1 Overview and main theorems

A lot of attention has been paid to so called sharp Gagliardo–Nirenberg inequalities. Such inequalities play a key role in the study of qualitative properties of some evolution PDEs (see, for example, [1, 6, 8, 14, 21, 32, 33]).

Let \(1 < p < n\) and \(1 \le q < r < p^*\), where \(p^* = \frac{np}{n-p}\) denotes the Sobolev critical exponent. Denote by \(D^{p,q}({\mathbb {R}}^n)\) the completion of \(C_0^\infty ({\mathbb {R}}^n)\) under the norm

$$\begin{aligned} ||u||_{D^{p,q}({\mathbb {R}}^n)} = \left( \,\, \int _{{\mathbb {R}}^n} |\nabla u|^p\; \mathrm{d}x \right) ^{\frac{1}{p}} + \left( \,\, \int _{{\mathbb {R}}^n} |u|^q\; \mathrm{d}x \right) ^{\frac{1}{q}}. \end{aligned}$$

The sharp Euclidean Gagliardo–Nirenberg inequality states that, for any function \(u \in D^{p,q}({\mathbb {R}}^n)\),

$$\begin{aligned} \left( \,\,\int _{{\mathbb {R}}^n} |u|^r\; \mathrm{d}x \right) ^{\frac{p}{r \theta }} \le A_0(p,q,r) \left( \,\,\int _{{\mathbb {R}}^n} |\nabla u|^p\; \mathrm{d}x \right) \left( \,\, \int _{{\mathbb {R}}^n} |u|^q\; \mathrm{d}x \right) ^{\frac{p(1 - \theta )}{\theta q}}, \end{aligned}$$
(1)

where \(\theta = \frac{np(r - q)}{r(q(p - n) + np)} \in (0,1)\) and \(A_0(p,q,r)\) is the best possible constant in this inequality, which is well defined thanks to the Euclidean Sobolev inequality.

The inequality (1) was introduced independently by Gagliardo and Nirenberg in [20] and [27]. Some particular cases are quite known. Indeed, in the limit case \(r = p^*\), (1) yields the well-known Euclidean Sobolev inequality introduced by Sobolev in [29]. The famous Nash inequality, introduced by Nash in [26], corresponds to \(p = 2, q = 1\) and \(\theta = n/(n +2)\). At last, the Moser inequality, introduced by Moser in [25], arises when \(p = 2, q = 2\) and \(\theta = n/(n + 2)\). According to Bakry et al. [5], non-sharp inequalities of type (1) are all equivalent for \(p \ge 1\) fixed and similar versions still hold when \(p \ge n\), whereas the Sobolev embedding is not valid in this case.

Over the past years, Some studies have been devoted to the search for extremal functions of (1). Different methods have been employed in this endeavor for certain parameters \(p, q\) and \(r\). Namely, Aubin [3] and Talenti [30] found extremal functions for Euclidean optimal Sobolev inequalities. Extremal functions to the sharp Nash inequality were found by Carlen and Loss in [9]. Besides, Cordero et al. [13] and Del Pino and Dolbeault [15] independently obtained extremal functions for the family of parameters \(p < q \le \frac{p(n-1)}{n-p}\) and \(r = \frac{p(q-1)}{p-1}\). In this case, the extremal functions are explicitly given by

$$\begin{aligned} u(x) = a \left( 1 + b |x|^{\frac{p}{p-1}}\right) ^{- \frac{p-1}{q-p}}, \end{aligned}$$

where \(a\) and \(b\) are positive constants. In particular, one easily sees that the set of extremals of (1) is not \(C^0\)-compact. The knowledge of extremal functions is open for several values of \(p, q\) and \(r\).

Let \((M,g)\) be a closed Riemannian manifold of dimension \(n \ge 2\) and let \(1 < p < n\) and \(1 \le q < r < p^*\). Denote by \(H^{1,p}(M)\) the Riemannian–Sobolev space defined as the completion of \(C^{\infty }(M)\) under the norm

$$\begin{aligned} ||u||_{H^{1,p}(M)} := \left( \,\, \int _{M} |\nabla _g u|^p\; \mathrm{d}v_g + \int _{M} |u|^p\; \mathrm{d}v_g \right) ^{1/p}. \end{aligned}$$

In [10], assuming \(1 < p \le 2\) and \(p < r\), it is proved the existence of a constant \(B\) such that the Riemannian Gagliardo–Nirenberg inequality

$$\begin{aligned} \left( \,\,\int _M |u|^r\; \mathrm{d}v_g \right) ^{\frac{p}{r \theta }}&\le \left( A_0(p,q,r,g) \int _M |\nabla _g u|^p\; \mathrm{d}v_g + B \int _M |u|^p\; \mathrm{d}v_g \right) \nonumber \\&\times \, \left( \,\,\int _M |u|^q\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}. \end{aligned}$$
(2)

holds for all \(u \in H^{1,p}(M)\), where \(\mathrm{d}v_g\) and \(\nabla _g\) denote, respectively, the Riemannian volume element and the gradient operator of \(g\) and \(A_0(p,q,r,g)\) stands for the first best possible constant in this inequality.

The case \(r=p^*\) and \(p=2\) was proved to be valid for some \(B\) by Hebey and Vaugon [22] and, independently, by Aubin and Li [4] and Druet [16] when \(1<p<2\), and generally non-valid for any \(B\) by Druet [17] when \(p>2\). The optimal Nash inequality, with \(p=2, q=1\) and \(\theta =n/(n+2)\), was obtained for some \(B\) by Humbert in [23] (see also [19]). Later, Brouttelande [7] extended its validity to \(p=2, 1\le q<r\) and \(q\le 2\le r<2+\frac{2}{n}q\). Closely related inequalities has been recently investigated by Chen and Sun in [12] for \(p>2\).

In a natural way, one then considers the sharp inequality

$$\begin{aligned} \left( \,\,\int _M |u|^r\; \mathrm{d}v_g \right) ^{\frac{p}{r \theta }}&\le \left( A_0(p,q,r,g) \int _M |\nabla _g u|^p\; \mathrm{d}v_g\right. \nonumber \\&\quad \left. + B_0(p,q,r,g) \int _M |u|^p\; \mathrm{d}v_g \right) \left( \,\,\int _M |u|^q\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}} \end{aligned}$$
(3)

which is also valid for all \(u \in H^{1,p}(M)\), where

$$\begin{aligned} B_0(p,q,r,g) := \min \{ B \in {\mathbb {R}}: (2)\ \text {is valid for all}\, u \in H^{1,p}(M) \}, \end{aligned}$$

and also the notion of extremal function as a non-zero function in \(H^{1,p}(M)\) which satisfies (3) with equality.

For \(r = p^*\), we refer the reader to the Druet and Hebey’s book [18] which is an excellent survey concerning the whole program of sharp Sobolev inequalities such as validity of saturated inequalities, existence of extremals, among others.

Let \(\mathcal{E}(p,q,r,g)\) be the set of all extremal functions \(u \in H^{1,p}(M)\) such that \(||u||_{L^r(M)} = 1\). A simple computation guarantees that each extremal function \(u_0 \in \mathcal{E}(p,q,r,g)\) satisfies an equation of kind

$$\begin{aligned} - \Delta _{p,g} u_0 + a |u_0|^{p - 2} u_0 + b |u_0|^{q - 2} u_0 = c |u_0|^{r - 2} u_0\quad \mathrm{on} \quad M\, \end{aligned}$$

where \(\Delta _{p,g} = -\mathrm{div}_g(|\nabla _g|^{p-2} \nabla _g)\) denotes the \(p\)-Laplace operator of \(g\) and \(a, b\) and \(c\) are positive constants. in particular, the elliptic regularity theory applied to this equation gives \(\mathcal{E}(p,q,r,g) \subset C^0(M)\). Note also that, by the strong maximum principle, extremal functions can be assumed positive on \(M\).

A first question is to know if \(\mathcal{E}(p,q,r,g)\) is non-empty. Another important one concerns with topological properties satisfied by \(\mathcal{E}(p,q,r,g)\) as, for example, if or not it is compact in the \(C^0\)-topology. This work answers positively both questions. The compactness is discussed into an uniform view point on the parameters \(p, q\) and \(r\). Two ingredients are essential in order this: results on continuity of \(A_0(p,q,r,g)\) and local boundedness of \(B_0(p,q,r,g)\) with respect to \(p, q\) and \(r\).

Namely, our main results are:

Theorem 1.1

Let \((M,g)\) be a closed Riemannian manifold of dimension \(n \ge 2\) and let \(1 \le q < r < p^*\). The set \(\mathcal{E}(p,q,r,g)\) is non-empty whenever \(1 < p < 2, p < r\) and \(1 \le q \le \frac{r}{r-p}\).

Theorem 1.2

Let \((M,g)\) be a closed Riemannian manifold of dimension \(n \ge 2\). For fixed parameters \(1 < p_1 \le p_2 < 2\) and \(1 \le q_1 \le q_2 < r_1 \le r_2 < p_1^*\) with \(p_2 < r_1, p_1 < r_2\) and \(q_2 \le \frac{r_2}{r_2 - p_1}\), the set \(\{ u \in \mathcal{E}(p,q,r,g) : \ p_1 \le p \le p_2,\ q_1 \le q \le q_2\) and \(r_1 \le r \le r_2 \}\) is compact in the \(C^0\)-topology. In particular, the same conclusion holds for each set \(\mathcal{E}(p,q,r,g)\), where \(1 < p < 2, p < r\) and \(q \le \frac{r}{r-p}\).

The study of the continuity of \(A_0(p,q,r,g)\) with respect to the triple \((p, q, r)\) can be translated in terms of the continuity of \(A_0(p,q,r)\) once these two best constants are equal whenever \(p \le r\).

When \(p < r\), it is natural to hope that \(A_0(p,q,r)\) continuously depends on \((p, q, r)\). Indeed, according to [14],

$$\begin{aligned} A_0(p,q,r)&= \frac{q - p}{p \sqrt{\pi }} \left( \frac{pq}{n(q - p)}\right) ^{\frac{1}{p}} \left( \frac{np - q(n - p)}{pq}\right) ^{\frac{1}{r}}\\&\times \left( \frac{\Gamma \left( \frac{q(p - 1)}{q - p}\right) \Gamma (\frac{n}{2} + 1)}{\Gamma \left( \frac{p - 1}{p(q - p)}(np - qn + qp)\right) \Gamma \left( \frac{n(p - 1)}{p} + 1\right) }\right) ^{\frac{1}{n}} \end{aligned}$$

for all \(p < q < \frac{p(n - 1)}{n - p}\) and \(r = \frac{p(q - 1)}{p - 1}\).

In [2], Agueh showed that \(A_0(p,q,r)\) can generally be splitted as

$$\begin{aligned} A_0(p,q,r) = D_0(p,q,r) m(p,q,r)^{\frac{nq-np-rp}{n(r-q)}}, \end{aligned}$$

where \(D_0(p,q,r)\) is explicitly given in terms of Gamma functions and \(m(p,q,r)\) is defined by

$$\begin{aligned} m(p,q,r) := \left\{ E_{p,q}(u) :\ u \in D^{p,q}({\mathbb {R}}^n) \quad \mathrm{and}\quad ||u||_{L^r({\mathbb {R}}^n)} = 1\right\} , \end{aligned}$$
(4)

where

$$\begin{aligned} E_{p,q}(u):= \frac{1}{p} \int _{{\mathbb {R}}^n} |\nabla u|^p\; \mathrm{d}x + \frac{1}{q} \int _{{\mathbb {R}}^n} |u|^q\; \mathrm{d}x. \end{aligned}$$

Using Corollary II.3 of [24], one concludes that the constant \(m(p,q,r)\) is attained for a positive function, which is radially symmetric, non-increasing, tends to \(0\) as \(|x| \rightarrow + \infty \) and satisfies the equation

$$\begin{aligned} - \Delta _p u + u^{q-1} = l(p,q,r) u^{r - 1} \quad \mathrm{in} \quad {\mathbb {R}}^n, \end{aligned}$$
(5)

where \(l(p,q,r)\) is a Lagrange multiplier. By using decaying properties for solutions of the above equation, we just establish the continuity of \(m(p,q,r)\) for the range \(1 < p < n\) and \(1 \le q < r < p^*\).

The proof of Theorem 2.1 and also of the local boundedness of \(B_0(p,q,r,g)\) are done by contradiction and are based on blow-up and concentration analyzes of minimizers associated to suitable functionals. Important additional difficulties arise in the concentration part when we seek to establish the desired contradiction. The ideas used for surrounding them are inspired in the recent paper [11]. Furthermore, our approach greatly simplifies that one made in the paper [10] devoted to the validity question for \(p < r\).

The complete proof of Theorems 1.1 and 1.2 will be carried out into four sections. Section 2 is dedicated to the proof of a result on continuity of \(A_0(p,q,r)\) which is stated as Theorem 2.1. In Sect. 3, we prove the bound of \(B_0(p,q,r,g)\) under the assumptions of Theorem 1.2 which is stated as Theorem 3.1. Finally, the proofs of existence of extremals and of compactness are done in Sects. 4 and 5, respectively.

