Existence and non-existence of maximizers for the Moser–Trudinger type inequalities under inhomogeneous constraints

Abstract

In this paper, we study the existence and non-existence of maximizers for the Moser–Trudinger type inequalities in $\mathbb {R}^N$ of the form

$$\begin{aligned} D_{N,\alpha }(a,b):= \sup _{u\in W^{1,N}(\mathbb {R}^N),\,\Vert \nabla u\Vert _{L^N(\mathbb {R}^N)}^a+\Vert u\Vert _{L^N(\mathbb {R}^N)}^b=1} \int _{\mathbb {R}^N}\Phi _N\left( \alpha |u|^{N'}\right) dx. \end{aligned}$$

Here $N\ge 2, N'=\frac{N}{N-1}, a,b>0, \alpha \in (0,\alpha _N]$ and $\Phi _N(t):=e^t-\sum _{j=0}^{N-2}\frac{t^j}{j!}$ where $\alpha _N:= N \omega _{N-1}^{1/(N-1)}$ and $\omega _{N-1}$ denotes the surface area of the unit ball in ${\mathbb {R}}^{N}$. We show the existence of the threshold $\alpha _*= \alpha _*(a,b,N) \in [0,\alpha _N]$ such that $D_{N,\alpha }(a,b)$ is not attained if $\alpha \in (0,\alpha _*)$ and is attained if $\alpha \in (\alpha _*, \alpha _N)$. We also provide the conditions on (a, b) in order that the inequality $\alpha _*< \alpha _N$ holds.

A Proof of \(N^2/(\alpha _N B_{GN}) < N\)

The purpose of this appendix is to prove

$$\begin{aligned} \frac{N^2}{\alpha _N B_{GN}} < N \end{aligned}$$

(A.1)

for all $N \ge 2$. Notice that when $N =2$, (A.1) is equivalent to $2/ B_{GN} < \alpha _2 = 4 \pi$ and this is already known (see [2, 12, 34]). However, we also provide a simple proof of this fact below.

We first rewrite (A.1). Using the Schwarz rearrangement, it follows that

$$\begin{aligned} B_{GN} = \sup _{u \in W^{1,N}_\mathrm{r} ({\mathbb {R}}^{N}) {\setminus } \{0\} } \frac{ \Vert u \Vert _{NN'}^{NN'} }{ \Vert u \Vert _{N}^N \Vert \nabla u \Vert _{N}^{NN'-N} }, \end{aligned}$$

where $W^{1,N}_\mathrm{r}({\mathbb {R}}^{N})$ is a set consisting of radial functions in $W^{1,N}({\mathbb {R}}^{N})$. For $u \in W^{1,N}_\mathrm{r}({\mathbb {R}}^{N})$, one has

$$\begin{aligned} \Vert \nabla u \Vert _N^N = \omega _{N-1} \int _0^\infty r^{N-1} |u'(r) |^{N} d r, \quad \Vert u \Vert _p^p = \omega _{N-1} \int _0^\infty r^{N-1} |u(r)|^p dr, \end{aligned}$$

which implies

$$\begin{aligned} B_{GN} = \omega _{N-1}^{-1/(N-1)} \sup _{u \in W^{1,N}_\mathrm{r} ({\mathbb {R}}^{N}) {\setminus } \{0\} } \frac{\int _0^\infty r^{N-1}|u(r)|^{NN'} dr }{\int _0^\infty r^{N-1} |u(r)|^N dr \left( \int _0^\infty r^{N-1} |u'(r)|^N dr \right) ^{1/(N-1)} }. \end{aligned}$$

Recalling $\alpha _N = N \omega _{N-1}^{1/(N-1)}$, we see that (A.1) is equivalent to

$$\begin{aligned} 1 - \inf _{ u \in W^{1,N}_\mathrm{r} ({\mathbb {R}}^{N}) {\setminus } \{0\} } \frac{ \int _0^\infty r^{N-1} |u(r)|^N dr \left( \int _0^\infty r^{N-1} |u'(r)|^N dr \right) ^{1/(N-1)} }{ \int _0^\infty r^{N-1}|u(r)|^{NN'} dr } >0. \end{aligned}$$

(A.2)

Therefore, instead of (A.1), we shall prove (A.2) for every $N \ge 2$.

