Let $ \Omega\subset \mathbb{R}^n (n\ge2) $ be a bounded domain and $ 1\le q\le\frac{np}{n-p} $. The Sobolev embedding theorem tells us the embedding $ W^{1, p}_0(\Omega)\subset L^q(\Omega) $ is continuous when $ 1<p<n $ but $ W^{1, n}_0(\Omega)\nsubseteq L^{\infty}(\Omega) $. Trudinger [1] established in the borderline case that $ W^{1, n}_0(\Omega)\subset L_{\varphi_n}(\Omega) $, where $ L_{\varphi_n}(\Omega) $ is the Orlicz space associated with the Young function $ \varphi_n(t)=\exp(\beta|t|^{\frac{n}{n-1}})-1 $ for some $ \beta>0 $. In the celebrated paper[2], Moser found the optimal $ \beta $. In fact, he proved that it holds

$ \begin{align*} \frac{1}{|\Omega|}\int_{\Omega}\exp(\beta_n|u|^{\frac{n}{n-1}})dx\le C, \end{align*} $

where $ \Omega $ be a domain with finite measure in Euclidean n-space $ \mathbb{R}^n $, $ \beta_{n}=n\left( \frac{n\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}\right) ^{\frac{1}{n-1}} $ and $ u\in W^{1, n}_0(\Omega) $ with $ \int_{\Omega}|\nabla u|^ndx\le1 $. Furthermore, this constant $ \beta_n $ is sharp. Similar exponential inequalities on bounded domain have been considered firstly by A. Alvino, V. Ferone and G. Trombetti [3] in Lorentz-Sobolev spaces of $ \mathbb{R}^{n} $.

There has also been substantial progress for Moser-Trudinger inequalities on Riemannian manifolds. In the case of compact Riemannian manifolds, the study of Trudinger-Moser inequalities can be traced back to Aubin [4], Cherrier [5, 6], and Fontana [7]. In the case of Cartan-Hadamard manifold Kong et al [8] and Bertrand et al [9] established the sharp Trudinger-Moser inequality.

One of the aim of this paper is to establish Trudinger-Moser type inequalities on $ M $ for Lorentz-Sobolev spaces. For simplicity, we also denote by $ L^{n, q}(\Omega) $ the Lorentz-Sobolev space on domain $ \Omega\subset M $. One of the main results is the following:

Theorem 1.1  Let $ \Omega $ be a bounded domain of $ M $, $ n\geq2 $, and let $ u\in C^{\infty}_{0}(\Omega) $ be such that

$ \begin{equation*} \|\nabla_{M} u\|_{L^{n, q}(\Omega)}\leq1, \;\;\;\;1\leq q\leq\infty. \end{equation*} $

(1) If $ 1<q<\infty $ and $ Ric_{g}\geq -(n-1)b^{2} $ for some $ b>0 $, then there exists a constant $ C=C(n, q) $ such that

$ \begin{equation} \frac{1}{|\Omega|}\int_{\Omega}e^{\beta_{q}|u(x)|^{q'}}dV\leq C. \end{equation} $

(1.1)

Furthermore, this inequality is sharp in the sense that if $ \beta_{q} $ is replaced by any $ \beta>\beta_{q} $, then inequality (1.1) can no longer hold with some $ C $ independent of $ u $.

(2) If $ q=\infty $, then for every $ \beta<\beta_{\infty} $, there exists a constant $ C=C(n, q, \beta) $ such that

$ \begin{equation} \frac{1}{|\Omega|}\int_{\Omega}e^{\beta|u(x)|}dV\leq C. \end{equation} $

(1.2)

Furthermore, this inequality is sharp in the sense that if $ \beta $ is replaced by any $ \beta\geq\beta_{\infty} $, then inequality (1.2) can no longer hold with some $ C $ independent of $ u $.

We begin by quoting some preliminary facts which will be needed in the sequel and refer to [8, 13] for more precise information about this subject.

Let $ M $ be an $ n $-dimensional complete Riemannian manifold with Riemannian metric $ ds^{2} $. If $ \{x^{i}\}_{1\leq i\leq n} $ is a local coordinate system, then we can write $ ds^{2}=\sum g_{ij}dx^{i}dx^{j} $ so that the Laplace-Beltrami operator $ \Delta_{M} $ in this local coordinate system is

$ \Delta_{M}=\sum\frac{1}{\sqrt{g}}\frac{\partial}{\partial x^{i}}\left(\sqrt{g}g^{ij}\frac{\partial}{\partial x^{j}}\right), $

where $ g=\det(g_{ij}) $ and $ (g^{ij})=(g_{ij})^{-1} $. We denote by $ \nabla_{M} $ the corresponding gradient.