2 Continuous dependence of \(A_0(p,q,r)\)

In this section, it is proved the following theorem:

Theorem 2.1

For each dimension \(n \ge 2\), the best constant \(A_0(p, q, r)\) is continuous on the set of parameters

$$\begin{aligned} 1 < p < n,\quad 1 \le q < r < p^*. \end{aligned}$$
(6)

In other words, given triples \((p_\alpha , q_\alpha , r_\alpha )\) converging to \((p_0, q_0, r_0)\) as \(\alpha \rightarrow + \infty \), if all these triples satisfy (6), then \(A_0(p_\alpha ,q_\alpha ,r_\alpha )\) converges to \(A_0(p_0,q_0,r_0)\) as \(\alpha \rightarrow + \infty \).

Let \(m(p,q,r)\) and \(l(p,q,r)\) be defined as in (4) and (5), respectively. Given \(\delta > 0\), one easily checks that these constants are bounded on all \((p,q,r)\) satisfying (6) with \(p \le n - \delta \). Indeed, fixed a nonzero function \(v \in C^{\infty }_0({\mathbb {R}}^n)\), we have

$$\begin{aligned} 0 \le m(p,q,r) \le E_{p,q}\left( \frac{v}{||v||_{L^r}}\right) \le C_1(n, \delta ) \end{aligned}$$
(7)

for all triple \((p,q,r)\) satisfying (6), where \(C_1(n, \delta )\) is a positive constant depending only on \(n\) and \(\delta \). In particular, the claim follows from

$$\begin{aligned} 0 \le l(p,q,r) \le \max \{p,q\} m(p,q,r) \le \max \left\{ n,\frac{n^2}{\delta }\right\} C_1(n, \delta ). \end{aligned}$$

Let us now describe a \(L^r\) decaying property satisfied by solutions of the problem (5).

Lemma 2.1

Let \(p_0, q_0\) and \(r_0\) be fixed numbers satisfying (6). Then, for any \(\delta _0 > 0\) small enough, there exist positive constants \(C_0\) and \(\zeta _0\), depending only on \(n\) and \(\delta _0\) such that, for any \((p,q,r)\) satisfying (6), \(p \in [p_0 - \delta _0, n - \delta _0]\) and \(q \in [1, p_0^*-\delta _0]\) and any positive radial minimizer \(u \in D^{p,q}({\mathbb {R}}^n)\) of \(m(p,q,r)\), one has

$$\begin{aligned} \int _{|x|>\rho } |u|^r\; \mathrm{d}x \le C_0 \rho ^{-\zeta _0} \end{aligned}$$

for all \(\rho \ge 1\). In particular, the above decaying holds for \((p,q,r)\) close enough to \((p_0,q_0,r_0)\).

Proof of Lemma 2.1

Let \(u \in D^{p,q}({\mathbb {R}}^n)\) be a positive radial minimizer of \(m(p,q,r)\). We next consider two distinct cases.

Assume first that \(q > p\). By Hölder’s inequality, one has

$$\begin{aligned} u^p(\rho )&= -p \int _{\rho }^{+ \infty } u^{p - 1} u' \mathrm{d}s = -p \int _{\rho }^{+ \infty } (u s^{\frac{n - 1}{q}})^{p - 1}u' s^{\frac{n-1}{p}} s^{(n - 1)( -\frac{p - 1}{q} - \frac{1}{p})} \mathrm{d}s\\&\le n ||\nabla u||_{L^p({\mathbb {R}}^n)} ||u||_{L^q({\mathbb {R}}^n)}^{p-1} \left( \int _\rho ^{+ \infty } s^{(n - 1)(- \frac{p - 1}{q} - \frac{1}{p}) t} \mathrm{d}s \right) ^{\frac{1}{t}}, \end{aligned}$$

where \(t=pq/[(q-p)(p-1)]\). By (7), we then derive

$$\begin{aligned} u^p(\rho ) \le C_1(n, \delta _0)\left( \,\,\int _\rho ^{+ \infty } s^{(n - 1)(-t + 1)} \mathrm{d}s \right) ^{\frac{1}{t}}. \end{aligned}$$

Because \(q \le p_0^* - \delta _0 < (p_0 - \delta _0)^* \le p^*\), the above inequality yields

$$\begin{aligned} u(\rho ) \le C_2(n, \delta _0) \rho ^{- \frac{ n - 1}{p} + \frac{n}{tp}} \end{aligned}$$
(8)

for all \(\rho > 0\) and \((p,q,r)\) as in the statement of lemma, where \(C_i(n, \delta _0), i=1,2\), are positive constants depending only on \(n\) and \(\delta _0\).

On the other hand, by Hölder’s inequality,

$$\begin{aligned} \int _{|x| > \rho } u^r \mathrm{d}x \le \left( \,\,\int _{|x| > \rho } u^q \mathrm{d}x\right) ^{\frac{p^*-r}{p^*-q}} \left( \,\,\int _{|x|>\rho } u^{p^*} \mathrm{d}x \right) ^{\frac{r-q}{p^*-q}}. \end{aligned}$$

By (7), the first right-hand side integral is bounded by a constant depending on \(n\) and \(\delta _0\). So, estimating the last integral with the aid (8), one obtains

$$\begin{aligned} \int _{|x|>\rho } u^r \mathrm{d}x \le C^*_0 \rho ^{- \zeta ^*_0} \end{aligned}$$

for all \(\rho \ge 1\), where \(C^*_0\) and \(\zeta ^*_0\) are positive constants depending only on \(n\) and \(\delta _0\).

In the case that \(q \le p\), Hölder’s inequality gives

$$\begin{aligned} u(\rho )^p = -p \int _\rho ^{+ \infty } (u s^{\frac{n - 1}{p}})^{p - 1} u' s^{\frac{n - 1}{p}} s^{- (n - 1)} \mathrm{d}s \le n ||\nabla u||_{L^p({\mathbb {R}}^n)} ||u||_{L^p({\mathbb {R}}^n)}^{p-1} \rho ^{-(n - 1)}. \end{aligned}$$

Applying an interpolation with respect to \(q\) and \(p^*\) and also (7), one derives

$$\begin{aligned} u(\rho ) \le C_3(n, \delta _0) \rho ^{- \frac{(n - 1)}{p}}, \end{aligned}$$

where \(C_3(n, \delta _0)\) is a positive constant depending only on \(n\) and \(\delta _0\).

Proceeding exactly as in the previous case, one gets

$$\begin{aligned} \int _{|x| > \rho } u^r \mathrm{d}x \le C^{**}_0 \rho ^{- \zeta ^{**}_0} \end{aligned}$$

for all \(\rho \ge 1\), where \(C^{**}_0\) and \(\zeta ^{**}_0\) are positive constants depending only on \(n\) and \(\delta _0\).

Finally, letting \(C_0 = \max \{C^*_0, C^{**}_0\}\) and \(\zeta _0 = \min \{\zeta ^*_0, \zeta ^{**}_0\}\), we conclude the proof.

We now are ready to prove the main result of this section.

Proof of Theorem 2.1

Let \((p_\alpha , q_\alpha , r_\alpha )\) and \((p_0, q_0, r_0)\) be triples satisfying (6) and such that \((p_\alpha , q_\alpha , r_\alpha )\) converges to \((p_0, q_0, r_0)\) as \(\alpha \rightarrow +\infty \). It suffices to show that there exists a subsequence, denoted also by \((p_\alpha , q_\alpha , r_\alpha )\), such that \(A_0(p_\alpha ,q_\alpha ,r_\alpha )\) converges to \(A_0(p_0,q_0,r_0)\) as \(\alpha \rightarrow +\infty \).

Let \(u_\alpha \in D^{p,q}({\mathbb {R}}^n)\) be a positive radial minimizer for \(m_\alpha = m(p_\alpha ,q_\alpha ,r_\alpha )\) such that \(||u_\alpha ||_{L^r({\mathbb {R}}^n)} = 1\). Thanks to the boundedness of \(m_\alpha \), we can apply the Moser iterative scheme to the Eq. (5) on concentric balls of radii \(R\). In particular, we find a positive constant \(C_0(R)\) depending on \(R\), so that

$$\begin{aligned} \Vert u_\alpha \Vert _{L^\infty (B_R)} \le C_0(R) \end{aligned}$$

for \(\alpha > 0\) large enough.

From the above estimate and elliptic regularity theory, one easily checks that \((u_\alpha )\) converges to \(u_0\) in \(C^1_{loc}({\mathbb {R}}^n)\), modulo a subsequence. This fact and Lemma 2.1 readily yield

$$\begin{aligned} 1=\int _{{\mathbb {R}}^n} u_\alpha ^{r_\alpha }\; \mathrm{d}x = \int _{B(0,\rho )} u_\alpha ^{r_\alpha }\; \mathrm{d}x + \int _{{\mathbb {R}}^n \setminus B(0,\rho )} u_\alpha ^{r_\alpha }\; \mathrm{d}x \quad \le \int _{B(0,\rho )} u_\alpha ^{r_\alpha }\; \mathrm{d}x + C_0 \rho ^{-\zeta _0}\;. \end{aligned}$$

Then, letting \(\alpha \rightarrow +\infty \), one obtains

$$\begin{aligned} 1 = \int _{{\mathbb {R}}^n} u_\alpha ^{r_\alpha }\; \mathrm{d}x \le \int _{B(0,\rho )} u_0^{r_0}\; \mathrm{d}x + C_0 \rho ^{-\zeta _0} \le \int _{{\mathbb {R}}^n} u_0^{r_0}\; \mathrm{d}x + C_0 \rho ^{-\zeta _0} \end{aligned}$$

for all \(\rho \ge 1\), so that

$$\begin{aligned} \int _{{\mathbb {R}}^n} u_0^{r_0}\; \mathrm{d}x \ge 1. \end{aligned}$$

Conversely,

$$\begin{aligned} \int _{B(0,\rho )} u_0^{r_0}\; \mathrm{}\mathrm{d}x = \lim _{\alpha \rightarrow + \infty } \int _{B(0,\rho )} u_\alpha ^r\; \mathrm{d}x \le 1, \end{aligned}$$

so that

$$\begin{aligned} \int _{{\mathbb {R}}^n} u_0^{r_0}\; \mathrm{d}x = 1. \end{aligned}$$

Let now \(\varphi \) be any function in \(C^{\infty }_0({\mathbb {R}}^n)\). One knows that

$$\begin{aligned} \left( \,\,\int _{{\mathbb {R}}^n} |\varphi |^{r_\alpha }\; \mathrm{d}x\right) ^{\frac{p_\alpha }{r_\alpha \theta _\alpha }} \le A_0(p_\alpha ,q_\alpha ,r_\alpha ) \left( \,\,\int _{{\mathbb {R}}^n} |\nabla \varphi |^{p_\alpha }\; \mathrm{d}x \right) \left( \,\,\int _{{\mathbb {R}}^n} |\varphi |^{q_\alpha }\; \mathrm{d}x \right) ^{\frac{p_\alpha (1 - \theta _\alpha )}{\theta _\alpha q_\alpha }}. \end{aligned}$$
(9)

Letting \(\alpha \rightarrow +\infty \), it follows that

$$\begin{aligned} \left( \,\,\int _{{\mathbb {R}}^n} |\varphi |^{r_0}\; \mathrm{d}x\right) ^{\frac{p_0}{r_0\theta _0}} \le \liminf _{\alpha \rightarrow + \infty } A_0(p_\alpha ,q_\alpha ,r_\alpha ) \left( \,\, \int _{{\mathbb {R}}^n} |\nabla \varphi |^{p_0}\; \mathrm{d}x \right) \left( \,\,\int _{{\mathbb {R}}^n} |\varphi |^{q_0}\; \mathrm{d}x \right) ^{\frac{p_0(1 - \theta _0)}{\theta _0 q_0}}\; , \end{aligned}$$

so that

$$\begin{aligned} A_0(p_0,q_0,r_0) \le \liminf _{\alpha \rightarrow + \infty } A_0(p_\alpha ,q_\alpha ,r_\alpha ). \end{aligned}$$
(10)

On the other hand, as proved in Theorem 2.1 of [2], \(u_\alpha \) is an extremal function for the inequality (9). Therefore,

$$\begin{aligned}&\left( \,\,\int _{B(0,\rho )} |\nabla u_0|^{p_0}\; \mathrm{d}x \right) \left( \int _{B(0,\rho )} u_0^{q_0}\; \mathrm{d}x \right) ^{\frac{p_0(1 - \theta _0)}{\theta _0 q_0}}\\&\quad = \lim _{\alpha \rightarrow + \infty } \left( \,\,\int _{B(0,\rho )} |\nabla u_\alpha |^{p_\alpha }\; \mathrm{d}x \right) \left( \,\,\int _{B(0,\rho )} u_\alpha ^{q_\alpha }\; \mathrm{d}x \right) ^{\frac{p_\alpha (1 - \theta _\alpha )}{\theta _\alpha q_\alpha }}\\&\quad \le \liminf _{\alpha \rightarrow + \infty } \left( \,\,\int _{{\mathbb {R}}^n} |\nabla u_\alpha |^{p_\alpha }\; \mathrm{d}x\right) \left( \,\,\int _{{\mathbb {R}}^n} u_\alpha ^{q_\alpha }\; \mathrm{d}x \right) ^{\frac{p_\alpha (1 - \theta _\alpha )}{\theta _\alpha q_\alpha }}\\&\quad =\liminf _{\alpha \rightarrow + \infty } A_0(p_\alpha ,q_\alpha ,r_\alpha )^{-1}\\&\quad = \left( \limsup _{\alpha \rightarrow + \infty } A_0(p_\alpha ,q_\alpha ,r_\alpha ) \right) ^{-1}, \end{aligned}$$

so that

$$\begin{aligned} \left( \,\,\int _{{\mathbb {R}}^n} |\nabla u_0|^{p_0}\; \mathrm{d}x \right) \left( \,\,\int _{{\mathbb {R}}^n} u_0^{q_0}\; \mathrm{d}x \right) ^{\frac{p_0(1 - \theta _0)}{\theta _0 q_0}} \le \left( \limsup _{\alpha \rightarrow + \infty } A_0(p_\alpha ,q_\alpha ,r_\alpha ) \right) ^{-1}. \end{aligned}$$

