Extremals for fractional Moser-Trudinger inequalities in dimension 1 via harmonic extensions and commutator estimates

TL;DR

This paper establishes the existence of extremal functions for fractional Moser-Trudinger inequalities in one dimension using blow-up analysis, harmonic extensions, and commutator estimates, advancing understanding in fractional Sobolev spaces.

Contribution

It introduces new sharp commutator estimates and applies blow-up analysis to prove extremal existence for fractional Moser-Trudinger inequalities in 1D.

Findings

Existence of extremals in interval and on the real line.

Development of new commutator estimates for fractional operators.

Application of blow-up analysis to fractional inequalities.

Abstract

We prove the existence of extremals for fractional Moser-Trudinger inequalities in an interval and on the whole real line. In both cases we use blow-up analysis for the corresponding Euler-Lagrange equation, which requires new sharp estimates obtained via commutator techniques.

Equations

$$\underset{u\in W_0^{1,n}(\Omega ),\Vert \nabla u{\Vert }_{L^n(\Omega )}\le 1}{\sup }{\int }_{\Omega }{e}^{{\alpha }_n|u{|}^{\frac{n}{n-1}}}dx\le C|\Omega |,{\alpha }_n:=n{\omega }_{n-1}^{\frac{1}{n-1}},$$

$$\underset{u\in W^{1,n}({\mathbb{R}}^n),\Vert u{\Vert }_{L^n({\mathbb{R}}^n)}^n+\Vert \nabla u{\Vert }_{L^n({\mathbb{R}}^n)}^n\le 1}{\sup }{\int }_{{\mathbb{R}}^n}\left(e^{{\alpha }_n|u{|}^{\frac{n}{n-1}}}-1\right)dx<\infty .$$

$$\underset{u\in {\tilde{H}}^{\frac{1}{2},2}(I),\Vert (-\Delta {)}^{\frac{1}{4}}u{\Vert }_{L^2(I)}\le 1}{\sup }{\int }_I\left(e^{\alpha u^2}-1\right)dx=C_{\alpha }<\infty ,\text{for }\alpha \le \pi ,$$

$$\underset{u\in H^{\frac{1}{2},2}(\mathbb{R}),\Vert u{\Vert }_{H^{\frac{1}{2},2}(\mathbb{R})}\le 1}{\sup }{\int }_{\mathbb{R}}\left(e^{\alpha u^2}-1\right)dx=D_{\alpha }<\infty ,\text{for }\alpha \le \pi ,$$

$${\int }_I\left(e^{\alpha u_{\alpha }^2}-1\right)dx=C_{\alpha }.$$

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

$$(-\Delta {)}^{\frac{1}{2}}{u}_k+u_k={\lambda }_k{u}_k{e}^{{\alpha }_k{u}_k^2}\text{in }\mathbb{R},$$

$$\Lambda :=\underset{k\to \infty }{\mathrm{limsup}}\Vert u_k{\Vert }_H^2<\infty .$$

$$(-\Delta {)}^{\frac{1}{2}}{u}_{\infty }+u_{\infty }={\lambda }_{\infty }{u}_{\infty }{e}^{\pi u_{\infty }^2}\text{in }\mathbb{R},$$

$${\lambda }_k{r}_k{\mu }_k^2{e}^{{\alpha }_k{\mu }_k^2}=\frac{1}{{\alpha }_k},$$

$${\eta }_k(x):=2{\alpha }_k{\mu }_k(u_k(r_{k}x)-{\mu }_k),{\eta }_{\infty }(x):=-\log \left(1+|x{|}^2\right),$$

$$(-\Delta {)}^{\frac{1}{2}}{u}_k={\lambda }_k{u}_k{e}^{{\alpha }_k{u}_k^2},$$

$$\underset{k\to \infty }{\lim }{\int }_I\left(e^{{\alpha }_k{u}_k^2}-1\right)dx=\underset{k\to \infty }{\lim }{C}_{{\alpha }_k}=C_{\pi },$$

$$\underset{k\to \infty }{\lim }{\mu }_k=\infty .$$

$$0<{\lambda }_k<{\lambda }_1(I).$$

$$(-\Delta {)}^{\frac{1}{2}}{u}_{\infty }={\lambda }_{\infty }{u}_{\infty }{e}^{\pi u_{\infty }^2}.$$

$${\lambda }_1(I){\int }_I{u}_k{\phi }_{1}dx={\lambda }_k{\int }_I{u}_k{e}^{{\alpha }_k{u}_k^2}{\phi }_{1}dx>{\lambda }_k{\int }_I{u}_k{\phi }_{1}dx,$$

$$\begin{array}{rl}{\int }_I{u}_{\infty }(-\Delta {)}^{\frac{1}{2}}\phi dx& =\underset{k\to \infty }{\lim }{\int }_I{u}_k(-\Delta {)}^{\frac{1}{2}}\phi dx\\ & =\underset{k\to \infty }{\lim }{\int }_I{\lambda }_k{u}_k{e}^{{\alpha }_k{u}_k^2}\phi dx\\ & ={\int }_I{\lambda }_{\infty }{u}_{\infty }{e}^{\pi u_{\infty }^2}\phi dx,\end{array}$$

$$\begin{array}{rl}{\int }_{I_2}{\lambda }_k{u}_k{e}^{{\alpha }_k{u}_k^2}\phi dx& \le \frac{{\sup }_I|\phi |}{L}{\int }_I{\lambda }_k{u}_k^2{e}^{{\alpha }_k{u}_k^2}dx\\ & =\frac{{\sup }_I|\phi |}{L}{\int }_I{u}_k(-\Delta {)}^{\frac{1}{2}}{u}_{k}dx\\ & =\frac{{\sup }_I|\phi |}{L}\Vert (-\Delta {)}^{\frac{1}{4}}{u}_k{\Vert }_{L^2(\mathbb{R})}^2,\end{array}$$

$${\int }_I{\lambda }_k{u}_k^2{e}^{{\alpha }_k{u}_k^2}dx=\Vert (-\Delta {)}^{\frac{1}{4}}{u}_k{\Vert }_{L^2(\mathbb{R})}^2=\Vert \nabla {\tilde{u}}_k{\Vert }_{L^2({\mathbb{R}}_{+}^2)}^2=1.$$

$${\tilde{\eta }}_k(x,y):=2{\alpha }_k{\mu }_k({\tilde{u}}_k(r_{k}x,r_{k}y)-{\mu }_k).$$

$${\tilde{\eta }}_{\infty }(x,y)=-\log \left((1+y{)}^2+x^2\right)$$

$$(-\Delta {)}^{\frac{1}{2}}{\eta }_{\infty }=2e^{{\eta }_{\infty }},{\int }_{\mathbb{R}}{e}^{{\eta }_{\infty }}dx=\pi .$$

$$\underset{k\to \infty }{\lim }{\int }_{-R{r}_k}^{R{r}_k}{\lambda }_k{\mu }_k^i{u}_k^{2-i}{e}^{{\alpha }_k{u}_k^2}dx=\frac{1}{\pi }{\int }_{-R}^R{e}^{{\eta }_{\infty }}dx.$$

$${\int }_{-R{r}_k}^{R{r}_k}{\lambda }_k{\mu }_k^i{u}_k^{2-i}{e}^{{\alpha }_k{u}_k^2}dx=\underset{=\frac{1}{{\alpha }_k}}{\underbrace{r_k{\lambda }_k{\mu }_k^2{e}^{{\alpha }_k{\mu }_k^2}}}{\int }_{-R}^R{\left(1+\frac{{\eta }_k}{2{\alpha }_k{\mu }_k^2}\right)}^{2-i}{e}^{{\eta }_k+\frac{{\eta }_k}{4{\alpha }_k{\mu }_k^2}}d\xi \to \frac{1}{\pi }{\int }_{-R}^R{e}^{{\eta }_{\infty }}d\xi ,$$

$$1={\int }_{-R{r}_k}^{R{r}_k}{\lambda }_k{u}_k^2{e}^{{\alpha }_k{u}_k^2}dx+{\int }_{I\setminus (-R{r}_k,R{r}_k)}{\lambda }_k{u}_k^2{e}^{{\alpha }_k{u}_k^2}dx=:(I{)}_k+(II{)}_k.$$

$$\begin{array}{r}\underset{k\to \infty }{\lim }(I{)}_k=\frac{1}{\pi }{\int }_{-R}^R{e}^{{\eta }_{\infty }}dx=1+o(1),\end{array}$$

$$\underset{R\to \infty }{\lim }\underset{k\to \infty }{\lim }(II{)}_k=0,$$

$$\underset{k\to \infty }{\mathrm{limsup}}\Vert (-\Delta {)}^{\frac{1}{4}}{u}_k^A{\Vert }_{L^2(\mathbb{R})}^2\le \frac{1}{A}.$$

$$\Vert (-\Delta {)}^{\frac{1}{4}}{u}_k^A{\Vert }_{L^2(\mathbb{R})}^2\le {\int }_{{\mathbb{R}}_{+}^2}|\nabla {\bar{u}}_k^A{|}^{2}dxdy.$$

Full text

1 Introduction

The celebrated Moser-Trudinger inequality [28] states that for $\Omega\subset\mathbb{R}^{n}$ with finite measure $|\Omega|$ we have

[TABLE]

where $\omega_{n-1}$ is the volume of the unit sphere in $\mathbb{R}^{n}$. The constant $\alpha_{n}$ is sharp in the sense that the supremum in (1) becomes infinite if $\alpha_{n}$ is replaced by any $\alpha>\alpha_{n}$. In the case $\Omega=\mathbb{R}^{2}$, B. Ruf [34] proved a similar inequality, using the full $W^{1,2}$-norm instead of the $L^{2}$-norm of the gradient, then generalized to $\mathbb{R}^{n}$, $n\geq 2$ by Li-Ruf [20] as

[TABLE]

Higher-order versions of (1) were proven by Adams [2] on the space $W^{k,\frac{n}{k}}(\Omega)$ for $n>k\in\mathbb{N}$.

