Show Summary Details
More options …

# Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco

IMPACT FACTOR 2018: 6.636

CiteScore 2018: 5.03

SCImago Journal Rank (SJR) 2018: 3.215
Source Normalized Impact per Paper (SNIP) 2018: 3.225

Mathematical Citation Quotient (MCQ) 2018: 3.18

Open Access
Online
ISSN
2191-950X
See all formats and pricing
More options …
Volume 8, Issue 1

# Critical and subcritical fractional Trudinger–Moser-type inequalities on $ℝ$

Futoshi Takahashi
• Corresponding author
• Department of Mathematics, Osaka City University & OCAMI, Sumiyoshi-ku, Osaka, 558-8585, Japan
• Email
• Other articles by this author:
• De Gruyter OnlineGoogle Scholar
Published Online: 2017-11-16 | DOI: https://doi.org/10.1515/anona-2017-0116

## Abstract

In this paper, we are concerned with the critical and subcritical Trudinger–Moser-type inequalities for functions in a fractional Sobolev space ${H}^{1/2,2}$ on the whole real line. We prove the relation between two inequalities and discuss the attainability of the suprema.

MSC 2010: 35A23; 26D10

## 1 Introduction

Let $\mathrm{\Omega }\subset {ℝ}^{N}$, with $N\ge 2$, be a domain with finite volume. Then the Sobolev embedding theorem assures that ${W}_{0}^{1,N}\left(\mathrm{\Omega }\right)↪{L}^{q}\left(\mathrm{\Omega }\right)$ for any $q\in \left[1,+\mathrm{\infty }\right)$. However, by a simple example we see that the embedding ${W}_{0}^{1,N}\left(\mathrm{\Omega }\right)↪{L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ does not hold. Instead, functions in ${W}_{0}^{1,N}\left(\mathrm{\Omega }\right)$ enjoy the exponential summability:

see Yudovich [19], Pohozaev [33] and Trudinger [37]. Later, Moser [26] improved the above embedding, and obtained the following inequality, now known as the Trudinger–Moser inequality:

$TM\left(\mathrm{\Omega },\alpha \right)=\underset{\begin{array}{c}u\in {W}_{0}^{1,N}\left(\mathrm{\Omega }\right)\\ {\parallel \nabla u\parallel }_{{L}^{N}\left(\mathrm{\Omega }\right)}\le 1\end{array}}{sup}\frac{1}{|\mathrm{\Omega }|}{\int }_{\mathrm{\Omega }}\mathrm{exp}\left(\alpha {|u|}^{N/\left(N-1\right)}\right)𝑑x\left\{\begin{array}{cc}<\mathrm{\infty },\hfill & \alpha \le {\alpha }_{N},\hfill \\ =\mathrm{\infty },\hfill & \alpha >{\alpha }_{N},\hfill \end{array}$

where

${\alpha }_{N}=N{\omega }_{N-1}^{1/\left(N-1\right)},$

and ${\omega }_{N-1}=|{S}^{N-1}|$ denotes the area of the unit sphere in ${ℝ}^{N}$. On the attainability of $TM\left(\mathrm{\Omega },\alpha \right)$, Carleson and Chang [5], Struwe [36], Flucher [10] and Lin [24] proved that $TM\left(\mathrm{\Omega },\alpha \right)$ is attained for any $0<\alpha \le {\alpha }_{N}$.

On domains with infinite volume, for example on the whole space ${ℝ}^{N}$, the Trudinger–Moser inequality does not hold as it is. However, several variants are known on the whole space. In the following, let

${\mathrm{\Phi }}_{N}\left(t\right)={e}^{t}-\sum _{j=0}^{N-2}\frac{{t}^{j}}{j!}$

denote the truncated exponential function.

First, Ogawa [27], Ogawa and Ozawa [28], Cao [4], Ozawa [29] (for small $\alpha >0$) and finally Adachi and Tanaka [1] proved that the following inequality holds true, which we call Adachi–Tanaka-type Trudinger–Moser inequality:

$A\left(N,\alpha \right)=\underset{\begin{array}{c}u\in {W}^{1,N}\left({ℝ}^{N}\right)\setminus \left\{0\right\}\\ {\parallel \nabla u\parallel }_{{L}^{N}\left({ℝ}^{N}\right)}\le 1\end{array}}{sup}\frac{1}{{\parallel u\parallel }_{{L}^{N}\left({ℝ}^{N}\right)}^{N}}{\int }_{{ℝ}^{N}}{\mathrm{\Phi }}_{N}\left(\alpha {|u|}^{N/\left(N-1\right)}\right)𝑑x\left\{\begin{array}{cc}<\mathrm{\infty },\hfill & \alpha <{\alpha }_{N},\hfill \\ =\mathrm{\infty },\hfill & \alpha \ge {\alpha }_{N}\hfill \end{array}$

(see also do Ó [8] and Cassani, Sani and Tarsi [6] for further information). This inequality enjoys the scale invariance under the scaling $u\left(x\right)↦{u}_{\lambda }\left(x\right)=u\left(\lambda x\right)$ for $\lambda >0$. Note that the critical exponent $\alpha ={\alpha }_{N}$ is not allowed for the finiteness of the supremum. Recently, it was proved by Ishiwata, Nakamura and Wadade [16] and Dong and Lu [9] that $A\left(N,\alpha \right)$ is attained for any $\alpha \in \left(0,{\alpha }_{N}\right)$. In this sense, the Adachi–Tanaka-type Trudinger–Moser inequality can be considered as a subcritical inequality.

On the other hand, Ruf [34] and Li and Ruf [22] proved that the following inequality holds true:

$B\left(N,\alpha \right)=\underset{\begin{array}{c}u\in {W}^{1,N}\left({ℝ}^{N}\right)\\ {\parallel u\parallel }_{{W}^{1,N}\left({ℝ}^{N}\right)}\le 1\end{array}}{sup}{\int }_{{ℝ}^{N}}{\mathrm{\Phi }}_{N}\left(\alpha {|u|}^{N/\left(N-1\right)}\right)𝑑x\left\{\begin{array}{cc}<\mathrm{\infty },\hfill & \alpha \le {\alpha }_{N},\hfill \\ =\mathrm{\infty },\hfill & \alpha >{\alpha }_{N}.\hfill \end{array}$

Here,

${\parallel u\parallel }_{{W}^{1,N}\left({ℝ}^{N}\right)}={\left({\parallel \nabla u\parallel }_{{L}^{N}\left({ℝ}^{N}\right)}^{N}+{\parallel u\parallel }_{{L}^{N}\left({ℝ}^{N}\right)}^{N}\right)}^{1/N}$

is the full Sobolev norm. Note that the scale invariance ($u↦{u}_{\lambda }$) does not hold for this inequality. Also note that the critical exponent $\alpha ={\alpha }_{N}$ is permitted to the finiteness. Later, Adimurthi and Yang [2] proved that for all $\beta \in \left[0,1\right)$ and all $\tau >0$ there holds

${A}_{N,\beta ,\tau }\left(\alpha \right)=\underset{\begin{array}{c}u\in {W}^{1,N}\left({ℝ}^{N}\right)\\ {\int }_{{ℝ}^{N}}\left({|\nabla u|}^{N}+\tau {|u|}^{N}\right)𝑑x\le 1\end{array}}{sup}{\int }_{{ℝ}^{N}}\frac{{\mathrm{\Phi }}_{N}\left(\alpha \left(1-\beta \right){|u|}^{N/\left(N-1\right)}\right)}{{|x|}^{N\beta }}𝑑x\left\{\begin{array}{cc}<\mathrm{\infty },\hfill & \alpha \le {\alpha }_{N},\hfill \\ =\mathrm{\infty },\hfill & \alpha >{\alpha }_{N},\hfill \end{array}$

by a different method. Clearly, the case $\beta =0$ and $\tau =1$ reduces to that of Ruf [34] and Li and Ruf [22].

Concerning the attainability of $B\left(N,\alpha \right)$, the following facts have been proved:

• If $N\ge 3$, then $B\left(N,\alpha \right)$ is attained for $0<\alpha \le {\alpha }_{N}$; see [22].

• If $N=2$, then there exists ${\alpha }_{*}>0$ such that $B\left(2,\alpha \right)$ is attained for ${\alpha }_{*}<\alpha \le {\alpha }_{2}$ ($=4\pi$); see [34] (for $\alpha ={\alpha }_{2}$, see [15]).

• If $N=2$ and $\alpha >0$ is sufficiently small, then $B\left(2,\alpha \right)$ is not attained; see [15].

The non-attainability of $B\left(2,\alpha \right)$ for α sufficiently small is attributed to the non-compactness of “vanishing” maximizing sequences, as described in [15]. Concerning the attainability of ${A}_{N,\beta ,\tau }\left(\alpha \right)$, recently Li and Yang [21] proved that ${A}_{N,\beta ,\tau }\left(\alpha \right)$ is attained when $0<\beta <1$, $\tau >0$ and $\alpha \le {\alpha }_{N}$. This complements the results by Li and Ruf [22] and Ishiwata [15].

In the following, we focus our attention on the fractional Sobolev spaces.

Let $s\in \left(0,1\right)$, $p\in \left[1,+\mathrm{\infty }\right)$ and let $\mathrm{\Omega }\subset {ℝ}^{N}$ be a bounded Lipschitz domain. For $s>0$, let us consider the space

${L}_{s}\left({ℝ}^{N}\right)=\left\{u\in {L}_{\mathrm{loc}}^{1}\left({ℝ}^{N}\right):{\int }_{{ℝ}^{N}}\frac{|u|}{1+{|x|}^{N+s}}𝑑x<\mathrm{\infty }\right\}.$

For $u\in {L}_{s}\left({ℝ}^{N}\right)$, we define the fractional Laplacian ${\left(-\mathrm{\Delta }\right)}^{s/2}u$ as follows: First, for $\varphi \in \mathcal{𝒮}\left({ℝ}^{N}\right)$, the rapidly decreasing function on ${ℝ}^{N}$ is defined via the normalized Fourier transform $\mathcal{ℱ}$ as

${\left(-\mathrm{\Delta }\right)}^{s/2}\varphi \left(x\right)={\mathcal{ℱ}}^{-1}\left({|\xi |}^{s}\mathcal{ℱ}\varphi \left(\xi \right)\right)\left(x\right)$

for $x\in {ℝ}^{N}$. Then for $u\in {L}_{s}\left({ℝ}^{N}\right)$, the fractional Laplacian ${\left(-\mathrm{\Delta }\right)}^{s/2}u$ is defined as an element of ${\mathcal{𝒮}}^{\prime }\left({ℝ}^{N}\right)$, the tempered distributions on ${ℝ}^{N}$, by the relation

$〈\varphi ,{\left(-\mathrm{\Delta }\right)}^{s/2}u〉=〈{\left(-\mathrm{\Delta }\right)}^{s/2}\varphi ,u〉={\int }_{ℝ}{\left(-\mathrm{\Delta }\right)}^{s/2}\varphi \cdot u𝑑x,\varphi \in \mathcal{𝒮}\left({ℝ}^{N}\right).$

Note that ${L}^{p}\left({ℝ}^{N}\right)\subset {L}_{s}\left({ℝ}^{N}\right)$ for any $p\ge 1$. Also note that it could happen that $\mathrm{supp}\left({\left(-\mathrm{\Delta }\right)}^{s/2}u\right)\not\subset \mathrm{\Omega }$ even if $\mathrm{supp}\left(u\right)\subset \mathrm{\Omega }$ for some open set Ω in ${ℝ}^{N}$.