Since \(u_0 \in D^{p_0,q_0}({\mathbb {R}}^n)\) and \(\Vert u_0\Vert _{L^r} = 1\), one has

$$\begin{aligned} \limsup _{\alpha \rightarrow + \infty } A_0(p_\alpha ,q_\alpha ,r_\alpha ) \le A_0(p_0,q_0,r_0). \end{aligned}$$
(11)

Finally, from (10) and (11), we conclude that

$$\begin{aligned} \lim _{\alpha \rightarrow + \infty } A_0(p_\alpha ,q_\alpha ,r_\alpha ) = A_0(p_0,q_0,r_0). \end{aligned}$$

3 Boundedness of \(B_0(p,q,r,g)\)

Our goal in this section is to establish the following result on bound of \(B_0(p,q,r,g)\):

Theorem 3.1

Let \((M,g)\) be a closed Riemannian manifold of dimension \(n \!\ge \! 2\). For fixed parameters \(1 < p_1 < p_2 < 2\) and \(1 \le q_1 < q_2 < r_1 < r_2 < p_1^*\) with \(p_2 < r_1\), there exists a constant \(K > 0\) such that \(B_0(p,q,r,g) \le K\) for all \(p_1 \le p \le p_2, q_1 \le q \le q_2\) and \(r_1 \le r \le r_2\).

The proof of this theorem is done into several claims and, in order to make the simpler notations, we denote \(\alpha = (p,q,r), \alpha _0 = (p_0,q_0,r_0), \theta = \theta (p,q,r)\) and \(\theta _0 = \theta (p_0,q_0,r_0)\). Here we assume \(\alpha \) converges to \(\alpha _0\).

From now on, several possibly different positive constants independent of \(\alpha \) will be denoted by \(c\) or \(c_i, i = 1, 2, \ldots \)

Let \(\kappa \in (0,1)\) be a fixed number. From the definition of \(B_0(p,q,r,g)\), we have

$$\begin{aligned} \nu _\alpha = \inf _{u \in E} J_\alpha (u) < A_0(p,q,r)^{-1}, \end{aligned}$$
(12)

where \(E = \{ u \in H^{1,p}(M):\; ||u||_{L^r(M)} = 1 \}\) and

$$\begin{aligned} J_\alpha (u) = \left( \,\,\int _M |\nabla _g u|^p\; \mathrm{d}v_g + C_\alpha \int _M |u|^p\; \mathrm{d}v_g \right) \left( \,\,\int _M |u|^q\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}} \end{aligned}$$

with \(C_\alpha = \frac{B_0(p,q,r,g)}{A_0(p,q,r)} \kappa \).

Since \(J_\alpha \) is of class \(C^1\), by using standard variational arguments, we find a minimizer \(u_\alpha \in E\) of \(J_\alpha \), i.e.

$$\begin{aligned} J_\alpha (u_\alpha ) = \nu _\alpha = \inf _{u \in E} J_\alpha (u). \end{aligned}$$
(13)

One may assume \(u_\alpha \ge 0\), since \(\nabla _g |u_\alpha | = \pm \nabla _g u_\alpha \). Each minimizer \(u_\alpha \) satisfies the Euler–Lagrange equation

$$\begin{aligned} A_\alpha \Delta _{p,g} u_\alpha + C_\alpha A_\alpha u_\alpha ^{p - 1} + \frac{1 - \theta }{\theta } \nu _\alpha ||u_\alpha ||_{L^q(M)}^{-q} u_\alpha ^{q - 1} = \frac{\nu _\alpha }{\theta } u_\alpha ^{r - 1}\ \ \mathrm{on}\ \ M, \end{aligned}$$
(14)

where \(\Delta _{p,g} = -\mathrm{div}_g(|\nabla _g|^{p-2} \nabla _g)\) is the \(p\)-Laplace operator of \(g\) and

$$\begin{aligned} A_\alpha = \left( \,\,\int _M u_\alpha ^q\; \mathrm{d}v_g \right) ^{\frac{p( 1 - \theta )}{\theta q}}. \end{aligned}$$

By the elliptic regularity theory [31], it follows that \(u_\alpha \) is of class \(C^1(M)\).

The proof is now carried out by contradiction. namely, assume \(B_0(p,q,r,g)\) is not bounded as \(\alpha \rightarrow \alpha _0\).

Thanks to Theorem 2.1, up to a subsequence, we have

$$\begin{aligned} \lim _{\alpha \rightarrow \alpha _0} C_\alpha = + \infty , \end{aligned}$$

where \(\alpha _0 = (p_0, q_0, r_0)\) with \(p_1 \le p_0 \le p_2, q_1 \le q_0 \le q_2\) and \(r_1 \le r_0 \le r_2\).

From (12) and (13), one gets

$$\begin{aligned} C_\alpha A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g < A_0(p,q,r)^{-1}, \end{aligned}$$

so that

$$\begin{aligned} A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g \rightarrow 0. \end{aligned}$$
(15)

One also knows that

$$\begin{aligned} A_0(p,q,r)^{-1} \le A_\alpha \left( \,\,\int _M |\nabla _g u_\alpha |^p\; \mathrm{d}v_g + C_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g \right) + \frac{\kappa }{A_0(p,q,r)} A_\alpha \int _M u_\alpha ^p \mathrm{d}v_g. \end{aligned}$$

Letting \(\alpha \rightarrow \alpha _0\) and evoking again Theorem 2.1, one obtains

$$\begin{aligned} \liminf _{\alpha \rightarrow \alpha _0} J_\alpha (u_\alpha ) \ge A(p_0,q_0,r_0)^{-1}. \end{aligned}$$

So, by (12), one has

$$\begin{aligned} \lim _{\alpha \rightarrow \alpha _0} \nu _\alpha = \lim _{\alpha \rightarrow \alpha _0} J_\alpha (u_\alpha ) = A(p_0,q_0,r_0)^{-1}. \end{aligned}$$
(16)

Finally, we assert that

$$\begin{aligned} \lim _{\alpha \rightarrow \alpha _0} A_\alpha = 0. \end{aligned}$$
(17)

Otherwise, if \(\limsup _{\alpha \rightarrow \alpha _0} A_\alpha > 0\), up to a subsequence, we can assume \(\lim _{\alpha \rightarrow \alpha _0} A_\alpha > 0\). Then, by (12) and (15) (instead of using that \(p \le r\)), there exists a constant \(c > 0\) such that

$$\begin{aligned} ||u_\alpha ||_{H^{1,p}(M)} \le c \end{aligned}$$

for \(\alpha \) close enough to \(\alpha _0\).

Because \(p_0 < r_0 < p_0^*\) and \(p\) and \(r\) tend respectively to \(p_0\) and \(r_0\), we can choose \(t < p_0\) and \(s\) so that \(p, r < s < t^*\). So, one easily deduces that \((u_\alpha )\) is bounded in \(H^{1,t}(M)\) for \(\alpha \) close enough to \(\alpha _0\) and, by compactness, \(u_\alpha \rightarrow u\) in \(L^s(M)\). Therefore,

$$\begin{aligned} ||u_\alpha - u||_{L^p(M)} \rightarrow 0 \end{aligned}$$

and

$$\begin{aligned} ||u_\alpha - u||_{L^r(M)} \rightarrow 0 \end{aligned}$$

as \(\alpha \rightarrow \alpha _0\).

From the first above limit and (15),

$$\begin{aligned} ||u_\alpha ||_{L^p(M)} \rightarrow ||u||_{L^{p_0}(M)} = 0 \end{aligned}$$

and from the second one,

$$\begin{aligned} 1 = ||u_\alpha ||_{L^r(M)} \rightarrow ||u||_{L^{r_0}(M)} \end{aligned}$$

as \(\alpha \rightarrow \alpha _0\). This contradiction concludes the claim (17).

Let \(x_\alpha \in M\) be a maximum point of \(u_\alpha \), i.e

$$\begin{aligned} u_\alpha (x_\alpha ) = ||u_\alpha ||_{L^\infty (M)}. \end{aligned}$$
(18)

Claim 1

We assert that

$$\begin{aligned} \lim _{\sigma \rightarrow + \infty } \lim _{\alpha \rightarrow \alpha _0} \int _{B(x_\alpha ,\sigma a_\alpha )} u_\alpha ^r\; \mathrm{d}v_g = 1, \end{aligned}$$
(19)

where

$$\begin{aligned} a_\alpha = A_\alpha ^{\frac{r}{np - nr + pr}}. \end{aligned}$$
(20)

Proof of Claim 1

By (17), it is clear that \(a_\alpha \rightarrow 0\) as \(\alpha \rightarrow \alpha _0\).

For \(x \in B(0, \sigma )\), set

$$\begin{aligned} \begin{array}{l} h_\alpha (x) = g(\exp _{x_\alpha } (a_\alpha x)),\\ \varphi _\alpha (x) = a_\alpha ^{\frac{n}{r}} u_\alpha (\exp _{x_\alpha } (a_\alpha x)). \end{array} \end{aligned}$$
(21)

Joining (14) and the definition of \(\theta \), one easily checks that

$$\begin{aligned} \Delta _{p,h_\alpha } \varphi _\alpha + C_\alpha a_\alpha ^p \varphi _\alpha ^{p - 1} + \frac{1 - \theta }{\theta } \nu _\alpha \; \varphi _\alpha ^{q - 1} = \frac{\nu _\alpha }{\theta } \varphi _\alpha ^{r - 1} \quad \mathrm{on} \quad B(0, \sigma ). \end{aligned}$$
(22)

A Moser’s iterative scheme applied to (22) (see [28]) produces

$$\begin{aligned} a_\alpha ^{n} ||u_\alpha ||_{L^\infty (M)}^r = \sup _{B(0,\frac{\sigma }{2})} \varphi _\alpha ^r \le c \int _{B(0,\sigma )} \varphi _\alpha ^r\; \mathrm{d}h_\alpha = c \int _{B(x_\alpha , \sigma a_\alpha )} u_\alpha ^r\; \mathrm{d}v_g \le c \end{aligned}$$

for \(\alpha \) close enough to \(\alpha _0\). This estimate together with

$$\begin{aligned} 1 = \int _M u_\alpha ^r\; \mathrm{d}v_g \le ||u_\alpha ||_{L^\infty (M)}^{r - q} \int _M u_\alpha ^q \mathrm{d}v_g = \left( ||u_\alpha ||_{L^\infty (M)} \; a_\alpha ^{\frac{n}{r}}\right) ^{r-q} \end{aligned}$$

yield

$$\begin{aligned} 1 \le ||u_\alpha ||_{L^\infty (M)} \; a_\alpha ^{\frac{n}{r}} \le c. \end{aligned}$$
(23)

In particular, there exists a constant \(c > 0\) such that

$$\begin{aligned} \int _{B(0,\sigma )} \varphi _\alpha ^r\; \mathrm{d}h_\alpha \ge c \end{aligned}$$
(24)

for \(\alpha \) close enough to \(\alpha _0\).

On the other hand, we have

$$\begin{aligned} \int _{B(0,\sigma )} \varphi _\alpha ^p\; \mathrm{}\mathrm{d}x \le c \int _{B(0,\sigma )} \varphi _\alpha ^p\; \mathrm{d}h_\alpha = a_\alpha ^{\frac{np}{r} - n} \int _{B(x_\alpha ,\sigma a_\alpha )} u_\alpha ^p\; \mathrm{d}v_g \le c(\sigma ) a_\alpha ^{\frac{np}{r}} ||u_\alpha ||_{L^\infty (M)}^p \le c(\sigma ), \end{aligned}$$

with \(c(\sigma ) \rightarrow +\infty \) as \(\sigma \rightarrow + \infty \).

