Critical and subcritical fractional Trudinger-Moser type inequalities on $\mathbb{R}$

TL;DR

This paper investigates critical and subcritical Trudinger-Moser inequalities within fractional Sobolev spaces on the real line, establishing their relationships and exploring whether the supremum values are attained.

Contribution

It provides new insights into the connection between critical and subcritical inequalities and analyzes the attainability of supremum in fractional Sobolev spaces.

Findings

Established the relation between critical and subcritical inequalities.

Proved results on the attainability of supremum values.

Extended the understanding of Trudinger-Moser inequalities in fractional settings.

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.

Equations

$$W_0^{1,N}(\Omega )\hookrightarrow \{u\in L^N(\Omega ):{\int }_{\Omega }\exp \left(\alpha |u{|}^{\frac{N}{N-1}}\right)dx<\infty \text{for any}\alpha >0\},$$

$$TM(\Omega ,\alpha )=\underset{\genfrac{}{}{0ex}{}{u\in W_0^{1,N}(\Omega )}{\Vert \nabla u{\Vert }_{L^N(\Omega )}\le 1}}{\sup }\frac{1}{|\Omega |}{\int }_{\Omega }\exp (\alpha |u{|}^{\frac{N}{N-1}})dx\{\begin{array}{ll}& <\infty ,\alpha \le {\alpha }_N,\\ & =\infty ,\alpha >{\alpha }_N,\end{array}$$

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

$$A(N,\alpha )=\underset{\genfrac{}{}{0ex}{}{u\in W^{1,N}({\mathbb{R}}^N)\setminus \{0\}}{\Vert \nabla u{\Vert }_{L^N({\mathbb{R}}^N)}\le 1}}{\sup }\frac{1}{\Vert u{\Vert }_{L^N({\mathbb{R}}^N)}^N}{\int }_{{\mathbb{R}}^N}{\Phi }_N(\alpha |u{|}^{\frac{N}{N-1}})dx\{\begin{array}{ll}& <\infty ,\alpha <{\alpha }_N,\\ & =\infty ,\alpha \ge {\alpha }_N.\end{array}$$

$$B(N,\alpha )=\underset{\genfrac{}{}{0ex}{}{u\in W^{1,N}({\mathbb{R}}^N)}{\Vert u{\Vert }_{W^{1,N}({\mathbb{R}}^N)}\le 1}}{\sup }{\int }_{{\mathbb{R}}^N}{\Phi }_N(\alpha |u{|}^{\frac{N}{N-1}})dx\{\begin{array}{ll}& <\infty ,\alpha \le {\alpha }_N,\\ & =\infty ,\alpha >{\alpha }_N.\end{array}$$

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

$$⟨\varphi ,(-\Delta {)}^{s/2}u⟩=⟨(-\Delta {)}^{s/2}\varphi ,u⟩={\int }_{\mathbb{R}}(-\Delta {)}^{s/2}\varphi \cdot udx,\varphi \in \mathcal{S}({\mathbb{R}}^N).$$

$$H^{s,p}({\mathbb{R}}^N)$$

$${\tilde{H}}^{s,p}(\Omega )$$

$$W^{s,p}({\mathbb{R}}^N)$$

$$[u{]}_{W^{s,p}({\mathbb{R}}^N)}^p$$

$${\tilde{W}}^{s,p}(\Omega )={\overline{C_c^{\infty }(\Omega )}}^{\Vert \cdot {\Vert }_{W^{s,p}({\mathbb{R}}^N)}}$$

$${\tilde{W}}^{s,p}(\Omega )=\left\{u\in W^{s,p}({\mathbb{R}}^N):u\equiv 0\text{on}{\mathbb{R}}^N\setminus \Omega \right\}$$

$$\underset{\genfrac{}{}{0ex}{}{u\in {\tilde{H}}^{s,p}(\Omega )}{\Vert (-\Delta {)}^{s/2}u{\Vert }_{L^p(\Omega )}\le 1}}{\sup }\frac{1}{|\Omega |}{\int }_{\Omega }\exp (\alpha |u{|}^{\frac{p}{p-1}})dx\{\begin{array}{ll}& <\infty ,\alpha \le {\alpha }_{N,p},\\ & =\infty ,\alpha >{\alpha }_{N,p}.\end{array}$$

$$\Vert (-\Delta {)}^{1/4}u{\Vert }_{L^2(\mathbb{R})}^2=\frac{1}{2\pi }[u{]}_{W^{1/2,2}(\mathbb{R})}^2,$$

