\left( x\right) ||{\widetilde{\Psi }}\left( y\right) | \mathrm {d}y\\&+\int _{\left| y\right| >1}|\Delta _{2^{-j}y}^{M}f\left( x\right) ||{\widetilde{\Psi }}\left( y\right) |\mathrm {d}y. \end{aligned}$$

Obviously we need only to estimate the second term. We have

$$\begin{aligned} \int _{\left| y\right| >1}|\Delta _{2^{-j}y}^{M}f\left( x\right) || {\widetilde{\Psi }}\left( y\right) |\mathrm {d}y\leqslant & {} \sum \limits _{k=0}^{\infty }\int _{{\overline{C}}_{k}}|\Delta _{2^{-j}y}^{M}f\left( x\right) ||{\widetilde{\Psi }}\left( y\right) |\mathrm {d} y \nonumber \\\leqslant & {} c\sum \limits _{k=0}^{\infty }2^{nj-Nk}\int _{{\overline{C}}_{k-j}}|\Delta _{h}^{M}f\left( x\right) |\mathrm {d}h, \end{aligned}$$

(5.5)

where ${\overline{C}}_{i}=\{x\in \mathbb {R}^{n}:2^{i}<\left| x\right| \leqslant 2^{i+1}\}$, $i\in \mathbb {N}_{0}$ and $N>0$ is at our disposal, and we have used the properties of the function ${\widetilde{\Psi }}$,

$$\begin{aligned} |{\widetilde{\Psi }}\left( y\right) |\leqslant c(1+\left| y\right| )^{-N} \end{aligned}$$

for any $y\in \mathbb {R}^{n}$ and any $N>n$. Now the right-hand side of (5.5) in $\ell _{\beta }^{s}({\dot{K}}_{q}^{\alpha ,p})$-norm is bounded by

$$\begin{aligned} c\sum \limits _{k=0}^{\infty }2^{-Nk}\Bigg (\sum \limits _{j=0}^{\infty }2^{(s+n)j\beta }(\omega _{j,k})^{\beta }\Bigg )^{1/\beta }, \end{aligned}$$

(5.6)

where

$$\begin{aligned} \omega _{j,k}=\Big \Vert \int _{{\overline{C}}_{k-j}}|\Delta _{h}^{M}f(\cdot )| \mathrm {d}h\Big |{\dot{K}}_{q}^{\alpha ,p}\Big \Vert ,\quad j,k\in \mathbb {N}_{0}. \end{aligned}$$

The sum (5.6) can be rewritten as

$$\begin{aligned}&c\sum \limits _{k=0}^{\infty }2^{-Nk}\Bigg (\sum \limits _{j=0}^{k+1}\cdot \cdot \cdot +\sum \limits _{j=k+2}^{\infty }\cdot \cdot \cdot \Bigg )^{1/\beta } \\&\quad \leqslant c\sum \limits _{k=0}^{\infty }2^{-Nk}\Bigg (\Bigg (\sum \limits _{j=0}^{k+1} \cdot \cdot \cdot \Bigg )^{1/\beta }+\Bigg (\sum \limits _{j=k+2}^{\infty }\cdot \cdot \cdot \Bigg )^{1/\beta }\Bigg ) \\&\quad =c\sum \limits _{k=0}^{\infty }2^{-Nk}(S_{k}^{_{1}}+S_{k}^{_{2}}). \end{aligned}$$

Clearly

$$\begin{aligned} (S_{k}^{_{1}})^{\beta }\leqslant c 2^{kn\beta }\big \Vert {\mathcal {M}}(f)| {\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{\beta }\sum \limits _{j=0}^{k+1}2^{sj\beta }\leqslant c 2^{(s+n)k\beta }\big \Vert f|{\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{\beta }, \end{aligned}$$

where $c>0$ is independent of k. Now let us estimate $S_{k}^{_{2}}$. We obtain

$$\begin{aligned} (S_{k}^{_{2}})^{\beta }\lesssim & {} 2^{\left( n+s\right) k\beta }\sum \limits _{v=1}^{\infty }2^{(s+n)v\beta }\Big \Vert \int _{\left| h\right| \leqslant 2^{-v}}|\Delta _{h}^{M}f|dh\Big |{\dot{K}}_{q}^{\alpha ,p} \Big \Vert ^{\beta } \\\lesssim & {} 2^{\left( n+s\right) k\beta }\sum \limits _{v=1}^{\infty }2^{sv\beta }\big \Vert d_{2^{-v}}^{M}(f)|{\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{\beta } \\\lesssim & {} 2^{\left( n+s\right) k\beta }\int \limits _{0}^{1}t^{-s\beta } \big \Vert d_{t}^{M}(f)|{\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{\beta }\frac{dt}{t}. \end{aligned}$$