Moreover,

$$\begin{aligned} \int _{B(0,\sigma )} |\nabla \varphi _\alpha |^p\; \mathrm{d}x \le c \int _{B(0,\sigma )} |\nabla _{h_\alpha } \varphi _\alpha |^p\; \mathrm{d}h_\alpha = A_\alpha \int _{B(x_\alpha ,\sigma a_\alpha )} |\nabla _g u_\alpha |^p\; \mathrm{d}v_g \le A_0(p,q,r)^{-1}. \end{aligned}$$
(25)

Let \(1 < t < p_0\). For \(\alpha \) close enough to \(\alpha _0\), the above inequalities imply that \((\varphi _\alpha )\) is bounded in \(H^{1,t}(B(0,\sigma ))\) for each \(\sigma >0\). So, modulo a subsequence, we derive the pointwise convergence \(\varphi _\alpha \rightarrow \varphi \) almost everywhere in \({\mathbb {R}}^n\). By Fatou’s Lemma,

$$\begin{aligned} \int _{B(0,\sigma )} \varphi ^{q_0}\; \mathrm{d}x&= \liminf _{\alpha \rightarrow \alpha _0} \int _{B(0,\sigma )} \varphi _\alpha ^q\; \mathrm{d}h_\alpha = \liminf _{\alpha \rightarrow \alpha _0} \frac{\int _{B(x_\alpha ,\sigma a_\alpha )} u_\alpha ^q\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} \le 1,\end{aligned}$$
(26)
$$\begin{aligned} \int _{B(0,\sigma )} \varphi ^{r_0}\; \mathrm{d}x&= \liminf _{\alpha \rightarrow \alpha _0} \int _{B(0,\sigma )} \varphi _\alpha ^r\; \mathrm{d}h_\alpha = \liminf _{\alpha \rightarrow \alpha _0} \int _{B(x_\alpha ,\sigma a_\alpha )} u_\alpha ^r\; \mathrm{d}v_g \le 1. \end{aligned}$$
(27)

In particular,

$$\begin{aligned} \varphi \in L^{q_0}({\mathbb {R}}^n) \cap L^{r_0}({\mathbb {R}}^n). \end{aligned}$$

In addition, proceeding as before, it is possible to choose \(t < p_0\) and \(s\) so that \(q, r < s < t^*\). Thus, for any \(\sigma >0\), we can assume

$$\begin{aligned} ||\varphi _\alpha - \varphi ||_{L^q(B(0,\sigma ))} \rightarrow 0 \end{aligned}$$
(28)

and

$$\begin{aligned} ||\varphi _\alpha - \varphi ||_{L^r(B(0,\sigma ))} \rightarrow 0 \end{aligned}$$
(29)

as \(\alpha \rightarrow \alpha _0\).

Let \(\eta \in C_0^1({\mathbb {R}})\) be a cutoff function such that \(\eta = 1\) on \([0,\frac{1}{2}], \eta = 0\) on \([1,\infty )\) and \(0 \le \eta \le 1\). Set now \(\eta _{\alpha ,\sigma }(x) = \eta ((\sigma a_\alpha )^{-1} d_g(x,x_\alpha ))\). Taking \(u_\alpha \eta _{\alpha ,\sigma }^p\) as a test function in (14), one gets

$$\begin{aligned}&A_\alpha \int _M |\nabla _g u_\alpha |^p \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g {+} A_\alpha \int _M |\nabla _g u_\alpha |^{p {-} 2} \nabla _g u_\alpha \cdot \nabla _g (\eta _{\alpha ,\sigma }^p) u_\alpha \; \mathrm{d}v_g {+} C_\alpha A_\alpha \int _M u_\alpha ^p \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g \nonumber \\&\quad + \frac{1 - \theta }{\theta } \nu _\alpha ||u_\alpha ||_{L^q(M)}^{-q} \int _M u_\alpha ^q \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g = \frac{\nu _\alpha }{\theta } \int _M u_\alpha ^r \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g. \end{aligned}$$
(30)

next we show that

$$\begin{aligned} \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} A_\alpha \int _M |\nabla _g u_\alpha |^{p - 2} \nabla _g u_\alpha \cdot \nabla _g (\eta _{\alpha ,\sigma }^p) u_\alpha \; \mathrm{d}v_g = 0. \end{aligned}$$
(31)

Indeed, it suffices to guarantee that

$$\begin{aligned} \lim _{\sigma \rightarrow \infty } \lim _{\alpha \rightarrow \alpha _0} A_\alpha \int _M u_\alpha ^p |\nabla _g \eta _{\alpha ,\sigma }|^p\; \mathrm{d}v_g = 0. \end{aligned}$$
(32)

Thanks to the inequality \(|\nabla _g \eta _{\alpha ,\sigma }| \le \frac{c}{\sigma a_\alpha }\) and (20), one obtains

$$\begin{aligned} A_\alpha \int _{M} u_\alpha ^p |\nabla _g \eta _{\alpha ,\sigma }|^p\; \mathrm{d}v_g&\le c \frac{A_\alpha }{\sigma ^p a_\alpha ^p} \int _{B(x_\alpha ,\sigma a_\alpha )} u_\alpha ^p\; \mathrm{d}v_g\\&\le c \frac{A_\alpha }{\sigma ^p a_\alpha ^p} \left( \,\,\int _M u_\alpha ^r\; \mathrm{d}v_g \right) ^{\frac{p}{r}} \left( \,\,\int _{B(x_\alpha , \sigma a_\alpha )}\; \mathrm{d}v_g\right) ^{1 - \frac{p}{r}}\\&= c \sigma ^{\frac{nr -np -pr}{r}} \end{aligned}$$

which clearly converges to \(0\) as \(\alpha \rightarrow \alpha _0\) and \(\sigma \rightarrow + \infty \).

Replacing (31) in (30), one arrives at

$$\begin{aligned}&\theta _0 A(p_0,q_0,r_0) \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \left( A_\alpha \int _M |\nabla _g u_\alpha |^p \eta _{\alpha , \sigma }^p\; \mathrm{d}v_g \right) \nonumber \\&\qquad + (1 - \theta _0) \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \frac{\int _M u_\alpha ^q \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g}\le \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \int _M u_\alpha ^r \eta _{\alpha , \sigma }^p\; \mathrm{d}v_g, \end{aligned}$$
(33)

where \(\theta _0 = \theta (p_0, q_0, r_0)\). In order to rewrite this inequality in a more suitable format, we first remark that

$$\begin{aligned} \left| \frac{\int _M u_\alpha ^q \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} - \frac{\int _M u_\alpha ^q \eta _{\alpha ,\sigma }^q\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} \right| \le \frac{\int _{B(x_\alpha ,\sigma a_\alpha ) \setminus B(x_\alpha , \sigma a_\alpha /2)} u_\alpha ^q\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} {=}\!\!\!\! \int _{B(0,\sigma ) \setminus B(0,\sigma /2)} \varphi _\alpha ^q\; \mathrm{d}h_\alpha . \end{aligned}$$

So, thanks to (28) and the fact that \(\varphi \in L^{q_0}({\mathbb {R}}^n)\), one has

$$\begin{aligned} \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \frac{\int _M u_\alpha ^q \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} = \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \frac{\int _M u_\alpha ^q \eta _{\alpha ,\sigma }^q\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g}. \end{aligned}$$

Estimating

$$\begin{aligned} \left| \int _M u_\alpha ^r \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g - \int _M u_\alpha ^r\eta _{\alpha ,\sigma }^r\; \mathrm{d}v_g \right| \le \int _{B(x_\alpha ,\sigma a_\alpha ) \setminus B(x_\alpha ,(\sigma a_\alpha )/2)} u_\alpha ^r\; \mathrm{d}v_g = \int _{B(0,\sigma )\setminus B(0,\sigma /2)} \varphi _\alpha ^r\; \mathrm{d}h_\alpha \end{aligned}$$

and arguing in a similar way, by (29), one gets

$$\begin{aligned} \lim _{\sigma \rightarrow + \infty } \lim _{\alpha \rightarrow \alpha _0} \int _M u_\alpha ^r \eta _{\alpha ,\sigma }^r\; \mathrm{d}v_g = \lim _{\sigma \rightarrow + \infty } \lim _{\alpha \rightarrow \alpha _0} \int _M u_\alpha ^r \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g. \end{aligned}$$

Consequently, (33) can be rewritten as

$$\begin{aligned}&\theta _0 A(p_0,q_0,r_0) \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \left( A_\alpha \int _M |\nabla _g u_\alpha |^p \eta _{\alpha , \sigma }^p\; \mathrm{d}v_g \right) \nonumber \\&\qquad + (1 - \theta _0) \lim _{\sigma +\rightarrow \infty } \lim _{\alpha \rightarrow \alpha _0} \frac{\int _M u_\alpha ^q \eta _{\alpha ,\sigma }^q\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g}\nonumber \\&\quad \le \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0}\int _M u_\alpha ^r \eta _{\alpha , \sigma }^r\; \mathrm{d}v_g. \end{aligned}$$
(34)

On the other hand,

$$\begin{aligned} \left( \int _M u_\alpha ^r \eta ^r_{\alpha ,\sigma }\; \mathrm{d}v_g \right) ^{\frac{p}{r \theta }}&\le \left( A_0(p,q,r) \int _M |\nabla _g(u_\alpha \eta _{\alpha ,\sigma })|^p\; \mathrm{d}v_g \right. \\&\left. + B_0(p,q,r,g) \int _M u_\alpha ^p \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g \right) \left( \int _M u_\alpha ^q \eta ^q_{\alpha ,\sigma }\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}} \end{aligned}$$

and the definition of \(A_\alpha \) lead to

$$\begin{aligned} \left( \int _M u_\alpha ^r \eta ^r_{\alpha ,\sigma }\; \mathrm{d}v_g \right) ^{\frac{p}{r \theta }}&\le (A_0(p,q,r) + \varepsilon ) \left( \int _M |\nabla _g u_\alpha |^p \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g\right) \left( \int _M u_\alpha ^q \eta _{\alpha ,\sigma }^q\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}\nonumber \\&+ c(\varepsilon ) \; A_\alpha \int _M u_\alpha ^p |\nabla _g \eta _{\alpha ,\sigma }|^p\; \mathrm{d}v_g + C_\alpha A_\alpha \int _M u_\alpha ^p \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g. \end{aligned}$$
(35)

Using then (14) and (32) and letting \(\alpha \rightarrow \alpha _0, \sigma \rightarrow +\infty \) and \(\varepsilon \rightarrow 0\), one gets

$$\begin{aligned}&\left( \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \left( \int _M u_\alpha ^r \eta _{\alpha ,\sigma }^r\; \mathrm{d}v_g \right) \right) ^{\frac{p_0}{r_0 \theta _0}}\nonumber \\&\quad \le A(p_0,q_0,r_0) \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \left( A_\alpha \int _M |\nabla _g u_\alpha |^p \eta _{\alpha ,\sigma }^p\; \mathrm{d}v_g \right) \nonumber \\&\qquad \times \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \left( \frac{\int _M u_\alpha ^q \eta _{\alpha ,\sigma }^q\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} \right) ^{\frac{p_0(1 - \theta _0)}{\theta _0 q_0}}. \end{aligned}$$
(36)

Let

$$\begin{aligned} X&= A(p_0,q_0,r_0) \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \left( A_\alpha \; \int _M |\nabla _g u_\alpha |^p \eta _{\alpha ,\sigma }^p \mathrm{d}v_g \right) , \\ Y&= \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \frac{\int _M u_\alpha ^q \eta _{\alpha ,\sigma }^q \mathrm{d}v_g}{\int _M u_\alpha ^q \mathrm{d}v_g}, \end{aligned}$$

and

$$\begin{aligned} Z = \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \int _M u_\alpha ^r \eta _{\alpha ,\sigma }^r \mathrm{d}v_g. \end{aligned}$$

It is clear that \(X, Y, Z \le 1\) and (34) and (36) take the form

$$\begin{aligned} \left\{ \begin{array}{l} \theta _0 X + (1 - \theta _0) Y \le Z \\ Z \le X^{\frac{r_0 \theta _0}{p_0}} Y^{\frac{r_0(1 - \theta _0)}{q_0}} \end{array} \right. \end{aligned}$$
(37)

By (24), we have \(Z > 0\), so that \(X, Y > 0\).

In order to end the proof of (19), it suffices to show that \(Z = 1\). By Young’s inequality, (37) immediately yields

$$\begin{aligned} \left\{ \begin{array}{l} X^{\theta _0} Y^{1 - \theta _0} \le Z\\ Z \le X^{\frac{r_0 \theta _0}{p_0}} Y^{\frac{r_0(1 - \theta _0)}{q_0}} \end{array} \right. \end{aligned}$$

But these two inequalities give

$$\begin{aligned} X^{\theta _0} Y^{1 - \theta _0} \le X^{\frac{r_0 \theta _0}{p_0}} Y^{\frac{r_0(1 - \theta _0)}{q_0}} \le X^{\theta _0} Y^{\frac{r_0(1 - \theta _0)}{q_0}}, \end{aligned}$$

so that \(Y = 1\). Therefore, by (20) and (23),

$$\begin{aligned} \int _{M \setminus B(x_\alpha ,\sigma a_\alpha )} u_\alpha ^r\; \mathrm{d}v_g \le ||u_\alpha ||_{L^\infty (M)}^{r - q} a_\alpha ^{\frac{n(r - q)}{r}} \frac{\int _{M\setminus B(x_\alpha ,\sigma a_\alpha )} u_\alpha ^q\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} \le c \frac{\int _{M\setminus B(x_\alpha ,\sigma a_\alpha )} u_\alpha ^q\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g}, \end{aligned}$$

which implies that

$$\begin{aligned} \lim _{\sigma \rightarrow +\infty } \lim _{\alpha \rightarrow \alpha _0} \int _{M \setminus B(x_\alpha , \sigma a_\alpha )} u_\alpha ^r\; \mathrm{d}v_g = 0. \end{aligned}$$

Thus, it follows that \(Z = 1\).