By using the above notion, we define the Bessel potential space ${H}^{s,p}\left(\mathrm{\Omega }\right)$ for a (possibly unbounded) set $\mathrm{\Omega }\subset {ℝ}^{N}$ as

${H}^{s,p}\left({ℝ}^{N}\right)=\left\{u\in {L}^{p}\left({ℝ}^{N}\right):{\left(-\mathrm{\Delta }\right)}^{s/2}u\in {L}^{p}\left({ℝ}^{N}\right)\right\},$

On the other hand, the Sobolev–Slobodeckij space ${W}^{s,p}\left({ℝ}^{N}\right)$ is defined as

${W}^{s,p}\left({ℝ}^{N}\right)=\left\{u\in {L}^{p}\left({ℝ}^{N}\right):{\left[u\right]}_{{W}^{s,p}\left({ℝ}^{N}\right)}<\mathrm{\infty }\right\},$${}_{{W}^{s,p}\left({ℝ}^{N}\right)}{}^{p}={\int }_{{ℝ}^{N}}{\int }_{{ℝ}^{N}}\frac{{|u\left(x\right)-u\left(y\right)|}^{p}}{{|x-y|}^{N+sp}}dxdy,$

and for a bounded domain $\mathrm{\Omega }\subset {ℝ}^{N}$, we define

${\stackrel{~}{W}}^{s,p}\left(\mathrm{\Omega }\right)={\overline{{C}_{c}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)}}^{{\parallel \cdot \parallel }_{{W}^{s,p}\left({ℝ}^{N}\right)}}$

where

${\parallel u\parallel }_{{W}^{s,p}\left({ℝ}^{N}\right)}={\left({\parallel u\parallel }_{{L}^{p}\left({ℝ}^{N}\right)}^{p}+{\left[u\right]}_{{W}^{s,p}\left({ℝ}^{N}\right)}^{p}\right)}^{1/p}.$

It is known that

if Ω is a Lipschitz domain, ${H}^{s,p}\left({ℝ}^{N}\right)={F}_{p,2}^{s}\left({ℝ}^{N}\right)$ (the Triebel–Lizorkin space), and ${W}^{s,p}\left({ℝ}^{N}\right)={B}_{p,p}^{s}\left({ℝ}^{N}\right)$ (the Besov space). Thus ${H}^{s,2}\left({ℝ}^{N}\right)={W}^{s,2}\left({ℝ}^{N}\right)$. However in general, ${H}^{s,p}\left({ℝ}^{N}\right)\ne {W}^{s,p}\left({ℝ}^{N}\right)$ for $p\ne 2$; see [31, 17] and the references therein.

Recently, Martinazzi [25] (see also [18]) proved a fractional Trudinger–Moser-type inequality on ${\stackrel{~}{H}}^{s,p}\left(\mathrm{\Omega }\right)$ as follows: Let $p\in \left(1,\mathrm{\infty }\right)$ and $s=\frac{N}{p}$ for $N\in ℕ$. Then for any open $\mathrm{\Omega }\subset {ℝ}^{N}$ with $|\mathrm{\Omega }|<\mathrm{\infty }$, it holds that

$\underset{\begin{array}{c}u\in {\stackrel{~}{H}}^{s,p}\left(\mathrm{\Omega }\right)\\ {\parallel {\left(-\mathrm{\Delta }\right)}^{s/2}u\parallel }_{{L}^{p}\left(\mathrm{\Omega }\right)}\le 1\end{array}}{sup}\frac{1}{|\mathrm{\Omega }|}{\int }_{\mathrm{\Omega }}\mathrm{exp}\left(\alpha {|u|}^{p/\left(p-1\right)}\right)𝑑x\left\{\begin{array}{cc}<\mathrm{\infty },\hfill & \alpha \le {\alpha }_{N,p},\hfill \\ =\mathrm{\infty },\hfill & \alpha >{\alpha }_{N,p}.\hfill \end{array}$

Here,

${\alpha }_{N,p}=\frac{N}{{\omega }_{N-1}}{\left(\frac{\mathrm{\Gamma }\left(\left(N-s\right)/2\right)}{\mathrm{\Gamma }\left(s/2\right){2}^{s}{\pi }^{N/2}}\right)}^{-p/\left(p-1\right)}.$

We note that, differently from the classical case, the attainability of the supremum is not known even for $N=1$ and $p=2$.

On the Sobolev–Slobodeckij spaces ${\stackrel{~}{W}}^{s,p}\left(\mathrm{\Omega }\right)$ with $sp=N$, a similar fractional Trudinger–Moser inequality was also proved by Parini and Ruf [31] when $N\ge 2$, and Iula [17] when $N=1$. They proved the validity of the inequality for sufficiently small values of $\alpha >0$, and the problem of the sharp exponent is still open.

In the following, we are interested in the simplest one-dimensional case, that is, we put $N=1$, $s=\frac{1}{2}$ and $p=2$. In this case, the Bessel potential space ${H}^{1/2,2}\left(ℝ\right)$ coincides with the Sobolev–Slobodeckij space ${W}^{1/2,2}\left(ℝ\right)$, and both seminorms are related as

${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}=\frac{1}{2\pi }{\left[u\right]}_{{W}^{1/2,2}\left(ℝ\right)}^{2};$

see [7, Proposition 3.6.]. Then the fractional Trudinger–Moser inequality in [25, 18] can be read as in the following proposition.

#### Proposition 1.1 (A fractional Trudinger–Moser inequality on ${\stackrel{\mathrm{~}}{H}}^{\mathrm{1}\mathrm{/}\mathrm{2}\mathrm{,}\mathrm{2}}\mathbf{}\mathrm{\left(}\mathrm{I}\mathrm{\right)}$).

Let $I\mathrm{\subset }\mathrm{R}$ be an open bounded interval. Then it holds that

$\underset{\begin{array}{c}u\in {\stackrel{~}{H}}^{1/2,2}\left(\mathrm{I}\right)\\ {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(\mathrm{I}\right)}\le 1\end{array}}{sup}\frac{1}{|I|}{\int }_{I}{e}^{\alpha {|u|}^{2}}𝑑x\left\{\begin{array}{cc}<\mathrm{\infty },\hfill & \alpha \le {\alpha }_{1,2}=\pi ,\hfill \\ =\mathrm{\infty },\hfill & \alpha >\pi .\hfill \end{array}$

For the fractional Adachi–Tanaka-type Trudinger–Moser inequality on the whole line, put

$A\left(\alpha \right)=\underset{\begin{array}{c}u\in {H}^{1/2,2}\left(ℝ\right)\setminus \left\{0\right\}\\ {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}\le 1\end{array}}{sup}\frac{1}{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1\right)𝑑x.$(1.1)

Then by the precedent results by Ogawa and Ozawa [28] and Ozawa [29] it is known that $A\left(\alpha \right)<\mathrm{\infty }$ for small exponent α.

On the other hand, a fractional Li–Ruf-type Trudinger–Moser inequality on ${H}^{1/2,2}\left(ℝ\right)$ is already known as follows.

#### Proposition 1.2 ([18]).

We have

$B\left(\alpha \right)=\underset{\begin{array}{c}u\in {H}^{1/2,2}\left(ℝ\right)\\ {\parallel u\parallel }_{{H}^{1/2,2}\left(ℝ\right)}\le 1\end{array}}{sup}{\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1\right)𝑑x\left\{\begin{array}{cc}<\mathrm{\infty },\hfill & \alpha \le \pi ,\hfill \\ =\mathrm{\infty },\hfill & \alpha >\pi .\hfill \end{array}$(1.2)

Here,

${\parallel u\parallel }_{{H}^{1/2,2}\left(ℝ\right)}={\left({\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}+{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\right)}^{1/2}$

is the full Sobolev norm on ${H}^{\mathrm{1}\mathrm{/}\mathrm{2}\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}\mathrm{R}\mathrm{\right)}$.

Concerning $A\left(\alpha \right)$ in (1.1), a natural question is to determine the range of the exponent $\alpha >0$ for which $A\left(\alpha \right)$ is finite. As pointed out in [14], this remained an open problem for a while. In this paper, we first prove the finiteness of the supremum in the full range of values of the exponent.

#### Theorem 1.3 (Full range Adachi–Tanaka-type on ${H}^{\mathrm{1}\mathrm{/}\mathrm{2}\mathrm{,}\mathrm{2}}\mathbf{}\mathrm{\left(}\mathrm{R}\mathrm{\right)}$).

We have

$A\left(\alpha \right)=\underset{\begin{array}{c}u\in {H}^{1/2,2}\left(ℝ\right)\setminus \left\{0\right\}\\ {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}\le 1\end{array}}{sup}\frac{1}{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1\right)𝑑x\left\{\begin{array}{cc}<\mathrm{\infty },\hfill & \alpha <\pi ,\hfill \\ =\mathrm{\infty },\hfill & \alpha \ge \pi .\hfill \end{array}$

Ozawa [30] proved that the Adachi–Tanaka-type Trudinger–Moser inequality is equivalent to the Gagliardo–Nirenberg-type inequality, and he obtained an exact relation between the best constants of both inequalities. Actually, he proved the result for general $1, and if $p=2$, the main result in [30] can be read as follows: Put ${\alpha }_{0}=sup\left\{\alpha >0:A\left(\alpha \right)<\mathrm{\infty }\right\}$ and

${\beta }_{0}=\underset{q\to \mathrm{\infty }}{lim sup}\underset{u\in {H}^{1/2,2}\left(ℝ\right),u\ne 0}{sup}\frac{{\parallel u\parallel }_{{L}^{q}\left(ℝ\right)}}{{q}^{1/2}{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}^{1-2/q}{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2/q}}.$

Then it is shown that $1/{\alpha }_{0}=2e{\beta }_{0}^{2}$; see [30, Theorem 1]. Thus, by a direct consequence of Theorem 1.3, we have the next corollary.