$$\underset{\genfrac{}{}{0ex}{}{u\in {\tilde{H}}^{1/2,2}(I)}{\Vert (-\Delta {)}^{1/4}u{\Vert }_{L^2(I)}\le 1}}{\sup }\frac{1}{|I|}{\int }_I{e}^{\alpha |u{|}^2}dx\{\begin{array}{ll}& <\infty ,\alpha \le {\alpha }_{1,2}=\pi ,\\ & =\infty ,\alpha >\pi \end{array}$$

$$A(\alpha )=\underset{\genfrac{}{}{0ex}{}{u\in H^{1/2,2}(\mathbb{R})\setminus \{0\}}{\Vert (-\Delta {)}^{1/4}u{\Vert }_{L^2(\mathbb{R})}\le 1}}{\sup }\frac{1}{\Vert u{\Vert }_{L^2(\mathbb{R})}^2}{\int }_{\mathbb{R}}\left(e^{\alpha u^2}-1\right)dx.$$

$$B(\alpha )=\underset{\genfrac{}{}{0ex}{}{u\in H^{1/2,2}(\mathbb{R})}{\Vert u{\Vert }_{H^{1/2,2}(\mathbb{R})}\le 1}}{\sup }{\int }_{\mathbb{R}}\left(e^{\alpha u^2}-1\right)dx\{\begin{array}{ll}& <\infty ,\alpha \le \pi ,\\ & =\infty ,\alpha >\pi .\end{array}$$

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

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

$${\beta }_0=\underset{q\to \infty }{\mathrm{limsup}}\underset{u\in H^{1/2,2}(\mathbb{R}),u\ne 0}{\sup }\frac{\Vert u{\Vert }_{L^q(\mathbb{R})}}{q^{1/2}\Vert (-\Delta {)}^{1/4}u{\Vert }_{L^2(\mathbb{R})}^{1-2/q}\Vert u{\Vert }_{L^2(\mathbb{R})}^{2/q}}.$$

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

$$\frac{C_1}{1-\alpha /\pi }\le A(\alpha )\le \frac{C_2}{1-\alpha /\pi }.$$

$$E(\alpha )=\underset{\genfrac{}{}{0ex}{}{u\in H^{1/2,2}(\mathbb{R})\setminus \{0\}}{\Vert (-\Delta {)}^{1/4}u{\Vert }_{L^2(\mathbb{R})}\le 1}}{\sup }\frac{1}{\Vert u{\Vert }_{L^2(\mathbb{R})}^2}{\int }_{\mathbb{R}}{u}^2{e}^{\alpha u^2}dx.$$

$$E(\alpha )\{\begin{array}{ll}& <\infty ,\alpha <\pi ,\\ & =\infty ,\alpha \ge \pi .\end{array}$$

$$\tilde{A}(\alpha )=\underset{\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{u\in H^{1/2,2}(\mathbb{R})\setminus \{0\}}{\Vert (-\Delta {)}^{1/4}u{\Vert }_{L^2(\mathbb{R})}\le 1}}{\Vert u{\Vert }_{L^2(\mathbb{R})}=1}}{\sup }{\int }_{\mathbb{R}}\left(e^{\alpha u^2}-1\right)dx.$$

$$\{\begin{array}{ll}& \Vert (-\Delta {)}^{1/4}{u}_{\lambda }{\Vert }_{L^2(\mathbb{R})}=\Vert (-\Delta {)}^{1/4}u{\Vert }_{L^2(\mathbb{R})},\\ & \Vert u_{\lambda }{\Vert }_{L^2(\mathbb{R})}^2={\lambda }^{-1}\Vert u{\Vert }_{L^2(\mathbb{R})}^2,\end{array}$$

$$2\pi \Vert (-\Delta {)}^{1/4}{u}_{\lambda }{\Vert }_{L^2(\mathbb{R})}^2$$

$$={\int }_{\mathbb{R}}{\int }_{\mathbb{R}}\frac{|u(\lambda x)-u(\lambda y){|}^2}{|x-y{|}^2}dxdy$$

$$={\int }_{\mathbb{R}}{\int }_{\mathbb{R}}\frac{|u(\lambda x)-u(\lambda y){|}^2}{|\lambda x-\lambda y{|}^2}d(\lambda x)d(\lambda y)$$

$$=[u{]}_{W^{1/2,2}(\mathbb{R})}^2=2\pi \Vert (-\Delta {)}^{1/4}u{\Vert }_{L^2(\mathbb{R})}^2.$$

$$\Vert (-\Delta {)}^{1/4}{u}_{\lambda }{\Vert }_{L^2(\mathbb{R})}\le 1\text{and}\Vert u_{\lambda }{\Vert }_{L^2(\mathbb{R})}^2=1.$$

$$\frac{1}{\Vert u{\Vert }_{L^2(\mathbb{R})}^2}{\int }_{\mathbb{R}}\left(e^{\alpha u^2}-1\right)dx={\int }_{\mathbb{R}}\left(e^{\alpha u_{\lambda }^2}-1\right)dx\le \tilde{A}(\alpha ),$$