A key tool in the proof of Theorem 3.1 consists of the following uniform estimate:

Claim 2

There exists a constant \(c > 0\), independent of \(p, q\) and \(r\), such that

$$\begin{aligned} d_g(x,x_\alpha )^p u_\alpha (x)^{r - p} \le c a_\alpha ^{\frac{np - nr + pr}{r}} \end{aligned}$$

for \(x \in M\) and \(\alpha \) close enough to \(\alpha _0\).

Proof of Claim 2

Suppose, by contradiction, that the above assertion is false.

Set

$$\begin{aligned} f_\alpha (x) = d_g(x,x_\alpha )^p u_\alpha (x)^{r - p} a_\alpha ^{\frac{nr - np - pr}{r}}. \end{aligned}$$

If \(y_\alpha \in M\) is a maximum point of \(f_\alpha \), then \(f_\alpha (y_\alpha ) = ||f_\alpha ||_{L^\infty (M)} \rightarrow + \infty \) when \(\alpha \rightarrow \alpha _0\). By (23), we have

$$\begin{aligned} f_\alpha (y_\alpha ) \le c \left( \frac{u_\alpha (y_\alpha )}{||u_\alpha ||_{L^\infty (M)}}\right) ^{r - p} d_g(x_\alpha ,y_\alpha )^p ||u_\alpha ||_{L^\infty (M)}^{\frac{pr}{n}} \le c d_g(x_\alpha ,y_\alpha )^p ||u_\alpha ||_{L^\infty (M)}^{\frac{pr}{n}}, \end{aligned}$$

so that

$$\begin{aligned} d_g(x_\alpha ,y_\alpha ) ||u_\alpha ||_{L^\infty (M)}^{\frac{r}{n}} \rightarrow +\infty . \end{aligned}$$
(38)

For any fixed \(\sigma > 0\) and \(\varepsilon \in (0,1)\), we next show that

$$\begin{aligned} B(y_\alpha ,\varepsilon \; d_g(x_\alpha ,y_\alpha )) \cap B\left( x_\alpha , \sigma ||u_\alpha ||_{L^\infty (M)}^{-\frac{r}{n}}\right) = \emptyset \end{aligned}$$
(39)

for \(\alpha \) close enough to \(\alpha _0\). Note that this claim follows readily from

$$\begin{aligned} d_g(x_\alpha ,y_\alpha ) \ge \sigma ||u_\alpha ||_{L^\infty (M)}^{-\frac{r}{n}} + \varepsilon d_g(x_\alpha ,y_\alpha ). \end{aligned}$$

On the other hand, the above inequality is equivalent to

$$\begin{aligned} d_g(x_\alpha , y_\alpha )(1 - \varepsilon ) ||u_\alpha ||_{L^\infty (M)}^{\frac{r}{n}} \ge \sigma , \end{aligned}$$

which is clearly satisfied since \(d_g(x_\alpha ,y_\alpha ) ||u_\alpha ||_{L^\infty (M)}^{\frac{r}{n}} \rightarrow + \infty \) and \(1 - \varepsilon > 0\).

We assert that exists a constant \(c > 0\) such that

$$\begin{aligned} u_\alpha (x) \le c u_\alpha (y_\alpha ) \end{aligned}$$
(40)

for \(x \in B(y_\alpha , \varepsilon d_g(x_\alpha ,y_\alpha ))\) and \(\alpha \) close enough to \(\alpha _0\). In fact, for \(x \in B(y_\alpha , \varepsilon d_g(x_\alpha ,y_\alpha ))\), we have

$$\begin{aligned} d_g(x,x_\alpha ) \ge d_g(x_\alpha ,y_\alpha ) - d_g(x,y_\alpha ) \ge (1 - \varepsilon ) d_g(x_\alpha ,y_\alpha ). \end{aligned}$$

Thus,

$$\begin{aligned} d_g(y_\alpha ,x_\alpha )^p u_\alpha (y_\alpha )^{r - p} a_\alpha ^{\frac{nr - np - pr}{r}}&= f_\alpha (y_\alpha ) \ge f_\alpha (x) = d_g(x,x_\alpha )^p u_\alpha (x)^{r - p} a_\alpha ^{\frac{nr - np - pr}{r}}\\&\ge (1 - \varepsilon )^p d_g(y_\alpha ,x_\alpha )^p u_\alpha (x)^{r - p} a_\alpha ^{\frac{nr - np - pr}{r}}, \end{aligned}$$

so that

$$\begin{aligned} u_\alpha (x) \le \left( \frac{1}{1 - \varepsilon }\right) ^{\frac{p}{r - p}} u_\alpha (y_\alpha ) \end{aligned}$$

for \(x \in B(y_\alpha , \varepsilon d_g(x_\alpha ,y_\alpha ))\) and \(\alpha \) close enough to \(\alpha _0\). This proves our claim.

Since \(f(y_\alpha ) \rightarrow + \infty \), one has

$$\begin{aligned} A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p-r}{p}} \rightarrow 0. \end{aligned}$$

So, we can define

$$\begin{aligned} \begin{array}{l} h_\alpha (x) = g(\exp _{y_\alpha }(A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p-r}{p}} x)) \\ \psi _\alpha (x) = u_\alpha (y_\alpha )^{-1} u_\alpha (\exp _{y_\alpha }(A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p-r}{p}} x)) \end{array} \end{aligned}$$

for each \(x \in B(0,2)\) and \(\alpha \) close enough to \(\alpha _0\).

By (14), one easily checks that

$$\begin{aligned}&\Delta _{p,h_\alpha } \psi _\alpha + C_\alpha A_\alpha u_\alpha (y_\alpha )^{p - r} \psi _\alpha ^{p - 1} + \frac{1 - \theta }{\theta } \nu _\alpha ||u_\alpha ||_{L^q(M)}^{-q} u_\alpha (y_\alpha )^{q - r} \psi _\alpha ^{q - 1}\nonumber \\&\quad = \frac{\nu _\alpha }{\theta } \psi _\alpha ^{r - 1} \ \ \mathrm{on}\ \ B(0,2). \end{aligned}$$
(41)

In particular,

$$\begin{aligned} \int _{B(0,2)} |\nabla _{h_\alpha } \psi _p|^{p - 2} \nabla _{h_\alpha } \psi _\alpha \cdot \nabla _{h_\alpha } \phi \; \mathrm{d}v_{h_\alpha } \le c \int _{B(0,2)} \psi _\alpha ^{r - 1} \phi \; \mathrm{d}v_{h_\alpha } \end{aligned}$$

for all positive test function \(\phi \in C_0^1(B(0,2))\). So, a Moser’s iterative scheme combined with (23) furnishes

$$\begin{aligned} 1&= \sup _{B\left( 0,\frac{1}{4}\right) } \psi _\alpha ^r \le c \int _{B\left( 0,\frac{1}{2}\right) } \psi _\alpha ^r\; \mathrm{d}v_{h_\alpha }\\&= c \left( A_\alpha ^{\frac{\theta q}{p(1 - \theta )}} u_\alpha (y_\alpha )^{r - q} \right) ^{-\frac{n(1 - \theta )}{\theta q}} \int _{B\left( y_\alpha , \frac{1}{2} A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p-r}{p}}\right) } u_\alpha ^r\; \mathrm{d}v_g\\&\le c \left( \frac{||u_\alpha ||_{L^\infty (M)}}{u_\alpha (y_\alpha )} \right) ^{\frac{np - rn + pr}{p}} \int _{B\left( y_\alpha , \frac{1}{2} A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p-r}{p}}\right) } u_\alpha ^r\; \mathrm{d}v_g. \end{aligned}$$

For simplicity, rewrite this last inequality as

$$\begin{aligned} 0 < c \le m_\alpha ^\varrho \int _{B\left( y_\alpha ,\frac{1}{2} A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p-r}{p}}\right) } u_\alpha ^r\; \mathrm{d}v_g, \end{aligned}$$
(42)

where \(m_\alpha = \frac{||u_\alpha ||_{L^\infty (M)}}{u_\alpha (y_\alpha )}\) and \(\varrho = \frac{np - rn + pr}{p}\).

By (20), (23) and (38), one has \(B(y_\alpha , \frac{1}{2} A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p - r}{p}}) \subset B(y_\alpha ,\varepsilon \mathrm{d}(x_\alpha ,y_\alpha ))\) for \(\alpha \) close enough to \(\alpha _0\). Therefore, (19) and (39) imply

$$\begin{aligned} \int _{B\left( y_\alpha , \frac{1}{2} A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p - r}{p}}\right) } u_\alpha ^r\; \mathrm{d}v_g \rightarrow 0, \end{aligned}$$

so that \(m_\alpha \rightarrow + \infty \) as \(\alpha \rightarrow \alpha _0\).

Our main goal now is to establish a contradiction to (42).

At first, by (23) and (40), one has

$$\begin{aligned} m_\alpha ^\varrho \int _{D_\alpha } u_\alpha ^r\; \mathrm{d}v_g \le m_\alpha ^\varrho ||u_\alpha ||^r_{L^\infty (D_\alpha )} (A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p - r}{p}})^n \le c m_\alpha ^\varrho u_\alpha (y_\alpha )^r (A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p - r}{p}})^n \le c, \end{aligned}$$
(43)

where \(D_\alpha = B(y_\alpha , A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p - r}{p}})\).

Consider the function \(\eta _\alpha (x) = \eta ( A_\alpha ^{-\frac{1}{p}} d_g(x,y_\alpha ) u_\alpha (y_\alpha )^{\frac{r - p}{p}})\), where \(\eta \in C_0^1({\mathbb {R}})\) is a cutoff function satisfying \(\eta = 1\) on \([0,\frac{1}{2}], \eta = 0 \) on \([1,\infty )\) and \(0 \le \eta \le 1\). Taking \(u_\alpha \eta _\alpha ^p\) as a test function in (14), one has

$$\begin{aligned}&A_\alpha \int _M |\nabla _g u_\alpha |^p \eta _\alpha ^p\; \mathrm{d}v_g + p A_\alpha \int _M |\nabla _g u_\alpha |^{p - 2} u_\alpha \eta _p^{p - 1} \nabla _g u_\alpha \cdot \nabla _g \eta _\alpha \; \mathrm{d}v_g + C_\alpha A_\alpha \\&\quad \times \,\int _M u_\alpha ^p \eta _\alpha ^p\; \mathrm{d}v_g+ \frac{1 - \theta }{\theta } \nu _\alpha ||u_\alpha ||_{L^q(M)}^{-q} \int _M u_\alpha ^q \eta _\alpha ^p\; \mathrm{d}v_g = \frac{\nu _\alpha }{\theta } \int _M u_\alpha ^r \eta _\alpha ^p\; \mathrm{d}v_g. \end{aligned}$$

From Hölder and Young inequalities, the above second term can be estimated as

$$\begin{aligned} \left| \int _M |\nabla _g u_\alpha |^{p - 2} u_\alpha \eta _\alpha ^{p - 1} \nabla _g u_\alpha \cdot \nabla _g \eta _\alpha \; \mathrm{d}v_g \right| \le \varepsilon \int _M |\nabla _g u_\alpha |^p \eta _\alpha ^p\; \mathrm{d}v_g + c_\varepsilon \int _M |\nabla _g \eta _\alpha |^p u_\alpha ^p\; \mathrm{d}v_g. \end{aligned}$$

Also, by (23) and (40), we have

$$\begin{aligned} A_\alpha \int _M|\nabla _g \eta _\alpha |^p u_\alpha ^p\; \mathrm{d}v_g&\le A_\alpha (A_\alpha ^{-\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{r - p}{p}})^p \int _{D_\alpha } u_\alpha ^p\; \mathrm{d}v_g\nonumber \\&\le c u_\alpha (y_\alpha )^r (A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p - r}{p}})^n \le c m_\alpha ^{- \varrho }. \end{aligned}$$
(44)

Putting these inequalities into (43), one gets

$$\begin{aligned} A_\alpha \int _M |\nabla _g u_\alpha |^p \eta _\alpha ^p\; \mathrm{d}v_g + c\; C_\alpha A_\alpha \int _M u_\alpha ^p \eta _\alpha ^p\; \mathrm{d}v_g + c \nu _\alpha ||u_\alpha ||_{L^q(M)}^{-q} \int _M u_\alpha ^q \eta _\alpha ^p\; \mathrm{d}v_g \le c m_\alpha ^{-\varrho }.\nonumber \\ \end{aligned}$$
(45)