#### Corollary 1.4.

We have ${\beta }_{\mathrm{0}}\mathrm{=}{\mathrm{\left(}\mathrm{2}\mathit{}\pi \mathit{}e\mathrm{\right)}}^{\mathrm{-}\mathrm{1}\mathrm{/}\mathrm{2}}$.

Furthermore, we obtain the relation between the suprema of both critical and subcritical Trudinger–Moser-type inequalities along the line of [20].

#### Theorem 1.5 (Relation).

We have

$B\left(\pi \right)=\underset{\alpha \in \left(0,\pi \right)}{sup}\frac{1-\left(\alpha /\pi \right)}{\left(\alpha /\pi \right)}A\left(\alpha \right).$

Also we obtain how the Adachi–Tanaka-type supremum $A\left(\alpha \right)$ behaves when α tends to π.

#### Theorem 1.6 (Asymptotic behavior).

There exist ${C}_{\mathrm{1}}\mathrm{,}{C}_{\mathrm{2}}\mathrm{>}\mathrm{0}$ such that for any $\alpha \mathrm{<}\pi$ which is close enough to π it holds that

$\frac{{C}_{1}}{1-\alpha /\pi }\le A\left(\alpha \right)\le \frac{{C}_{2}}{1-\alpha /\pi }.$

Note that the estimate from above follows from Theorem 1.5 and Proposition 1.2. On the other hand, we will see that the estimate from below follows from a computation using the Moser sequence.

Concerning the existence of maximizers of the Adachi–Tanaka-type supremum $A\left(\alpha \right)$ in (1.1), we have the following theorem.

#### Theorem 1.7 (Attainability of $A\mathbf{}\mathrm{\left(}\alpha \mathrm{\right)}$).

$A\left(\alpha \right)$ is attained for any $\alpha \mathrm{\in }\mathrm{\left(}\mathrm{0}\mathrm{,}\pi \mathrm{\right)}$.

On the other hand, as for $B\left(\alpha \right)$ in (1.2), we have the following result.

#### Theorem 1.8 (Non-attainability of $B\mathbf{}\mathrm{\left(}\alpha \mathrm{\right)}$).

$B\left(\alpha \right)$ is not attained for $\mathrm{0}\mathrm{<}\alpha \mathrm{\ll }\mathrm{1}$.

It is plausible that there exists ${\alpha }_{*}>0$ such that $B\left(\alpha \right)$ is attained for ${\alpha }_{*}<\alpha \le \pi$, but we do not have a proof up to now.

Finally, we improve the subcritical Adachi–Tanaka-type inequality along the line of [9].

#### Theorem 1.9.

For $\alpha \mathrm{>}\mathrm{0}$, set

$E\left(\alpha \right)=\underset{\begin{array}{c}u\in {H}^{1/2,2}\left(ℝ\right)\setminus \left\{0\right\}\\ {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}\le 1\end{array}}{sup}\frac{1}{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{ℝ}{u}^{2}{e}^{\alpha {u}^{2}}𝑑x.$

Then we have

$E\left(\alpha \right)\left\{\begin{array}{cc}<\mathrm{\infty },\hfill & \alpha <\pi ,\hfill \\ =\mathrm{\infty },\hfill & \alpha \ge \pi .\hfill \end{array}$

Furthermore, $E\mathit{}\mathrm{\left(}\alpha \mathrm{\right)}$ is attained for all $\alpha \mathrm{\in }\mathrm{\left(}\mathrm{0}\mathrm{,}\pi \mathrm{\right)}$.

Since

${e}^{\alpha {t}^{2}}-1\le \alpha {t}^{2}{e}^{\alpha {t}^{2}}$

for $t\in ℝ$, Theorem 1.9 extends Theorem 1.3. In the classical case, Dong and Lu [9] used a rearrangement technique to reduce the problem to one dimension and obtained a similar inequality by estimating a one-dimensional integral. In the fractional setting ${H}^{1/2,2}$, we cannot follow this argument and we need a new idea.

The organization of the paper is as follows: In Section 2, we prove Theorems 1.3, 1.5 and 1.6. In Section 3, we prove Theorems 1.7 and 1.8. In Section 4, we prove Theorem 1.9.

## Note.

After this work was completed, the author was informed by the anonymous referee that the full range Adachi–Tanaka-type inequality is proven, among other relevant results, in the recent preprints [12, 13] by different methods.

## 2 Proofs of Theorems 1.3, 1.5 and 1.6

For the proofs of Theorems 1.3, 1.5 and 1.6, we prepare several lemmas.

#### Lemma 2.1.

Set

$\stackrel{~}{A}\left(\alpha \right)=\underset{\begin{array}{c}u\in {H}^{1/2,2}\left(ℝ\right)\setminus \left\{0\right\}\\ {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}\le 1\\ {\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}=1\end{array}}{sup}{\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1\right)𝑑x.$

Then $\stackrel{\mathrm{~}}{A}\mathit{}\mathrm{\left(}\alpha \mathrm{\right)}\mathrm{=}A\mathit{}\mathrm{\left(}\alpha \mathrm{\right)}$ for any $\alpha \mathrm{>}\mathrm{0}$.

#### Proof.

For any $u\in {H}^{1/2,2}\left(ℝ\right)\setminus \left\{0\right\}$ and $\lambda >0$, we put ${u}_{\lambda }\left(x\right)=u\left(\lambda x\right)$ for $x\in ℝ$. Then we have

$\left\{\begin{array}{cc}\hfill {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\lambda }\parallel }_{{L}^{2}\left(ℝ\right)}& ={\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)},\hfill \\ \hfill {\parallel {u}_{\lambda }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}& ={\lambda }^{-1}{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\hfill \end{array}$(2.1)

since

$2\pi {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\lambda }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}={\left[{u}_{\lambda }\right]}_{{W}^{1/2,2}\left(ℝ\right)}^{2}$$={\int }_{ℝ}{\int }_{ℝ}\frac{{|u\left(\lambda x\right)-u\left(\lambda y\right)|}^{2}}{{|x-y|}^{2}}𝑑x𝑑y$$={\int }_{ℝ}{\int }_{ℝ}\frac{{|u\left(\lambda x\right)-u\left(\lambda y\right)|}^{2}}{{|\lambda x-\lambda y|}^{2}}d\left(\lambda x\right)d\left(\lambda y\right)$$={\left[u\right]}_{{W}^{1/2,2}\left(ℝ\right)}^{2}$$=2\pi {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}.$

Thus for any $u\in {H}^{1/2,2}\left(ℝ\right)\setminus \left\{0\right\}$ with ${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}\le 1$, if we choose $\lambda ={\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}$, then ${u}_{\lambda }\in {H}^{1/2,2}\left(ℝ\right)$ satisfies

${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\lambda }\parallel }_{{L}^{2}\left(ℝ\right)}\le 1\mathit{ }\text{and}\mathit{ }{\parallel {u}_{\lambda }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}=1.$

Thus

$\frac{1}{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1\right)𝑑x={\int }_{ℝ}\left({e}^{\alpha {u}_{\lambda }^{2}}-1\right)𝑑x\le \stackrel{~}{A}\left(\alpha \right),$

which implies $A\left(\alpha \right)\le \stackrel{~}{A}\left(\alpha \right)$. The opposite inequality is trivial. ∎

#### Lemma 2.2.

For any $\mathrm{0}\mathrm{<}\alpha \mathrm{<}\pi$, it holds that

$A\left(\alpha \right)\le \frac{\left(\alpha /\pi \right)}{1-\left(\alpha /\pi \right)}B\left(\pi \right).$

#### Proof.

Choose any $u\in {H}^{1/2,2}\left(ℝ\right)$ with ${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}\le 1$ and ${\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}=1$. Further, put $v\left(x\right)=Cu\left(\lambda x\right)$, where ${C}^{2}=\frac{\alpha }{\pi }\in \left(0,1\right)$ and $\lambda ={C}^{2}/\left(1-{C}^{2}\right)$. Then by the scaling rules (2.1) we see

${\parallel v\parallel }_{{H}^{1/2,2}\left(ℝ\right)}^{2}={\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{{L}^{2}\left(ℝ\right)}^{2}+{\parallel v\parallel }_{{L}^{2}\left(ℝ\right)}^{2}$$={C}^{2}{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}+{\lambda }^{-1}{C}^{2}{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}$$\le {C}^{2}+{\lambda }^{-1}{C}^{2}=1.$

Also we have

${\int }_{ℝ}\left({e}^{\pi {v}^{2}}-1\right)𝑑x={\int }_{ℝ}\left({e}^{\pi {C}^{2}{u}^{2}\left(\lambda x\right)}-1\right)𝑑x$$={\lambda }^{-1}{\int }_{ℝ}\left({e}^{\pi {C}^{2}{u}^{2}\left(y\right)}-1\right)𝑑y$$=\frac{1-{C}^{2}}{{C}^{2}}{\int }_{ℝ}\left({e}^{\alpha {u}^{2}\left(y\right)}-1\right)𝑑y$$=\frac{1-\left(\alpha /\pi \right)}{\left(\alpha /\pi \right)}{\int }_{ℝ}\left({e}^{\alpha {u}^{2}\left(y\right)}-1\right)𝑑y.$

Thus, testing $B\left(\pi \right)$ by v, we see

$B\left(\pi \right)\ge {\int }_{ℝ}\left({e}^{\pi {v}^{2}}-1\right)𝑑x\ge \frac{1-\left(\alpha /\pi \right)}{\left(\alpha /\pi \right)}{\int }_{ℝ}\left({e}^{\alpha {u}^{2}\left(y\right)}-1\right)𝑑y.$

By taking the supremum for $u\in {H}^{1/2,2}\left(ℝ\right)$ with ${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}\le 1$ and ${\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}=1$, we have

$B\left(\pi \right)\ge \frac{1-\left(\alpha /\pi \right)}{\left(\alpha /\pi \right)}\stackrel{~}{A}\left(\alpha \right).$

Finally, Lemma 2.1 implies the result. ∎

#### Proof of Theorem 1.3.

The assertion that $A\left(\alpha \right)<\mathrm{\infty }$ for $\alpha <\pi$ follows from Lemma 2.2 and the fact that $B\left(\pi \right)<\mathrm{\infty }$ by Proposition 1.2.