Full text

1. Introduction

Let $\Omega\subset\mathbb{R}^{N}$, $N\geq 2$ be a domain with finite volume. Then the Sobolev embedding theorem assures that $W^{1,N}_{0}(\Omega)\hookrightarrow L^{q}(\Omega)$ for any $q\in[1,+\infty)$, however, a simple example shows that the embedding $W^{1,N}_{0}(\Omega)\hookrightarrow L^{\infty}(\Omega)$ does not hold. Instead, functions in $W^{1,N}_{0}(\Omega)$ enjoy the exponential summability:

[TABLE]

see Yudovich [29], Pohozaev [24], and Trudinger [28]. Later, Moser [18] improved the embedding above as follows, now known as the Trudinger-Moser inequality:

[TABLE]

here $\alpha_{N}=N\omega_{N-1}^{\frac{1}{N-1}}$ and $\omega_{N-1}=|S^{N-1}|$ denotes the area of the unit sphere in $\mathbb{R}^{N}$. On the attainability of $TM(\Omega,\alpha)$, Carleson-Chang [4], Flucher [6], and Lin [13] proved that $TM(\Omega,\alpha)$ is attained for any $0<\alpha\leq\alpha_{N}$.

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

[TABLE]

denote the truncated exponential function.

First, Ogawa [20], Ogawa-Ozawa [21], Cao [3], Ozawa [23], and Adachi-Tanaka [1] proved that the following inequality holds true, which we call Adachi-Tanaka type Trudinger-Moser inequality:

[TABLE]

The inequality enjoys the scale invariance under the scaling $u(x)\mapsto u_{\lambda}(x)=u(\lambda x)$ for $\lambda>0$. Note that the critical exponent $\alpha=\alpha_{N}$ is not allowed for the finiteness of the supremum. Recently, it is proved that $A(N,\alpha)$ is attained for any $\alpha\in(0,\alpha_{N})$ by Ishiwata-Nakamura-Wadade [10] and Dong-Lu [5]. In this sense, Adachi-Tanaka type Trudinger-Moser inequality has a subcritical nature of the problem.

On the other hand, Ruf [26] and Li-Ruf [15] proved that the following inequality holds true:

[TABLE]

Here $\|u\|_{W^{1,N}(\mathbb{R}^{N})}=\left(\|\nabla u\|_{L^{N}(\mathbb{R}^{N})}^{N}+\|u\|_{L^{N}(\mathbb{R}^{N})}^{N}\right)^{1/N}$ is the full Sobolev norm. Note that the scale invariance $(u\mapsto u_{\lambda})$ does not hold for this inequality. Also note that the critical exponent $\alpha=\alpha_{N}$ is permitted to the finiteness.

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

• •

If $N\geq 3$, then $B(N,\alpha)$ is attained for $0<\alpha\leq\alpha_{N}$ [26].

• •

If $N=2$, then there exists $\alpha_{*}>0$ such that $B(2,\alpha)$ is attained for $\alpha_{*}<\alpha\leq\alpha_{2}(=4\pi)$ [26], [9].

• •

If $N=2$ and $\alpha>0$ is sufficiently small, then $B(2,\alpha)$ is not attained. [9].

The non-attainability of $B(2,\alpha)$ for $\alpha$ sufficiently small is attributed to the non-compactness of “vanishing” maximizing sequences, as described in [9].

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

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

[TABLE]

For $u\in L_{s}(\mathbb{R}^{N})$, we define the fractional Laplacian $(-\Delta)^{s/2}u$ as follows: First, for $\phi\in\mathcal{S}(\mathbb{R}^{N})$, the rapidly decreasing functions on $\mathbb{R}^{N}$, $(-\Delta)^{s/2}\phi$ is defined via the normalized Fourier transform $\mathcal{F}$ as $(-\Delta)^{s/2}\phi(x)=\mathcal{F}^{-1}(|\xi|^{s}\mathcal{F}\phi(\xi))(x)$ for $x\in\mathbb{R}^{N}$. Then for $u\in L_{s}(\mathbb{R}^{N})$, $(-\Delta)^{s/2}u$ is defined as the element of $\mathcal{S}^{\prime}(\mathbb{R}^{N})$, the tempered distributions on $\mathbb{R}^{N}$, by the relation

[TABLE]

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

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

[TABLE]

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

[TABLE]

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

[TABLE]

where $\|u\|_{W^{s,p}(\mathbb{R}^{N})}=\left(\|u\|_{L^{p}(\mathbb{R}^{N})}^{p}+[u]_{W^{s,p}(\mathbb{R}^{N})}^{p}\right)^{1/p}$. It is known that