Taking N large enough such that $N>s+n$, we estimate the right-hand side of (5.5) in $\ell _{\beta }^{s}({\dot{K}}_{q}^{\alpha ,p})$ -norm by

$$\begin{aligned} c\big \Vert f|{\dot{K}}_{q}^{\alpha ,p}B_{\beta }^{s}\big \Vert _{M}. \end{aligned}$$

This finishes the proof. $\square$

Proof of Proposition 4.4

The proof is based on [1]. Observe that $g_{\tau ,\alpha }\in {\dot{K}}_{q}^{\alpha ,p}$, because of $\tau +\frac{n}{q}>0$. We need to prove that

$$\begin{aligned} \sup _{0<t\leqslant 1}t^{-(\tau +\frac{n}{q})}\sup _{\left| h\right|<t} \big \Vert \Delta _{h}^{m}g|{\dot{K}}_{q}^{\alpha ,p}\big \Vert<\infty ,\quad 0<\tau +\frac{n}{q}<m,\quad m\in \mathbb {N}, \end{aligned}$$

since

$$\begin{aligned} \big \Vert d_{t}^{m}(g)|{\dot{K}}_{q}^{\alpha ,p}\big \Vert \leqslant \sup _{\left| h\right|<t}\big \Vert \Delta _{h}^{m}g|{\dot{K}}_{q}^{\alpha ,p}\big \Vert ,\quad 0<t\leqslant 1 \end{aligned}$$

We will present the proof into several steps.

Step 1. $\alpha =0$. Since $\tau >-\frac{n}{q}$, for any $1<p\leqslant q$ we have $\tau >-\frac{n}{p}$. From Theorem 2.6 we obtain

$$\begin{aligned} g\in B{_{p,\infty }^{\tau +\frac{n}{p}}}\hookrightarrow {\dot{K}} _{q}^{0,p}B_{\infty }^{\tau +\frac{n}{q}}, \end{aligned}$$

if $p\leqslant q$. Now if $p>q$ we have

$$\begin{aligned} g\in B{_{q,\infty }^{\tau +\frac{n}{q}}=}{\dot{K}}_{q}^{0,q}B_{\infty }^{\tau + \frac{n}{q}}\hookrightarrow {\dot{K}}_{q}^{0,p}B_{\infty }^{\tau +\frac{n}{q}}. \end{aligned}$$

Step 2. $-\frac{n}{q}<\alpha <0$. Can be covered by the similar arguments used in Step 1.

Step 3. $0<\alpha <\tau +\frac{n}{q}$.

Substep 3.1. We prove that $g\in {\dot{K}}_{q}^{\alpha ,p}B{ _{\infty }^{\tau +\frac{n}{q}}}$ if $0<\tau +\frac{n}{q}\leqslant 1$.

Substep 3.1.1. We consider the case of $0<\tau +\frac{n}{q }<1$. Clearly,

$$\begin{aligned} \big \Vert \triangle _{h}^{1}g|{\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{p}= & {} \sum \limits _{k=-\infty }^{\infty }2^{k\alpha p}\big \Vert \triangle _{h}^{1}g\,\chi _{k}\big \Vert _{q}^{p} \\= & {} \sum \limits _{k\in \mathbb {Z},2^{k}<2\left| h\right| }\cdot \cdot \cdot +\sum \limits _{k\in \mathbb {Z},2^{k}\geqslant 2\left| h\right| }\cdot \cdot \cdot \\= & {} H_{1}+H_{2}. \end{aligned}$$

We estimate each term on the right separately. First notice that

$$\begin{aligned} \big \Vert \triangle _{h}^{1}g\,\chi _{k}\big \Vert _{q}^{q}= & {} \int _{\{x:\left| x\right| \leqslant 2\left| h\right| \}\cap C_{k}}\left| \triangle _{h}^{1}g\left( x\right) \right| ^{q}\,\mathrm {d}x\,+\int _{\{x:\left| x\right| >2\left| h\right| \}\cap C_{k}}\left| \triangle _{h}^{1}g\left( x\right) \right| ^{q}\,\mathrm {d}x \\= & {} T_{1,k}(h)+T_{2,k}(h),\quad h\in \mathbb {R}^{n},k\in \mathbb {Z}.\\ \end{aligned}$$