When $N=2$, we can check (A.2) by setting $u(r) := \max \{ 0 , (1-r)^3 \}$. In fact, since

$$\begin{aligned} \begin{aligned}&\int _0^1 r (1-r)^6 dr = B(2,7) = \frac{\Gamma (2) \Gamma (7)}{\Gamma (9)}, \quad 9 \int _0^1 r (1-r)^4 dr = 9 B(2,5) = 9 \frac{\Gamma (2) \Gamma (5)}{\Gamma (7)},\\&\int _0^1 r (1-r)^{12} dr = B(2,13) = \frac{\Gamma (2)\Gamma (13)}{\Gamma (15)} \end{aligned} \end{aligned}$$

where B(x, y) and $\Gamma (z)$ are the beta function and the gamma function, we see

$$\begin{aligned} \begin{aligned}&\frac{ \int _0^\infty r^{N-1} |u(r)|^N dr \left( \int _0^\infty r^{N-1} |u'(r)|^N dr \right) ^{1/(N-1)} }{ \int _0^\infty r^{N-1}|u(r)|^{NN'} dr }\\&\quad = \frac{\Gamma (7)}{\Gamma (9)} 9 \frac{\Gamma (5)}{\Gamma (7)} \frac{\Gamma (15)}{\Gamma (13)} = 9 \frac{1}{8\cdot 7 \cdot 6 \cdot 5} 14 \cdot 13 = \frac{39}{40} < 1. \end{aligned} \end{aligned}$$

For the case $N \ge 3$, we choose $u(r) = e^{-r}$ as a test function. Observe that

$$\begin{aligned} \begin{aligned}&\int _0^\infty r^{N-1} e^{-Nr} dr = \frac{\Gamma (N)}{N^N} = \int _0^\infty r^{N-1} | (e^{-r})' |^N d r, \\&\int _0^\infty r^{N-1} e^{-N^2 r/ (N-1)} dr = \left( \frac{N-1}{N^2} \right) ^N \Gamma (N). \end{aligned} \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} \begin{aligned}&\frac{ \int _0^\infty r^{N-1} |u(r)|^N dr \left( \int _0^\infty r^{N-1} |u'(r)|^N dr \right) ^{1/(N-1)} }{ \int _0^\infty r^{N-1}|u(r)|^{NN'} dr}\\&\quad = \Gamma (N)^{1/(N-1)} (N-1)^{-N} N^{(N^2-2N)/(N-1)} =: C_N. \end{aligned} \end{aligned}$$

(A.3)

Thus to prove (A.2), it suffices to show $C_N^{N-1} < 1$ for $N\ge 3$, which is equivalent to $(N-1) \log (C_N) < 0$.

From

$$\begin{aligned} C_{N}^{N-1} = (N-1) ! N^{N^2 - 2 N} (N-1)^{-N^2+N}, \end{aligned}$$

it follows that

$$\begin{aligned} \begin{aligned} \log (C_N^{N-1})&= \sum _{k=1}^{N-1} \log k + (N^2 - 2N) \log N - (N^2-N) \log (N-1)\\&=\sum _{k=1}^{N-1} \log k + (N^2 - N) \left( \log N - \log (N-1) \right) - N \log N\\&=\left\{ \sum _{k=1}^{N-1} \log k - N \left( \log N - 1\right) \right\} + \left\{ - N + N(N-1) \log \left( 1 + \frac{1}{N-1} \right) \right\} \\&=: d_N+e_N \end{aligned} \end{aligned}$$

