Extremizers for the Airy–Strichartz inequality

Abstract

We identify the compactness threshold for optimizing sequences of the Airy–Strichartz inequality as an explicit multiple of the sharp constant in the Strichartz inequality. In particular, if the sharp constant in the Airy–Strichartz inequality is strictly smaller than this multiple of the sharp constant in the Strichartz inequality, then there is an optimizer for the former inequality. Our result is valid for the full range of Airy–Strichartz inequalities (except the endpoints) both in the diagonal and off-diagonal cases.

Appendices

Appendix A: A generalized Brézis–Lieb lemma for mixed Lebesgue spaces

Let us review some basics. We assume that (X, dx) and (Y, dy) are measure spaces and consider a sequence $(f_n)$ of non-negative measurable functions on $X\times Y$ which converges almost everywhere to some function f. Moreover, we fix an exponent $r>0.$ Our first remark is that the monotone convergence theorem remains true, in the sense that, if for each n one has $f_{n+1}\ge f_n$ almost everywhere, then

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _Y \left( \int _X f_n\,dx \right) ^r dy = \int _Y \left( \int _X f\,dx \right) ^r dy . \end{aligned}$$

To see this, we first apply for almost every fixed $y\in Y$ the usual monotone convergence theorem in X to see that $g_n := \left( \int _X f_n\,dx \right) ^r$ converges to $g := \left( \int _X f\,dx \right) ^r.$ Indeed, by Fubini’s theorem, for a.e. $y\in Y, f_n(\cdot ,y)$ converges to $f(\cdot ,y)$ a.e. on X. Then, we apply the monotone convergence theorem in Y to $(g_n)$ and we obtain the claim.

Our second remark is that Fatou’s lemma remains true, in the sense that

$$\begin{aligned} \liminf _{n\rightarrow \infty } \int _Y \left( \int _X f_n\,dx \right) ^r dy \ge \int _Y \left( \int _X f\,dx \right) ^r dy. \end{aligned}$$

This follows, as usual, by applying the monotone convergence theorem to $F_n:=\inf _{m\ge n} f_m.$

Our third remark is that the dominated convergence theorem remains true, in the sense that, if $f_n\le F$ with $\int _Y \left( \int _X F \right) ^r dy <\infty ,$ then

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _Y \left( \int _X f_n\,dx \right) ^r dy = \int _Y \left( \int _X f\,dx \right) ^r dy . \end{aligned}$$

To see it, just apply the usual dominated convergence theorem, first to the sequence $f_n(\cdot ,y)$ for a.e. $y\in Y,$ and then to the sequence $(\int _X f_n\,dx)^r.$

In mixed Lebesgue space, we have the following version of the triangle inequality.

Lemma A.1

Let (X, dx) and (Y, dy) be measure space,  let $0<p,q<{\infty },$ and let $f,g\in L^p_x L^q_y(X\times Y).$ Then,  we have

$$\begin{aligned} \left| \! \left| f+g \right| \! \right| _{L^p_xL^q_y}^\beta \leqslant \left| \! \left| f \right| \! \right| _{L^p_xL^q_y}^\beta + \left| \! \left| g \right| \! \right| _{L^p_xL^q_y}^\beta , \end{aligned}$$

where $\beta =\beta (p,q)=\min (p,q,1).$

Proof

Throughout the proof, we will use the inequality $(a+b)^r\leqslant a^r+b^r$ for all $a,b\geqslant 0,r\in (0,1].$ We distinguish 4 cases. First, if $p,q\geqslant 1,$ then it follows from the triangle inequality in $L^p_xL^q_y.$ Secondly, if $p<1\leqslant q,$ we have

$$\begin{aligned} \left| \! \left| f+g \right| \! \right| _{L^p_xL^q_y}^p= & {} \int _X \left| \! \left| f+g \right| \! \right| _{L^q_y}^p\,dx\leqslant \int _X\left( \left| \! \left| f \right| \! \right| _{L^q_y}+ \left| \! \left| g \right| \! \right| _{L^q_y}\right) ^p\,dx\\\leqslant & {} \int _X\left( \left| \! \left| f \right| \! \right| _{L^q_y}^p+ \left| \! \left| g \right| \! \right| _{L^q_y}^p\right) \,dx. \end{aligned}$$

Thirdly, if $q<1$ and $p\geqslant q,$ then

$$\begin{aligned} \left| \! \left| f+g \right| \! \right| _{L^p_xL^q_y}^q= & {} \left| \! \left| \int _Y|f+g|^q\,dy \right| \! \right| _{L^{p/q}_x}\leqslant \left| \! \left| \int _Y|f|^q\,dy+\int _Y|g|^q\,dy \right| \! \right| _{L^{p/q}_x}\\\leqslant & {} \left| \! \left| \int _Y|f|^q\,dy \right| \! \right| _{L^{p/q}_x}+ \left| \! \left| \int _Y|g|^q\,dy \right| \! \right| _{L^{p/q}_x}. \end{aligned}$$

Finally, if $q<1$ and $p<q,$ then

$$\begin{aligned} \left| \! \left| f+g \right| \! \right| _{L^p_xL^q_y}^p= & {} \int _X\left( \int _Y|f+g|^q\,dy\right) ^{p/q}\,dx\\\leqslant & {} \int _X\left( \int _Y|f|^q\,dy+\int _Y|f|^q\,dy\right) ^{p/q}\,dx\\\leqslant & {} \int _X\left( \int _Y|f|^q\,dy\right) ^{p/q}\,dx+\int _X\left( \int _Y|g|^q\,dy\right) ^{p/q}\,dx. \end{aligned}$$

$\square$

After these preliminaries, we can state and prove the one-sided analogue of the Brézis–Lieb lemma, which is originally due to [6, 35]. In [20, Lem. 3.1] we have obtained a two-fold generalization of this lemma, namely, we allow the leading term to depend on n and we allow for a remainder that converges strongly to zero. The following proposition is a generalization of this generalization to the case of mixed Lebesgue spaces. We emphasize that instead of equality we only have an asymptotic inequality.