From now on, we let $ M $ be a Cartan-Hadamard manifold. That is, $ M $ is a complete, simply connected Riemannian manifold with negative curvature. Let $ p\in M $ and denote by $ \rho_{p}(x)= \rm{dist}(x, p) $ for all $ x\in M $, where $ \rm{dist}(\cdot, \cdot) $ denotes the geodesic distance. For simplicity, we denote by $ \rho(x)=\rho_{p}(x) $. Then $ \rho(x) $ is smooth on $ M\setminus\{p\} $ and it satisfies

$ |\nabla \rho(x)|=1, \;\;x\in M\setminus\{p\}. $

For any $ \delta>0 $, denote by $ B_{\delta}(p)=\{x\in M: \rho(x)<\delta\} $ the geodesic ball in $ M $. We introduce the density function $ J_{p}(\theta, t) $ of the volume form in normal coordinates as follows (see e.g. [10], page 166-167). Choose an orthonormal basis $ \{\theta, e_{2}, \cdots, e_{n}\} $ on $ T_{p}M $ and let $ c(t)= \rm{Exp}_{p}t\theta $ be a geodesic. $ \{Y_{i}(t)\}_{2\leq i\leq n} $ are Jacobi fields satisfying the initial conditions

$ Y_{i}(0)=0, \;\; Y'_{i}(0)=e_{i}, \;\;2\leq i\leq n, $

so that the density function can be given by

$ J_{p}(t, \theta)=t^{-n+1}\sqrt{\det(\langle Y_{i}(t), Y_{j}(t)\rangle)}, \;\;t>0. $

We note that $ J_{p}(t, \theta) $ does not depend on $ \{e_{2}, \cdots, e_{n}\} $ and $ J_{p}(t, \theta)\in C^{\infty}(T_{p}M\backslash\{p\}) $ by the definition of $ J_{p}(t, \theta) $. Furthermore, if we set $ J_{p}(\theta, 0)\equiv1 $, then $ J_{p}(t, \theta)\in C(T_{p}M) $ and

$ \begin{equation} J_{p}(t, \theta)=1+O(t^{2}) \;\; \rm{as}\;\; t\rightarrow 0, \end{equation} $

(2.1)

since $ Y_{i}(t) $ has the asymptotic expansion (see e.g. [10], page 169)

$ Y_{i}(t)=te_{i}-\frac{t^{3}}{6}R(c'(t), e_{i})c'(t)+o(t^{3}), $

where $ R(\cdot, \cdot) $ is the curvature tensor on $ M $. By the definition of $ J_{p}(t, \theta) $, we have the following formula in polar coordinates on $ M $:

$ \int_{M}f(x)dV=\int^{\infty}_{0}\int_{\mathbb{S}^{n-1}}f(\rho, \theta)\rho^{n-1}J_{p}(\rho, \theta)d\rho d\sigma, \;\;f\in L^{1}(M), $

where $ d\sigma $ denotes the canonical measure of the unit sphere of $ T_{p}(M) $. Finally, we recall a useful fact of $ J_{p}(t, \theta) $ which plays an important role in the study of Moser-Trudinger inequalities. If the sectional curvature $ K $ on $ M $ satisfies $ K\leq -b $, then (see [10], page 172, line -2, the proof of Bishop-Gunther comparison theorem)

$ \begin{equation} \frac{1}{J_{p}(t, \theta)}\cdot\frac{\partial J_{p}(t, \theta)}{\partial t}\geq\frac{J_{b}'(t)}{J_{b}(t)}, \;\; t>0. \end{equation} $

(2.2)

In particular, since $ M $ is with negative curvature, we have

$ \begin{equation*} \frac{1}{J_{p}(t, \theta)}\cdot\frac{\partial J_{p}(t, \theta)}{\partial t}\geq\frac{J_{0}'(t)}{J_{0}(t)}=0, \end{equation*} $

which means $ J_{p}(t, \theta) $, as a function of $ t $ on $ [0, +\infty) $, is monotonically increasing.