For the proof of $A\left(\pi \right)=\mathrm{\infty }$, we use the Moser sequence

(2.2)

and its estimates

${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}=\pi +o\left(1\right),$(2.3)${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\le \pi \left(1+{\left(C\mathrm{log}\left(1/\epsilon \right)\right)}^{-1}\right),$(2.4)${\parallel {u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}=O\left({\left(\mathrm{log}\left(1/\epsilon \right)\right)}^{-1}\right)$(2.5)

as $\epsilon \to 0$ for some $C>0$. Note that ${u}_{\epsilon }\in {\stackrel{~}{W}}^{1/2,2}\left(\left(-1,1\right)\right)\subset {W}^{1/2,2}\left(ℝ\right)={H}^{1/2,2}\left(ℝ\right)$. For estimate (2.3), we refer to [17, Proposition 2.2]. For estimate (2.4), we refer to [17, (35)]. Actually, after a careful look at the proof of [17, Proposition 2.2], we confirm that

$\underset{\epsilon \to 0}{lim}\left(\mathrm{log}\left(1/\epsilon \right)\right)\left({\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}-\pi \right)\le C$

for a positive $C>0$, which implies (2.4). For (2.5), we compute

${\parallel {u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}={\int }_{|x|\le \epsilon }\left(\mathrm{log}\left(1/\epsilon \right)\right)𝑑x+{\int }_{\epsilon <|x|\le 1}{\left(\frac{\mathrm{log}\left(1/|x|\right)}{{\left(\mathrm{log}\left(1/\epsilon \right)\right)}^{1/2}}\right)}^{2}𝑑x$$=2\epsilon \mathrm{log}\left(1/\epsilon \right)+\frac{2}{\mathrm{log}\left(1/\epsilon \right)}{\int }_{\mathrm{log}\left(1/\epsilon \right)}^{0}{t}^{2}\left(-{e}^{t}\right)𝑑x$$=2\epsilon \mathrm{log}\left(1/\epsilon \right)+\frac{2}{\mathrm{log}\left(1/\epsilon \right)}\left(\mathrm{\Gamma }\left(3\right)+o\left(1\right)\right)$

as $\epsilon \to 0$. Thus we obtain (2.5).

By testing $A\left(\pi \right)$ by ${v}_{\epsilon }={u}_{\epsilon }/{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}$, we have

$A\left(\pi \right)\ge \frac{1}{{\parallel {v}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{ℝ}\left({e}^{\pi {v}_{\epsilon }^{2}}-1\right)𝑑x$$\ge \frac{{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{{\parallel {u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{|x|\le \epsilon }\left({e}^{\pi {v}_{\epsilon }^{2}}-1\right)𝑑x$$\ge \frac{{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{{\parallel {u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}\epsilon \mathrm{exp}\left(\pi \frac{\mathrm{log}\left(1/\epsilon \right)}{{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}\right)$$\ge \frac{{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{{\parallel {u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}\epsilon \mathrm{exp}\left(\frac{\mathrm{log}\left(1/\epsilon \right)}{1+{\left(C\mathrm{log}\left(1/\epsilon \right)\right)}^{-1}}\right)$

since ${e}^{t}-1\ge \left(1/2\right){e}^{t}$ for t large and by (2.4). Also since

we see

$\frac{t}{1+1/Ct}=t-\frac{1}{C}+o\left(1\right)$

as $t\to \mathrm{\infty }$. Put $t=\mathrm{log}\left(1/\epsilon \right)$. We see

$\mathrm{exp}\left(\frac{\mathrm{log}\left(1/\epsilon \right)}{1+{\left(C\mathrm{log}\left(1/\epsilon \right)\right)}^{-1}}\right)=\mathrm{exp}\left(\mathrm{log}\left(1/\epsilon \right)-1/C+o\left(1\right)\right)=\left(1/\epsilon \right){e}^{-1/C+o\left(1\right)},$

$\epsilon \mathrm{exp}\left(\frac{\mathrm{log}\left(1/\epsilon \right)}{1+{\left(C\mathrm{log}\left(1/\epsilon \right)\right)}^{-1}}\right)\ge {e}^{-1/C+o\left(1\right)}\ge \delta >0$

for some $\delta >0$ independent of $\epsilon \to 0$. Therefore, by (2.3), (2.4) and (2.5) we have for ${\delta }^{\prime }>0$,

$A\left(\pi \right)\ge \frac{\pi +o\left(1\right)}{{\left(C\mathrm{log}\left(1/\epsilon \right)\right)}^{-1}}\delta \ge {\delta }^{\prime }\left(\mathrm{log}\left(1/\epsilon \right)\right)\to \mathrm{\infty }$

as $\epsilon \to 0$. This proves $A\left(\pi \right)=\mathrm{\infty }$. ∎

#### Proof of Theorem 1.5.

By Lemma 2.2, we have

$B\left(\pi \right)\ge \underset{\alpha \in \left(0,\pi \right)}{sup}\frac{1-\left(\alpha /\pi \right)}{\left(\alpha /\pi \right)}A\left(\alpha \right).$

Let us prove the opposite inequality. Let

$\left\{{u}_{n}\right\}\subset {H}^{1/2,2}\left(ℝ\right),{u}_{n}\ne 0,{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}+{\parallel {u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\le 1$

be a maximizing sequence of $B\left(\pi \right)$. We may assume ${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}<1$ for any $n\in ℕ$. Put

${v}_{n}\left(x\right)=\frac{{u}_{n}\left({\lambda }_{n}x\right)}{{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}} \left(x\in ℝ\right),$${\lambda }_{n}=\frac{1-{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}>0.$

Thus by (2.1) we see

${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{v}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}=1,$${\parallel {v}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}=\frac{{\lambda }_{n}^{-1}}{{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\parallel {u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}=\frac{{\parallel {u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{1-{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}\le 1$

since ${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}+{\parallel {u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\le 1$. Thus, setting ${\alpha }_{n}=\pi {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}<\pi$ for any $n\in ℕ$, we may test $A\left({\alpha }_{n}\right)$ by $\left\{{v}_{n}\right\}$, which results in

$B\left(\pi \right)+o\left(1\right)={\int }_{ℝ}\left({e}^{\pi {u}_{n}^{2}\left(y\right)}-1\right)𝑑y$$={\lambda }_{n}{\int }_{ℝ}\left({e}^{\pi {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}{v}_{n}^{2}\left(x\right)}-1\right)𝑑x$$\le {\lambda }_{n}\frac{1}{{\parallel {v}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{ℝ}\left({e}^{{\alpha }_{n}{v}_{n}^{2}\left(x\right)}-1\right)𝑑x$$\le {\lambda }_{n}A\left({\alpha }_{n}\right)$$=\frac{1-\left({\alpha }_{n}/\pi \right)}{\left({\alpha }_{n}/\pi \right)}A\left({\alpha }_{n}\right)$$\le \underset{\alpha \in \left(0,\pi \right)}{sup}\frac{1-\left(\alpha /\pi \right)}{\left(\alpha /\pi \right)}A\left(\alpha \right).$

Here we have used a change of variables $y={\lambda }_{n}x$ for the second equality, and ${\parallel {v}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\le 1$ for the first inequality. Letting $n\to \mathrm{\infty }$, we have the desired result. ∎

#### Proof of Theorem 1.6.

We need to prove that there exists ${C}_{1}>0$ such that for any $\alpha <\pi$ which is sufficiently close to π it holds that

$A\left(\alpha \right)\ge \frac{{C}_{1}}{1-\alpha /\pi }.$

Again we use the Moser sequence (2.2) and we test $A\left(\alpha \right)$ by ${v}_{\epsilon }={u}_{\epsilon }/{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}$. As in the similar calculations in the proof of Theorem 1.3, we have

$A\left(\alpha \right)\ge \frac{1}{{\parallel {v}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{ℝ}\left({e}^{\alpha {v}_{\epsilon }^{2}}-1\right)𝑑x$$\ge \frac{\left(1/2\right)}{{\parallel {v}_{\epsilon }\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{|x|\le \epsilon }{e}^{\alpha {v}_{\epsilon }^{2}}𝑑x$$\ge C\epsilon \left(\mathrm{log}\left(1/\epsilon \right)\right)\mathrm{exp}\left(\frac{\alpha }{\pi }\frac{\mathrm{log}\left(1/\epsilon \right)}{1+{\left(C\mathrm{log}\left(1/\epsilon \right)\right)}^{-1}}\right)$$=C\epsilon \left(\mathrm{log}\left(1/\epsilon \right)\right)\mathrm{exp}\left({\delta }_{\epsilon }\mathrm{log}\left(1/\epsilon \right)\right),$

where we put

${\delta }_{\epsilon }=\frac{\alpha }{\pi }\frac{1}{1+{\left(C\mathrm{log}\left(1/\epsilon \right)\right)}^{-1}}\in \left(0,1\right).$

Now, for $\alpha <\pi$ which is sufficiently close to π, we fix $\epsilon >0$ small such that

$\frac{1}{1-\alpha /\pi }\le \mathrm{log}\left(1/\epsilon \right)\le \frac{2}{1-\alpha /\pi },$(2.6)

which implies

$\mathrm{exp}\left(-\frac{2}{1-\alpha /\pi }\right)\le \epsilon \le \mathrm{exp}\left(-\frac{1}{1-\alpha /\pi }\right).$

With this choice of $\epsilon >0$, we have

$A\left(\alpha \right)\ge C\epsilon \left(\mathrm{log}\left(1/\epsilon \right)\right)\mathrm{exp}\left({\delta }_{\epsilon }\mathrm{log}\left(1/\epsilon \right)\right)$$=C\epsilon \left(\mathrm{log}\left(1/\epsilon \right)\right){\left(1/\epsilon \right)}^{{\delta }_{\epsilon }}$(2.7)$=C{\epsilon }^{1-{\delta }_{\epsilon }}\left(\mathrm{log}\left(1/\epsilon \right)\right).$(2.8)

Now, we estimate that

${\epsilon }^{1-{\delta }_{\epsilon }}\ge {\left(\mathrm{exp}\left(-\frac{2}{1-\alpha /\pi }\right)\right)}^{1-{\delta }_{\epsilon }}$$=\mathrm{exp}\left(-\frac{2}{1-\alpha /\pi }\left(1-{\delta }_{\epsilon }\right)\right)$$=\mathrm{exp}\left(-\left(\frac{2}{1-\alpha /\pi }\right)\left\{\left(1-\frac{\alpha }{\pi }\right)+\left(\frac{\alpha }{\pi }\right)\left(1-\frac{1}{1+{\left(C\mathrm{log}1/\epsilon \right)}^{-1}}\right)\right\}\right)$$=\mathrm{exp}\left(-2-\left(\frac{2\left(\alpha /\pi \right)}{1-\alpha /\pi }\right)\left(\frac{1}{1+C\mathrm{log}1/\epsilon }\right)\right)$$\ge \mathrm{exp}\left(-2-\left(\frac{2\left(\alpha /\pi \right)}{1-\alpha /\pi }\right)\left(\frac{1}{1+\frac{C}{1-\alpha /\pi }}\right)\right)$$={e}^{-2}\cdot {e}^{-\frac{2\left(\alpha /\pi \right)}{C+1-\alpha /\pi }}$$={e}^{-2}\cdot {e}^{-f\left(\alpha /\pi \right)},$

where $f\left(t\right)=2t/\left(C+1-t\right)$ for $t\in \left[0,1\right]$ and we have used (2.6) in the last inequality. We easily see that $f\left(0\right)=0$ and ${f}^{\prime }\left(t\right)=2\left(C+1\right)/{\left(C+1-t\right)}^{2}>0$ for $t>0$, thus $f\left(t\right)$ is strictly increasing in t and ${\mathrm{max}}_{t\in \left[0,1\right]}f\left(t\right)=f\left(1\right)=\frac{2}{C}$. Thus we have