In [17] the authors proved the following $1$-dimensional fractional extension of the previous results (for the definition of $H^{\frac{1}{2},2}(\mathbb{R})$ and $(-\Delta)^{\frac{1}{4}}$ see (65) in the Appendix).

Theorem A

Set $I:=(-1,1)\subset\mathbb{R}$ and $\tilde{H}^{\frac{1}{2},2}(I):=\{u\in H^{\frac{1}{2},2}(\mathbb{R}):u\equiv 0\text{ on }\mathbb{R}\setminus I\}$. Then we have

[TABLE]

and

[TABLE]

where $\|u\|_{H^{\frac{1}{2},2}(\mathbb{R}{})}^{2}:=\|(-\Delta)^{\frac{1}{4}}u\|_{L^{2}{(\mathbb{R})}}^{2}+\|u\|_{L^{2}(\mathbb{R})}^{2}$. The constant $\pi$ is sharp in (3) and (4).

More general results have recently appeared, see e.g. [1, 12, 18, 27, 35, 38], in which both the dimension and the (fractional) order of differentiability have been generalized. For instance, (3) and (4) can be seen as $1$-dimensional cases of the more general results of [18, 27, 12] that hold in arbitrary dimension $n$.

The existence of extremals for this kind of inequalities is a challenging question. Existence of extremals for (1) was originally proven by L. Carleson and A. Chang [5] in the case of the unit ball, a fundamental result later extended by Struwe [37] and Flucher [11] to the case of general bounded domains in $\mathbb{R}^{2}$ and by K. Lin [22] to the case of bounded domains in $\mathbb{R}^{n}$. In the case of the Li-Ruf inequality (2), the existence of extremals appears in [20] when $n\geq 3$ and was proven by Ishiwata [16] when $n=2$. For the higher-order Adams inequality the existence of extremals has been proven in various cases, e.g. by Li-Ndiaye [21] on a $4$ dimensional closed manyfold, by Lu-Yang [23] for a $4$ dimensional bounded domain and by DelaTorre-Mancini [7] for a bounded domain in $\mathbb{R}^{2m}$, $m\geq 1$ arbitrary.

On the other hand, the existence of extremals for the fractional Moser-Trudinger inequality has remained open until now, with the exception of Takahashi [38] considering a subcritical version of (4) of Adachi-Tanaka type [1], and Li-Liu [19] treating the case of a fractional Moser-Trudinger on $H^{\frac{1}{2},2}(\partial M)$ with $M$ a compact Riemann surface with boundary. The idea of Li and Liu is that working on the boundary of a compact manifold, one can localize the $H^{\frac{1}{2},2}$-norm.

Applying the same method for an interval $I\subset\mathbb{R}$ creates problems near $\partial I$, which require additional care in the estimate, and the problem becomes even more challenging when working on the whole $\mathbb{R}$. The main purpose of this paper is to handle these two cases and prove that the suprema in (3) and (4) are attained.

Theorem 1.1

For any $0<\alpha\leq\pi$, the inequality (3) has an extremal i.e. there exists $u_{\alpha}\in\tilde{H}^{\frac{1}{2},2}(I)$ such that $\|(-\Delta)^{\frac{1}{4}}u_{\alpha}\|_{L^{2}(\mathbb{R})}\leq 1$ and

[TABLE]

Theorem 1.1 is rather simple to prove for $\alpha\in(0,\pi)$, while the case $\alpha=\pi$ relies on a delicate blow-up analysis for subcritical extremals.

A similar analysis can be carried out for the Ruf-type inequality (4). However, working on the whole real line we need to face additional difficulties due to the lack of compactness of the embedding of $H=H^{\frac{1}{2},2}(\mathbb{R})$ into $L^{2}(\mathbb{R})$: vanishing at infinity might occur for maximizing sequences, even in the sub-critical case $\alpha\in(0,\pi)$. This issue is not merely technical indeed Takahashi [38] proved that (4) has no extremal when $\alpha$ is small enough. Here, in analogy with the results in dimension $n\geq 2$, we prove that the supremum in (4) is attained if $\alpha$ sufficiently close to $\pi$.

Theorem 1.2

There exists $\alpha^{*}\in(0,\pi)$ such that for $\alpha^{*}\leq\alpha\leq\pi$ the inequality (4) has an extremal, namely, there exists $\bar{u}_{\alpha}\in H^{\frac{1}{2},2}(\mathbb{R})$ such that $\|\bar{u}_{\alpha}\|_{H^{\frac{1}{2},2}(\mathbb{R})}\leq 1$ and

[TABLE]

As for Theorem 1.1, the proof of Theorem 1.2 for $\alpha=\pi$ is based on blow-up analysis. In fact we need to study the blow-up of a non-local equation on the whole real line (no boundary conditions), as done in the following theorem.

Theorem 1.3

Let $(u_{k})\subset H=H^{\frac{1}{2},2}(\mathbb{R})$ be a sequence of non-negative solutions to

[TABLE]

where $\alpha_{k}\to\pi$ and $\lambda_{k}\to\lambda_{\infty}\geq 0$. Assume $u_{k}$ even and decreasing ($u_{k}(-x)=u_{k}(x)\leq u_{k}(y)$ for $x\geq y\geq 0$) for every $k$ and set $\mu_{k}:=\sup_{\mathbb{R}}u_{k}=u_{k}(0)$. Assume also that

[TABLE]

Then up to extracting a subsequence we have that either

• (i)

$\mu_{k}\leq C$, $u_{k}\to u_{\infty}$ in $C^{\ell}_{\operatorname{loc}}(\mathbb{R})$ for every $\ell\geq 0$, where $u_{\infty}\in C^{\ell}_{\operatorname{loc}}(\mathbb{R})\cap H$ solves

[TABLE]

or

• (ii)

$\mu_{k}\to\infty$, $u_{k}\to u_{\infty}$ weakly in $H$ and strongly in $C^{0}_{\operatorname{loc}}(\bar{\mathbb{R}}\setminus\{0\})$ where $u_{\infty}$ is a solution to (7). Moreover, setting $r_{k}$ such that

[TABLE]

and

[TABLE]

one has $\eta_{k}\to\eta_{\infty}$ in $C^{\ell}_{\operatorname{loc}}(\mathbb{R}{})$ for every $\ell\geq 0$, $\sup_{k}\|\eta_{k}\|_{L_{s}(\mathbb{R})}<\infty$ for any $s>0$ (cfr. (63)), and $\Lambda\geq\|u_{\infty}\|_{H}^{2}+1$.

The proof of Theorem 1.3 is quite delicate because local elliptic estimates of a nonlocal equation depend on global bounds as we shall prove in Lemma 3.6. This will be based on sharp commutator estimates (Lemma 3.3), as developed in [24] for the case of a bounded domain in $\mathbb{R}^{n}$, extending to the fractional case the approach of [26].

We expect similar existence results to hold for a perturbed version of inequalities (3)-(4), as in [25] and [39] (see also the recent results in [15]), but we will not investigate this issue here.

2 Proof of Theorem 1.1

2.1 Strategy of the proof

We will focus on the case $\alpha=\pi$, since the existence of extremals for (3) with $\alpha\in(0,\pi)$ follows easily by Vitali’s convergence theorem, see e.g. the argument in [25, Proposition 6].

Let $u_{k}$ be an extremal of (3) for $\alpha=\alpha_{k}=\pi-\frac{1}{k}$. By replacing $u_{k}$ with $|u_{k}|$ we can assume that $u_{k}\geq 0$. Moreover $\|(-\Delta)^{\frac{1}{4}}u_{k}\|_{L^{2}(\mathbb{R})}=1$, and $u_{k}$ satisfies the Euler-Lagrange equation

[TABLE]

with bounds on the Lagrange multipliers $\lambda_{k}$ (see (13)).

Using the monotone convergence theorem we also get

[TABLE]

where $C_{\alpha_{k}}$ and $C_{\pi}$ are as in (3).

If $\mu_{k}:=\max_{I}u_{k}=O(1)$ as $k\to\infty$, then up to a subsequence $u_{k}\to u_{\infty}$ locally uniformly, where by (11) $u_{\infty}$ maximizes (3) with $\alpha=\pi$. Therefore we will work by contradiction, assuming

[TABLE]

By studying the blow-up behavior of $u_{k}$, see in particular Propositions 2.2 and 2.8, we will show that (12) implies $C_{\pi}\leq 4\pi$ (Proposition 2.9), but with suitable test functions we will also prove that $C_{\pi}>4\pi$ (Proposition 2.10), hence contradicting (12) and completing the proof of Theorem 1.1.

2.2 The blow-up analysis

The following proposition is well known in the local case, and its proof in the present setting is similar to the local one. We give it for completeness.

Proposition 2.1

We have $u_{k}\in C^{\infty}(I)\cap C^{0,\frac{1}{2}}(\bar{I})$, $u_{k}>0$ in $I$, and $u_{k}$ is symmetric with respect to [math] and decreasing with respect to $|x|$. Moreover,

[TABLE]

Up to a subsequence we have $\lambda_{k}\to\lambda_{\infty}$ and $u_{k}\to u_{\infty}$ weakly in $\tilde{H}^{\frac{1}{2},2}(I)$ and strongly in $L^{2}(I)$, where $u_{\infty}$ solves

[TABLE]

Proof.