We firstly recall the rearrangement of functions on $ M $. Suppose $ f $ is a nonnegative function on $ M $. The non-increasing rearrangement of $ f $ is defined by

$ \begin{equation} f^{\ast}(t)=\inf\{s>0: \lambda_{f}(s)\leq t\}, \end{equation} $

(3.1)

where $ \lambda_{f}(s)=|\{x\in M: f(x)>s\}| $. Here we use the notation $ |\Sigma| $ for the measure of a measurable set $ \Sigma\subset M $. Set

$ f^{\ast\ast}(t)=\frac{1}{t}\int^{t}_{0}f^{\ast}(s)ds. $

Denote by$ L^{p, q}(\Omega) $ the Lorentz space of those function $ f $ on a domain $ \Omega\subset M $ satisfying

$\|f\|_{L^{p, q}(\Omega)}= \begin{cases}\left\|t^{\frac{1}{p}-\frac{1}{q}} f^*(t)\right\|_{L^q(0,|\Omega|)}, & 1 \leq q＜\infty ; \\ \sup\limits_{t>0} f^*(t) t^{1 / p}, & q=\infty\end{cases}$

is finite.Similarly, denote by

$\|f\|_{L^{p, q}(\Omega)}^*= \begin{cases}\left\|t^{\frac{1}{p}-\frac{1}{q}} f^{* *}(t)\right\|_{L^q(0,|\Omega|)}, & 1 \leq q＜\infty; \\ \sup\limits_{t>0} f^{* *}(t) t^{1 / p}, & q=\infty.\end{cases}$

We remark that for $ 1<q<+\infty $ and $ 0< r\leq +\infty $, there holds (see [11], Theorem 3.4)

$ \begin{equation} \|f\|_{L^{q, r}} \leq\|f\|^{\ast}_{L^{q, r}}\leq C(p, q)\|f\|_{L^{q, r}}, \end{equation} $

(3.2)

where $ C(p, q) $ is a constant depending on $ p $ and $ r $. Moreover, we have the following generalization of Young's inequality for convolution (see [12]):

Proposition 3.1  Let $ 1<r, p_{1}, p_{2}<+\infty $ and $ 1\leq s, q_{1}, q_{2}\leq\infty $. If

$ \frac{1}{p_{1}}+\frac{1}{p_{2}}-1=\frac{1}{r}\;\; \rm{and}\;\;\frac{1}{q_{1}}+\frac{1}{q_{2}}\geq\frac{1}{s}, $

then there exists $ C>0 $ such that

$ \begin{equation} \begin{split} \|f\ast g\|_{L^{r, s}}\leq&C\|f\|_{L^{p_{1}, q_{1}}}\|g\|_{L^{p_{2}, q_{2}}}, \;\;f\in L(p_{1}, q_{1}), \;\;g\in L(p_{2}, q_{2}). \end{split} \end{equation} $

(3.3)

We also need the following inequality (see [12], Corollary 1.8)

$ \begin{equation} \begin{split} \|f\ast g\|_{\infty}\leq \int^{\infty}_{0}f^{\ast}(t)g^{\ast}(t)dt. \end{split} \end{equation} $

(3.4)

Lemma 3.2  Let $ 0<\alpha<n $ and set $ \phi_{\alpha}=\frac{1}{\rho^{n-\alpha}J_{p}(\rho, \theta)} $. Then

$ \phi_{\alpha}^{\ast}(t)\leq \left(\frac{t}{\omega_{n}}\right)^{-(n-\alpha)/n}, \;\;t>0, $

where $ \omega_{n}=\pi^{n/2}/\Gamma(1+n/2) $ is the volume of $ \mathbb{S}^{n} $.

Proof  As in the proof of [8], Lemma 3.2, we set, for any $ s>0 $,

$ \begin{equation} \lambda_{\phi_{\alpha}}(s)=\int_{\{x\in M: \phi_{\alpha}(x)>s\}}dV=\int_{\{(\rho, \theta)\in M: \rho^{n-\alpha}J_{p}(\rho, \theta)<s^{-1}\}}dV. \end{equation} $

(3.5)

We denote by $ \rho_{\theta}(s) $ the solution of $ \rho^{n-\alpha}J_{p}(\rho, \theta)=s^{-1} $. Then $ \rho_{\theta}(s)^{n-\alpha}J_{p}(\rho_{\theta}(s), \theta)=s^{-1} $ and