[TABLE]

if $\Omega$ is a Lipschitz domain and $H^{s,p}(\mathbb{R}^{N})=F^{s}_{p,2}(\mathbb{R}^{N})$ (Triebel-Lizorkin space), $W^{s,p}(\mathbb{R}^{N})=B^{s}_{p,p}(\mathbb{R}^{N})$ (Besov space). Thus $H^{s,2}(\mathbb{R}^{N})=W^{s,2}(\mathbb{R}^{N})$, however in general, $H^{s,p}(\mathbb{R}^{N})\neq W^{s,p}(\mathbb{R}^{N})$ for $p\neq 2$. See [25], [11] and the references therein.

Recently, Martinazzi [17] (see also [12]) proved a fractional Trudinger-Moser type inequality on $\tilde{H}^{s,p}(\Omega)$ as follows: Let $p\in(1,\infty)$ and $s=N/p$ for $N\in\mathbb{N}$. Then for any open $\Omega\subset\mathbb{R}^{N}$ with $|\Omega|<\infty$, it holds

[TABLE]

Here $\alpha_{N,p}=\frac{N}{\omega_{N-1}}\left(\frac{\Gamma((N-s)/2)}{\Gamma(s/2)2^{s}\pi^{N/2}}\right)^{-p/(p-1)}$.

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 $\tilde{W}^{s,p}(\Omega)$ with $sp=N$, similar fractional Trudinger-Moser inequality is also proved by Parini-Ruf [25] when $N\geq 2$ and Iula [11] when $N=1$. In this case, the result is weaker and the inequality holds true only for $0\leq\alpha<\alpha^{*}_{N,p}$ for some (explicit) value $\alpha^{*}_{N,p}$. Also, it is not known the inequality holds or not when $\alpha=\alpha^{*}_{N,p}$.

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

[TABLE]

see Proposition 3.6. in [19]. Then the fractional Trudinger-Moser inequality in [17], [12] can be read as

Proposition 1**.**

*(A fractional Trudinger-Moser inequality on $\tilde{H}^{1/2,2}(I)$)

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

[TABLE]

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

[TABLE]

Then by the precedent results by Ogawa-Ozawa [21] and Ozawa [23], it is known that $A(\alpha)<\infty$ for small exponent $\alpha$.

On the other hand, it is already known a fractional Li-Ruf type Trudinger-Moser inequality on $H^{1/2,2}(\mathbb{R})$:

Proposition 2**.**

(Iula-Maalaoui-Martinazzi [12])

[TABLE]

Here

[TABLE]

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

Concerning $A(\alpha)$ in (1.1), a natural question is that to what range of the exponent $\alpha$ the supremum is finite. As pointed out in [8], it remained an open problem for a while. In this paper, first we prove the finiteness of supremum in the full range of values of exponent.

Theorem 1**.**

*(Full range Adachi-Tanaka type on $H^{1/2,2}(\mathbb{R})$)

We have*

[TABLE]

Ozawa [22] proved that the Adachi-Tanaka type Trudinger-Moser inequality is equivalent to the Gagliardo-Nirenberg type inequality, and he also proved an exact relation between the best constants of both inequalities. As a result, we have the next corollary.

Corollary 1**.**

Set

[TABLE]

Then $\beta_{0}=(2\pi e)^{-1/2}$.

Furthermore, we obtain the relation between the suprema of both critical and subcritical Trudinger-Moser type inequalities along the line of Lam-Lu-Zhang [14].

Theorem 2**.**

(Relation) We have

[TABLE]

Also we obtain how Adachi-Tanaka type supremum $A(\alpha)$ behaves when $\alpha$ tends to $\pi$.

Theorem 3**.**

*(Asymptotic behavior)

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

[TABLE]

Note that the estimate from the above follows from Theorem 2 and Proposition 2. On the other hand, we will see that that the estimate from the below follows from a computation using the Moser sequence.

Concerning the existence of maximizers of Adachi-Tanaka type supremum $A(\alpha)$ in (1.1), we see

Theorem 4**.**

*(Attainability of $A(\alpha)$)

$A(\alpha)$ is attained for any $\alpha\in(0,\pi)$.*

On the other hand, as for $B(\alpha)$ in (1.2), we have

Theorem 5**.**

*(Non-attainability of $B(\alpha)$)

For $0<\alpha<<1$, $B(\alpha)$ is not attained.*

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

Finally, we improve the subcritical Adachi-Tanaka type inequality along the line of Dong-Lu [5]:

Theorem 6**.**

For $\alpha>0$, set

[TABLE]

Then we have

[TABLE]

Furthermore, $E(\alpha)$ is attained for all $\alpha\in(0,\pi)$.