For the first claim see Remark 1.4 in [24]. The positivity follows from the maximum principle, and symmetry and monotonicity follow from the moving point technique, see e.g. [8, Theorem 11].

Now testing (10) with $\varphi_{1}$, the first eigenfunction of $(-\Delta)^{\frac{1}{2}}$ in $\tilde{H}^{\frac{1}{2},2}(I)$, positive and with eigenvalue $\lambda_{1}(I)>0$, we obtain

[TABLE]

hence proving (13). By the theorem of Banach-Alaoglu and the compactness of the Sobolev embedding of $\tilde{H}^{\frac{1}{2},2}(I)\hookrightarrow L^{2}(I)$, we obtain the claimed convergence of $u_{k}$ to $u_{\infty}$. Finally, to show that $u_{\infty}$ solves (14), test with $\varphi\in C^{\infty}_{c}(I)$:

[TABLE]

where the convergence of the last integral is justified by splitting $I$ into $I_{1}:=\{x\in I:u_{k}(x)\leq L\}$ and $I_{2}:=\{x\in I:u_{k}(x)>L\}$, applying the dominated convergence on $I_{1}$ and bounding

[TABLE]

and letting $L\to\infty$. ∎

Let $\tilde{u}_{k}$ be the harmonic extension of $u_{k}$ to $\mathbb{R}^{2}_{+}$ given by the Poisson integral, see (66) in the appendix. Notice that

[TABLE]

Let $r_{k}=\frac{1}{\alpha_{k}\lambda_{k}\mu_{k}^{2}e^{\alpha_{k}\mu_{k}^{2}}}$ and $\eta_{k}(x):=2\alpha_{k}\mu_{k}(u_{k}(r_{k}x)-\mu_{k})$ be as in (8) and (9), and set

[TABLE]

Note that $\tilde{\eta}_{k}$ is the Poisson integral of $\eta_{k}$.

Proposition 2.2

We have $r_{k}\to 0$ and $\tilde{\eta}_{k}\to\tilde{\eta}_{\infty}$ in $C^{\ell}_{\operatorname{loc}}(\overline{\mathbb{R}^{2}_{+}})$ for every $\ell\geq 0$, where

[TABLE]

is the Poisson integral (compare to (66)) of $\eta_{\infty}:=-\log\left({1+x^{2}}\right)$, and

[TABLE]

Proof.

According to Lemma 2.2, Theorem 1.5 and Proposition 2.7 in [24], we have $r_{k}\to 0$, $\eta_{k}\to\eta_{\infty}$ in $C^{\ell}_{\operatorname{loc}}(\mathbb{R})$ for every $\ell\geq 0$ and $(\eta_{k})$ is uniformly bounded in $L_{\frac{1}{2}}(\mathbb{R})$ (see (63)).

To obtain the local convergence of $\tilde{\eta}_{k}$, fix $R>0$ and split the integral in the Poisson integral (66) of $\tilde{\eta}_{k}$ into an integral over $(-R,R)$ and an integral over $\mathbb{R}\setminus(-R,R)$, for $R$ large. The former is bounded by the convergence of $\eta_{k}$ locally, the latter by the boundedness of $\eta_{k}$ in $L_{\frac{1}{2}}(\mathbb{R})$, provided $(x,y)\in B_{\frac{R}{2}}\cap\mathbb{R}_{+}^{2}$. As a consequence we get that $\tilde{\eta}_{k}$ is locally uniformly bounded in $\overline{\mathbb{R}^{2}_{+}}$. Since $\tilde{\eta}_{k}$ is harmonic, we conclude by elliptic estimates. ∎

Corollary 2.3

For $R>0$ and $i=0,1,2$, we have

[TABLE]

Moreover, $u_{\infty}\equiv 0$, i.e. up to a subsequence $u_{k}\to 0$ in $L^{2}(I)$, weakly in $\tilde{H}^{\frac{1}{2},2}(I)$, and a.e in $I$.

Proof.

With the change of variables $\xi=\frac{x}{r_{k}}$, writing $u_{k}(r_{k}\cdot)=\mu_{k}+\frac{\eta_{k}}{2\alpha_{k}\mu_{k}}$ and using (8) and Proposition 2.2, we see that

[TABLE]

as $k\to\infty$, as claimed in (17).

In order to prove the last statement, recalling that $\|(-\Delta)^{\frac{1}{4}}u_{k}\|_{L^{2}}=1$, we write

[TABLE]

By (17) and (16) we get

[TABLE]

with $o(1)\to 0$ as $R\to\infty$. This in turn implies that

[TABLE]

which is possible only if $u_{\infty}\equiv 0$, or $\lambda_{\infty}=0$ (by Fatou’s lemma). But on account of (14), also in the latter case we have $u_{\infty}\equiv 0$. ∎

Lemma 2.4

For $A>1$, set $u_{k}^{A}:=\min\left\{u_{k},\frac{\mu_{k}}{A}\right\}$. Then we have

[TABLE]

Proof.

We set $\bar{u}_{k}^{A}:=\min\left\{\tilde{u}_{k},\frac{\mu_{k}}{A}\right\}$. Since $\bar{u}_{k}^{A}$ is an extension (in general not harmonic) of $u_{k}^{A}$, we have

[TABLE]

Using integration by parts and the harmonicity of $\tilde{u}_{k}$ we get

[TABLE]

Proposition 2.2 implies that $u_{k}^{A}(r_{k}x)=\frac{\mu_{k}}{A}$ for $|x|\leq R$ and $k\geq k_{0}(R)$. Then, with (16) and (17) we obtain

[TABLE]

Set now $v_{k}^{A}:=\left(u_{k}-\frac{\mu_{k}}{A}\right)^{+}=u_{k}-u_{k}^{A}$. With similar computations we get

[TABLE]

Since

[TABLE]

we get that

[TABLE]

Then, we conclude using (19), and (20). ∎

Proposition 2.5

We have

[TABLE]

Moreover

[TABLE]

Proof.

Fix $A>1$ and let $u_{k}^{A}$ be defined as in Lemma 2.4. We split

[TABLE]

Using Corollary 2.3 and Vitali’s theorem, we see that

[TABLE]

since $e^{\alpha_{k}(u_{k}^{A})^{2}}$ is uniformly bounded in $L^{A}(I)$ by Lemma 2.4 together with Theorem A.

By (15) and Corollary 2.3, we now estimate

[TABLE]

with $o(1)\to 0$ as $k\to\infty$.Together with (11), and letting $A\downarrow 1$, this gives

[TABLE]

The converse inequality follows from (17) as follows:

[TABLE]

with $o(1)\to 0$ as $k\to\infty$. Letting $R\to\infty$ and recalling (16) we obtain (21).

Finally, (22) follows at once from (21), because otherwise we would have $C_{\pi}=0$, which is clearly impossible. ∎

Proposition 2.6

Let us set $f_{k}:=\lambda_{k}\mu_{k}u_{k}e^{\alpha_{k}u_{k}^{2}}$. Then we have

[TABLE]

as $k\to\infty$, for any $\varphi\in C(\bar{I})$. In particular, $f_{k}\rightharpoonup\delta_{0}$ in the sense of Radon measures in $I$.

Proof.

Take $\varphi\in{C}(\bar{I})$. For given $R>0$, $A>1$, we split

[TABLE]

On $\{u_{k}\leq\frac{\mu_{k}}{A}\}$ we have $u_{k}=u_{k}^{A}$ and Lemma 2.4 and Theorem A imply that $u_{k}e^{\alpha_{k}u_{k}^{2}}$ is uniformly bounded in $L^{1}$ (depending on $A$). Thus using (22) we get $I_{3}\to 0$.

With (15) and (17) we also get

[TABLE]

with $o(1)\to 0$ as $k\to\infty$. Thanks to (16), we conclude that $I_{2}\to 0$ as $k\to\infty$ and $R\to\infty$.

As for $I_{1}$, again with (17) we compute

[TABLE]

so that $I_{1}\to\varphi(0)$ as $k\to\infty$ and $R\to\infty$. ∎

Given $x\in I$, let $G_{x}:\mathbb{R}\setminus\{0\}\to\mathbb{R}$ be the Green’s function of $(-\Delta)^{\frac{1}{2}}$ on $I$ with singularity at $x$. We recall that we have the explicit formula (see e.g. [3])

[TABLE]

In the following we further denote

[TABLE]

Lemma 2.7

We have $\mu_{k}u_{k}\rightarrow G:=G_{0}$ in $L^{\infty}_{\operatorname{loc}}(\overline{I}\setminus\{0\})\cap L^{1}(I)$ as $k\to+\infty$.

Proof.