On the other hand, the sharp Riemannian Gagliardo–Nirenberg inequality gives

$$\begin{aligned}&\left( \,\, \int _{B(y_\alpha , \frac{1}{2} A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p-r}{p}})} u_\alpha ^r\; \mathrm{d}v_g \right) ^{\frac{p}{r \theta }} \le \left( \,\,\int _M (u_\alpha \eta _\alpha ^p)^r\; \mathrm{d}v_g \right) ^{\frac{p}{r \theta }}\nonumber \\&\quad \le c \left( \,\,\int _M |\nabla _g u_\alpha |^p \eta _\alpha ^{p^2}\; \mathrm{d}v_g\right) \left( \,\,\int _M (u_\alpha \eta _\alpha ^p)^q\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}\nonumber \\&\qquad + \; c \left( \,\,\int _M |\nabla _g \eta _\alpha |^p u_\alpha ^p\; \mathrm{d}v_g + c\; C_\alpha \int _M (u_\alpha \eta _\alpha ^p)^p\; \mathrm{d}v_g\right) \left( \,\, \int _M (u_\alpha \eta _\alpha ^p)^q\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}.\nonumber \\ \end{aligned}$$
(46)

Thanks to (44) and (45), we can estimate each term of the right-hand side of (46). Indeed,

$$\begin{aligned}&\left( \int _M |\nabla _g u_\alpha |^p \eta _\alpha ^{p^2}\; \mathrm{d}v_g\right) \left( \int _M (u_\alpha \eta _\alpha ^p)^q\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}\\&\quad \le \left( A_\alpha \int _M |\nabla _g u_\alpha |^p \eta _\alpha ^p\; \mathrm{d}v_g \right) \left( ||u_\alpha ||_{L^q(M)}^{-q} \int _M u_\alpha ^q \eta _\alpha ^p\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}\\&\quad \le c m_\alpha ^{-\varrho \left( 1 + \frac{p( 1 - \theta )}{\theta q}\right) },\\&\left( \int _M |\nabla _g \eta _\alpha |^p u_\alpha ^p\; \mathrm{d}v_g\right) \left( \int _M (u_\alpha \eta _\alpha ^p)^q\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}\\&\quad \le A_\alpha \int _M |\nabla _g \eta _\alpha |^p u_\alpha ^p\; \mathrm{d}v_g \left( ||u_\alpha ||_{L^q(M)}^{-q} \int _M u_\alpha ^q \eta _\alpha ^p\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}\\&\quad \le c m_\alpha ^{-\varrho \left( 1 + \frac{p(1 - \theta )}{\theta q}\right) } \end{aligned}$$

and

$$\begin{aligned}&C_\alpha \left( \int _M (u_\alpha \eta _\alpha ^p)^p\; \mathrm{d}v_g\right) \left( \int _M (u_\alpha \eta _\alpha ^p)^q\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}\\&\quad \le C_\alpha A_\alpha \int _M u_\alpha ^p \eta _\alpha ^p\; \mathrm{d}v_g \left( ||u_\alpha ||_{L^q(M)}^{-q} \int _M u_\alpha ^q \eta _\alpha ^p \; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}\\&\quad \le c m_\alpha ^{-\varrho \left( 1 + \frac{p(1 - \theta )}{\theta q}\right) }. \end{aligned}$$

Replacing these three estimates in (46), one gets

$$\begin{aligned} \left( \,\,\int _{B(y_p, \frac{1}{2} A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p-r}{p}})} u_\alpha ^r\; \mathrm{d}v_g \right) ^{\frac{p}{r \theta }} \le c \; m_\alpha ^{-\varrho \left( 1 + \frac{p(1 - \theta )}{\theta q}\right) }, \end{aligned}$$

so that

$$\begin{aligned} m_\alpha ^\varrho \int _{B\left( y_\alpha , \frac{1}{2} A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p-r}{p}}\right) } u_\alpha ^r\; \mathrm{d}v_g \le c \; m_\alpha ^{\varrho \left( 1 - \frac{r \theta }{p} - \frac{r (1 - \theta )}{q}\right) }. \end{aligned}$$

Since \(m_\alpha \rightarrow + \infty \) and

$$\begin{aligned} \lim _{\alpha \rightarrow \alpha _0} \left( 1 - \frac{r \theta }{p} - \frac{r (1 - \theta )}{q}\right) < c < 0, \end{aligned}$$

we derive

$$\begin{aligned} m_\alpha ^\varrho \int _{B\left( y_\alpha , \frac{1}{2} A_\alpha ^{\frac{1}{p}} u_\alpha (y_\alpha )^{\frac{p-r}{p}}\right) } u_\alpha ^r\; \mathrm{d}v_g \rightarrow 0. \end{aligned}$$

But this notedly contradicts (42).

Proof of Theorem 3.1

In order to establish the desired contradiction, we will perform several integral estimates by using the Claim 2. Assume, without loss of generality, that the radius of injectivity of \(M\) is \({>}1\).

Let \(\eta \in C^1_0({\mathbb {R}})\) be a cutoff function as in the above proof and define \(\eta _{\alpha ,\delta }(x) =\eta (\frac{d_g(x,x_\alpha )}{\delta })\) for \(0 < \delta \le 1\). In normal coordinates around \(x_\alpha \), the sharp Euclidean Gagliardo–Nirenberg inequality furnishes

$$\begin{aligned} \left( \,\,\int _{B(0,\delta )} u_\alpha ^r \eta _{\alpha ,\delta }^r\; \mathrm{d}x \right) ^{\frac{p}{r \theta }} \le A_0(p,q,r) \left( \,\, \int _{B(0,\delta )} |\nabla (u_\alpha \eta _{\alpha ,\delta })|^p\; \mathrm{d}x\right) \left( \,\,\int _{B(0,\delta )} u_\alpha ^q \eta _{\alpha ,\delta }^q\; \mathrm{d}x \right) ^{\frac{p(1 - \theta )}{\theta q}}. \end{aligned}$$

Expanding the metric \(g\) on these same coordinates, one locally gets

$$\begin{aligned} (1 - c d_g(x,x_\alpha )^2)\; \mathrm{d}v_g \le \mathrm{d}x \le (1 + c d_g(x,x_\alpha )^2)\; \mathrm{d}v_g \end{aligned}$$
(47)

and

$$\begin{aligned} |\nabla (u_\alpha \eta _{\alpha ,\delta })|^p \le |\nabla _g(u_\alpha \eta _{\alpha ,\delta })|^p(1 + c d_g(x,x_\alpha )^2). \end{aligned}$$
(48)

Thanks to these expansions, one arrives at

$$\begin{aligned} \left( \,\,\int _{B(0,\delta )} u_\alpha ^r \eta _{\alpha ,\delta }^r\; \mathrm{d}x \right) ^{\frac{p}{r \theta }}&\le \left( A_0(p,q,r) \; A_\alpha \int _M |\nabla _g (u_\alpha \eta _{\alpha ,\delta })|^p\; \mathrm{d}v_g\right. \\&\left. +\,\, c A_\alpha \int _M |\nabla _g (u_\alpha \eta _{\alpha ,\delta })|^p d_g(x,x_\alpha )^2\; \mathrm{d}v_g \right) \\&\times \left( \frac{\int _{B(0,\delta )} u_\alpha ^q \eta _{\alpha ,\delta }^q\; \mathrm{d}x}{\int _M u_\alpha ^q\; \mathrm{d}v_g}\right) ^{\frac{p(1 - \theta )}{\theta q}}. \end{aligned}$$

Using now the inequalities

$$\begin{aligned} |\nabla _g (u_\alpha \eta _{\alpha ,\delta })|^p \le |\nabla _g u_\alpha |^p \eta _{\alpha ,\delta }^p + c |\eta _{\alpha ,\delta } \nabla _g u_p|^{p - 1} |u_\alpha \nabla _g \eta _{\alpha ,\delta }| + c |u_\alpha \nabla _g \eta _{\alpha ,\delta }|^p \end{aligned}$$

and

$$\begin{aligned} A_0(p,q,r) \left( A_\alpha \int _M |\nabla _g u_\alpha |^p\; \mathrm{d}v_g \right) \le 1 - A_0(p,q,r) \left( C_\alpha A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g \right) , \end{aligned}$$

we then derive

$$\begin{aligned} \left( \int _{B(0,\delta )} u_\alpha ^r \eta _{\alpha ,\delta }^r\; \mathrm{d}x \right) ^{\frac{p}{r \theta }}&\le \left( 1 - c_1 C_\alpha A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g\right. \nonumber \\&\left. + c_2 F_\alpha + c_2 G_\alpha + c_3 \delta ^{-p} A_\alpha \int _{M \setminus B(x_\alpha ,\frac{\delta }{2})} u_\alpha ^p\; \mathrm{d}v_g \right) \nonumber \\&\times \left( \frac{\int _{B(0,\delta )} u_\alpha ^q \eta _{\alpha ,\delta }^q\; \mathrm{d}x}{\int _M u_\alpha ^q\; \mathrm{d}v_g}\right) ^{\frac{p(1 - \theta )}{\theta q}}, \end{aligned}$$
(49)

where

$$\begin{aligned} F_\alpha = A_\alpha \int _M |\nabla _g u_\alpha |^p \eta _{\alpha ,\delta }^p d_g(x,x_\alpha )^2\; \mathrm{d}v_g \end{aligned}$$

and

$$\begin{aligned} G_\alpha = A_\alpha \int _M |\nabla _g u_\alpha |^{p-1} \eta _{\alpha ,\delta }^{p-1} u_\alpha |\nabla _g \eta _{\alpha ,\delta }|\; \mathrm{d}v_g. \end{aligned}$$

In order to estimate \(F_\alpha \) and \(G_\alpha \), let \(\zeta _{\alpha ,\delta }(x) = 1 - \eta (\frac{2}{\delta } d_g(x,x_\alpha ))\), where \(\eta \) is a cutoff function as above. Taking \(u_\alpha \zeta _{\alpha ,\delta }^p\) as a test function in (14), one gets

$$\begin{aligned} A_\alpha \int _M |\nabla _g u_\alpha |^p \zeta _{\alpha ,\delta }^p \; \mathrm{d}v_g \le c \int _M u_\alpha ^r \zeta _{\alpha ,\delta }^p\; \mathrm{d}v_g + c A_\alpha \int _M |\nabla _g u_\alpha |^{p-1} \zeta _{\alpha ,\delta }^{p - 1} |\nabla _g \zeta _{\alpha ,\delta }| u_\alpha \; \mathrm{d}v_g. \end{aligned}$$

By Young’s inequality, one has

$$\begin{aligned} A_\alpha \int _M |\nabla _g u_\alpha |^p \zeta _{\alpha ,\delta }^p\; \mathrm{d}v_g \le c \delta ^{-p} A_\alpha \int _{M \setminus B(x_\alpha ,\frac{\delta }{2})} u_\alpha ^p\; \mathrm{d}v_g + c \int _{M \backslash B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^r\; \mathrm{d}v_g, \end{aligned}$$

so that

$$\begin{aligned} G_\alpha \le A_\alpha \int _M |\nabla _g u_\alpha |^{p - 1} \zeta _{\alpha ,\delta }^p u_\alpha \; \mathrm{d}v_g \le c \delta ^{-p} A_\alpha \int _{M \setminus B(x_\alpha ,\frac{\delta }{2})} u_\alpha ^p\; \mathrm{d}v_g + c \int _{M \setminus B\left( x_\alpha , \frac{\delta }{4}\right) } u_\alpha ^r\; \mathrm{d}v_g. \end{aligned}$$
(50)

Using further the fact that \(p < 2\), one has

$$\begin{aligned} \int _M |\nabla _g u_\alpha |^{p - 1} \eta _{\alpha ,\delta }^p u_\alpha d_g(x,x_\alpha )\; \mathrm{d}v_g&\le \varepsilon \int _M |\nabla _g u_\alpha |^p \eta _{\alpha ,\delta }^p d_g(x,x_\alpha )^2\; \mathrm{d}v_g\nonumber \\&+ c_\varepsilon \int _M u_\alpha ^p d_g(x,x_\alpha )^{2 - p}\; \mathrm{d}v_g. \end{aligned}$$
(51)

Besides, taking \(u_\alpha d_g(\cdot , x_\alpha )^2 \eta _{\alpha ,\delta }^p\) as a test function in (14), one gets

$$\begin{aligned}&A_\alpha \int _M |\nabla _g u_\alpha |^p \eta _{\alpha ,\delta }^p d_g(x,x_\alpha )^2\; \mathrm{d}v_g + \frac{ \int _M u_\alpha ^q \eta _{\alpha ,\delta }^p d_g(x,x_\alpha )^2\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g}\\&\quad \le c \int _{B(x_\alpha , \delta )} u_\alpha ^r d_g(x,x_\alpha )^2\; \mathrm{d}v_g + c A_\alpha \int _M |\nabla _g u_\alpha |^{p - 1} \eta _{\alpha ,\delta }^p u_\alpha d_g(x,x_\alpha )\; \mathrm{d}v_g + c G_\alpha .\nonumber \end{aligned}$$
(52)