Since $e^{\alpha t^{2}}-1\leq\alpha t^{2}e^{\alpha t^{2}}$ for $t\in\mathbb{R}$, Theorem 6 extends Theorem 1. In the classical case, Dong-Lu used a rearrangement technique to reduce the problem to one-dimension and obtained the similar inequality by estimating a one-dimensional integral. The method is similar to [4]. 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 Theorem 1, 2, and 3. In section 3, we prove Theorem 4 and 5. In section 4, we prove Theorem 6.

2. Proof of Theorem 1, 2, and 3

For the proofs of Theorem 1, 2, and 3, we prepare several lemmas.

Lemma 1**.**

Set

[TABLE]

Then $\tilde{A}(\alpha)=A(\alpha)$ for any $\alpha>0$.

Proof.

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

[TABLE]

since

[TABLE]

Thus for any $u\in H^{1/2,2}(\mathbb{R})\setminus\{0\}$ with $\|(-\Delta)^{1/4}u\|_{L^{2}(\mathbb{R})}\leq 1$, if we choose $\lambda=\|u\|^{2}_{L^{2}(\mathbb{R})}$, then $u_{\lambda}\in H^{1/2,2}(\mathbb{R})$ satisfies

[TABLE]

Thus

[TABLE]

which implies $A(\alpha)\leq\tilde{A}(\alpha)$. The opposite inequality is trivial. ∎

Lemma 2**.**

For any $0<\alpha<\pi$, it holds

[TABLE]

Proof.

Choose any $u\in H^{1/2,2}(\mathbb{R})$ with $\|(-\Delta)^{1/4}u\|_{L^{2}(\mathbb{R})}\leq 1$ and $\|u\|_{L^{2}(\mathbb{R})}=1$. Put $v(x)=Cu(\lambda x)$ where $C^{2}=\alpha/\pi\in(0,1)$ and $\lambda=\frac{C^{2}}{1-C^{2}}$. Then by scaling rules (2.2), we see

[TABLE]

Also we have

[TABLE]

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

[TABLE]

By taking the supremum for $u\in H^{1/2,2}(\mathbb{R})$ with $\|(-\Delta)^{1/4}u\|_{L^{2}(\mathbb{R})}\leq 1$ and $\|u\|_{L^{2}(\mathbb{R})}=1$, we have

[TABLE]

Finally, Lemma 1 implies the result. ∎

Proof of Theorem 1: The assertion that $A(\alpha)<\infty$ for $\alpha<\pi$ follows from Lemma 2 and the fact $B(\pi)<\infty$ by Proposition 2.

For the proof of $A(\pi)=\infty$, we use the Moser sequence

[TABLE]

and its estimates

[TABLE]

as $\varepsilon\to 0$ for some $C>0$. Note $u_{\varepsilon}\in\tilde{W}^{1/2,2}((-1,1))\subset W^{1/2,2}(\mathbb{R})=H^{1/2,2}(\mathbb{R})$. For the estimate (2.4), we refer to Iula [11] Proposition 2.2. For the estimate (2.5), we refer to [11] equation (35). Actually, after a careful look of the proof of Proposition 2.2 in [11], we confirm that

[TABLE]

for a positive $C>0$, which implies (2.5). For (2.6), we compute

[TABLE]

as $\varepsilon\to 0$. Thus we obtain (2.6).

By testing $A(\pi)$ by $v_{\varepsilon}=u_{\varepsilon}/\|(-\Delta)^{1/4}u_{\varepsilon}\|_{L^{2}(\mathbb{R})}$, we have

[TABLE]

since $e^{t}-1\geq(1/2)e^{t}$ for $t$ large and (2.5). Also since

[TABLE]

we see $\frac{t}{1+\frac{1}{Ct}}=t-1/C+o(1)$ as $t\to\infty$. Put $t=\log(1/\varepsilon)$, we see

[TABLE]

which leads to

[TABLE]

for some $\delta>0$ independent of $\varepsilon\to 0$. Therefore, by (2.4), (2.5), (2.6), we have for $\delta^{\prime}>0$

[TABLE]

as $\varepsilon\to 0$. This proves $A(\pi)=\infty$. ∎

Proof of Theorem 2: By Lemma 2, we have

[TABLE]

Let us prove the opposite inequality. Let $\{u_{n}\}\subset H^{1/2,2}(\mathbb{R})$, $u_{n}\neq 0$, $\|(-\Delta)^{1/4}u_{n}\|^{2}_{L^{2}(\mathbb{R})}+\|u_{n}\|^{2}_{L^{2}(\mathbb{R})}\leq 1$, be a maximizing sequence of $B(\pi)$. We may assume $\|(-\Delta)^{1/4}u_{n}\|^{2}_{L^{2}(\mathbb{R})}<1$ for any $n\in\mathbb{N}$. Put