Let us set $v_{k}:=\mu_{k}u_{k}-G$ and $f_{k}=\mu_{k}\lambda_{k}u_{k}e^{\alpha_{k}u_{k}^{2}}$. Arguing as in Proposition 2.6, we show that $\|f_{k}\|_{L^{1}(\mathbb{R})}\to 1$ as $k\to\infty$. Moreover, since $u_{k}$ is decreasing with respect to $|x|$, we get that $u_{k}\to 0$ and $f_{k}\to 0$ locally uniformly in $\overline{I}\setminus\{0\}$ as $k\to\infty$. By Green’s representation formula, we have

[TABLE]

Fix $\sigma\in(0,1)$. If we assume $|x|\geq\sigma$, $|y|\leq\frac{\sigma}{2}$, then we have

[TABLE]

where $C$ is a constant depending only on $\sigma$. Then, for any $\varepsilon\in(0,\frac{\sigma}{2})$, we can write

[TABLE]

where $o(1)\to 0$ uniformly in $I\setminus(-\sigma,\sigma)$ as $k\to\infty$. Clearly, (27) implies

[TABLE]

Since $\varepsilon$ and $\sigma$ can be arbitrarily small, this shows that $v_{k}\to 0$ in $L^{\infty}_{\operatorname{loc}}(\overline{I}\setminus\{0\})$. With a similar argument, we prove the $L^{1}$ convergence. Indeed, integrating (25), for $\varepsilon\in(0,1)$ we get

[TABLE]

Since

[TABLE]

we get

[TABLE]

Moreover, using the change of variables $x=yz$, we obtain

[TABLE]

Then, we have

[TABLE]

Clearly (28) and (29) yield $\limsup_{k\to+\infty}\|v_{k}-G\|_{L^{1}(I)}=O(\varepsilon\log\frac{1}{\varepsilon})$. Since $\varepsilon$ can be arbitrarily small we get the conclusion. ∎

Proposition 2.8

We have $\mu_{k}\tilde{u}_{k}\rightarrow\tilde{G}$ in $C^{0}_{\operatorname{loc}}(\overline{\mathbb{R}^{2}_{+}}\setminus\{(0,0)\})\cap C^{1}_{\operatorname{loc}}(\mathbb{R}^{2}_{+})$, where $\tilde{G}$ is the Poisson extension of $G$.

Proof.

As in the proof of Lemma 2.7 we denote $v_{k}:=\mu_{k}u_{k}-G$. Let us consider the Poisson extension $\widetilde{v}_{k}=\mu_{k}\widetilde{u}_{k}-\widetilde{G}$. For any fixed $\varepsilon>0$, we can split

[TABLE]

By Lemma 2.7, we have

[TABLE]

as $k\to\infty$. Moreover, assuming $(x,y)\in\mathbb{R}^{2}_{+}\setminus B_{2\varepsilon}(0,0)$, we get

[TABLE]

Hence $\widetilde{v}_{k}\to 0$ in $C^{0}_{\operatorname{loc}}(\mathbb{R}^{2}_{+}\setminus B_{2\varepsilon}(0,0))$. Finally, since can $\varepsilon$ be arbitrarily small and $\tilde{v}_{k}$ is harmonic in $\mathbb{R}^{2}_{+}$, we get $\widetilde{v}_{k}\to 0$ in $C^{0}_{\operatorname{loc}}(\overline{\mathbb{R}^{2}_{+}}\setminus\{(0,0)\})\cap C^{1}_{\operatorname{loc}}(\mathbb{R}^{2}_{+})$. ∎

2.3 The two main estimates and completion of the proof

We shall now conclude our contradiction argument by showing the incompatibility of (12) with (11) and the the definition of $C_{\pi}$. In this final part of the proof, we will use the precise asymptotic of $\widetilde{G}$ near $(0,0)$. Since $\log|(x,y)|$ is the Poisson integral of $\log|x|$ (see Proposition A.3), and since $S\in C(\mathbb{R})$, (24) guarantees the existence of the limit

[TABLE]

In fact, using (23) we get $S_{0}=\frac{\log 2}{\pi}$. More precisely, noting that $S\in C^{\infty}(I)$, we can write

[TABLE]

with $h\in C^{\infty}(\overline{\mathbb{R}^{2}_{+}}\cap B_{1}(0,0))\cap C(\overline{\mathbb{R}^{2}_{+}})$ and $h(0,0)=0$.

Proposition 2.9

*If (12) holds, then $C_{\pi}\leq 2\pi e^{\pi S_{0}}=4\pi.$ *

Proof.

For a fixed large $L>0$ and a fixed and small $\delta>0$ set

[TABLE]

Recalling that $\|\nabla\tilde{u}_{k}\|_{L^{2}}^{2}=1$, we have

[TABLE]

Clearly the left-hand side bounds

[TABLE]

where the function $\tilde{\Phi}_{k}$ is the unique solution to

[TABLE]

given explicitly by

[TABLE]

Using Proposition 2.2 we obtain

[TABLE]

where for fixed $L>0$ we have $o(1)\to 0$ as $k\to\infty$, and $|O(L^{-1})|\leq\frac{C}{L}$ uniformly for $L$ and $k$ large. Moreover, using Proposition 2.8 and (30), we obtain

[TABLE]

where for fixed $\delta>0$ we have $o(1)\to 0$ as $k\to\infty$, and $|O(\delta)|\leq C\delta$ uniformly for $\delta$ small and $k$ large.

Still with Proposition 2.2 we get

[TABLE]

Similarly with Proposition 2.8 we get

[TABLE]

where we used the expansion in (30) and the boundary conditions

[TABLE]

We then get

[TABLE]

or

[TABLE]

Rearranging gives

[TABLE]

with $o(1)\to 0$ as $k\to\infty$. Then, recalling that $\alpha_{k}\uparrow\pi$, letting $k\to\infty$ first and then $L\to\infty$, $\delta\to 0$, we obtain

[TABLE]

and using Proposition 2.5 we conclude. ∎

Proposition 2.10

There exists a function $u\in\tilde{H}^{\frac{1}{2},2}(I)$ with $\|(-\Delta)^{\frac{1}{4}}u\|_{L^{2}(\mathbb{R})}\leq 1$ such that

[TABLE]

Proof.

For $\varepsilon>0$ choose $L=L(\varepsilon)>0$ such that as $\varepsilon\to 0$ we have $L\to\infty$ and $L\varepsilon\to 0$. Fix

[TABLE]

and

[TABLE]

By the maximum principle we have $\mathbb{R}^{2}_{+}\cap B_{L\varepsilon}\subset\Omega_{L\varepsilon}$. Indeed, $\tilde{G}$ is harmonic in $\mathbb{R}^{2}_{+}$, $\tilde{G}\geq\gamma_{L\varepsilon}$ on $\partial(\mathbb{R}^{2}_{+}\cap B_{L\varepsilon})\setminus\{(0,0)\}$, and $\tilde{G}\to+\infty$ as $(x,y)\to(0,0)$. Notice also that (30) gives

[TABLE]

For some constants $B$ and $c$ to be fixed we set

[TABLE]

Observe that $\mathbb{R}^{2}_{+}\cap B_{L\varepsilon}(0,-\varepsilon)\subseteq\mathbb{R}^{2}_{+}\cap B_{L\varepsilon}\subseteq\Omega_{L\varepsilon}$. To have continuity on $\mathbb{R}^{2}_{+}\cap\partial B_{L\varepsilon}(0,-\varepsilon)$ we impose

[TABLE]

which, together with (32), gives the relation

[TABLE]

Moreover

[TABLE]

and

[TABLE]

where the last equality follows from (30). We now impose $\|\nabla U_{\varepsilon}\|_{L^{2}(\mathbb{R}^{2}_{+})}=1$, obtaining

[TABLE]

which, together with (33), implies

[TABLE]

Let now $I_{L,\varepsilon}^{1}=(-\varepsilon\sqrt{L^{2}-1},\varepsilon\sqrt{L^{2}-1})$ and $I_{L\varepsilon}^{2}$ be the disjoint sub-intervals of $I$ obtained by intersecting $I\times\{0\}$ respectively with $B_{L\varepsilon}(0,-\varepsilon)$ and $\overline{\mathbb{R}^{2}_{+}\setminus\Omega_{L\varepsilon}}$. Then, for $u_{\varepsilon}(x):=U_{\varepsilon}(x,0)$, using a change of variables and (34)-(35) we get

[TABLE]

Moreover

[TABLE]

with

[TABLE]

Now observe that $c^{2}=-\frac{\log\varepsilon}{\pi}+O(1)$ by (34), and choose $L=\log^{2}\varepsilon$ to obtain

[TABLE]

so that

[TABLE]

for $\varepsilon$ small enough.

Finally notice that

[TABLE]

since the Poisson extension $\tilde{u}_{\varepsilon}$ minimizes the Dirichlet energy among extensions with finite energy. ∎

3 Proof of Theorem 1.3

Let $u_{k}\in H\cap C^{\infty}(\mathbb{R})$ be a sequence of positive even and decreasing solutions to (5) satisfying the energy bound (6) and with $\lambda_{k}\to\lambda_{\infty}\geq 0$ as $k\to\infty$.

First we show that case (i) holds when $\mu_{k}\leq C$.

Lemma 3.1

If $\mu_{k}\leq C$ then (i) holds.

Proof.

By assumption we know that $u_{k}$ and $f_{k}:=(-\Delta)^{\frac{1}{2}}u_{k}=\lambda_{k}u_{k}e^{\alpha_{k}u_{k}^{2}}-u_{k}$ are uniformly bounded in $L^{\infty}(\mathbb{R})$. Then, by elliptic estimates and a bootstrap argument, we can find $u_{\infty}\in C^{\infty}(\mathbb{R})$ such that up to a subsequence $u_{k}\to u_{\infty}$ in $C^{\ell}_{\operatorname{loc}}(\mathbb{R})$ for every $\ell\geq 0$. To prove that $u_{\infty}$ satisfies (7), note that $f_{k}\to f_{\infty}:=\lambda_{\infty}u_{\infty}e^{\pi u_{\infty}^{2}}-u_{\infty}$ locally uniformly on $\mathbb{R}$ and set $M=\sup_{k}(\|f_{k}\|_{L^{\infty}(\mathbb{R})}+\mu_{k})$. For any $\varphi\in\mathcal{S}(\mathbb{R})$ (the Schwarz space of rapidly decreasing functions) and any $R>0$, we have that

[TABLE]

Similarly, recalling that $(-\Delta)^{\frac{1}{2}}\varphi$ has quadratic decay at infinity (see e.g. [14, Prop. 2.1]), we get

[TABLE]

Hence $u$ is a weak solution of (7). ∎

From now on we will assume that $\mu_{k}\to+\infty$ and prove that (ii) of Theorem 1.3 holds.