$ \begin{equation*} \lambda_{\phi_{\alpha}}(s)=\int_{\{(\rho, \theta)\in M: \rho^{n-1}J_{p}(\rho, \theta)<s^{-1}\}}dV=\int_{\mathbb{S}^{n-1}}\int^{\rho_{\theta}(s)}_{0}\rho^{n-1}J_{p}(\rho, \theta)d\sigma d\rho. \end{equation*} $

Therefore, since $ \phi_{\alpha}^{\ast}(t)=\inf\{s>0: \lambda_{\phi_{\alpha}}(s)\leq t\} $, we have

$ \begin{equation} t=\lambda_{\phi_{\alpha}}(\phi_{\alpha}^{\ast}(t))=\int_{\mathbb{S}^{n-1}}\int^{\rho_{\theta}(\phi_{\alpha}^{\ast}(t))}_{0}\rho^{n-1}J_{p}(\rho, \theta)d\sigma d\rho, \end{equation} $

(3.6)

where $ \rho_{\theta}(\phi_{\alpha}^{\ast}(t)) $ satisfies

$ \begin{equation} \rho_{\theta}(\phi_{\alpha}^{\ast}(t))^{n-\alpha}J_{p}(\rho_{\theta}(\phi_{\alpha}^{\ast}(t)), \theta)=\frac{1}{\phi_{\alpha}^{\ast}(t)}. \end{equation} $

(3.7)

For simplicity, we set $ \rho_{\theta}(t)=\rho_{\theta}(\phi_{\alpha}^{\ast}(t)) $ in the rest of proof. Then,

$ t=\lambda_{\phi_{\alpha}}(\phi_{\alpha}^{\ast}(t))=\int_{\mathbb{S}^{n-1}}\int^{\rho_{\theta}(t)}_{0}\rho^{n-1}J_{p}(\rho, \theta)d\sigma d\rho $

and $ \rho_{\theta}(t) $ satisfies

$ \begin{equation*} \rho_{\theta}(t)^{n-\alpha}J_{p}(\theta, \rho_{\theta}(t))=\frac{1}{\phi_{\alpha}^{\ast}(t)}. \end{equation*} $

Thus, since $ J_{p}(\rho, \theta) $, as a function of $ \rho $ on $ [0, +\infty) $, is monotonically increasing and $ J_{p}(\rho, \theta)\geq J_{p}(0, \theta)=1 $, we have

$ \begin{equation*} \begin{split} t=&\int_{\mathbb{S}^{n-1}}\int^{\rho_{\theta}(t)}_{0}\rho^{n-1}J_{p}(\rho, \theta)d\sigma d\rho\\ \leq&\int_{\mathbb{S}^{n-1}}\int^{\rho_{\theta}(t)}_{0}\rho^{n-1}J_{p}(\theta, \rho_{\theta}(t))d\sigma d\rho\\ =&\int_{\mathbb{S}^{n-1}}J_{p}(\theta, \rho_{\theta}(t))\left(\int^{\rho_{\theta}(t)}_{0}\rho^{n-1}d\rho\right)d\sigma\\ =&\frac{1}{n}\int_{\mathbb{S}^{n-1}}J_{p}(\theta, \rho_{\theta}(t))\rho^{n}_{\theta}(t)d\sigma\\ \leq& \frac{1}{n}\int_{\mathbb{S}^{n-1}}J^{n/(n-\alpha)}_{p}(\theta, \rho_{\theta}(t))\rho^{n}_{\theta}(t)d\sigma\\ =&\frac{1}{n}\int_{\mathbb{S}^{n-1}}\left[J_{p}(\theta, \rho_{\theta}(t))\rho^{n-\alpha}_{\theta}(t)\right]^{n/(n-\alpha)}d\sigma\\ =&\frac{1}{n}[\phi_{\alpha}^{\ast}(t)]^{-n/(n-\alpha)}|\mathbb{S}^{n-1}|=[\phi_{\alpha}^{\ast}(t)]^{-n/(n-\alpha)}\omega_{n}. \end{split} \end{equation*} $

Therefore, $ \phi_{\alpha}^{\ast}(t)\leq \left(\frac{t}{\omega_{n}}\right)^{-(n-\alpha)/n}, \;\;t>0. $ This completes the proof of Lemma 3.2.