$\bullet$ Estimation of $H_{1}$. Obviously $T_{2,k}(h)=0$ if $2^{k}<2\left| h\right| ,h\in \mathbb {R}^{n}\$and $k\in \mathbb {Z}$. Therefore, we need only to estimate $T_{1,k}(h)$. Observe that

$$\begin{aligned} \triangle _{h}^{1}g\left( x\right) =g\left( x+h\right) -g\left( x\right) \end{aligned}$$

and

$$\begin{aligned} |x+h|\leqslant |x|+|h|<3\left| h\right| ,\quad \text {if}\quad 2^{k}<2\left| h\right| \quad \text {and}\quad x\in C_{k}, \end{aligned}$$

which yields

$$\begin{aligned} T_{1,k}(h)\leqslant & {} 2^{q-1}\int _{\{x:|x+h|\leqslant 3\left| h\right| \}\cap C_{k}}\left| g\left( x+h\right) \right| ^{q}\,\mathrm {d} x\\&+2^{q}\int _{\{x:\left| x\right| \leqslant 2\left| h\right| \}\cap C_{k}}\left| g\left( x\right) \right| ^{q}\,\mathrm {d}x \\\leqslant & {} 2^{q}\int _{\left| x\right| \leqslant 3\left| h\right| }\left| g\left( x\right) \right| ^{q}\,\mathrm {d}x \\\leqslant & {} c\int _{0}^{3\left| h\right| }r^{\left( \tau -\alpha \right) q+n-1} \mathrm {d}r \\\leqslant & {} c\left| h\right| ^{(\tau +\frac{n}{q}-\alpha )q}, \end{aligned}$$

since $\alpha <\tau +\frac{n}{q}$ where $c>0$ is independent of k and h. Therefore

$$\begin{aligned} H_{1}\leqslant c\left| h\right| ^{(\tau +\frac{n}{q})p}\sum \limits _{k\in \mathbb {Z},2^{k}<2\left| h\right| }\Big (\frac{2^{k}}{|h|}\Big ) ^{\alpha p}\leqslant c\left| h\right| ^{(\tau +\frac{n}{q})p}, \end{aligned}$$

because of $\alpha >0$.

$\bullet$ Estimation of $H_{2}$. As before one finds that

$$\begin{aligned}&T_{1,k}(h)\leqslant \int _{\{x:|x+h|\leqslant 3\left| h\right| \}}\left| \triangle _{h}^{1}g\left( x\right) \right| ^{q}\,\mathrm {d}x\\&\quad \leqslant c\int _{\left| x\right| \leqslant 3\left| h\right| }\left| g\left( x\right) \right| ^{q}\,\mathrm {d}x\leqslant c\left| h\right| ^{(\tau +\frac{n}{q}-\alpha )q}, \end{aligned}$$

since again $\alpha <\tau +\frac{n}{q}$ where $c>0$ is independent of k and h. By the mean value theorem;

$$\begin{aligned} \left| \triangle _{h}^{1}g\left( x\right) \right| \leqslant c|h||x|^{\tau -\alpha -1},\quad \left| x\right| >2\left| h\right| , \end{aligned}$$

we obtain

$$\begin{aligned} T_{2,k}(h)\leqslant & {} \left| h\right| ^{q}\int _{C_{k}}\left| x\right| ^{\left( \tau -\alpha -1\right) q}\,\mathrm {d}x \\\leqslant & {} c\left| h\right| ^{q}2^{k(q\left( \tau -\alpha -1\right) +n)}\\= & {} c\left| h\right| ^{q}2^{kq(\tau +\frac{n}{q}-\alpha )}2^{-kq}. \end{aligned}$$

This yields that

$$\begin{aligned} H_{2}\leqslant & {} c\left| h\right| ^{p}\sum \limits _{k\in \mathbb {Z} ,2^{k}\geqslant 2\left| h\right| }2^{kp(\tau +\frac{n}{q}-1)} \\= & {} c\left| h\right| ^{(\tau +\frac{n}{q})p}\sum \limits _{k\in \mathbb { Z},2^{k}\geqslant 2\left| h\right| }\Big (\frac{2^{k}}{\left| h\right| }\Big )^{p(\tau +\frac{n}{q}-1)} \\\leqslant & {} c\left| h\right| ^{(\tau +\frac{n}{q})p}, \end{aligned}$$

since $\tau +\frac{n}{q}<1$. Therefore,

$$\begin{aligned} \big \Vert \triangle _{h}^{1}g|{\dot{K}}_{q}^{\alpha ,p}\big \Vert \leqslant c\left| h\right| ^{\tau +\frac{n}{q}},\quad \text { if }\quad 0<\alpha<\tau +\frac{ n}{q}<1. \end{aligned}$$