Lemma 3.2

Let $\eta_{k}$ be defined as in Theorem 1.3. Then $\eta_{k}$ is bounded in $C^{0,\alpha}_{\operatorname{loc}}(\mathbb{R})$ for $\alpha\in(0,1)$.

Proof.

Note that

[TABLE]

Moreover we have that

[TABLE]

is bounded in $L^{\infty}$. Since $\eta_{k}\leq 0$, and $\eta_{k}(0)=0$ this implies that $\eta_{k}$ is bounded in $L^{\infty}_{\operatorname{loc}}(\mathbb{R})$ and then in $C^{\alpha}_{\operatorname{loc}}(\mathbb{R})$ for any $\alpha\in(0,1)$. ∎

The bound of Lemma 3.2 implies that, up to a subsequence $\eta_{k}\to\eta_{\infty}$ in $C^{0,\alpha}_{\operatorname{loc}}(\mathbb{R})$ for some function $\eta_{\infty}$. However, it does not provide a limit equation for $\eta_{\infty}$. In order to prove that $\eta_{\infty}$ solves

[TABLE]

we will prove that that $\eta_{k}$ is bounded in $L_{s}(\mathbb{R})$ for any $s>0$. This bound can be obtained thanks to the commutator estimates proved in [24]. Part of the argument must be modified since the ${u_{k}}^{\prime}s$ are not compactly supported. We start by recalling the following technical lemma, which is a consequence of the estimates in [24].

Lemma 3.3

For any $s\in(0,1)$, there exists a constant $C=C(s)$ such that, for any $\varphi,\psi\in C^{\infty}_{c}(\mathbb{R}^{n})$, $\rho\in\mathbb{R}^{+}$, we have

[TABLE]

where

[TABLE]

Proof.

Let $\theta\in C^{\infty}_{c}((-2,2))$ be a cut-off function such that $\theta\equiv 1$ on $(-1,1)$ and $0\leq\theta\leq 1$. Let us denote $\theta_{\rho}=\theta(\frac{\cdot}{\rho})$. Let us also introduce the Riesz operators

[TABLE]

where the constant $\kappa_{s}$ is defined by the identity $\widehat{\kappa_{s}|\cdot|^{-s}}={|\cdot|^{s-1}}$. With this definition $I_{1-s}$ is the inverse of $(-\Delta)^{\frac{1-s}{2}}$. Then we can split

[TABLE]

where we use the commutator notation $[u,I_{1-s}](v)=uI_{1-s}v-I_{1-s}(uv)$ for any $u,v\in C^{\infty}_{c}(\mathbb{R})$. Applying respectively Proposition 3.2, Proposition 3.4 and Proposition A.3. in [24], we get that

[TABLE]

that

[TABLE]

and that

[TABLE]

∎

As a consequence of Lemma 3.3 we obtain the following crucial estimate.

Lemma 3.4

For any $s\in(0,1)$ there exists a constant $C=C(s)$ such that

[TABLE]

for any $\rho>0$, and $u\in H\cap C^{\infty}(\mathbb{R})$. Here $E_{1}$ and $E_{2,2\rho}$ are defined as in Lemma 3.3.

Proof.

By the Hölder inequality for Lorentz spaces (see e.g. [31, Theorem 3.5]), we have

[TABLE]

We shall bound the RHS of (36) by approximating $u$ with compactly supported functions and applying Lemma 3.3. To this purpose, we take a sequence of cut-off function $(\tau_{j})_{j\in\mathbb{N}}\subseteq C^{\infty}_{c}(\mathbb{R})$ such that $\tau_{j}(x)=1$ for $|x|\leq j$, $\tau_{j}(x)=0$ for $|x|\geq j+1$, $0\leq\tau_{j}\leq 1$ and $|\tau_{j}^{\prime}|\leq 2$. We define $u_{j}:=\tau_{j}u$. We claim that

[TABLE]

and

[TABLE]

The first claim is proved in [10, Lemma 12]. We shall prove the second claim. Set $v_{j}=u_{j}-u$. Then, for any fixed $R_{0}>0$ and $x\in(-R_{0},R_{0})$, if $j>2R_{0}$ we have

[TABLE]

with $C$ depending only on $s$. As $j\to\infty$, we get (38).

Now, By Lemma 3.3, we know that, for any $j$,

[TABLE]

where $C$ depends only on $s$. Clearly, (37) yields

[TABLE]

Moreover

[TABLE]

Finally, (37) and (38) imply that $u_{j}(-\Delta)^{\frac{s}{2}}u_{j}\rightarrow u(-\Delta)^{\frac{s}{2}}u$ in $L^{q}_{\operatorname{loc}}(\mathbb{R})$ for every $q\in[1,\infty)$, and therefore in $L^{(\frac{1}{s},\infty)}(-\rho,\rho)$. Then, passing to the limit in (39) we get

[TABLE]

and together with (36) we conclude. ∎

We can now apply Lemma 3.4 to $u_{k}$. After scaling, we get the following bound on $\eta_{k}$.

Lemma 3.5

For any $s\in(0,1)$, there exists a constant $C=C(s)>0$ such that

[TABLE]

Proof.

First we observe that $f_{k}:=(-\Delta)^{\frac{1}{2}}u_{k}=\lambda_{k}u_{k}e^{\alpha_{k}u_{k}^{2}}-u_{k}$ is bounded in $L\log^{\frac{1}{2}}L_{\operatorname{loc}}(\mathbb{R})$. Indeed, we have

[TABLE]

so that

[TABLE]

Since $|f_{k}|u_{k}$ is bounded in $L^{1}(\mathbb{R})$ by (5) and (6), we get that $f_{k}$ is bounded in $L\log^{\frac{1}{2}}L_{\operatorname{loc}}(\mathbb{R})$.

Then Lemma 3.4 and (6) imply the existence of $C=C(s)$ such that

[TABLE]

For any $R>0$, we can apply this with $\rho=Rr_{k}$ and rewrite it in terms of $\eta_{k}$. Then, we obtain

[TABLE]

Since, by Lemma 3.2, $\eta_{k}$ is locally bounded, if $k$ is sufficiently large we get $1+\frac{\eta_{k}}{\mu_{k}^{2}}\geq\frac{1}{2}$ and the proof is complete. ∎

Lemma 3.6

The sequence $(\eta_{k})$ is bounded in $L_{s}(\mathbb{R})$ for any $s>0$.

Proof.

It is sufficient to prove the statement for $s\in(0,1/2)$. Since $\eta_{k}\leq 0$, Lemma 3.5 gives

[TABLE]

Take $2s<\alpha<1$. Since $\eta_{k}$ is bounded in $C^{\alpha}_{\operatorname{loc}}(\mathbb{R})$ by Lemma 3.2, we have that

[TABLE]

Similarly

[TABLE]

Therefore, we obtain that

[TABLE]

But for $x\in(-1,1)$ and $y\notin(-2,2)$ we have $|x-y|\leq|y|+|x|\leq 2|y|\leq 2(1+|y|^{1+2s})^{\frac{1}{1+2s}}$. Hence

[TABLE]

This and Lemma 3.2 imply that $\eta_{k}$ is bounded in $L_{s}(\mathbb{R})$. ∎

Proof of Theorem 1.3 (completed). By Lemma 3.2, up to a subsequence we can assume that $\eta_{k}\to\eta_{\infty}$ in $C^{\alpha}_{\operatorname{loc}}(\mathbb{R})$ for any $\alpha\in(0,1)$, with $\eta_{\infty}\in C^{\alpha}_{\operatorname{loc}}(\mathbb{R})$. Let us denote

[TABLE]

As observed in the proof of Lemma 3.2, we have $r_{k}\mu_{k}^{2}\to 0$ as $k\to\infty$ and thus $f_{k}\to 2e^{\eta_{\infty}}$ locally uniformly on $\mathbb{R}$. Moreover $f_{k}$ is bounded in $L^{\infty}(\mathbb{R})$. Then, for any Schwarz function $\varphi\in\mathcal{S}(\mathbb{R})$ we have

[TABLE]

as $k,R\to+\infty$. On the other hand, we know by Lemma 3.6 that $\eta_{k}$ is bounded in $L_{s}(\mathbb{R})$ and, consequently, $\eta_{\infty}\in L_{s}(\mathbb{R})$, $s>0$. In particular, for $s\in(0,\frac{1}{2})$, letting $k\to\infty$ first, and then $R\to\infty$ we get

[TABLE]

Then $\eta_{\infty}$ is a weak solution $(-\Delta)^{\frac{1}{2}}\eta_{\infty}=2e^{\eta_{\infty}}$ and $\eta_{\infty}\in L_{s}(\mathbb{R})$ for any $s$. Moreover, repeating the argument of Corollary 2.3 and using (6), we get

[TABLE]

which implies $e^{\eta_{\infty}}\in L^{1}(\mathbb{R})$. Then $\eta_{\infty}(x)=-\log(1+x^{2})$, see e.g. [6, Theorem 1.8].

To complete the proof, we shall study the properties of the weak limit $u_{\infty}$ of $u_{k}$ in $H$. First, we show that $u_{\infty}$ is a weak solution of (7). Let us denote

[TABLE]

Take any function $\varphi\in\mathcal{S}(\mathbb{R})$. On the one hand, since $(-\Delta)^{\frac{1}{2}}\varphi\in L^{2}(\mathbb{R})$ and $u_{k}\rightharpoonup u_{\infty}$ weakly in $L^{2}(\mathbb{R})$, we have

[TABLE]

as $k\to\infty$. On the other hand, for any large $t>0$ we get

[TABLE]

as $k,t\to\infty$, where we used that $g_{\infty}\in L^{2}(\mathbb{R})$ by Theorem A (see e.g. Lemma 2.3 of [17]) together with the dominated convergence theorem and the bounds $\|u_{k}g_{k}\|_{L^{1}(\mathbb{R})}\leq\Lambda$ and $\|u_{k}\|_{L^{2}(\mathbb{R})}\leq\Lambda$. Then, $u_{\infty}$ is a weak solution of (7).