${\epsilon }^{1-{\delta }_{\epsilon }}\ge {e}^{-2}\cdot {e}^{-2/C}=:{C}_{0},$

which is independent of α. Going back to (2.7) with (2.6), we observe that

$A\left(\alpha \right)\ge C{\epsilon }^{1-{\delta }_{\epsilon }}\left(\mathrm{log}\left(1/\epsilon \right)\right)\ge C{C}_{0}\left(\mathrm{log}\left(1/\epsilon \right)\right)\ge \frac{C{C}_{0}}{1-\alpha /\pi },$

which proves the result. ∎

## 3 Proofs of Theorems 1.7 and 1.8

For $u\in {H}^{1/2,2}\left(ℝ\right)$, we denote by ${u}^{*}$ its symmetric decreasing rearrangement defined as follows: For a measurable set $A\subset ℝ$, let ${A}^{*}$ denote an open interval ${A}^{*}=\left(-|A|/2,|A|/2\right)$. We define ${u}^{*}$ by

${u}^{*}\left(x\right)={\int }_{0}^{\mathrm{\infty }}{\chi }_{{\left\{y\in ℝ:|u\left(y\right)|>t\right\}}^{*}}\left(x\right)𝑑t,$

where ${\chi }_{A}$ denote the indicator function of a measurable set $A\subset ℝ$. Note that ${u}^{*}$ is nonnegative, even and decreasing on the positive line ${ℝ}_{+}=\left[0,+\mathrm{\infty }\right)$. It is known that

${\int }_{ℝ}F\left({u}^{*}\right)𝑑x={\int }_{ℝ}F\left(|u|\right)𝑑x$

for any nonnegative measurable function $F:{ℝ}_{+}\to {ℝ}_{+}$, which is the difference of two monotone increasing functions ${F}_{1}$, ${F}_{2}$ with ${F}_{1}\left(0\right)={F}_{2}\left(0\right)=0$ such that either ${F}_{1}\circ |u|$ or ${F}_{2}\circ |u|$ is integrable. Also the inequality of Pólya–Szegő type

${\int }_{ℝ}{|{\left(-\mathrm{\Delta }{u}^{*}\right)}^{1/4}|}^{2}𝑑x\le {\int }_{ℝ}{|{\left(-\mathrm{\Delta }u\right)}^{1/4}|}^{2}𝑑x$

holds true for $u\in {H}^{1/2,2}\left(ℝ\right)$; see for example [3, 32, 23].

#### Remark 3.1.

Note that the radial compactness lemma by Strauss [35] is violated on $ℝ$. More precisely, let

${H}_{\mathrm{rad}}^{1/2,2}\left(ℝ\right)=\left\{u\in {H}^{1/2,2}\left(ℝ\right):u\left(x\right)=u\left(-x\right),x\ge 0\right\};$

then ${H}_{\mathrm{rad}}^{1/2,2}\left(ℝ\right)$ cannot be embedded compactly in ${L}^{q}\left(ℝ\right)$ for any $q>0$. To see this, let $\psi \ne 0$ be an even function in ${C}_{c}^{\mathrm{\infty }}\left(ℝ\right)$ with $\mathrm{supp}\left(\psi \right)\subset \left(-1,1\right)$, and put ${u}_{n}\left(x\right)=\psi \left(x-n\right)+\psi \left(x+n\right)$. Then we see that ${u}_{n}$ is an even, compactly supported smooth function, and ${u}_{n}⇀0$ weakly in ${H}^{1/2,2}\left(ℝ\right)$ as $n\to \mathrm{\infty }$. But $\left\{{u}_{n}\right\}$ does not have any strong convergent subsequence in ${L}^{q}\left(ℝ\right)$ because ${\parallel {u}_{n}\parallel }_{{L}^{q}\left(ℝ\right)}^{q}=2{\parallel \psi \parallel }_{{L}^{q}\left(ℝ\right)}^{q}>0$ for any n sufficient large.

However, for a sequence ${\left\{{u}_{n}\right\}}_{n\in ℕ}\subset {H}^{1/2,2}\left(ℝ\right)$ with ${u}_{n}$ even, nonnegative and decreasing on ${ℝ}_{+}$, we have the following compactness result.

#### Proposition 3.2.

Assume $\mathrm{\left\{}{u}_{n}\mathrm{\right\}}\mathrm{\subset }{H}^{\mathrm{1}\mathrm{/}\mathrm{2}\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}\mathrm{R}\mathrm{\right)}$ to be a sequence such that ${u}_{n}$ is even, nonnegative and decreasing on ${\mathrm{R}}_{\mathrm{+}}$. Let ${u}_{n}\mathrm{⇀}u$ weakly in ${H}^{\mathrm{1}\mathrm{/}\mathrm{2}\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}\mathrm{R}\mathrm{\right)}$. Then ${u}_{n}\mathrm{\to }u$ strongly in ${L}^{q}\mathit{}\mathrm{\left(}\mathrm{R}\mathrm{\right)}$ for any $q\mathrm{\in }\mathrm{\left(}\mathrm{2}\mathrm{,}\mathrm{+}\mathrm{\infty }\mathrm{\right)}$ for a subsequence.

#### Proof.

Since $\left\{{u}_{n}\right\}\subset {H}^{1/2,2}\left(ℝ\right)$ is a weakly convergent sequence, we have

$\underset{n\in ℕ}{sup}{\parallel {u}_{n}\parallel }_{{H}^{1/2,2}\left(ℝ\right)}\le C$

for some $C>0$. We also have ${u}_{n}\left(x\right)\to u\left(x\right)$ a.e. $x\in ℝ$ for a subsequence, thus u is even, nonnegative and decreasing on ${ℝ}_{+}$. Now, we use the estimate below, which is referred to a simple radial lemma: If $u\in {L}^{2}\left(ℝ\right)$ is even, nonnegative and decreasing on ${ℝ}_{+}$, then it holds that

${u}^{2}\left(x\right)\le \frac{1}{2|x|}{\int }_{-|x|}^{|x|}{u}^{2}\left(y\right)dy\le \frac{1}{2|x|}{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\mathit{ }\left(x\ne 0\right).$(3.1)

Thus ${u}_{n}^{2}\left(x\right)\le \frac{C}{2|x|}$ for $x\ne 0$ by

$\underset{n\in ℕ}{sup}{\parallel {u}_{n}\parallel }_{{H}^{1/2,2}\left(ℝ\right)}\le C,$

and ${u}^{2}\left(x\right)\le \frac{C}{2|x|}$ by the pointwise convergence. Now, set ${v}_{n}={|{u}_{n}-u|}^{q}$ for $q>2$. Then we see ${v}_{n}\left(x\right)\to 0$ a.e. $x\in ℝ$. Moreover,

${\int }_{|x|\ge R}{|{u}_{n}-u|}^{q}𝑑x=2{\int }_{R}^{\mathrm{\infty }}{|{u}_{n}-u|}^{q}𝑑x$$\le {2}^{q}\left({\int }_{R}^{\mathrm{\infty }}{|{u}_{n}|}^{q}𝑑x+{\int }_{R}^{\mathrm{\infty }}{|u|}^{q}𝑑x\right)$$\le C{\int }_{R}^{\mathrm{\infty }}\frac{dx}{{|x|}^{q/2}}=\frac{C{R}^{1-q/2}}{\left(q/2\right)-1}\to 0$

as $R\to \mathrm{\infty }$ since $q>2$. Thus ${\left\{{v}_{n}\right\}}_{n\in ℕ}$ is uniformly integrable. Also, by [7, Theorem 6.9] we know that

For any $q>2$, take ${q}_{0}$ such that $2. Since ${u}_{n}-u$ is uniformly bounded in ${H}^{1/2,2}\left(ℝ\right)$, we have ${\parallel {u}_{n}-u\parallel }_{{L}^{{q}_{0}}\left(ℝ\right)}\le C$, and

${\int }_{I}{v}_{n}𝑑x={\int }_{I}{|{u}_{n}-u|}^{q}𝑑x\le {\left({\int }_{I}{|{u}_{n}-u|}^{{q}_{0}}𝑑x\right)}^{q/{q}_{0}}{|I|}^{1-q/{q}_{0}}$

for any bounded measurable set $I\subset ℝ$. Therefore, ${\int }_{I}{v}_{n}𝑑x\to 0$ if $|I|\to 0$, which implies that $\left\{{v}_{n}\right\}$ is uniformly absolutely continuous. Thus by Vitali’s convergence theorem (see, for example, [11, p. 187]) we obtain ${v}_{n}={|{u}_{n}-u|}^{q}\to 0$ strongly in ${L}^{1}\left(ℝ\right)$, which is the desired conclusion. ∎

#### Proposition 3.3.

Assume $\mathrm{\left\{}{u}_{n}\mathrm{\right\}}\mathrm{\subset }{H}^{\mathrm{1}\mathrm{/}\mathrm{2}\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}\mathrm{R}\mathrm{\right)}$ to be a sequence with ${\mathrm{\parallel }{\mathrm{\left(}\mathrm{-}\mathrm{\Delta }\mathrm{\right)}}^{\mathrm{1}\mathrm{/}\mathrm{4}}\mathit{}{u}_{n}\mathrm{\parallel }}_{{L}^{\mathrm{2}}\mathit{}\mathrm{\left(}\mathrm{R}\mathrm{\right)}}\mathrm{\le }\mathrm{1}$. Let ${u}_{n}\mathrm{⇀}u$ weakly in ${H}^{\mathrm{1}\mathrm{/}\mathrm{2}\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}\mathrm{R}\mathrm{\right)}$ for some u and assume ${u}_{n}$ is even, nonnegative and decreasing on ${\mathrm{R}}_{\mathrm{+}}$. Then we have

${\int }_{ℝ}\left({e}^{\alpha {u}_{n}^{2}}-1-\alpha {u}_{n}^{2}\right)𝑑x\to {\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1-\alpha {u}^{2}\right)𝑑x$

for any $\alpha \mathrm{\in }\mathrm{\left(}\mathrm{0}\mathrm{,}\pi \mathrm{\right)}$.

#### Proof.

A similar proposition has already appeared; see [16, Lemma 3.1] and [9, Lemma 5.5]. We prove it here for the reader’s convenience.