[TABLE]

Thus by (2.2), we see

[TABLE]

since $\|(-\Delta)^{1/4}u_{n}\|^{2}_{L^{2}(\mathbb{R})}+\|u_{n}\|^{2}_{L^{2}(\mathbb{R})}\leq 1$. Thus, setting $\alpha_{n}=\pi\|(-\Delta)^{1/4}u_{n}\|^{2}_{L^{2}(\mathbb{R})}<\pi$ for any $n\in\mathbb{N}$, we may test $A(\alpha_{n})$ by $\{v_{n}\}$, which results in

[TABLE]

Here we have used a change of variables $y=\lambda_{n}x$ for the second equality, and $\|v_{n}\|^{2}_{L^{2}(\mathbb{R})}\leq 1$ for the first inequality. Letting $n\to\infty$, we have the desired result. ∎

Proof of Theorem 3:

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

[TABLE]

Again we use the Moser sequence (2.3) and we test $A(\alpha)$ by $v_{\varepsilon}=u_{\varepsilon}/\|(-\Delta)^{1/4}u_{\varepsilon}\|_{L^{2}(\mathbb{R})}$. As in the similar calculations in the proof of Theorem 1, we have

[TABLE]

where we put $\delta_{\varepsilon}=(\frac{\alpha}{\pi})\frac{1}{1+(C\log(1/\varepsilon))^{-1}}\in(0,1)$.

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

[TABLE]

which implies

[TABLE]

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

[TABLE]

Now, we estimate that

[TABLE]

where $f(t)=\frac{2t}{C+1-t}$ for $t\in[0,1]$ and we have used (2.7) in the last inequality. We easily see that $f(0)=0$, $f^{\prime}(t)=\frac{2(C+1)}{(C+1-t)^{2}}>0$ for $t>0$, thus $f(t)$ is strictly increasing in $t$ and $\max_{t\in[0,1]}f(t)=f(1)=2/C$. Thus we have

[TABLE]

which is independent of $\alpha$. Backing to (2.8) with (2.7), we observe that

[TABLE]

which proves the result. ∎

3. Proof of Theorem 4 and 5

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

[TABLE]

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

[TABLE]

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

[TABLE]

holds true for $u\in H^{1/2,2}(\mathbb{R})$, see for example, [2] and [16].

Remark 1*.*

Note that Radial Compactness Lemma by Strauss [27] is violated on $\mathbb{R}$. More precisely, let

[TABLE]

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

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

Proposition 3**.**

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

Proof.

Since $\{u_{n}\}\subset H^{1/2,2}(\mathbb{R})$ is a weakly convergent sequence, we have $\sup_{n\in\mathbb{N}}\|u_{n}\|_{H^{1/2,2}(\mathbb{R})}\leq C$ for some $C>0$. We also have $u_{n}(x)\to u(x)$ a.e $x\in\mathbb{R}$ for a subsequence, thus $u$ is even, nonnegative and decreasing on $\mathbb{R}_{+}$. Now, we use the estimate below, which is referred to a Simple Radial Lemma: If $u\in L^{2}(\mathbb{R})$ is even, nonnegative and decreasing on $\mathbb{R}_{+}$, then it holds

[TABLE]

Thus $u_{n}^{2}(x)\leq\frac{C}{2|x|}$ for $x\neq 0$ by $\sup_{n\in\mathbb{N}}\|u_{n}\|_{H^{1/2,2}(\mathbb{R})}\leq C$ and $u^{2}(x)\leq\frac{C}{2|x|}$ by the pointwise convergence. Now, set $v_{n}=|u_{n}-u|^{q}$ for $q>2$. Then we see $v_{n}(x)\to 0$ a.e. $x\in\mathbb{R}$. Moreover,

[TABLE]

as $R\to\infty$ since $q>2$. Thus $\{v_{n}\}_{n\in\mathbb{N}}$ is uniformly integrable. Also by [19] Theorem 6.9, we know that

[TABLE]

For any $q>2$, take $q_{0}$ such that $2<q<q_{0}<\infty$. Since $u_{n}-u$ is uniformly bounded in $H^{1/2,2}(\mathbb{R})$, we have $\|u_{n}-u\|_{L^{q_{0}}(\mathbb{R})}\leq C$, and

[TABLE]

for any bounded measurable set $I\subset\mathbb{R}$. Therefore $\int_{I}v_{n}dx\to 0$ if $|I|\to 0$, which implies $\{v_{n}\}$ is uniformly absolutely continuous. Thus by Vitali’s Convergence Theorem (see for example, [7] p.187), we obtain $v_{n}=|u_{n}-u|^{q}\to 0$ strongly in $L^{1}(\mathbb{R})$, which is the desired conclusion. ∎