Now, observe that

[TABLE]

with

[TABLE]

as $R\to\infty$, and

[TABLE]

for any $R>1$, by Fatou’s lemma. Thus we conclude that

[TABLE]

Finally, to prove that $u_{k}\to u_{\infty}$ in $C^{\ell}_{\operatorname{loc}}(\mathbb{R}\setminus\{0\})$ for every $\ell\geq 0$, we use the monotonicity of $u_{k}$, which implies that $u_{k}$ is locally bounded away from [math], hence we can conclude by elliptic estimates, as in Lemma 3.1.

4 Proof of Theorem 1.2

Let us denote

[TABLE]

The proof of Theorem 1.2 is organized as follows. First, we prove that $D_{\alpha}$ is attained for $\alpha\in(0,\pi)$ sufficiently close to $\pi$. Then, we fix a sequence $(\alpha_{k})_{k\in N}$ such that $\alpha_{k}\nearrow\pi$ as $k\to+\infty$, and for any large $k$ we take a positive extremal $u_{k}\in H$ for $D_{\alpha_{k}}$. With a contradiction argument similar to the one of Section 2, we show that $\mu_{k}:=\sup_{\mathbb{R}}u_{k}\leq C$. Finally, we show that $u_{k}\to u_{\infty}$ in $L^{\infty}_{\operatorname{loc}}(\mathbb{R})\cap L^{2}(\mathbb{R})$, where $u_{\infty}$ is a maximizer for $D_{\pi}$.

4.1 Subcritical extremals: Ruling out vanishing

The following lemma describes the effect of the lack of compactness of the embedding ${H\subseteq L^{2}(\mathbb{R})}$ on $E_{\alpha}$, and holds uniformly for $\alpha\in[0,\pi]$.

Lemma 4.1

Let $(\alpha_{k})\subseteq[0,\pi]$ and $(u_{k})\subseteq H$ be two sequences such that:

$\alpha_{k}\to\alpha_{\infty}\in[0,\pi]$* as $k\to\infty$.* 2. 2.

$\|u_{k}\|_{H}\leq 1$, $u_{k}\rightharpoonup u_{\infty}$ weakly in $H$, $u_{k}\to u_{\infty}$ a.e. in $\mathbb{R}$, and $e^{\alpha_{k}u_{k}^{2}}\to e^{\alpha_{\infty}u_{\infty}^{2}}$ in $L^{1}_{\operatorname{loc}}(\mathbb{R})$ as $k\to\infty$. 3. 3.

The $u_{k}$’s are even and monotone decreasing i.e. $u_{k}(-x)=u_{k}(x)\geq u_{k}(y)$ for $0\leq x\leq y$.

Then we have

[TABLE]

as $k\to\infty$.

Proof.

Since $u_{k}$ is even and decreasing, we know that

[TABLE]

for any $x\in\mathbb{R}\setminus\{0\}$. In particular, there exists a constant $C>0$, such that

[TABLE]

for $|x|\geq 1$. Applying the dominated convergence theorem for $|x|\geq 1$, using the assumption that $e^{\alpha_{k}u_{k}^{2}}\to e^{\alpha_{\infty}u_{\infty}^{2}}$ in $L^{1}_{\operatorname{loc}}(\mathbb{R})$, and recalling that $(u_{k})$ is precompact in $L^{1}_{\operatorname{loc}}(\mathbb{R})$, we find that

[TABLE]

and the Lemma follows. ∎

Lemma 4.2

Take $\alpha\in(0,\pi)$. If $D_{\alpha}>\alpha$, then $D_{\alpha}$ is attained by an even an decreasing function, i.e. there exists $u_{\alpha}\in H$ even and decreasing s.t. $\|u_{\alpha}\|_{H}=1$ and $E_{\alpha}(u_{\alpha})=D_{\alpha}$.

Proof.

Let $(u_{k})\subset H$ be a maximizing sequence for $E_{\alpha}$. W.l.o.g. we can assume $u_{k}\to u_{\infty}\in H$ weakly in $H$ and a.e. on $\mathbb{R}$. Moreover, up to replacing $u_{k}$ with its symmetric decreasing rearrangement, we can assume that $u_{k}$ is even and decreasing (see [30]). Since $\alpha\in(0,\pi)$ the sequence $e^{\alpha u_{k}^{2}}-1$ is bounded in $L^{\frac{\pi}{\alpha}}(\mathbb{R})$, with $\frac{\pi}{\alpha}>1$. Then, by Vitali’s theorem, we get $e^{\alpha u_{k}^{2}}\to e^{\alpha u_{\infty}^{2}}$ in $L^{1}_{\operatorname{loc}}(\mathbb{R})$, and Lemma 4.1 yields

[TABLE]

This implies that $u_{\infty}\not\equiv 0$, since otherwise we have $E_{\alpha}(u_{k})=\alpha\|u_{k}\|_{L^{2}(\mathbb{R})}^{2}+o(1)\leq\alpha+o(1),$ which contradicts the assumption $D_{\alpha}>\alpha$. Let us denote

[TABLE]

Observe that $L,\tau\in(0,1]$. Let us consider the sequence $v_{k}(x)=u_{k}(\tau x)$. Clearly, we have $v_{k}\rightharpoonup v_{\infty}$ weakly in $H$, where $v_{\infty}(x):=u_{\infty}(\tau x)$. Moreover, since

[TABLE]

we get $\|v_{\infty}\|_{H}\leq 1$. By (42) we have

[TABLE]

If $\tau<1$, this implies $D_{\alpha}\leq\alpha L\leq\alpha$, contradicting the assumptions. Hence $\tau=1$ and (43) gives $D_{\alpha}=E_{\alpha}(u_{\infty})$. Finally we have $\|u_{0}\|_{H}=1$, otherwise $E_{\alpha}(\frac{u_{\infty}}{\|u_{\infty}\|_{H}})>E_{\alpha}(u_{\infty})=D_{\alpha}$. ∎

Lemma 4.3

There exists $\alpha^{*}\in(0,\pi)$ such that $D_{\alpha}>\alpha$ for any $\alpha\in(\alpha^{*},\pi]$. In particular $D_{\alpha}$ is attained by an even and decreasing function $u_{\alpha}$ for any $\alpha\in(\alpha^{*},\pi)$ by Lemma 4.2.

Proof.

This follows from Proposition 4.14 by continuity. Indeed Proposition 4.14 gives $D_{\pi}>2\pi e^{-\gamma}>\pi$. ∎

4.2 The critical case

Next, we take a sequence $\alpha_{k}$ such that $\alpha_{k}\nearrow\pi$ as $k\to\infty$. For any large $k$, Lemma 4.3 yields the existence $u_{k}\in H$ even and decreasing such that $D_{\alpha_{k}}=E_{\alpha_{k}}(u_{k})$. Each $u_{k}$ satisfies

[TABLE]

and $\|u_{k}\|_{H}=1$. Note that $u_{k}\in C^{\infty}(\mathbb{R})$ by elliptic estimates. Multiplying the equation by $u_{k}$ and using the basic inequality $te^{t}\geq e^{t}-1$, for $t\geq 0$, we infer

[TABLE]

Since $D_{\alpha_{k}}\to D_{\pi}>0$, we get that $\lambda_{k}$ is uniformly bounded.

Then the sequence $u_{k}$ satisfies the alternative of Theorem 1.3. If case (i) holds, then we can argue as in Lemma 4.2 and Lemma 4.3 and prove that $D_{\pi}$ is attained. Therefore, we shall assume by contradiction that case (ii) occurs.

Let $r_{k}$ and $\eta_{k}$ be as in Theorem 1.3. Let $\tilde{\eta}_{k}$ denote the Poisson integral of $\eta_{k}$.

Proposition 4.4

We have $\tilde{\eta}_{k}\to\tilde{\eta}_{\infty}$ in $C^{\ell}_{\operatorname{loc}}(\overline{\mathbb{R}^{2}_{+}})$ for every $\ell\geq 0$, where

[TABLE]

is the Poisson integral (compare to (66)) of $\eta_{\infty}:=-\log\left({1+x^{2}}\right)$.

Proof.

By Theorem 1.3 we know that $\eta_{k}\to\eta_{\infty}$ in $C^{\ell}_{\operatorname{loc}}(\mathbb{R})$ and that $\eta_{k}$ is bounded in $L_{\frac{1}{2}}$. Then, we can repeat the argument of the proof of Proposition 2.2. ∎

Remark 4.5

As in (17), the convergence $\eta_{k}\to\eta_{\infty}$ in $L^{\infty}_{\operatorname{loc}}(\mathbb{R})$ implies

[TABLE]

for $i=0,1,2$ and for any $R>0$.

Lemma 4.6

We have $u_{k}\to 0$ in $L^{2}(\mathbb{R})$.

Proof.