Lemma 3.3  Let $ 0<\alpha<n $ and set $ \widetilde{\phi}_{\alpha}=\frac{1}{\rho^{n-\alpha}f_{\mathbb{S}^{n-1}}J_{p}(\rho, \theta)d\theta} $, where $ f_{\mathbb{S}^{n-1}}J_{p}(\rho, \theta)d\theta=\frac{1}{|\mathbb{S}^{n-1}|}\int_{\mathbb{S}^{n-1}}J_{p}(\rho, \theta)d\theta $. Then

$ \widetilde{\phi}_{\alpha}^{\ast}(t)\leq \left(\frac{t}{\omega_{n}}\right)^{-(n-\alpha)/n}, \;\;t>0, $

Proof  The proof is similar to that given in Lemma 3.2. Set

$ \begin{equation*} \lambda_{\widetilde{\phi}_{\alpha}}(s)=\int_{\{x\in M: \widetilde{\phi}_{\alpha}(x)>s\}}dV=\int_{\{(\rho, \theta)\in M: \rho^{n-\alpha}\int_{\mathbb{S}^{n-1}}J_{p}(\rho, \theta)d\theta<s^{-1}\}}dV. \end{equation*} $

Denote by $ \rho(s) $ the solution of $ \rho^{n-\alpha}\int_{\mathbb{S}^{n-1}}J_{p}(\rho, \theta)d\theta=s^{-1} $. Then $ \rho(s) $ satisfies

$ \begin{equation*} \rho(s)^{n-\alpha}\int_{\mathbb{S}^{n-1}}J_{p}(\rho(s), \theta)d\theta=s^{-1} \end{equation*} $

and

$ \begin{equation*} \lambda_{\widetilde{\phi}_{\alpha}}(s)=\int_{\{(\rho, \theta)\in M: \rho^{n-1}J_{p}(\rho, \theta)<s^{-1}\}}dV=\int_{\mathbb{S}^{n-1}}\int^{\rho(s)}_{0}\rho^{n-1}J_{p}(\rho, \theta) d\sigma d\rho. \end{equation*} $

With the same argument as that in the proof of Lemma 3.2, we have

$ \begin{equation*} \begin{split} t=&\lambda_{\widetilde{\phi}_{\alpha}}(\widetilde{\phi}_{\alpha}^{\ast}(t)) =\int_{\mathbb{S}^{n-1}}\int^{\rho(\widetilde{\phi}_{\alpha}^{\ast}(t))}_{0}\rho^{n-1}J_{p}(\rho, \theta)d\sigma d\rho\\ \leq&\int_{\mathbb{S}^{n-1}}J_{p}(\theta, \rho(\widetilde{\phi}_{\alpha}^{\ast}(t)))d\sigma\int^{\rho(\widetilde{\phi}_{\alpha}^{\ast}(t))}_{0}\rho^{n-1} d\rho\\ =&\frac{|\mathbb{S}^{n-1}|}{n}\rho^{n}(\widetilde{\phi}_{\alpha}^{\ast}(t))f_{\mathbb{S}^{n-1}}J_{p}(\theta, \rho(\widetilde{\phi}_{\alpha}^{\ast}(t))d\sigma\\ \leq& \omega_{n}\left[\rho^{n-\alpha}(\widetilde{\phi}_{\alpha}^{\ast}(t))f_{\mathbb{S}^{n-1}}J_{p}(\theta, \rho(\widetilde{\phi}_{\alpha}^{\ast}(t))d\sigma\right] ^{\frac{n}{n-\alpha}}=\omega_{n}[\widetilde{\phi}_{\alpha}^{\ast}(t)]^{-n/(n-\alpha)}. \end{split} \end{equation*} $

The rest of proof is completely to that given in Lemma 3.2 and we omit it.

For simplicity, we set $ \phi=\phi_{1}=\frac{1}{\rho^{n-\alpha}J_{p}(\rho, \theta)}. $ Then by by Lemma 3.2,

$ \begin{equation} \phi^{\ast}(t)\leq \left(\frac{t}{\omega_{n}}\right)^{-(n-1)/n}, \;\;t>0. \end{equation} $

(3.8)

Now we can give the proof of the main result.