Proposition 4**.**

Assume $\{u_{n}\}\subset H^{1/2,2}(\mathbb{R})$ be a sequence with $\|(-\Delta)^{1/4}u_{n}\|_{L^{2}(\mathbb{R})}\leq 1$. Let $u_{n}\rightharpoonup u$ weakly in $H^{1/2,2}(\mathbb{R})$ for some $u$ and assume $u_{n}$ is even, nonnegative and decreasing on $\mathbb{R}_{+}$. Then we have

[TABLE]

for any $\alpha\in(0,\pi)$.

Proof.

The similar proposition above is already appeared, see [10] Lemma 3.1, and [5] Lemma 5.5. We prove it here for the reader’s convenience.

Put $\Phi_{\alpha}(t)=e^{\alpha t^{2}}-1$ and $\Psi_{\alpha}(t)=e^{\alpha t^{2}}-1-\alpha t^{2}$. Note that $\Phi_{\alpha}(t)$ is nonnegative, strictly convex and $\Psi_{\alpha}^{\prime}(t)=2\alpha t\Phi_{\alpha}(t)$. Thus by the mean value theorem, we have

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Thus we have

[TABLE]

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

First, direct calculation shows that

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

[TABLE]

here we used (3.4) for the third inequality and Theorem 1 for the last inequality, the use of which is valid since $\|(-\Delta)^{1/4}u_{n}\|_{L^{2}(\mathbb{R})}\leq 1$ and $\|(-\Delta)^{1/4}u\|_{L^{2}(\mathbb{R})}\leq 1$ by the weak lower semicontinuity. Note that $\{u_{n}\}$ satisfies $\sup_{n\in\mathbb{N}}\|u_{n}\|_{H^{1/2,2}(\mathbb{R})}\leq C$ for some $C>0$. Thus we have obtained $\|\Phi_{\alpha}(u_{n})+\Phi_{\alpha}(u)\|_{L^{b}(\mathbb{R})}=O(1)$ independent of $n$.

Next, we estimate the term $\||u_{n}|+|u|\|_{L^{a}(\mathbb{R})}$. Since $\{u_{n}\}$ is a bounded sequence in $H^{1/2,2}(\mathbb{R})$, we have by [19] Theorem 6.9 that $\|u\|_{L^{q}(\mathbb{R})}\leq C\|u_{n}\|_{H^{1/2,2}(\mathbb{R})}$ for any $q\geq 2$. Thus we see $\||u_{n}|+|u|\|_{L^{a}(\mathbb{R})}\leq C$ for some $C>0$ independent of $n$ for $a\geq 2$. Now, note that if we choose $1<b<\pi/\alpha$ and $a>2$ sufficiently large, then we can find $c>2$ such that $1/a+1/b+1/c=1$.

By these choices and Proposition 3, we conclude that $\|u_{n}-u\|_{L^{c}(\mathbb{R})}\to 0$ as $n\to\infty$. Backing to (3) with all together, we conclude that

[TABLE]

which is the desired conclusion. ∎

Now, we prove Theorem 4. We will show that $A(\alpha)$ in (1.1) is attained for any $0<\alpha<\pi$. Since $A(\alpha)=\tilde{A}(\alpha)$ by Lemma 1, we choose a maximizing sequence for $\tilde{A}(\alpha)$:

[TABLE]

Here $\{u_{n}\}_{n\in\mathbb{N}}\subset H^{1/2,2}(\mathbb{R})$ satisfies $\|(-\Delta)^{1/4}u_{n}\|_{L^{2}(\mathbb{R})}\leq 1$ and $\|u_{n}\|_{L^{2}(\mathbb{R})}=1$. By appealing to the use of rearrangement, we may furthermore assume that $u_{n}$ is nonnegative, even, and decreasing on $\mathbb{R}_{+}$. Since $\{u_{n}\}_{n\in\mathbb{N}}\subset H^{1/2,2}(\mathbb{R})$ is a bounded sequence, we have $u\in H^{1/2,2}(\mathbb{R})$ such that $u_{n}\rightharpoonup u$ in $H^{1/2,2}(\mathbb{R})$. By Proposition 4, we see

[TABLE]

as $n\to\infty$. Therefore, since $\|u_{n}\|_{L^{2}(\mathbb{R})}^{2}=1$, we have, letting $n\to\infty$,

[TABLE]

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

[TABLE]

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

[TABLE]