Substep 3.1.2. We consider the case of $\tau +\frac{n}{q}=1$. We have,

$$\begin{aligned} \big \Vert \triangle _{h}^{2}g|{\dot{K}}_{q}^{\alpha ,p}\big \Vert ^{p}= & {} \sum \limits _{k=-\infty }^{\infty }2^{k\alpha p}\big \Vert \triangle _{h}^{2}g\,\chi _{k}\big \Vert _{q}^{p} \\= & {} \sum \limits _{k\in \mathbb {Z},2^{k}<2\left| h\right| }\cdot \cdot \cdot +\sum \limits _{k\in \mathbb {Z},2^{k}\geqslant 2\left| h\right| }\cdot \cdot \cdot \\= & {} I_{1}+I_{2}. \end{aligned}$$

Observe that

$$\begin{aligned} \big \Vert \triangle _{h}^{2}g\,\chi _{k}\big \Vert _{q}^{q}= & {} \int _{\{x:\left| x\right| \leqslant 2\left| h\right| \}\cap C_{k}}\left| \triangle _{h}^{2}g\left( x\right) \right| ^{q}\,\mathrm {d}x\,+\int _{\{x:\left| x\right| >2\left| h\right| \}\cap C_{k}}\left| \triangle _{h}^{2}g\left( x\right) \right| ^{q}\,\mathrm {d}x \\= & {} V_{1,k}(h)+V_{2,k}(h),\quad h\in \mathbb {R}^{n},k\in \mathbb {Z}. \end{aligned}$$

$\bullet$ Estimation of $I_{1}$. Obviously $V_{2,k}(h)=0$ if $2^{k}<2\left| h\right| ,h\in \mathbb {R}^{n},k\in \mathbb {Z}$. Therefore, we need only to estimate $V_{1,k}(h)$. We have

$$\begin{aligned} \triangle _{h}^{2}g\left( x\right) =g\left( x+h\right) +g\left( x-h\right) -2g\left( x\right) \end{aligned}$$

and

$$\begin{aligned} |x-h|\leqslant |x|+|h|<3\left| h\right| ,\quad \text {if}\quad 2^{k}<2\left| h\right| \quad \text {and}\quad x\in C_{k}, \end{aligned}$$

which yields

$$\begin{aligned} V_{1,k}(h)\leqslant & {} 3^{q}\int _{\left| x\right| \leqslant 3\left| h\right| }\left| g\left( x\right) \right| ^{q}\,\mathrm {d}x \\= & {} c\int _{0}^{3\left| h\right| }r^{\left( \tau -\alpha \right) q+n-1} \mathrm {d}r \\= & {} c\left| h\right| ^{(1-\alpha )q}, \end{aligned}$$

since $\alpha <\tau +\frac{n}{q}=1$ where $c>0$ is independent of k and h . Therefore

$$\begin{aligned} I_{1}\leqslant c\left| h\right| ^{p}\sum \limits _{k\in \mathbb {Z} ,2^{k}<2\left| h\right| }\Big (\frac{2^{k}}{\left| h\right| } \Big )^{\alpha p}\leqslant c|h|^{p}, \end{aligned}$$

because of $\alpha >0$.

$\bullet$ Estimation of $I_{2}$. As before one finds

$$\begin{aligned} V_{1,k}(h)\leqslant c\int _{\left| x\right| \leqslant 3\left| h\right| }\left| g\left( x\right) \right| ^{q}\,\mathrm {d}x\leqslant c\left| h\right| ^{(1-\alpha )q}, \end{aligned}$$

where $c>0$ is independent of k and h. Again by the mean value theorem;

$$\begin{aligned} \left| \triangle _{h}^{2}g\left( x\right) \right| \leqslant c|h|^{2}|x|^{\tau -\alpha -2},\quad \left| x\right| >2\left| h\right| \end{aligned}$$

we obtain

$$\begin{aligned} V_{2,k}(h)\leqslant & {} \left| h\right| ^{2q}\int _{C_{k}}\left| x\right| ^{\left( \tau -\alpha -2\right) q}\,\mathrm {d}x \\\leqslant & {} c\left| h\right| ^{2q}2^{k(q\left( \tau -\alpha -2\right) +n)} \\= & {} c\left| h\right| ^{2q}2^{kq(\tau +\frac{n}{q}-\alpha )}2^{-2kq}. \end{aligned}$$