Put

${\mathrm{\Phi }}_{\alpha }\left(t\right)={e}^{\alpha {t}^{2}}-1\mathit{ }\text{and}\mathit{ }{\mathrm{\Psi }}_{\alpha }\left(t\right)={e}^{\alpha {t}^{2}}-1-\alpha {t}^{2}.$

Note that ${\mathrm{\Phi }}_{\alpha }\left(t\right)$ is nonnegative, strictly convex and ${\mathrm{\Psi }}_{\alpha }^{\prime }\left(t\right)=2\alpha t{\mathrm{\Phi }}_{\alpha }\left(t\right)$. Thus by the mean value theorem we have

$|{\mathrm{\Psi }}_{\alpha }\left({u}_{n}\right)-{\mathrm{\Psi }}_{\alpha }\left(u\right)|\le {\mathrm{\Psi }}_{\alpha }^{\prime }\left(\theta {u}_{n}+\left(1-\theta \right)u\right)|{u}_{n}-u|$$\le 2\alpha |\theta {u}_{n}+\left(1-\theta \right)u|{\mathrm{\Phi }}_{\alpha }\left(\theta {u}_{n}+\left(1-\theta \right)u\right)|{u}_{n}-u|$$\le 2\alpha \left(|{u}_{n}|+|u|\right)\left(\theta {\mathrm{\Phi }}_{\alpha }\left({u}_{n}\right)+\left(1-\theta \right){\mathrm{\Phi }}_{\alpha }\left(u\right)\right)|{u}_{n}-u|$$\le 2\alpha \left(|{u}_{n}|+|u|\right)\left({\mathrm{\Phi }}_{\alpha }\left({u}_{n}\right)+{\mathrm{\Phi }}_{\alpha }\left(u\right)\right)|{u}_{n}-u|.$

Thus we have

${\int }_{ℝ}|{\mathrm{\Psi }}_{\alpha }\left({u}_{n}\right)-{\mathrm{\Psi }}_{\alpha }\left(u\right)|𝑑x\le 2\alpha {\int }_{ℝ}\left(|{u}_{n}|+|u|\right)\left({\mathrm{\Phi }}_{\alpha }\left({u}_{n}\right)+{\mathrm{\Phi }}_{\alpha }\left(u\right)\right)|{u}_{n}-u|𝑑x$$\le 2\alpha {\parallel |{u}_{n}|+|u|\parallel }_{{L}^{a}\left(ℝ\right)}{\parallel {\mathrm{\Phi }}_{\alpha }\left({u}_{n}\right)+{\mathrm{\Phi }}_{\alpha }\left(u\right)\parallel }_{{L}^{b}\left(ℝ\right)}{\parallel {u}_{n}-u\parallel }_{{L}^{c}\left(ℝ\right)}$(3.2)

by Hölder’s inequality, where $a,b,c>1$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ are chosen later.

First, direct calculation shows that

${\left({\mathrm{\Phi }}_{\alpha }\left(t\right)\right)}^{b}<{e}^{b\alpha {t}^{2}}-1\mathit{ }\left(t\in ℝ\right)$(3.3)

for all $b>1$. Thus if we fix $1 so that $b\alpha <\pi$ is realized, then we have

${\parallel {\mathrm{\Phi }}_{\alpha }\left({u}_{n}\right)+{\mathrm{\Phi }}_{\alpha }\left(u\right)\parallel }_{{L}^{b}\left(ℝ\right)}^{b}\le {\left({\parallel {\mathrm{\Phi }}_{\alpha }\left({u}_{n}\right)\parallel }_{{L}^{b}\left(ℝ\right)}+{\parallel {\mathrm{\Phi }}_{\alpha }\left(u\right)\parallel }_{{L}^{b}\left(ℝ\right)}\right)}^{b}$$\le {2}^{b-1}\left({\int }_{ℝ}{\left({\mathrm{\Phi }}_{\alpha }\left({u}_{n}\right)\right)}^{b}𝑑x+{\int }_{ℝ}{\left({\mathrm{\Phi }}_{\alpha }\left(u\right)\right)}^{b}𝑑x\right)$$\le {2}^{b-1}\left({\int }_{ℝ}\left({e}^{b\alpha {u}_{n}^{2}}-1\right)𝑑x+{\int }_{ℝ}\left({e}^{b\alpha {u}^{2}}-1\right)𝑑x\right)$$\le {2}^{b-1}A\left(b\alpha \right)\left({\parallel {u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}+{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\right).$

Here we used (3.3) for the third inequality and Theorem 1.3 for the last inequality, the use of which is valid since ${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}\le 1$ and ${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}\le 1$ by the weak lower semicontinuity. Note that $\left\{{u}_{n}\right\}$ satisfies

$\underset{n\in ℕ}{sup}{\parallel {u}_{n}\parallel }_{{H}^{1/2,2}\left(ℝ\right)}\le C$

for some $C>0$. Thus we have obtained ${\parallel {\mathrm{\Phi }}_{\alpha }\left({u}_{n}\right)+{\mathrm{\Phi }}_{\alpha }\left(u\right)\parallel }_{{L}^{b}\left(ℝ\right)}=O\left(1\right)$ independent of n.

Next, we estimate the term ${\parallel |{u}_{n}|+|u|\parallel }_{{L}^{a}\left(ℝ\right)}$. Since $\left\{{u}_{n}\right\}$ is a bounded sequence in ${H}^{1/2,2}\left(ℝ\right)$, we have by [7, Theorem 6.9] that ${\parallel u\parallel }_{{L}^{q}\left(ℝ\right)}\le C{\parallel {u}_{n}\parallel }_{{H}^{1/2,2}\left(ℝ\right)}$ for any $q\ge 2$. Thus we see ${\parallel |{u}_{n}|+|u|\parallel }_{{L}^{a}\left(ℝ\right)}\le C$ for some $C>0$ independent of n for $a\ge 2$. Now, note that if we choose $1 and $a>2$ sufficiently large, then we can find $c>2$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$.

By these choices and Proposition 3.2, we conclude that ${\parallel {u}_{n}-u\parallel }_{{L}^{c}\left(ℝ\right)}\to 0$ as $n\to \mathrm{\infty }$. Going back to (3), we conclude that

${\int }_{ℝ}{\mathrm{\Psi }}_{\alpha }\left({u}_{n}\right)dx\to {\int }_{ℝ}{\mathrm{\Psi }}_{\alpha }\left(u\right)dx\mathit{ }\left(n\to \mathrm{\infty }\right),$

which is the desired conclusion. ∎

#### Proof of Theorem 1.7.

We will show that $A\left(\alpha \right)$ in (1.1) is attained for any $0<\alpha <\pi$. Since $A\left(\alpha \right)=\stackrel{~}{A}\left(\alpha \right)$ by Lemma 2.1, we choose a maximizing sequence for $\stackrel{~}{A}\left(\alpha \right)$:

${\int }_{ℝ}\left({e}^{\alpha {u}_{n}^{2}}-1\right)dx=A\left(\alpha \right)+o\left(1\right)\mathit{ }\left(n\to \mathrm{\infty }\right).$

Here ${\left\{{u}_{n}\right\}}_{n\in ℕ}\subset {H}^{1/2,2}\left(ℝ\right)$ satisfies ${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}\le 1$ and ${\parallel {u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}=1$. Appealing to the use of rearrangement, moreover we may assume that ${u}_{n}$ is nonnegative, even and decreasing on ${ℝ}_{+}$. Since ${\left\{{u}_{n}\right\}}_{n\in ℕ}\subset {H}^{1/2,2}\left(ℝ\right)$ is a bounded sequence, we have $u\in {H}^{1/2,2}\left(ℝ\right)$ such that ${u}_{n}⇀u$ in ${H}^{1/2,2}\left(ℝ\right)$. By Proposition 3.3, we see

${\int }_{ℝ}\left({e}^{\alpha {u}_{n}^{2}}-1-\alpha {u}_{n}^{2}\right)𝑑x={\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1-\alpha {u}^{2}\right)𝑑x$

as $n\to \mathrm{\infty }$. Therefore, since ${\parallel {u}_{n}\parallel }_{{L}^{2}\left(ℝ\right)}^{2}=1$, we have, letting $n\to \mathrm{\infty }$,

$A\left(\alpha \right)=\alpha +{\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1-\alpha {u}^{2}\right)𝑑x.$(3.4)

Next, we claim that $A\left(\alpha \right)>\alpha$ for any $0<\alpha <\pi$. Indeed, take any ${u}_{0}\in {H}^{1/2,2}\left(ℝ\right)$ such that ${u}_{0}\not\equiv 0$, ${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{u}_{0}\parallel }_{{L}^{2}\left(ℝ\right)}\le 1$ and ${\parallel {u}_{0}\parallel }_{{L}^{2}\left(ℝ\right)}=1$. Then we have

$A\left(\alpha \right)=\stackrel{~}{A}\left(\alpha \right)\ge {\int }_{ℝ}\left({e}^{\alpha {u}_{0}^{2}}-1\right)𝑑x=\alpha +{\int }_{ℝ}\left({e}^{\alpha {u}_{0}^{2}}-1-\alpha {u}_{0}^{2}\right)𝑑x.$

Now, since ${e}^{\alpha {t}^{2}}-1-\alpha {t}^{2}>0$ for any $t>0$, we have

${\int }_{ℝ}\left({e}^{\alpha {u}_{0}^{2}}-1-\alpha {u}_{0}^{2}\right)𝑑x>0$

for ${u}_{0}\not\equiv 0$, which results in $A\left(\alpha \right)>\alpha$, the claim.