Proof of Theorem 1.1

(1) Without loss of generality, we assume $ \Omega\subset B(x_0, R)=\{x\in M: d(x, x_0)< R\} $ for some $ x_0\in M $ and $ R>0 $. Then by (3.1) in [8] we have

$ \begin{align*} |u(p)|\leq\frac{1}{n\omega_{n}}\int_{M}|\nabla_{M} u|\frac{1}{\rho^{n-1}J_{p}(\rho, \theta)}dV\leq \frac{1}{n\omega_{n}}\int_{M}|\nabla_{M} u|\frac{1}{\rho^{n-1}}dV. \end{align*} $

With the same arguments as in [9], we have, for some $ A>0 $ and $ \epsilon>0 $,

$ \begin{align*} (\frac{1}{\rho^{n-1}})^{\ast}(t)\leq \left(\frac{t}{\omega_{n}}\right)^{-(n-1)/n}+ At^{-\frac{n-1-\epsilon}{n}}, \;\; t\rightarrow0+. \end{align*} $

Applying O'Neil's lemma (see [12], Lemma 1.5), we have, by (3.1) in [8],

$ \begin{equation*} \begin{split} u^{\ast}(t)\leq& \frac{1}{n\omega_{n}}\left(\frac{1}{t}\int^{t}_{0}|\nabla_{M}u|^{\ast}(s)ds\int^{t}_{0} \left[\left(\frac{s}{\omega_{n}}\right)^{-(n-1)/n}+ As^{-\frac{n-1-\epsilon}{n}}\right]ds+\right.\\ &\left. \int^{|\Omega|}_{t}|\nabla_{M}u|^{\ast}(s) \left[\left(\frac{s}{\omega_{n}}\right)^{-(n-1)/n}+ As^{-\frac{n-1-\epsilon}{n}}\right]ds\right), \end{split} \end{equation*} $

i.e.

$ \begin{equation} \begin{split} u^{\ast}(t)\leq&\frac{1}{n\omega^{1/n}_{n}}\left((nt^{-\frac{n-1}{n}}+A\frac{n}{1+\epsilon}t^{-\frac{n-1-\epsilon}{n}})\int^{t}_{0}|\nabla_{M}u|^{\ast}(s)ds+\right.\\ &\left. \int^{|\Omega|}_{t}|\nabla_{M}u|^{\ast}(s) \left[s^{-\frac{n-1}{n}}+A\left(\omega_{n}\right)^{-(n-1)/n}s^{-\frac{n-1-\epsilon}{n}}\right]ds\right). \end{split} \end{equation} $

(4.1)

With the same arguments as in [13], there exists a constant $ C>0 $ independent of $ u $ such that

$ \begin{equation*} \begin{split} &\frac{1}{|\Omega|}\int_{\Omega}e^{\beta_{q}|u|^{q'}}dV <C. \end{split} \end{equation*} $

Now we prove the sharpness of $ \beta_{q} $. Denote by $ B_{r}=\{x\in M:\rho(x)<r\} $. Set, for each $ \varepsilon\in(0, 1) $,

$f_{\varepsilon}(\rho)= \begin{cases}\int_\rho^1\left[t f_{\mathbb{S}^{n-1}} J_p(t, \theta) d \theta\right]^{-1} d t \cdot\left\{\int_{\varepsilon}^1\left[t f_{\mathbb{S}^{n-1}} J_p(t, \theta) d \theta\right]^{-1} d t\right\}^{-1}, & \varepsilon \leq \varrho \leq 1; \\ 1, & 0 \leq \rho \leq \varepsilon.\end{cases} $

We compute

$\left|\nabla_M f_{\varepsilon}(\rho)\right|= \begin{cases}{\left[\rho f_{\mathbb{S}^n-1} J_p(\rho, \theta) d \theta\right]^{-1}\left\{\int_{\varepsilon}^1\left[t f_{\mathbb{S}^{n-1}} J_p(t, \theta) d \theta\right]^{-1} d t\right\}^{-1},} & \varepsilon＜\varrho \leq 1; \\ 0, & 0 \leq \rho＜\varepsilon.\end{cases}$

Following the proof of Lemma 3.3, we have

$ \begin{equation*} \begin{split} |\nabla_{M}f_{\varepsilon}|^{\ast}(t)\leq& \left\{ \int^{1}_{\varepsilon}\left[tf_{\mathbb{S}^{n-1}}J_{p}(t, \theta)d\theta\right]^{-1}dt\right\}^{-1}\left(\frac {\omega_{n}}{t+|B_{\varepsilon}|}\right)^{\frac{1}{n}}, \;\; 0\leq t< |B_{1}|-|B_{\varepsilon}| \end{split} \end{equation*} $

and

$ |\nabla_{M}f_{\varepsilon}|^{\ast}(t)=0, \;\;|B_{1}|-|B_{\varepsilon}|\leq t\leq |B_{1}|. $