Joining now (50), (51) and (52), one obtains

$$\begin{aligned} F_\alpha&\le c \int _{B(x_\alpha , \delta )} u_\alpha ^r d_g(x,x_\alpha )^2\; \mathrm{d}v_g + c \int _{M \backslash B(x_\alpha , \frac{\delta }{4})} u_\alpha ^r\; \mathrm{d}v_g \nonumber \\&+ c \delta ^{-p} A_\alpha \int _{M \setminus B(x_\alpha ,\frac{\delta }{2})} u_\alpha ^p\; \mathrm{d}v_g + c \delta ^{2 - p} A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g\;. \end{aligned}$$

On the other hand, the Claim 2 gives

$$\begin{aligned} \int _{B(x_\alpha ,\delta )} u_\alpha ^r d_g(x,x_\alpha )^2\; \mathrm{d}v_g \le c \delta ^{2 - p} A_\alpha \int _M u_\alpha ^p \mathrm{d}v_g \end{aligned}$$
(53)

and

$$\begin{aligned} \int _{M \backslash B(x_\alpha , \frac{\delta }{4})} u_\alpha ^r\; \mathrm{d}v_g&\le 16 \int _{M\backslash B(x_\alpha , \frac{\delta }{4})} u_\alpha ^p u_\alpha ^{r - p} d_g(x,x_\alpha )^2\; \mathrm{d}v_g\nonumber \\&\le c \delta ^{p - 2} A_\alpha \int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p \mathrm{d}v_g. \end{aligned}$$
(54)

Consequently,

$$\begin{aligned} F_\alpha&\le c \delta ^{2 - p} A_\alpha \int _M u_\alpha ^p \; \mathrm{d}v_g + c \delta ^{-p} A_\alpha \int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p \mathrm{d}v_g \quad \hbox {and} \quad \nonumber \\ G_\alpha&\le c \delta ^{-p} A_\alpha \int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p \; \mathrm{d}v_g. \end{aligned}$$
(55)

Putting these two estimates in (49), one arrives at

$$\begin{aligned} \left( \int _{B(x_\alpha ,\delta )} u_\alpha ^r \eta _{\alpha ,\delta }^r\; \mathrm{d}x \right) ^{\frac{p}{r \theta }}&\le \left( 1 - \left( c_1\; C_\alpha + c \delta ^{2 - p}\right) A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g\right. \nonumber \\&\left. + c \delta ^{-p} A_\alpha \int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g \right) \left( \frac{\int _{B(x_\alpha ,\delta )} u_\alpha ^q \eta _{\alpha ,\delta }^q\; \mathrm{d}x}{\int _M u_\alpha ^q\; \mathrm{d}v_g} \right) ^{\frac{p(1 - \theta )}{\theta q}}.\nonumber \\ \end{aligned}$$
(56)

However, by (48), we have

$$\begin{aligned} \left( \int _M u_\alpha ^r \eta _{\alpha ,\delta }^r\; \mathrm{d}x \right) ^{\frac{p}{r \theta }}&\ge \left( \int _M u_\alpha ^r \eta _{\alpha ,\delta }^r\; \mathrm{d}v_g - c \int _M u_\alpha ^r \eta _{\alpha ,\delta }^r d_g(x,x_\alpha )^2\; \mathrm{d}v_g \right) ^{\frac{p}{r \theta }}\\&\ge 1 - c\int _{M \setminus B(x_\alpha ,\delta )} u_\alpha ^r\; \mathrm{d}v_g - c \int _M u_\alpha ^r \eta _{\alpha ,\delta }^r d_g(x,x_\alpha )^2\; \mathrm{d}v_g \end{aligned}$$

and

$$\begin{aligned} \left( \frac{\int _{B(x_\alpha ,\delta )} u_\alpha ^q \eta _{\alpha ,\delta }^q\; \mathrm{d}x}{\int _M u_\alpha ^q\; \mathrm{d}v_g} \right) ^{\frac{p(1 - \theta )}{\theta q}}&\le \left( \frac{\int _M u_\alpha ^q \eta _{\alpha ,\delta }^q\; \mathrm{d}v_g + c \int _M u_\alpha ^q \eta _{\alpha ,\delta }^q d_g(x,x_\alpha )^2\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} \right) ^{\frac{p(1 - \theta )}{\theta q}}\\&\le \left( \frac{\int _M u_\alpha ^q \eta _{\alpha ,\delta }^q\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} \right) ^{\frac{p(1 - \theta )}{\theta q}} + c \frac{\int _M u_\alpha ^q \eta _{\alpha ,\delta }^q d_g(x,x_\alpha )^2\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g}\\&\le 1 + c \frac{\int _M u_\alpha ^q \eta _{\alpha ,\delta }^q d_g(x,x_\alpha )^2\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g}. \end{aligned}$$

Replacing these two inequalities in (56) and using the fact that \(p < 2\), one gets

$$\begin{aligned} 0&\le - C_\alpha A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g + \frac{\int _M u_\alpha ^q \eta _{\alpha ,\delta }^q d_g(x,x_\alpha )^2\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} + c \int _M u_\alpha ^r \eta _{\alpha ,\delta }^r d_g(x,x_\alpha )^2\; \mathrm{d}v_g\\&\quad + c \int _{M \setminus B(x_\alpha ,\delta )} u_\alpha ^r\; \mathrm{d}v_g + c \delta ^{2 - p} A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g + c \delta ^{-p} A_\alpha \int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) }u_\alpha ^p\; \mathrm{d}v_g. \end{aligned}$$

By (53) and (54), we then derive

$$\begin{aligned} C_\alpha A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g&\le c \frac{\int _M u_\alpha ^q \eta _{p,q,\delta }^p d_g(x,x_\alpha )^2\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} + c \delta ^{2 - p} A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g\nonumber \\&\quad + c \delta ^{-p} A_\alpha \int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g. \end{aligned}$$
(57)

Plugging (50), (51), (53) and (54) in (52), one obtains

$$\begin{aligned} \frac{\int _M u_\alpha ^q \eta _{\alpha ,\delta }^p d_g(x,x_\alpha )^2\; \mathrm{d}v_g}{\int _M u_\alpha ^q\; \mathrm{d}v_g} \le c \delta ^{2 - p} A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g + c \delta ^{-p} A_\alpha \int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g. \end{aligned}$$

Introducing now this inequality in (57), one gets

$$\begin{aligned} C_\alpha \le c \delta ^{2 - p} + c(\delta ) \frac{\int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g}{\int _M u_\alpha ^p\; \mathrm{d}v_g} \le c \delta ^{2 - p} + c(\delta ), \end{aligned}$$
(58)

where \(c(\delta ) \rightarrow +\infty \) as \(\delta \rightarrow 0^+\). But this is a contradiction, since \(\lim _{\alpha \rightarrow \alpha _0} C_\alpha = + \infty \).

4 Proof of Theorem 1.1

In this section, we furnish the proof of the existence of an extremal function for parameters \(p, q\) and \(r\) as in Theorem 1.1.

Given \(\alpha \in (0,1)\), consider the functional

$$\begin{aligned} J_\alpha (u) = \left( \,\,\int _M |\nabla _g u|^p\; \mathrm{d}v_g + C_\alpha \int _M |u|^p\; \mathrm{d}v_g \right) \left( \,\,\int _M |u|^q\; \mathrm{d}v_g\right) ^{\frac{p(1 - \theta )}{\theta q}} \end{aligned}$$

constrained to \(E = \{ u \in H^{1,p}(M):\; ||u||_{L^r(M)} = 1 \}\), where \(C_\alpha = \frac{B_0(p,q,r,g)}{A_0(p,q,r)} \alpha \).

The definition of \(B_0(p,q,r,g)\) yields

$$\begin{aligned} \nu _\alpha = \inf _{u \in E} J_\alpha (u) < A_0(p,q,r)^{-1}. \end{aligned}$$
(59)

In a standard way, one knows that \(\nu _\alpha \) is attained by a nonnegative function \(u_\alpha \in E\) of \(C^1\) class. In particular, \(u_\alpha \) satisfies the Euler-Lagrange equation

$$\begin{aligned} A_\alpha \Delta _{p,g} u_\alpha + C_\alpha A_\alpha u_\alpha ^{p - 1} + \frac{1 - \theta }{\theta } \nu _\alpha ||u_\alpha ||_{L^q(M)}^{-q} u_\alpha ^{q - 1} = \frac{\nu _\alpha }{\theta } u_\alpha ^{r - 1} \quad \hbox {on} \quad M \end{aligned}$$
(60)

where

$$\begin{aligned} A_\alpha = \left( \,\,\int _M u_\alpha ^q\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}. \end{aligned}$$

We assert that

$$\begin{aligned} \lim _{\alpha \rightarrow 1^-} A_\alpha > 0. \end{aligned}$$

If so, the conclusion of Theorem 1.1 follows. In fact, the above claim and (59) imply that the sequence \((u_\alpha )\) is bounded in \(H^{1,p}(M)\). So, up to a subsequence, \((u_\alpha )\) converges weakly to \(u_0\) in \(H^{1,p}(M)\) and also strongly in \(L^p(M), L^q(M)\) and \(L^r(M)\), so that \(u_0 \in E\). Moreover, letting \(\alpha \rightarrow 1^-\) in the inequality

$$\begin{aligned} J_\alpha (u_\alpha ) < A_0(p,q,r)^{-1}, \end{aligned}$$

one readily concludes that \(u_0\) is extremal for (3).

Instead, assume

$$\begin{aligned} \lim _{\alpha \rightarrow 1^-} A_\alpha = 0. \end{aligned}$$

In this case, since \(p \le r\),

$$\begin{aligned} A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g \rightarrow 0, \end{aligned}$$
(61)

which in turn implies that

$$\begin{aligned} \lim _{\alpha \rightarrow 1^-} \nu _\alpha = A_0(p,q,r)^{-1}. \end{aligned}$$
(62)

Because (60) is quite similar to (14), proceeding in the same spirit of the proof of Theorem 3.1, we achieve the following conclusions:

Let \(x_\alpha \in M\) be a maximum point of \(u_\alpha \). Then, for any \(\sigma > 0\), the concentration property of \(u_\alpha \) around \(x_\alpha \) holds, namely

$$\begin{aligned} \lim _{\sigma \rightarrow \infty } \lim _{\alpha \rightarrow 1^-} \int _{B(x_\alpha ,\sigma a_\alpha )} u_\alpha ^r\; \mathrm{d}v_g = 1, \end{aligned}$$
(63)

where

$$\begin{aligned} a_\alpha = A_\alpha ^{\frac{r}{np - nr +pr}} \end{aligned}$$

with \(a_\alpha \rightarrow 0\) as \(\alpha \rightarrow 1^-\).

The above concentration leads to a uniform estimate referred as distance type lemma. namely, there exists a constant \(c > 0\), independent of \(\alpha \), such that

$$\begin{aligned} d_g(x,x_\alpha )^p u_\alpha (x)^{r - p} \le c a_\alpha ^{\frac{np - nr +pr}{r}} \end{aligned}$$
(64)

for all \(x \in M\) and \(\alpha \) close enough to \(1^-\).

As before, using (62), (63) and (64), with natural adaptations one arrives at [see (58)]

$$\begin{aligned} C_\alpha \le c \delta ^{2 - p} + c(\delta ) \frac{\int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g}{\int _M u_\alpha ^p\; \mathrm{d}v_g} \end{aligned}$$
(65)

for \(\delta > 0\) small enough, where \(c(\delta ) \rightarrow + \infty \) as \(\delta \rightarrow 0^+\).

We assert that

$$\begin{aligned} \lim _{\alpha \rightarrow 1^-} \frac{\int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g}{\int _M u_\alpha ^p\; \mathrm{d}v_g} = 0 \end{aligned}$$

whenever \(p, q < r < p^*\) and \(1 \le q \le \frac{r}{r-p}\).

At first, an integration of the Eq. (14) on \(M\) furnishes, for any nonnegative function \(h \in C^1(M)\),

$$\begin{aligned} A_\alpha \int _M |\nabla _g u_\alpha |^{p - 2} \nabla _g u_\alpha \cdot \nabla _g h\; \mathrm{d}v_g \le c \int _M u_\alpha ^{r - 1} h\; \mathrm{d}v_g. \end{aligned}$$

On the other hand, the claim 2 yields, for any nonnegative function \(h \in C^1(M \setminus B(x_\alpha , \lambda ))\),

$$\begin{aligned} \int _M u_\alpha ^{r - 1} h\; \mathrm{d}v_g \le c_\lambda A_\alpha \int _M u_\alpha ^{p - 1} h\; \mathrm{d}v_g \end{aligned}$$

for some constant \(c_\lambda > 0\). Thus,

$$\begin{aligned} \int _M |\nabla _g u_\alpha |^{p - 2} \nabla _g u_\alpha \cdot \nabla _g h\; \mathrm{d}v_g \le c \int _M u_\alpha ^{p - 1} h\; \mathrm{d}v_g \end{aligned}$$

for all nonnegative function \(h \in C^1(M \setminus B(x_\alpha , \lambda ))\). A Moser’s iteration then produces

$$\begin{aligned} ||u_\alpha ||_{L^{\infty }(M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) )} \le c ||u_\alpha ||_{L^p(M)}. \end{aligned}$$

We now analyze two distinct cases: \(q \le p < r\) and \(p < q < r\).