By the claim and (3.4), we conclude that the weak limit u satisfies $u\not\equiv 0$. By the weak lower semi-continuity, we have that $u\not\equiv 0$ satisfies ${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}\le 1$ and ${\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}\le 1$. Thus, by (3.4) again, we see

$A\left(\alpha \right)=\alpha +{\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1-\alpha {u}^{2}\right)𝑑x$$\le \alpha +\frac{1}{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1-\alpha {u}^{2}\right)𝑑x$$=\alpha +\frac{1}{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1\right)𝑑x-\alpha \frac{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}$$=\frac{1}{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}{\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1\right)𝑑x.$

Thus we have shown that $u\in {H}^{1/2,2}\left(ℝ\right)$ maximizes $A\left(\alpha \right)$. ∎

Next, we prove Theorem 1.8. We follow Ishiwata’s argument in [15]. Let

$M=\left\{u\in {H}^{1/2,2}\left(ℝ\right):{\parallel u\parallel }_{{H}^{1/2,2}\left(ℝ\right)}=1\right\},$${J}_{\alpha }:M\to ℝ,{J}_{\alpha }\left(u\right)={\int }_{ℝ}\left({e}^{\alpha {u}^{2}}-1\right)𝑑x.$

Actually, we will show a stronger claim that ${J}_{\alpha }$ has no critical point on M for sufficiently small $\alpha >0$. Assume to the contrary that there exists a critical point $v\in M$ of ${J}_{\alpha }$ for small $\alpha >0$. Then we define an orbit on M through v as

${v}_{\tau }\left(x\right)=\sqrt{\tau }v\left(\tau x\right),\tau \in \left(0,\mathrm{\infty }\right),{w}_{\tau }=\frac{{v}_{\tau }}{{\parallel {v}_{\tau }\parallel }_{{H}^{1/2}}}\in M.$

Note that ${w}_{1}=v$, thus it must hold that

$\frac{d}{d\tau }{|}_{\tau =1}{J}_{\alpha }\left({w}_{\tau }\right)=0.$

By the scaling rules (2.1), we see for any $p\ge 2$,

${\parallel {v}_{\tau }\parallel }_{{L}^{p}\left(ℝ\right)}^{p}={\tau }^{p/2-1}{\parallel v\parallel }_{{L}^{p}\left(ℝ\right)}^{p}\mathit{ }\text{and}\mathit{ }{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{v}_{\tau }\parallel }_{{L}^{2}\left(ℝ\right)}=\tau {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{{L}^{2}\left(ℝ\right)}.$

Now, we see

${J}_{\alpha }\left({w}_{\tau }\right)={\int }_{ℝ}\left({e}^{\alpha {w}_{\tau }^{2}}-1\right)𝑑x$$={\int }_{ℝ}\sum _{j=1}^{\mathrm{\infty }}\frac{{\alpha }^{j}}{j!}\frac{{v}_{\tau }^{2j}\left(x\right)}{{\left({\parallel {v}_{\tau }\parallel }_{2}^{2}+{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{v}_{\tau }\parallel }_{2}^{2}\right)}^{j}}$$=\sum _{j=1}^{\mathrm{\infty }}\frac{{\alpha }^{j}}{j!}\frac{{\parallel {v}_{\tau }\parallel }_{2j}^{2j}}{{\left({\parallel {v}_{\tau }\parallel }_{2}^{2}+{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}{v}_{\tau }\parallel }_{2}^{2}\right)}^{j}}$$=\sum _{j=1}^{\mathrm{\infty }}\frac{{\alpha }^{j}}{j!}\frac{{\tau }^{j-1}{\parallel v\parallel }_{2j}^{2j}}{{\left({\parallel v\parallel }_{2}^{2}+\tau {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{2}^{2}\right)}^{j}}$$=\sum _{j=1}^{\mathrm{\infty }}\frac{{\alpha }^{j}}{j!}{f}_{j}\left(\tau \right),$

where

${f}_{j}\left(\tau \right)=\frac{{\tau }^{j-1}c}{{\left(b+\tau a\right)}^{j}}$

with $a={\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{2}^{2}$, $b={\parallel v\parallel }_{2}^{2}$ and $c={\parallel v\parallel }_{2j}^{2j}$. Since

${f}_{j}^{\prime }\left(\tau \right)=\frac{{\tau }^{j-2}c}{{\left(b+\tau a\right)}^{j+1}}\left\{-\tau a+\left(j-1\right)b\right\}$

and ${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{2}^{2}+{\parallel v\parallel }_{2}^{2}=1$, we calculate

$\frac{d}{d\tau }{|}_{\tau =1}{J}_{\alpha }\left({w}_{\tau }\right)=\sum _{j=1}^{\mathrm{\infty }}{\left[\frac{{\alpha }^{j}}{j!}\frac{{\tau }^{j-2}{\parallel v\parallel }_{2j}^{2j}}{{\left({\parallel v\parallel }_{2}^{2}+\tau {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{2}^{2}\right)}^{j+1}}\left\{-\tau {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{2}^{2}+\left(j-1\right){\parallel v\parallel }_{2}^{2}\right\}\right]}_{\tau =1}$$\le -\alpha {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{2}^{2}{\parallel v\parallel }_{2}^{2}+\sum _{j=2}^{\mathrm{\infty }}\frac{{\alpha }^{j}}{\left(j-1\right)!}{\parallel v\parallel }_{2j}^{2j}$$=\alpha {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{2}^{2}{\parallel v\parallel }_{2}^{2}\left\{-1+\sum _{j=2}^{\mathrm{\infty }}\frac{{\alpha }^{j-1}}{\left(j-1\right)!}\frac{{\parallel v\parallel }_{2j}^{2j}}{{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{2}^{2}{\parallel v\parallel }_{2}^{2}}\right\}.$

Here, we need the following lemma.

#### Lemma 3.4 (Ogawa and Ozawa [28]).

There exists $C\mathrm{>}\mathrm{0}$ such that for any $u\mathrm{\in }{H}^{\mathrm{1}\mathrm{/}\mathrm{2}\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}\mathrm{R}\mathrm{\right)}$ and $p\mathrm{\ge }\mathrm{2}$, it holds that

${\parallel u\parallel }_{{L}^{p}\left(ℝ\right)}^{p}\le C{p}^{p/2}{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}^{p-2}{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}.$

#### Proof.

For $p=2j$, Lemma 3.4 implies

$\frac{{\parallel v\parallel }_{2j}^{2j}}{{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{2}^{2}{\parallel v\parallel }_{2}^{2}}\le C{\left(2j\right)}^{j}\underset{\le 1\left(j\ge 2\right)}{\underset{⏟}{{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{2}^{2j-4}}}\le C{\left(2j\right)}^{j}.$

Thus, for $0<\alpha \ll 1$ sufficiently small (it would be enough that $\alpha <\frac{1}{2e}$), Stirling’s formula $j!\sim {j}^{j}{e}^{-j}\sqrt{2\pi j}$ implies that

$\sum _{j=2}^{\mathrm{\infty }}\frac{{\alpha }^{j-1}}{\left(j-1\right)!}\frac{{\parallel v\parallel }_{2j}^{2j}}{{\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{2}^{2}{\parallel v\parallel }_{2}^{2}}\le \sum _{j=2}^{\mathrm{\infty }}\frac{{\alpha }^{j-1}}{\left(j-1\right)!}{\left(2j\right)}^{j}\le \alpha C$

for some $C>0$ independent of α. Therefore we have

${\frac{d}{d\tau }{J}_{\alpha }\left({w}_{\tau }\right)|}_{\tau =1}<0$

for small α, which is the desired contradiction. ∎

## 4 Proof of Theorem 1.9

In order to prove Theorem 1.9, first we set

$F\left(\beta \right)=\underset{\begin{array}{c}u\in {H}^{1/2,2}\left(ℝ\right)\\ {\parallel u\parallel }_{{H}^{1/2,2}\left(ℝ\right)}\le 1\end{array}}{sup}{\int }_{ℝ}{u}^{2}{e}^{\beta {u}^{2}}𝑑x$(4.1)

for $\beta >0$. Then we obtain the following result.

#### Proposition 4.1.

We have $F\mathit{}\mathrm{\left(}\beta \mathrm{\right)}\mathrm{<}\mathrm{\infty }$ for $\beta \mathrm{<}\pi$.

#### Proof.

We follow the proof of [18, Theorem 1.5]. Take any $u\in {H}^{1/2,2}\left(ℝ\right)$ with ${\parallel u\parallel }_{{H}^{1/2,2}\left(ℝ\right)}\le 1$ in the admissible sets for $F\left(\beta \right)$ in (4.1). By appealing to the rearrangement, we may assume that u is even, nonnegative and decreasing on ${ℝ}_{+}$. We divide the integral

${\int }_{ℝ}{u}^{2}{e}^{\beta {u}^{2}}𝑑x={\int }_{ℝ\setminus I}{u}^{2}{e}^{\beta {u}^{2}}𝑑x+{\int }_{I}{u}^{2}{e}^{\beta {u}^{2}}𝑑x=\left(\mathrm{I}\right)+\left(\mathrm{II}\right),$

where $I=\left(-\frac{1}{2},\frac{1}{2}\right)$.

First, we estimate (I). By the radial lemma (3.1), we see for any $k\in ℕ$, $k\ge 2$,

Thus,

${\int }_{ℝ\setminus I}{u}^{2k}\left(x\right)𝑑x\le \frac{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2k}}{{2}^{k}}{\int }_{ℝ\setminus I}\frac{dx}{{|x|}^{k}}=\frac{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2k}}{{2}^{k-1}}{\int }_{1/2}^{\mathrm{\infty }}\frac{dx}{{x}^{k}}=\frac{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2k}}{k-1}.$

Therefore, we have

$\left(\mathrm{I}\right)={\int }_{ℝ\setminus I}{u}^{2}{e}^{\beta {u}^{2}}𝑑x$$={\int }_{ℝ\setminus I}{u}^{2}\left(1+\sum _{k=1}^{\mathrm{\infty }}\frac{{\beta }^{k}{u}^{2k}}{k!}\right)𝑑x$$={\int }_{ℝ\setminus I}{u}^{2}𝑑x+\sum _{k=2}^{\mathrm{\infty }}\frac{{\beta }^{k-1}}{\left(k-1\right)!}{\int }_{ℝ\setminus I}{u}^{2k}𝑑x$$\le {\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}+\sum _{k=2}^{\mathrm{\infty }}\frac{{\beta }^{k-1}}{\left(k-1\right)!}\frac{{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2k}}{k-1}$$={\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\left(1+\sum _{k=2}^{\mathrm{\infty }}\frac{{\beta }^{k-1}}{\left(k-1\right)\left(k-1\right)!}{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2\left(k-1\right)}\right).$

Now by the constraint ${\parallel u\parallel }_{{H}^{1/2,2}\left(ℝ\right)}\le 1$, we have ${\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}\le 1$. Also if we put

${a}_{k}=\frac{{\beta }^{k-1}}{\left(k-1\right)\left(k-1\right)!},$

then ${\sum }_{k=2}^{\mathrm{\infty }}{a}_{k}$ converges since ${a}_{k+1}/{a}_{k}=\beta \left(k-1\right)/{k}^{2}\to 0$ as $k\to \mathrm{\infty }$. Thus we obtain

$\left(\mathrm{I}\right)\le 1+\sum _{k=2}^{\mathrm{\infty }}\frac{{\beta }^{k-1}}{\left(k-1\right)\left(k-1\right)!}\le C,$

where $C>0$ is independent of $u\in {H}^{1/2,2}\left(ℝ\right)$ with ${\parallel u\parallel }_{{H}^{1/2,2}\left(ℝ\right)}\le 1$.