Therefore,

$ \begin{equation*} \begin{split} \|\nabla_{M}f_{\varepsilon}\|^{q'}_{L^{n, q}(B_{1})}=&\left[\int^{|B_{1}|}_{0} \left(t^{\frac{1}{n}}|\nabla_{M}f|^{\ast}(t)\right)^{q}\frac{dt}{t}\right] ^{\frac{1}{q-1}}\\ \leq&\frac{\omega^{q'/n}_{n}}{\left\{ \int^{1}_{\varepsilon}\left[tf_{\mathbb{S}^{n-1}}J_{p}(t, \theta)d\theta\right]^{-1}dt\right\}^{q'}} \left[\int^{|B_{1}|-|B_{\varepsilon}|}_{0} \left(\frac {t}{t+|B_{\varepsilon}|}\right)^{\frac{q}{n}}\frac{dt}{t}\right]^{\frac{1}{q-1}}. \end{split} \end{equation*} $

Substituting $ s=1+\frac{t}{|B_{\varepsilon}|} $ in the integral we have

$ \begin{equation} \begin{split} \|\nabla_{M}f_{\varepsilon}\|^{q'}_{L^{n, q}(B_{1})} \leq&\frac{\omega^{q'/n}_{n}}{\left\{ \int^{1}_{\varepsilon}\left[tf_{\mathbb{S}^{n-1}}J_{p}(t, \theta)d\theta\right]^{-1}dt\right\}^{q'}} \left[\int^{\frac{|B_{1}|}{|B_{\varepsilon}|}}_{1} \left(1-\frac{1}{s}\right)^{\frac{q}{n}}\frac{ds}{s-1}\right]^{\frac{1}{q-1}}. \end{split} \end{equation} $

(4.2)

Now assume that

$ \frac{1}{|B_{1}|}\int_{B_{1}}\exp\left[\beta\left(\frac{|f_{\varepsilon}|}{\|\nabla_{M} f_{\varepsilon}\|_{L^{n, q}(B_{1})}}\right)^{q'}\right]dV\leq C $

for some $ \beta>0 $. Using the fact $ f_{\varepsilon}\equiv 1 $ on $ B_{\varepsilon} $, we have

$ \frac{|B_{\varepsilon}|}{|B_{1}|}\exp\left(\beta\frac{1}{\|\nabla_{M} f_{\varepsilon}\|^{q'}_{L^{n, q}(B_{1})}}\right)\leq C. $

By (4.2), we have

$ \begin{equation} \begin{split} \beta\leq& \left(\ln C+\ln |B|+\ln |B_{\varepsilon}|^{-1}\right)\|\nabla_{M} f_{\varepsilon}\|^{q'}_{L^{n, q}(B_{1})}\\ \leq&\omega^{q'/n}_{n} \frac{\ln C+\ln |B|+\ln |B_{\varepsilon}|^{-1}}{\left\{ \int^{1}_{\varepsilon}\left[tf_{\mathbb{S}^{n-1}}J_{p}(t, \theta)d\theta\right]^{-1}dt\right\}^{q'}} \left[\int^{\frac{|B_{1}|}{|B_{\varepsilon}|}}_{1} \left(1-\frac{1}{s}\right)^{\frac{q}{n}}\frac{ds}{s-1}\right]^{\frac{1}{q-1}}. \end{split} \end{equation} $

(4.3)

By (2.1), we have

$ |B_{\varepsilon}|=\int^{\varepsilon}_{0}\int_{\mathbb{S}^{n-1}}\rho^{n-1}J(\rho, \theta)d\rho d\theta=\omega_{n}\varepsilon^{n}(1+O(\varepsilon^{2})), \;\;\varepsilon\rightarrow0+. $