Assume the first above situation. From the Claim 2 and integration of (14), we have

$$\begin{aligned}&\int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g \le c A_\alpha ^{\frac{p - q}{r - p}} \int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^q\; \mathrm{d}v_g \le c ||u_\alpha ||_{L^\infty (M \setminus B(x_\alpha , \frac{\delta }{4})} \int _M u_\alpha ^{q - 1}\; \mathrm{d}v_g\\&\quad \le c \left( \int _M u_\alpha ^p\; \mathrm{d}v_g \right) ^{\frac{1}{p}} \left( \int _M u_\alpha ^q\; \mathrm{d}v_g \right) \left( \int _M u_\alpha ^{r-1}\; \mathrm{d}v_g \right) \le c \left( \int _M u_\alpha ^p \mathrm{d}v_g \right) ^{\frac{q + 1}{p}}\;. \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{\int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g}{\int _M u_\alpha ^p\; \mathrm{d}v_g} \le c \left( \int _M u_\alpha ^p\; \mathrm{d}v_g \right) ^{\frac{q - p +1}{p}} \rightarrow 0 \end{aligned}$$

as \(\alpha \rightarrow 1^-\), since \(p < 2\) and \(q \ge 1\) imply \(q - p + 1 > 0\).

Assume now the second case. Using Hölder’s inequality and arguing in a similar manner as above, one gets

$$\begin{aligned}&\int _{M\setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g \le c \left( \int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^q\; \mathrm{d}v_g \right) ^{\frac{p}{q}}\nonumber \\&\quad \le c \left( \left( \int _M u_\alpha ^p\; \mathrm{d}v_g \right) ^{\frac{1}{p}} \left( \int _M u_\alpha ^q\; \mathrm{d}v_g\right) \left( \int _M u_\alpha ^{r - 1}\; \mathrm{d}v_g\right) \right) ^{\frac{p}{q}}. \end{aligned}$$

If \(r - 1 < p\), by Hölder’s inequality,

$$\begin{aligned} \frac{\int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g}{\int _M u_\alpha ^p\; \mathrm{d}v_g} \le c \left( \int _M u_\alpha ^q\; \mathrm{d}v_g\right) ^{\frac{p}{q}} \left( \int _M u_\alpha ^p\; \mathrm{d}v_g \right) ^{\frac{r}{q} - 1} \rightarrow 0 \end{aligned}$$

Otherwise, if \(r - 1 \ge p\), then an interpolation argument combined the normalization \(||u_\alpha ||_r = 1\) yields

$$\begin{aligned} \int _M u_\alpha ^{r - 1} \mathrm{d}v_g \le c \left( \int _M u_\alpha ^p \mathrm{d}v_g \right) ^{\frac{1}{r - p}}. \end{aligned}$$

Thus,

$$\begin{aligned} \int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g \le c \left( \left( \int _M u_\alpha ^q\; \mathrm{d}v_g \right) \left( \int _M u_\alpha ^p\; \mathrm{d}v_g \right) ^{\frac{1}{p} + \frac{1}{r - p}} \right) ^{\frac{p}{q}}, \end{aligned}$$

so that

$$\begin{aligned} \frac{ \int _{M\setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g }{\int _M u_\alpha ^p\; \mathrm{d}v_g} \le c \left( \int _M u_\alpha ^q\; \mathrm{d}v_g \right) ^{\frac{p}{q}} \left( \int _M u_\alpha ^p \mathrm{d}v_g \right) ^{\left( \frac{1}{r - p} + \frac{1 - q}{p}\right) \frac{p}{q}} \rightarrow 0 \end{aligned}$$

as \(\alpha \rightarrow 1^-\), since the inequality \(q \le \frac{r}{r-p}\) is equivalent to \(\frac{1}{r - p} + \frac{1 - q}{p} \ge 0\).

So, taking the limit in (65), one obtains

$$\begin{aligned} \frac{B_0(p,q,r,g)}{A_0(p,q,r)} \le c \delta ^{2 - p} \end{aligned}$$

for all \(\delta > 0\) small enough.

Finally, the facts that \(p < 2\) and

$$\begin{aligned} B_0(p,q,r,g) \ge v_g(M)^{-\frac{p}{n}} > 0\; , \end{aligned}$$
(66)

which can be easily checked by replacing a constant function in (3), lead to the desired contradiction.

5 Proof of Theorem 1.2

In this last section, we present the proof of the compactness theorem.

Let \(\alpha = (p, q, r)\). Consider a sequence \((u_\alpha )\) formed by extremal functions \(u_\alpha \in \mathcal{E}(p,q,r,g)\) for parameters \(p_1 \le p \le p_2, q_1 \le q \le q_2\) and \(r_1 \le r \le r_2\). Without loss of generality, assume \((\alpha )\) converges to \(\alpha _0 = (p_0, q_0, r_0)\).

It is clear that \(u_\alpha \) satisfies

$$\begin{aligned} 1&= \left( \int _M |u_\alpha |^r\; \mathrm{d}v_g\right) ^{\frac{p}{\theta r}} = \left( A_0(p,q,r)\int _M |\nabla _g u_\alpha |^p\; \mathrm{d}v_g\right. \\&\quad \left. + B_0(p,q,r,g) \int _M |u_\alpha |^p\; \mathrm{d}v_g \right) \left( \int _M |u_\alpha |^q\; \mathrm{d}v_g\right) ^{\frac{p(1 -\theta )}{\theta q}} \end{aligned}$$

and is a \(C^1\) solution of the equation

$$\begin{aligned}&A_0(p,q,r) A_\alpha \Delta _{p,g} u_\alpha + B_0(p,q,r,g) A_\alpha u_\alpha ^{p - 1} + \frac{1 - \theta }{\theta } ||u_\alpha ||_{L^q(M)}^{-q} u_\alpha ^{q - 1} = \frac{1}{\theta } u_\alpha ^{r - 1}\, \mathrm{on}\, M,\nonumber \\ \end{aligned}$$
(67)

where

$$\begin{aligned} A_\alpha = \left( \int _M u_\alpha ^q\; \mathrm{d}v_g \right) ^{\frac{p( 1 - \theta )}{\theta q}}. \end{aligned}$$

As in the proof of Theorem 1.1, we show that

$$\begin{aligned} \lim _{\alpha \rightarrow \alpha _0} A_\alpha > 0. \end{aligned}$$
(68)

Assuming the above assertion is true, we prove that \((u_\alpha )\) is weakly compact in a certain sense. Precisely, consider \(t < p_0\) so that \(p, q, r \le s < t^*\). This chosen guarantees, up to a subsequence, that \(u_\alpha \rightharpoonup u_0\) in \(H^{1,t}(M)\) and \(u_\alpha \rightarrow u_0\) in \(L^s(M)\). In particular,

$$\begin{aligned}&||u_\alpha - u_0||_{L^p(M)} \rightarrow 0,\\&||u_\alpha - u_0||_{L^q(M)} \rightarrow 0 \end{aligned}$$

and

$$\begin{aligned} ||u_\alpha - u_0||_{L^r(M)} \rightarrow 0 \end{aligned}$$

as \(\alpha \rightarrow \alpha _0\), so that \(||u_0||_{L^{r_0}(M)} = 1\).

On the other hand, by Theorems 2.1 and 3.1,

$$\begin{aligned} \int _M |\nabla _g u_0|^t\; \mathrm{d}v_g&\le \liminf _{\alpha \rightarrow \alpha _0} \int _M |\nabla _g u_\alpha |^t\; \mathrm{d}v_g \le \liminf _{\alpha \rightarrow \alpha _0} \left( v_g(M)^{1 - \frac{t}{p}} \left( \int _M |\nabla _g u_\alpha |^p\; \mathrm{d}v_g\right) ^{\frac{t}{p}}\right) \\&= \left[ \left( \left( \int _M |u_0|^{q_0}\; \mathrm{d}v_g\right) ^{- \frac{p_0(1 - \theta _0)}{\theta _0 q_0}} - B_0 \int _M |u_0|^{p_0}\; \mathrm{d}v_g \right) A(p_0,q_0,r_0)^{-1}\right] ^{\frac{t}{p_0}}, \end{aligned}$$

where \(B_0 := \lim _{\alpha \rightarrow \alpha _0} B_0(p,q,r,g)\). Letting \(t \rightarrow p_0^-\), by Fatou’s Lemma, one has

$$\begin{aligned} \int _M |\nabla _g u_0|^{p_0}\; \mathrm{d}v_g \le \left( \left( \int _M |u_0|^{q_0}\; \mathrm{d}v_g\right) ^{- \frac{p_0(1 - \theta _0)}{\theta _0 q_0}} - B_0 \int _M |u_0|^{p_0}\; \mathrm{d}v_g \right) A(p_0,q_0,r_0)^{-1}. \end{aligned}$$

Thus,

$$\begin{aligned} \left( \int _M |u_0|^r\; \mathrm{d}v_g\right) ^{\frac{p}{\theta r}}&= 1 \ge \left( A(p_0,q_0,r_0) \int _M |\nabla _g u_0|^{p_0}\; \mathrm{d}v_g\right. \\&\quad \left. + B_0 \int _M u_0^{p_0} \mathrm{d}v_g \right) \left( \int _M u_0^{q_0} \mathrm{d}v_g \right) ^{\frac{p_0(1 - \theta _0)}{\theta _0 q_0}}, \end{aligned}$$

so that, by (3), one has \(B_0 \le B(p_0,q_0,r_0,g)\). On the other hand, for fixed \(u\), passing the limit in

$$\begin{aligned} \left( \int _M |u|^r\; \mathrm{d}v_g \right) ^{\frac{p}{r \theta }}&\le \left( A_0(p,q,r,g) \int _M |\nabla _g u|^p\; \mathrm{d}v_g\right. \\&\quad \left. + B_0(p,q,r,g) \int _M |u|^p\; \mathrm{d}v_g \right) \left( \int _M |u|^q\; \mathrm{d}v_g \right) ^{\frac{p(1 - \theta )}{\theta q}}, \end{aligned}$$

one gets

$$\begin{aligned} \left( \int _M |u|^{r_0}\; \mathrm{d}v_g \right) ^{\frac{p_0}{r_0 \theta _0}}&\le \left( A_0(p_0,q_0,r_0,g) \int _M |\nabla _g u|^{p_0}\; \mathrm{d}v_g\right. \\&\quad \left. + B_0 \int _M |u|^{p_0}\; \mathrm{d}v_g \right) \left( \int _M |u|^{q_0}\; \mathrm{d}v_g \right) ^{\frac{p_0(1 - \theta _0)}{\theta _0 q_0}}, \end{aligned}$$

so that \(B(p_0,q_0,r_0) \le B_0\). So, we conclude that \(B_0 = B(p_0,q_0,r_0)\) and \(u_0\) is a corresponding extremal function. This end the weak compactness.

In order to attain the \(C^0\) compactness, note that (67) and (68) yield

$$\begin{aligned} \int _M |\nabla _g u_\alpha |^{p - 2} \nabla _g u \cdot \nabla _g h\; \mathrm{d}v_g \le c \int _M u_\alpha ^{r - 1} h\; \mathrm{d}v_g \end{aligned}$$

for all nonnegative function \(h \in C^1(M)\). Evoking now a Moser’s iterative scheme, one obtains

$$\begin{aligned} ||u_\alpha ||_{L^\infty (M)} = \sup _{x \in M} u_\alpha (x) \le c ||u_\alpha ||_{L^r(M)} \le c, \end{aligned}$$

for some constant \(c > 0\), which is independent of \(\alpha \). The conclusion follows then from the classical elliptic theory.

Finally, it only remains to show (68). Suppose by contradiction that

$$\begin{aligned} \lim _{\alpha \rightarrow \alpha _0} A_\alpha = 0. \end{aligned}$$

Then,

$$\begin{aligned} A_\alpha \int _M u_\alpha ^p\; \mathrm{d}v_g \rightarrow 0. \end{aligned}$$

The assumptions imply that \(p \le p_2 < r_1 \le r \le r_2 < p^*\) and \(1 \le q \le q_2 \le \frac{r_2}{r_2 - p_1} \le \frac{r}{r - p}\). Thanks to these inequalities, the same strategy of proof of Theorem 3.1 yields the Claims 1 and 2. As before, these claims produce

$$\begin{aligned} B_0(p,q,r,g) \le c \delta ^{2 - p} + c(\delta ) \frac{\int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g}{\int _M u_\alpha ^p\; \mathrm{d}v_g}, \end{aligned}$$

with \(c(\delta ) \rightarrow + \infty \) as \(\delta \rightarrow 0^+\). Proceeding now in the same spirit of the proof of Theorem 1.1, one concludes that

$$\begin{aligned} \lim _{\alpha \rightarrow \alpha _0} \frac{\int _{M \setminus B\left( x_\alpha ,\frac{\delta }{4}\right) } u_\alpha ^p\; \mathrm{d}v_g}{\int _M u_\alpha ^p\; \mathrm{d}v_g} = 0. \end{aligned}$$

Using the facts that \(p \le p_2 < 2, \delta > 0\) can be taken small enough and the lower estimate (66) holds for \(B_0(p,q,r,g)\), we derive a clear contradiction as \(\alpha \rightarrow \alpha _0\).