Next, we estimate (II). Set

$v\left(x\right)=\left\{\begin{array}{cc}u\left(x\right)-u\left(1/2\right),\hfill & |x|\le \frac{1}{2},\hfill \\ 0,\hfill & |x|>\frac{1}{2}.\hfill \end{array}$

Then by the argument of [18] we know that

${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\le {\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}$

and

${u}^{2}\left(x\right)\le {v}^{2}\left(x\right)\left(1+{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\right)+2$

for $x\in I$. Put

$w=v\sqrt{1+{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}}.$

Then we have $w\in {\stackrel{~}{H}}^{1/2,2}\left(\mathrm{I}\right)$ since $v\equiv 0$ on $ℝ\setminus I$, and

${\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}w\parallel }_{{L}^{2}\left(ℝ\right)}^{2}=\left(1+{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\right){\parallel {\left(-\mathrm{\Delta }\right)}^{1/4}v\parallel }_{{L}^{2}\left(ℝ\right)}^{2}$$\le \left(1+{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\right)\left(1-{\parallel u\parallel }_{{L}^{2}\left(ℝ\right)}^{2}\right)\le 1.$

Thus we may use the fractional Trudinger–Moser inequality (Proposition 1.1) to w to obtain

${\int }_{I}{e}^{\pi {w}^{2}}𝑑x\le C$

for some $C>0$ independent of u. By ${u}^{2}\le {w}^{2}+2$ on I, we conclude that

${\int }_{I}{e}^{\pi {u}^{2}}𝑑x\le {\int }_{I}{e}^{\pi \left({w}^{2}+2\right)}𝑑x={e}^{2\pi }{\int }_{I}{e}^{\pi {w}^{2}}𝑑x\le {C}^{\prime }.$

Now, since $\beta <\pi$, there is an absolute constant ${C}_{0}$ such that

${t}^{2}{e}^{\beta {t}^{2}}\le {C}_{0}{e}^{\pi {t}^{2}}$

for any $t\in ℝ$. Finally, we obtain

$\left(\mathrm{II}\right)={\int }_{I}{u}^{2}{e}^{\beta {u}^{2}}𝑑x\le {C}_{0}{\int }_{I}{e}^{\pi {u}^{2}}𝑑x\le {C}_{0}{C}^{\prime }.$

Proposition 4.1 follows from the estimates (I) and (II). ∎

By using Proposition 4.1 and arguing as in the proof of Theorem 1.3 (after establishing claims similar to those in Lemma 2.1 and Lemma 2.2), it is easy to obtain the following proposition.

#### Proposition 4.2.

For any $\mathrm{0}\mathrm{<}\alpha \mathrm{<}\beta \mathrm{<}\pi$, we have

$E\left(\alpha \right)\le \left(\frac{1}{1-\alpha /\beta }\right)F\left(\beta \right).$

Since $F\left(\beta \right)<\mathrm{\infty }$ for any $\beta <\pi$, this proves the first part of Theorem 1.9. For the attainability of $E\left(\alpha \right)$ for $\alpha \in \left(0,\pi \right)$ it is enough to argue as in the proof of Theorem 1.7. We omit the details.

## References

• [1]

S. Adachi and K. Tanaka, A scale-invariant form of Trudinger–Moser inequality and its best exponent, Proc. Amer. Math. Soc. (1999), no. 1102, 148–153.  Google Scholar

• [2]

Adimurthi and Y. Yang, An interpolation of Hardy inequality and Trundinger–Moser inequality in ${ℝ}^{N}$ and its applications, Int. Math. Res. Not. IMRN (2010), no. 13, 2394–2426.  Google Scholar

• [3]

F. J. Almgren, Jr. and E. H. Lieb, Symmetric decreasing rearrangement is sometimes continuous, J. Amer. Math. Soc. 2 (1989), no. 4, 683–773.

• [4]

D. M. Cao, Nontrivial solution of semilinear elliptic equation with critical exponent in ${𝐑}^{2}$, Comm. Partial Differential Equations 17 (1992), no. 3–4, 407–435.  Google Scholar

• [5]

L. Carleson and S.-Y. A. Chang, On the existence of an extremal function for an inequality of J. Moser, Bull. Sci. Math. (2) 110 (1986), no. 2, 113–127.  Google Scholar

• [6]

D. Cassani, F. Sani and C. Tarsi, Equivalent Moser type inequalities in ${ℝ}^{2}$ and the zero mass case, J. Funct. Anal. 267 (2014), no. 11, 4236–4263.  Google Scholar

• [7]

E. Di Nezza, G. Palatucci and E. Valdinoci, Hitchhiker’s guide to the fractional Sobolev spaces, Bull. Sci. Math. 136 (2012), no. 5, 521–573.

• [8]

J. A. M. B. do Ó, N-Laplacian equations in ${𝐑}^{N}$ with critical growth, Abstr. Appl. Anal. 2 (1997), no. 3–4, 301–315.  Google Scholar

• [9]

M. Dong and G. Lu, Best constants and existence of maximizers for weighted Trudinger–Moser inequalities, Calc. Var. Partial Differential Equations 55 (2016), no. 4, Article ID 88.

• [10]

M. Flucher, Extremal functions for the Trudinger–Moser inequality in 2 dimensions, Comment. Math. Helv. 67 (1992), no. 3, 471–497.

• [11]

G. B. Folland, Real analysis, 2nd ed., Pure Appl. Math. (New York), John Wiley & Sons, New York, 1999.  Google Scholar

• [12]

L. Fontana and C. Morpurgo, Sharp Adams and Moser–Trudinger inequalities on ${ℝ}^{n}$ and other spaces of infinite measure, preprint (2015), https://arxiv.org/abs/1504.04678.

• [13]

L. Fontana and C. Morpurgo, Sharp exponential integrability for critical Riesz potentials and fractional Laplacians on ${ℝ}^{n}$, preprint (2017), https://arxiv.org/abs/1702.02708.

• [14]

A. Iannizzotto and M. Squassina, 1/2-Laplacian problems with exponential nonlinearity, J. Math. Anal. Appl. 414 (2014), no. 1, 372–385.

• [15]

M. Ishiwata, Existence and nonexistence of maximizers for variational problems associated with Trudinger–Moser type inequalities in ${ℝ}^{N}$, Math. Ann. 351 (2011), no. 4, 781–804.  Google Scholar

• [16]

M. Ishiwata, M. Nakamura and H. Wadade, On the sharp constant for the weighted Trudinger–Moser type inequality of the scaling invariant form, Ann. Inst. H. Poincaré Anal. Non Linéaire 31 (2014), no. 2, 297–314.

• [17]

S. Iula, A note on the Moser–Trudinger inequality in Sobolev–Slobodeckij spaces in dimension one, preprint (2016), https://arxiv.org/abs/1610.00933v1.

• [18]

S. Iula, A. Maalaoui and L. Martinazzi, A fractional Moser–Trudinger type inequality in one dimension and its critical points, Differential Integral Equations 29 (2016), no. 5–6, 455–492.  Google Scholar

• [19]

V. I. Judovič, Some estimates connected with integral operators and with solutions of elliptic equations, Dokl. Akad. Nauk SSSR 138 (1961), 805–808.  Google Scholar

• [20]

N. Lam, G. Lu and L. Zhang, Equivalence of critical and subcritical sharp Trudinger–Moser–Adams inequalities, preprint (2015), https://arxiv.org/abs/1504.04858v1.

• [21]

X. Li and Y. Yang, Extremal functions for singular Trudinger–Moser inequalities in the entire Euclidean space, preprint (2016), https://arxiv.org/abs/1612.08247.

• [22]

Y. Li and B. Ruf, A sharp Trudinger–Moser type inequality for unbounded domains in ${ℝ}^{n}$, Indiana Univ. Math. J. 57 (2008), no. 1, 451–480.  Google Scholar

• [23]

E. H. Lieb and M. Loss, Analysis, 2nd ed., Grad. Stud. Math. 14, American Mathematical Society, Providence, 2001.  Google Scholar

• [24]

K.-C. Lin, Extremal functions for Moser’s inequality, Trans. Amer. Math. Soc. 348 (1996), no. 7, 2663–2671.

• [25]

L. Martinazzi, Fractional Adams–Moser–Trudinger type inequalities, Nonlinear Anal. 127 (2015), 263–278.

• [26]

J. Moser, A sharp form of an inequality by N. Trudinger, Indiana Univ. Math. J. 20 (1970/71), 1077–1092.  Google Scholar

• [27]

T. Ogawa, A proof of Trudinger’s inequality and its application to nonlinear Schrödinger equations, Nonlinear Anal. 14 (1990), no. 9, 765–769.

• [28]

T. Ogawa and T. Ozawa, Trudinger type inequalities and uniqueness of weak solutions for the nonlinear Schrödinger mixed problem, J. Math. Anal. Appl. 155 (1991), no. 2, 531–540.

• [29]

T. Ozawa, On critical cases of Sobolev’s inequalities, J. Funct. Anal. 127 (1995), no. 2, 259–269.

• [30]

T. Ozawa, Characterization of Trudinger’s inequality, J. Inequal. Appl. 1 (1997), no. 4, 369–374.

• [31]

E. Parini and B. Ruf, On the Moser–Trudinger inequality in fractional Sobolev–Slobodeckij spaces, preprint (2016), https://arxiv.org/abs/1607.07681v1.

• [32]

Y. J. Park, Fractional Polya–Szegö inequality, J. Chungcheong Math. Soc. 24 (2011), no. 2, 267–271.  Google Scholar

• [33]

S. Pohozaev, The Sobolev embedding in the case $pl=n$, Proceedings of the Technical Scientic Conference on Advances of Scientic Research (1964/1965), Energetics Institute, Moscow (1965), 158–170.  Google Scholar

• [34]

B. Ruf, A sharp Trudinger–Moser type inequality for unbounded domains in ${ℝ}^{2}$, J. Funct. Anal. 219 (2005), no. 2, 340–367.  Google Scholar

• [35]

W. A. Strauss, Existence of solitary waves in higher dimensions, Comm. Math. Phys. 55 (1977), no. 2, 149–162.

• [36]

M. Struwe, Critical points of embeddings of ${H}_{0}^{1,n}$ into Orlicz spaces, Ann. Inst. H. Poincaré Anal. Non Linéaire 5 (1988), no. 5, 425–464.  Google Scholar

• [37]

N. S. Trudinger, On imbeddings into Orlicz spaces and some applications, J. Math. Mech. 17 (1967), 473–483.  Google Scholar

## About the article

Revised: 2017-08-10

Accepted: 2017-09-21

Published Online: 2017-11-16

Funding Source: Japan Society for the Promotion of Science

Award identifier / Grant number: 15H03631

Award identifier / Grant number: 26610030

Part of this work was supported by JSPS Grant-in-Aid for Scientific Research (B), no. 15H03631, and JSPS Grant-in-Aid for Challenging Exploratory Research, no. 26610030.

Citation Information: Advances in Nonlinear Analysis, Volume 8, Issue 1, Pages 868–884, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496,

Export Citation