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

By the claim and (3.5), we conclude that the weak limit $u$ satisfies $u\not\equiv 0$. By the weak lower semi continuity, we have $u\not\equiv 0$ satisfies $\|(-\Delta)^{1/4}u\|_{L^{2}(\mathbb{R})}\leq 1$ and $\|u\|_{L^{2}(\mathbb{R})}\leq 1$. Thus by (3.5) again, we see

[TABLE]

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

Next, we prove Theorem 5. We follow Ishiwata’s argument in [9]. Let

[TABLE]

Actually, we will show a stronger claim that $J_{\alpha}$ has no critical point on $M$ for sufficiently small $\alpha>0$. Assume 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

[TABLE]

Note that $w_{1}=v$ thus it must be \frac{d}{d\tau}\Big{|}_{\tau=1}J_{\alpha}(w_{\tau})=0. By scaling rules (2.2), we see for any $p\geq 2$,

[TABLE]

Now, we see

[TABLE]

where $f_{j}(\tau)=\frac{\tau^{j-1}c}{(b+\tau a)^{j}}$ with $a=\|(-\Delta)^{1/4}v\|_{2}^{2}$, $b=\|v\|_{2}^{2}$ and $c=\|v\|_{2j}^{2j}$. Since

[TABLE]

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

[TABLE]

Here, we need the following lemma:

Lemma 3**.**

*(Ogawa-Ozawa [21])

There exists $C>0$ such that for any $u\in H^{1/2,2}(\mathbb{R})$ and $p\geq 2$, it holds*

[TABLE]

For $p=2j$, Lemma 3 implies

[TABLE]

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

[TABLE]

for some $C>0$ independent of $\alpha$. Therefore we have \frac{d}{d\tau}J_{\alpha}(w_{\tau})\Big{|}_{\tau=1}<0 for small $\alpha$, which is a desired contradiction. ∎

4. Proof of Theorem 6.

In order to prove Theorem 6, first we set

[TABLE]

for $\beta>0$. Then we have

Proposition 5**.**

We have $F(\beta)<\infty$ for $\beta<\pi$

Proof.

We follow the proof of Theorem 1.5 in [12]. Take any $u\in H^{1/2,2}(\mathbb{R})$ with $\|u\|_{H^{1/2,2}(\mathbb{R})}\leq 1$ in the admissible sets for $F(\beta)$ in (4.1). By appealing to the rearrangement, we may assume that $u$ is even, nonnegative and decreasing on $\mathbb{R}_{+}$. We divide the integral

[TABLE]

where $I=(-1/2,1/2)$.

First, we estimate $(I)$. By the Radial Lemma (3.2), we see for any $k\in\mathbb{N}$, $k\geq 2$,

[TABLE]

Thus

[TABLE]

Therefore, we have

[TABLE]

Now by the constraint $\|u\|_{H^{1/2,2}(\mathbb{R})}\leq 1$, we have $\|u\|_{L^{2}(\mathbb{R})}\leq 1$. Also if we put $a_{k}=\frac{\beta^{k-1}}{(k-1)(k-1)!}$, then $\sum_{k=2}^{\infty}a_{k}$ converges since $a_{k+1}/a_{k}=\beta\frac{k-1}{k^{2}}\to 0$ as $k\to\infty$. Thus we obtain

[TABLE]

where $C>0$ is independent of $u\in H^{1/2,2}(\mathbb{R})$ with $\|u\|_{H^{1/2,2}(\mathbb{R})}\leq 1$.

Next, we estimate $(II)$. Set

[TABLE]

Then by the argument of [12], we know that

[TABLE]

for $x\in I$. Put $w=v\sqrt{1+\|u\|_{L^{2}(\mathbb{R})}^{2}}$. Then we have $w\in\tilde{H}^{1/2,2}(I)$ since $v\equiv 0$ on $\mathbb{R}\setminus I$, and

[TABLE]

Thus we may use the fractional Trudinger-Moser inequality (Proposition 1) to $w$ to obtain

[TABLE]

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

[TABLE]

Now, since $\beta<\pi$, there is an absolute constant $C_{0}$ such that $t^{2}e^{\beta t^{2}}\leq C_{0}e^{\pi t^{2}}$ for any $t\in\mathbb{R}$. Finally, we obtain

[TABLE]

Proposition 5 follows from the estimates $(I)$ and $(II)$. ∎

By using Proposition 5 and arguing as in the proof of Theorem 1 (after establishing the similar claims as in Lemma 1 and Lemma 2), it is easy to obtain the following Proposition:

Proposition 6**.**

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

[TABLE]

Since $F(\beta)<\infty$ for any $\beta<\pi$, this proves the first part of Theorem 6. For the attainability of $E(\alpha)$ for $\alpha\in(0,\pi)$, it is enough to argue as in the proof of Theorem 4. We omit the details. ∎

Acknowledgments.

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