Also by (2.1), one can easily check

$ \lim\limits_{\varepsilon\rightarrow0+}\frac{\ln C+\ln |B|+\ln |B_{\varepsilon}|^{-1}}{ \int^{1}_{\varepsilon}\left[tf_{\mathbb{S}^{n-1}}J_{p}(t, \theta)d\theta\right]^{-1}dt}=n $

and

$ \lim\limits_{\varepsilon\rightarrow0+}\frac{1}{ \int^{1}_{\varepsilon}\left[tf_{\mathbb{S}^{n-1}}J_{p}(t, \theta)d\theta\right]^{-1}dt}\int^{\frac{|B_{1}|}{|B_{\varepsilon}|}}_{1} \left(1-\frac{1}{s}\right)^{\frac{q}{n}}\frac{ds}{s-1}=n. $

Therefore, passing the limit $ \varepsilon\rightarrow 0+ $ in (4.3) yields

$ \beta\leq \omega^{q'/n}_{n}n^{1+\frac{1}{q-1}}=\beta_{q}. $

(2) Since $ \|\nabla_{M}u\|_{L^{n, \infty}(\Omega)}\leq1 $, we have

$ \begin{equation} \begin{split} |\nabla_{M}u|^{\ast}\leq t^{-\frac{1}{n}}, \;\;0\leq t\leq |\Omega|. \end{split} \end{equation} $

(4.4)

Therefore, by (4.1), we have for $ 0<t<|\Omega| $,

$ \begin{equation*} \begin{split} u^{\ast}(t)\leq& \frac{1}{n\omega^{1/n}_{n}}\left((nt^{-\frac{n-1}{n}}+A\frac{n}{1+\epsilon}t^{-\frac{n-1-\epsilon}{n}})\int^{t}_{0}|\nabla_{M}u|^{\ast}(s)ds+\right.\\ &\left. \int^{|\Omega|}_{t}|\nabla_{M}u|^{\ast}(s) \left[s^{-\frac{n-1}{n}}+A\left(\omega_{n}\right)^{-(n-1)/n}s^{-\frac{n-1-\epsilon}{n}}\right]ds\right)\\ \leq&\frac{1}{\beta_{\infty}}\left(\frac{n^{2}}{n-1}+\frac{n^{2}A}{(n-1)(1+\epsilon)}t^{\frac{\epsilon}{n}}+\frac{n}{\epsilon} (|\Omega|^{\frac{\epsilon}{n}}-t^{\frac{\epsilon}{n}})+\ln\frac{|\Omega|}{t}\right). \end{split} \end{equation*} $

Thus, for every $ \beta<\beta_{\infty} $, we have

$ \begin{equation*} \begin{split} \frac{1}{|\Omega|}\int_{\Omega}e^{\beta|u(x)|}dV=&\frac{1}{|\Omega|}\int^{|\Omega|}_{0}e^{\beta u^{\ast}(t)}dt\\ \leq&\frac{1}{|\Omega|}\int^{|\Omega|}_{0}\exp\frac{\beta}{\beta_{\infty}}\left(\frac{n^{2}}{n-1}+\frac{n^{2}A}{(n-1)(1+\epsilon)}t^{\frac{\epsilon}{n}}+\frac{n}{\epsilon} (|\Omega|^{\frac{\epsilon}{n}}-t^{\frac{\epsilon}{n}})+\ln\frac{|\Omega|}{t}\right)dt<\infty. \end{split} \end{equation*} $

To see the constant $ \beta_{\infty} $ is sharp, we consider the function

$ f(\rho)=\frac{1}{\beta_{\infty}}n\int^{1}_{\rho}\left[tf_{\mathbb{S}^{n-1}}J_{p}(t, \theta)d\theta\right]^{-1}dt, \;\; 0<\rho\leq 1. $

By (2.1), we have

$ f(\rho)=\frac{1}{\beta_{\infty}}n\ln\frac{1}{\rho}+O(\rho^{2}), \;\;\rho\rightarrow0+. $

Therefore,

$ \int_{B_{1}}e^{\beta_{\infty}|f(x)|}dV=\infty. $

On the other hand, since $ |\nabla_{M}f|=\frac{n}{\beta_{\infty}\rho f_{\mathbb{S}^{n-1}}J_{p}(\rho, \theta)d\theta}=\frac{1}{\omega^{1/n}_{n}\rho f_{\mathbb{S}^{n-1}}J_{p}(\rho, \theta)d\theta} $, we have, by Lemma 3.3,

$ |\nabla_{M}f|^{\ast}(t)\leq t^{-\frac{1}{n}} $

and thus

$ \|\nabla_{M}f\|_{L^{n, \infty}(B_{1})}\leq 1. $

The proof is thereby completed.