Indeed, otherwise up to a subsequence we would have $\|(-\Delta)^{\frac{1}{4}}u_{k}\|_{L^{2}(\mathbb{R})}\leq\frac{1}{A}$ for some $A>1$. Consider, the function $v_{k}=(u_{k}-u_{k}(1))^{+}$. Then, $v_{k}\in\tilde{H}^{\frac{1}{2},2}(I)$ and $\|(-\Delta)^{\frac{1}{4}}v_{k}\|_{L^{2}(\mathbb{R})}\leq\frac{1}{A}$. The Moser-Trudinger inequality (3) gives that $e^{\alpha_{k}v_{k}^{2}}$ is bounded in $L^{A}(\mathbb{R})$. Since

[TABLE]

and $|v_{k}-u_{k}|\leq u_{k}(1)\to 0$ as $k\to\infty$, we get that $e^{\alpha_{k}u_{k}^{2}}$ is uniformly bounded in $L^{p}(\mathbb{R})$ for every $1<p<A$. Therefore, we have

[TABLE]

as $k\to\infty$. But then, by Lemma 4.1 we find $D^{\pi}_{g}\leq\pi$, which contradicts Lemma 4.3. ∎

Lemma 4.7

For $A>1$, set $u_{k}^{A}:=\min\left\{u_{k},\frac{\mu_{k}}{A}\right\}$. Then we have

[TABLE]

Proof.

The proof is similar to the one of Lemma 2.4. We set $\bar{u}_{k}^{A}:=\min\left\{\tilde{u}_{k},\frac{\mu_{k}}{A}\right\}$. Since $\bar{u}_{k}^{A}$ is an extension of $u_{k}^{A}$, using integration by parts and the harmonicity of $\tilde{u}_{k}$ we get

[TABLE]

Proposition 4.4 implies that $u_{k}^{A}(r_{k}x)=\frac{\mu_{k}}{A}$ for $|x|\leq R$ and $k\geq k_{0}(R)$. Noting that $u_{k}^{A}\leq u_{k}$ and using Lemma 4.6, and Remark 4.5, we get

[TABLE]

Set now $v_{k}^{A}:=\left(u_{k}-\frac{\mu_{k}}{A}\right)^{+}$. With similar computations we get

[TABLE]

Since

[TABLE]

we get that

[TABLE]

Then, we conclude using (45). ∎

Proposition 4.8

We have

[TABLE]

Moreover

[TABLE]

Proof.

Fix $A>1$ and write

[TABLE]

Using Lemmas 4.11 and 4.7 together with Theorem A we see that

[TABLE]

since $e^{\alpha_{k}(u_{k}^{A})^{2}}-1$ is uniformly bounded in $L^{p}$, for any $1\leq p<A$. By (41) and Lemma 4.6, we find

[TABLE]

We now estimate

[TABLE]

with $o(1)\to 0$ as $k\to\infty$, where we used that

[TABLE]

Letting $A\downarrow 1$, this gives

[TABLE]

The converse inequality follows from Remark 4.5:

[TABLE]

with $o(1)\to 0$ as $k\to\infty$. Letting $R\to\infty$ we obtain (46).

Finally, (47) follows at once from (46), because otherwise we would have $D_{\pi}=0$, which is clearly impossible. ∎

Lemma 4.9

We have

[TABLE]

as $k\to\infty$, in the sense of Radon measures in $\mathbb{R}$.

Proof.

The proof follows step by step the one Proposition 2.6, with (4), Proposition 4.4, Remark 4.5 Lemma 4.6 and Lemma 4.7 used in place of (3), Proposition 2.2, (17), Lemma 2.3 and Lemma 2.4. We omit the details. ∎

For $x\in\mathbb{R}$, let $G_{x}$ be the Green function of $(-\Delta)^{\frac{1}{2}}+Id$ on $\mathbb{R}$ with singularity at $x$. In the following we denote $G:=G_{0}$. By translation invariance, we get $G_{x}(y)=G(y-x)$ for any $x,y\in\mathbb{R}$, $x\neq y$. Moreover, the inversion formula for the Fourier-transform implies that

[TABLE]

where

[TABLE]

We recall that the identity

[TABLE]

holds for any $x\in\mathbb{R}\setminus\{0\}$, where $\gamma$ denotes the Euler-Mascheroni constant see e.g. [13, Chapter 12.2].

Proposition 4.10

The function $G$ satisfies the following properties.

We have $G\in C^{\infty}(\mathbb{R}\setminus\{0\})$ and

[TABLE] 2. 2.

*We have $G(x)=O(|x|^{-2})$ and $G^{\prime}(x)=O(|x|^{-3})$ as $|x|\to\infty$. * 3. 3.

Let $\tilde{G}$ be the Poisson extension of $G$. There exists a function $f\in C^{1}(\overline{\mathbb{R}^{2}_{+}})$ such that $f(0,0)=0$ and

[TABLE]

Proof.

Property 1. follows directly by formula (48) and the identity in (49). Similarly, since

[TABLE]

as $t\to+\infty$, we get 2.

Given $R>0$, let $\psi\in C^{\infty}_{c}(\mathbb{R})$ be a cut-off function with $\psi\equiv 1$ on $(-R,R)$. Let us denote $g_{0}:=-\frac{1}{\pi}\log|\cdot|-\frac{\gamma}{\pi}$, $g_{1}:=\frac{1}{2}|\cdot|\psi$, $g_{2}:=G-g_{0}-g_{1}$. By Proposition A.3, we have

[TABLE]

Denoting $\theta(x,y):=\arctan{\frac{x}{y}}$ the angle between the $y$-axis and the segment connecting the origin to $(x,y)$, the function

[TABLE]

is harmonic in $\mathbb{R}^{2}_{+}$, continuous on $\overline{\mathbb{R}}^{2}_{+}$, and identically [math] on $(-R,R)\times\mathbb{R}$. By [33, Theorem C], we get that $h\in C^{\infty}(\overline{\mathbb{R}^{2}_{+}}\cap B_{R}(0,0))$. Finally, note that formula (48) implies $g_{2}\in C^{2}(\mathbb{R})$ and $g_{2}(0)=0$. Hence, standard elliptic regularity yields $\tilde{g}_{2}\in C^{1,\alpha}(\overline{\mathbb{R}^{2}_{+}}\cap B_{R}(0,0))$, for any $\alpha\in(0,1)$. In particular $\tilde{g}_{2}(0,0)=g_{2}(0)=0$. ∎

Lemma 4.11

We have $\mu_{k}u_{k}\to G$ in ${L^{2}(\mathbb{R})}\cap L^{\infty}(\mathbb{R}\setminus(-\varepsilon,\varepsilon))$, for any $\varepsilon>0$,

Proof.

Let us set $v_{k}:=\mu_{k}u_{k}-G$ and $f_{k}=\mu_{k}\lambda_{k}u_{k}e^{\alpha_{k}u_{k}^{2}}$. By Lemma 4.9 we have $\|f_{k}\|_{L^{1}(I)}\to 1$ as $k\to+\infty$, $I=(-1,1)$. Then, arguing as in Lemma 2.8, we get

[TABLE]

Using (41), Lemma 4.6 and (47), we get that $f_{k}\to 0$ in $L^{2}(\mathbb{R}\setminus I)$. In particular

[TABLE]

Fix $\sigma\in(0,1)$ and assume $|x|\geq\sigma$. If we further take $|y|\leq\frac{\sigma}{2}$, then Proposition 4.10 implies

[TABLE]

where $C$ is a constant depending only on $\sigma$. Thus, for any $\varepsilon\in(0,\frac{\sigma}{2})$, we can write

[TABLE]

where $o(1)\to 0$ as $k\to\infty$ (depending on $\varepsilon$ and $\sigma$). Here, we used that $f_{k}\to 0$ in $L^{\infty}(\mathbb{R}\setminus(-\varepsilon,\varepsilon))$ by (41) and (47). Since $\varepsilon$ is arbitrarily small, (53) shows that $v_{k}\to 0$ in $L^{\infty}(\mathbb{R}\setminus(-\sigma,\sigma))$.

Next, we prove the $L^{2}$ convergence. First, Hölder’s inequality and Fubini’s theorem give

[TABLE]

as $k\to\infty$. With a similar argument, after integrating (52) and using the triangular inequality in $L^{2}$, we find

[TABLE]

Since $G\in L^{2}(\mathbb{R})$, the function $\psi(y):=\int_{\mathbb{R}}|G(x-y)-G(x)|^{2}dx$ is continuous on $\mathbb{R}$ and $\psi(0)=0$. Let $\varphi\in C(\mathbb{R})$ be a compactly supported function such that $\varphi\equiv\psi$ on $I$. Then, Lemma 4.9 implies

[TABLE]

as $k\to\infty$, and the conclusion follows.

∎

Repeating the argument of Proposition 2.8, we get the following:

Lemma 4.12

We have $\mu_{k}\tilde{u}_{k}\rightarrow\tilde{G}$ in $C^{0}_{\operatorname{loc}}(\overline{\mathbb{R}^{2}_{+}}\setminus\{(0,0)\})\cap C^{1}_{\operatorname{loc}}(\mathbb{R}^{2}_{+})$, where $\tilde{G}$ is the Poisson extension of $G$.

With Proposition 4.4 and Lemma 4.12 we can give an upper bound on $D_{\pi}$.

Proposition 4.13

Under the assumption that $\mu_{k}\to\infty$ as $k\to\infty$, we have $D_{\pi}\leq 2\pi e^{-\gamma}$.

Proof.