Claim 1

For each$N \ge 3$, there holds

$$\begin{aligned} e_N < - \frac{1}{2}. \end{aligned}$$

In fact, by $\log (1+x) = \sum _{k=1}^\infty (-1)^{k-1} x^k / k$ for $0 \le x <1$ and $N \ge 3$, we have

$$\begin{aligned} \begin{aligned}&\log \left( 1 + \frac{1}{N-1} \right) \\&\quad = \frac{1}{N-1} - \frac{1}{2(N-1)^2} \\&\qquad + \frac{1}{3(N-1)^3} + \sum _{k=2}^\infty \left( -\frac{1}{2k} \frac{1}{(N-1)^{2k}} + \frac{1}{2k+1} \frac{1}{(N-1)^{2k+1}} \right) \\&\quad < \frac{1}{N-1} - \frac{1}{2(N-1)^2} + \frac{1}{3(N-1)^3}. \end{aligned} \end{aligned}$$

Hence, one sees that

$$\begin{aligned} \begin{aligned} e_N&< - N + N (N-1) \left\{ \frac{1}{N-1} - \frac{1}{2(N-1)^2} + \frac{1}{3(N-1)^3} \right\} \\&= - \frac{1}{2} \frac{N}{N-1} + \frac{N}{3(N-1)^2} = -\frac{1}{2} + \frac{3-N}{6(N-1)^2} \le -\frac{1}{2}. \end{aligned} \end{aligned}$$

Claim 2

There holds $d_{N+1}< d_{N}$for every$N \ge 3$.

Since

$$\begin{aligned} \begin{aligned} d_{N} - d_{N+1}&= \sum _{k=1}^{N-1} \log k - N \left( \log N - 1 \right) - \sum _{k=1}^{N} \log k + (N+1) \left\{ \log (N+1) - 1 \right\} \\&= (N+1) \left\{ \log (N+1) - \log N \right\} - 1, \end{aligned} \end{aligned}$$

set $f(x) := (x+1) \{ \log (x+1) - \log (x) \} - 1$. Noting that

$$\begin{aligned} \begin{aligned} f'(x)&= \log (x+1) - \log (x) + 1 - \frac{x+1}{x}\\&= \log (x+1) - \log (x) - \frac{1}{x} = \log (1+x^{-1}) - \frac{1}{x},\\ f''(x)&= \frac{1}{x+1} - \frac{1}{x} + \frac{1}{x^2} = \frac{1}{x^2(x+1)} > 0, \end{aligned} \end{aligned}$$

and that $f'(x) \rightarrow 0$ as $x \rightarrow \infty$, we infer that $f'(x) < 0$ for all $x \ge 1$. Since

$$\begin{aligned} f(x)= & {} (x+1) \log ( 1 + x^{-1} ) - 1 = (x+1) \left( x^{-1} - \frac{x^{-2}}{2} + O(x^{-3}) \right) -1 \\= & {} \frac{x^{-1}}{2} + O(x^{-2}), \end{aligned}$$

we have $f(x) \rightarrow 0$ as $x \rightarrow \infty$, and $f(x) > 0$ for all $x \ge 1$. Thus $f(N) > 0$ and Claim 2 holds.

Claim 3

There holds $\log (C_N^{N-1}) < 0$for every$N \ge 3$.

By Claims 1–2, it is enough to prove $d_3 < 1/2$. Since $d_3 = \log 2 -3 ( \log 3 - 1 )$, we see that $d_3<\frac{1}{2}$ is equivalent to $e^5<\frac{729}{4}$. Moreover, we observe that

$$\begin{aligned} e^5< (2.8)^5 = 172.10368 < 182.25 = \frac{729}{4}. \end{aligned}$$

Hence, Claim 3 is proved.

From (A.2), (A.3) and Claim 3, we get the desired inequality $N^2 / (\alpha _N B_{GN}) < N$.