This yields that

$$\begin{aligned} I_{2}\leqslant & {} c\left| h\right| ^{2p}\sum \limits _{k\in \mathbb {Z} ,2^{k}\geqslant 2\left| h\right| }2^{-kp} \\= & {} c\left| h\right| ^{p}\sum \limits _{k\in \mathbb {Z},2^{k}\geqslant 2\left| h\right| }2^{-kp}\left| h\right| ^{p} \\\leqslant & {} c\left| h\right| ^{p}. \end{aligned}$$

Therefore,

$$\begin{aligned} \big \Vert \triangle _{h}^{2}g|{\dot{K}}_{q}^{\alpha ,p}\big \Vert \leqslant c\left| h\right| ,\quad \text {if}\quad 0<\alpha <\tau +\frac{n}{q}=1. \end{aligned}$$

Step 4. $\tau +\frac{n}{q}>1$. Using the same arguments of [39, Theorem 4.5]

$$\begin{aligned} \big \Vert g|{\dot{K}}_{q}^{\alpha ,p}{B_{\infty }^{\tau +\frac{n}{q}}}\big \Vert =\sum _{|\lambda |\leqslant m}\big \Vert g^{(\lambda )}|{\dot{K}}_{q}^{\alpha ,p}{ B_{\infty }^{\tau +\frac{n}{q}-m}}\big \Vert , \end{aligned}$$

is an equivalent quasi-norms on ${\dot{K}}_{q}^{\alpha ,p}{B_{\infty }^{\tau + \frac{n}{q}}}$. Let us prove by induction that $g^{(\lambda )}\in {\dot{K}} _{q}^{\alpha ,p}{B_{\infty }^{\tau +\frac{n}{q}-m}}$ for any $|\lambda |\leqslant m$ with $m=-[-\tau -\frac{n}{q}]-1$. The case $m=0$ is covered by the above steps. Observe that

$$\begin{aligned} 1\leqslant m<\tau +\frac{n}{q}<m+1. \end{aligned}$$

and $g^{(\lambda )}$ is a linear combination of

$$\begin{aligned} g_{j}=\left| x\right| ^{\tau -\alpha -j}\Theta \left( x\right) \omega \left( x\right) ,\quad j=0,\ldots ,|\lambda |, \end{aligned}$$

where $\Theta \in C^{\infty }\left( \mathbb {R}^{n}\backslash \{0\}\right)$ is homogeneous of degree 0 and $\omega \in {\mathcal {D}}(\mathbb {R}^{n})$. Since $j<\tau +\frac{n}{q}$, it follows that $g_{j}\in {\dot{K}}_{q}^{\alpha ,p}$. If $j\ne 0$, then by induction

$$\begin{aligned} g_{j}\in {\dot{K}}_{q}^{\alpha ,p}{B_{\infty }^{\tau +\frac{n}{q}-j}} \hookrightarrow {\dot{K}}_{q}^{\alpha ,p}{B_{\infty }^{\tau +\frac{n}{q}-m}.} \end{aligned}$$

Now if $j=0$, then by the fact that $\tau +\frac{n}{q}>m\geqslant 1$,

$$\begin{aligned} \big \Vert g_{0}\big \Vert _{{\dot{W}}_{q,1}^{\alpha ,p}}=\sum \limits _{|\beta |\leqslant 1} \big \Vert \frac{\partial ^{\beta }g_{0}}{\partial ^{\beta }x}\big \Vert _{{\dot{K}} _{q}^{\alpha ,p}}<\infty . \end{aligned}$$

Since ${\dot{W}}_{q,1}^{\alpha ,p}={\dot{K}}_{q}^{\alpha ,p}F_{2}^{1}$, see [24], we obtain

$$\begin{aligned} g_{0}\in {\dot{K}}_{q}^{\alpha ,p}{F}_{2}^{{1}}\hookrightarrow {\dot{K}} _{q}^{\alpha ,p}{F}_{\infty }^{\tau +\frac{n}{q}}\hookrightarrow {\dot{K}} _{q}^{\alpha ,p}{B_{\infty }^{\tau +\frac{n}{q}-m}.} \end{aligned}$$

The proof of Proposition 4.4 is complete. $\square$