For a fixed and small $\delta>0$ set

[TABLE]

Recalling that $\|\nabla\tilde{u}_{k}\|_{L^{2}(\mathbb{R})}^{2}=\|(-\Delta)^{\frac{1}{4}}u_{k}\|^{2}_{L^{2}(\mathbb{R})}=1-\|u_{k}\|^{2}_{L^{2}(\mathbb{R})}$, we have

[TABLE]

Clearly the left-hand side bounds

[TABLE]

Using Proposition 4.4, Proposition 4.10 and Lemma 4.12 we obtain

[TABLE]

where $o(1)\to 0$ as $k\to\infty$ for fixed $L>0$, $\delta>0$, and $|O(L^{-1})|\leq CL^{-1}$, $|O(\delta|\log\delta|)|\leq C\delta|\log\delta|$, uniformly for $\delta$ small, and $L$, $k$ large. Still with Proposition 4.4 we get

[TABLE]

Similarly Lemma 4.12 and Proposition 4.10 yield

[TABLE]

where we used that

[TABLE]

From Lemma 4.11 we get that $\mu_{k}u_{k}\to G$ in $L^{2}(\mathbb{R})$, hence

[TABLE]

as $k\to+\infty$. We then get

[TABLE]

Using (55) and rearranging as in the proof of Proposition 2.9, we find

[TABLE]

with $o(1)\to 0$ as $k\to\infty$. Then, recalling that $\alpha_{k}\uparrow\pi$, letting $k\to\infty$ first and then $L\to\infty$, $\delta\to 0$, we obtain

[TABLE]

and using Proposition 4.8 we conclude. ∎

Proposition 4.14

There exists a function $u\in H^{\frac{1}{2},2}(\mathbb{R})$ such that $\|u\|_{H}\leq 1$ and $E_{\pi}(u)>2\pi e^{-\gamma}.$

Proof.

For $\varepsilon>0$ choose $L=L(\varepsilon)>0$ such that as $\varepsilon\to 0$ we have $L\to\infty$ and $L\varepsilon\to 0$. Fix

[TABLE]

and

[TABLE]

By the maximum principle we have $\mathbb{R}^{2}_{+}\cap B_{L\varepsilon}\subset\Omega_{L\varepsilon}$. Notice also that Proposition 4.10 gives

[TABLE]

and $\Omega_{L\varepsilon}\subseteq\mathbb{R}^{2}_{+}\cap B_{2L_{\varepsilon}}$. For suitable constants $B,c\in\mathbb{R}$ to be fixed we set

[TABLE]

Observe that $\mathbb{R}^{2}_{+}\cap B_{L\varepsilon}(0,-\varepsilon)\subseteq\mathbb{R}^{2}\cap B_{L\varepsilon}\subseteq\Omega_{L\varepsilon}$. We choose $B$ in order to have continuity on $\mathbb{R}^{2}_{+}\cap\partial B_{L\varepsilon}(0,-\varepsilon)$, i.e. we impose

[TABLE]

which gives the relation

[TABLE]

This choice of $B$ also implies that the function $cU_{\varepsilon}$ does not depend on the value of $c$. Then we can choose $c$ by imposing

[TABLE]

where we set $u_{\varepsilon}(x)=U_{\varepsilon}(x,0)$. Since the harmonic extension $\tilde{u}_{\varepsilon}$ minimizes the Dirichlet energy among extensions with finite energy, we have

[TABLE]

and (57) implies $\|u_{\varepsilon}\|_{H^{\frac{1}{2},2}(\mathbb{R})}^{2}\leq 1$.

In order to obtain a more precise expansion of $B$ and $c$ we compute

[TABLE]

and

[TABLE]

By the divergence theorem we have for $\tau<L\varepsilon$ and letting $\tau\to 0$,

[TABLE]

where in the third identity we used that $\Omega_{L\varepsilon}\subset B_{2L\varepsilon}$ for $L\varepsilon$ small enough. Observe also that

[TABLE]

Together with (57)-(59) this gives

[TABLE]

which, together with (56), implies

[TABLE]

Now, observe that $B_{L\varepsilon}(0,-\varepsilon)\cap(\mathbb{R}\times\{0\})=(-\varepsilon\sqrt{L^{2}-1},\varepsilon\sqrt{L^{2}-1})$ and that

[TABLE]

Moreover

[TABLE]

with

[TABLE]

Now choose $L=\log^{2}\varepsilon$ to obtain

[TABLE]

so that

[TABLE]

for $\varepsilon$ small enough. ∎

Proof of Theorem 1.2 (completed). By Propositions 2.10 and 4.14, we know that $\mu_{k}\leq C$. Then, by dominated convergence theorem we have $e^{\alpha_{k}u_{k}^{2}}\to e^{\pi u_{\infty}^{2}}$ in $L^{1}_{\operatorname{loc}}(\mathbb{R})$. Then, by Lemma 4.2, we infer

[TABLE]

This implies that $u_{\infty}\not\equiv 0$, otherwise we would have $E_{\alpha_{k}}(u_{k})\leq\pi\|u_{k}\|_{L^{2}(\mathbb{R})}^{2}+o(1)\leq\pi+o(1)$, which contradicts the strict inequality $D_{\pi}>2\pi e^{-\gamma}>\pi$, since $E_{\alpha_{k}}(u_{k})=D_{\alpha_{k}}\to D_{\pi}$ as $k\to\infty$.

Let us denote $L:=\limsup_{k\to\infty}\|u_{k}\|^{2}_{2}$, $\tau=\frac{\|u_{\infty}\|^{2}_{L^{2}(\mathbb{R})}}{L}$ and observe that $L,\tau\in(0,1]$. Let us consider the sequence $v_{k}(x)=u_{k}(\tau x)$. Clearly, we have $v_{k}\rightharpoonup v_{\infty}$ in $H$, where $v_{\infty}(x):=u_{\infty}(\tau x)$. Since

[TABLE]

we get $\|v_{\infty}\|_{H^{\frac{1}{2},2}}\leq 1$. By (60) we have

[TABLE]

If $\tau<1$, this implies $D_{\pi}\leq\pi L,$ which is not possible. Hence, we must have $\tau=1$ and $E_{\pi}(u_{\infty})=D_{\pi}$. $\square$

Appendix A Appendix: The half-Laplacian on $\mathbb{R}$

For $u\in\mathcal{S}$ (the Schwarz space of rapidly decaying functions) we set

[TABLE]

One can prove that it holds (see e.g.)

[TABLE]

from which it follows that

[TABLE]

Then one can set

[TABLE]

and for every $u\in L_{s}(\mathbb{R})$ one defines the tempered distribution $(-\Delta)^{s}u$ as

[TABLE]

Moreover we will define for $p\geq 1$ and $s\in(0,1)$

[TABLE]

In the case $s=\frac{1}{2}$ we have $K_{\frac{1}{2}}=\frac{1}{\pi}$ in (62) and a simple alternative definition of $(-\Delta)^{\frac{1}{2}}$ can be given via the Poisson integral. For $u\in L_{\frac{1}{2}}(\mathbb{R})$ define the Poisson integral

[TABLE]

which is harmonic in $\mathbb{R}^{2}_{+}=\mathbb{R}\times(0,\infty)$ and satisfies the boundary condition $\tilde{u}|_{\mathbb{R}\times\{0\}}=u$ in the following sense:

Proposition A.1

If $u\in L^{\frac{1}{2}}(\mathbb{R})$, then $\tilde{u}(\cdot,y)\in L^{1}_{\operatorname{loc}}(\mathbb{R})$ for $y\in(0,\infty)$ and $\tilde{u}(\cdot,y)\to u$ in the sense of distributions as $y\to 0^{+}$. If $u\in L^{\frac{1}{2}}(\mathbb{R})\cap C((a,b))$ for some interval $(a,b)\subseteq\mathbb{R}$, then $\tilde{u}$ extends continuously to $(a,b)\times\{0\}$ and $\tilde{u}(x,0)=u(x)$ for any $x\in(a,b)$. If $u\in H^{\frac{1}{2}}(\mathbb{R})$, then $\tilde{u}\in H^{1}(\mathbb{R}^{2}_{+})$, the identity $\|\nabla\tilde{u}\|_{L^{2}(\mathbb{R}^{2}_{+})}=\|(-\Delta)^{\frac{1}{4}}u\|_{L^{2}(\mathbb{R})}$ holds, and $\tilde{u}|_{\mathbb{R}\times\{0\}}=u$ in the sense of traces.

Then we have (see e.g [4])

[TABLE]

where the identity is pointwise if $u$ is regular enough (for instance $C^{1,\alpha}_{\operatorname{loc}}(\mathbb{R})$), and has to be read in the sense of tempered distributions in general, with

[TABLE]

More precisely:

Proposition A.2

If $u\in L_{\frac{1}{2}}(\mathbb{R})\cap C^{1,\alpha}_{\operatorname{loc}}((a,b))$ for some interval $(a,b)\subset\mathbb{R}$ and some $\alpha\in(0,1)$, then the tempered distribution $(-\Delta)^{\frac{1}{2}}u$ defined in (64) coincides on the interval $(a,b)$ with the functions given by (62) and (67). For general $u\in L_{\frac{1}{2}}(\mathbb{R})$ the definitions (64) and (67) are equivalent, where the right-hand side of (67) is defined by (68).

It is known that the Poisson integral of a function $u\in L^{\frac{1}{2}}(\mathbb{R})$ is the unique harmonic extension of $u$ under some growth constraints at infinity. In fact, combining [36, Theorem 2.1 and Corollary 3.1] and [33, Theorem C] we get:

Proposition A.3

For any $u\in L_{\frac{1}{2}}(\mathbb{R})$, the Poisson extension $\tilde{u}$ satisfies $\tilde{u}(x,y)=o(y^{-1}(x^{2}+y^{2}))$ as $|(x,y)|\to\infty$. Moreover, if $U$ is a harmonic function in $\mathbb{R}^{2}_{+}$ which satisfies $U(x,y)=o(y^{-1}(x^{2}+y^{2}))$ as $|(x,y)|\to\infty$ and $U(\cdot,y)\to u$ as $y\to 0^{+}$ in the sense of distributions, then $U=\tilde{u}$ in $\mathbb{R}^{2}_{+}$.