Proposition A.2

Let (X, dx) and (Y, dy) be measure spaces and $(f_n)$ be a sequence of measurable functions on $X\times Y,$ and let $0< p, q<\infty .$ Assume that

$$\begin{aligned} \sup _n \int _X \left( \int _Y |f_n|^q \,dy \right) ^{p/q} dx <\infty , \end{aligned}$$

and that $f_n$ may be split as

$$\begin{aligned} f_n=\pi _n+\rho _n+\sigma _n \end{aligned}$$

with $|\pi _n|\leqslant \Pi$ for some $\Pi \in L^p_xL^q_y(X\times Y), \rho _n\rightarrow 0$ a.e. in $(x,y)\in X\times Y,$ and $\sigma _n\rightarrow 0$ in $L^p_x L^q_y(X \times Y).$ Then,  as $n\rightarrow \infty ,$

$$\begin{aligned} \left| \! \left| f_n \right| \! \right| _{L^p_xL^q_y}^\alpha \leqslant \left| \! \left| \pi _n \right| \! \right| _{L^p_x L^q_y}^\alpha + \left| \! \left| \rho _n \right| \! \right| _{L^p_x L^q_y}^\alpha +o(1), \end{aligned}$$

where $\alpha =\alpha (p,q)=\min (p,q).$

Proof

We first show that we may get rid of $\sigma _n,$ that is,

$$\begin{aligned} \left| \! \left| f_n \right| \! \right| _{L^p_x L^q_y}= \left| \! \left| \pi _n+\rho _n \right| \! \right| _{L^p_x L^q_y}+o_{n\rightarrow {\infty }}(1). \end{aligned}$$

(A.1)

This follows from Lemma A.1, which implies that with $\beta =\min (\alpha ,1),$

$$\begin{aligned} \left| \left| \! \left| f_n \right| \! \right| _{L^p_xL^q_y}^\beta - \left| \! \left| \pi _n+\rho _n \right| \! \right| _{L^p_xL^q_y}^\beta \right| \leqslant \left| \! \left| \sigma _n \right| \! \right| _{L^p_xL^q_y}^\beta =o_{n\rightarrow {\infty }}(1) . \end{aligned}$$

For $\alpha \geqslant 1$ this immediately gives (A.1) and for $\alpha <1$ we use in addition the boundedness of $\Vert f_n\Vert _{L^p_x L^q_y}$ to deduce (A.1).

Next, we shall show that

$$\begin{aligned} \int _X \left( \int _Y \left| |\pi _n+\rho _n|^q-|\pi _n|^q -|\rho _n|^q \right| dy \right) ^{p/q} dx = o_{n\rightarrow {\infty }}(1) . \end{aligned}$$

(A.2)

Let us first argue that this implies the conclusion. When $p\leqslant q,$ we use the elementary inequality

$$\begin{aligned} A^\theta \leqslant B^\theta + C^\theta + |A-B-C|^\theta ,\quad A,B,C\geqslant 0,\ 0<\theta \leqslant 1 , \end{aligned}$$

with $\theta =p/q$ and $A= \int _Y |\pi _n+\rho _n|^q\,dy,B=\int _Y |\pi _n|^q\,dy$ and $C=\int _Y |\rho _n|^q\,dy.$ Then

$$\begin{aligned} \int _X |A-B-C|^\theta \,dx \le \int _X \left( \int _Y \left| |\pi _n+\rho _n|^q-|\pi _n|^q -|\rho _n|^q \right| dy \right) ^{p/q} dx , \end{aligned}$$

so the conclusion follows by integrating the elementary inequality with respect to x. In the other case $p>q,$ the inequality (A.2) implies that

$$\begin{aligned} \int _Y|\pi _n+\rho _n|^q\,dy=\int _Y|\pi _n|^q\,dy+\int _Y|\rho _n|^q\,dy+o_{L^{p/q}_x}(1), \end{aligned}$$

as $n\rightarrow {\infty },$ so that the result follows from the triangle inequality in $L^{p/q}_x.$ Thus, it remains to prove (A.2). As in the usual Brézis–Lieb proof, we use the fact that for any $\varepsilon >0$ there is a $C_\varepsilon$ such that for any $a,b\in {{\mathbb {C}} },$

$$\begin{aligned} \left| |a+b|^q - |b|^q \right| \le \varepsilon |b|^q + C_\varepsilon |a|^q . \end{aligned}$$

Let us define

$$\begin{aligned} h_n^{(\varepsilon )} := \left( \left| |\pi _n+\rho _n|^q - |\pi _n|^q -|\rho _n|^q\right| - \varepsilon |\rho _n|^q \right) _+ . \end{aligned}$$

On the full measure set $\{\Pi <{\infty }\}\cap \{\rho _n\rightarrow 0\}, h_n^{(\varepsilon )}\rightarrow 0$ since $\pi _n(x,y)$ is bounded there. Hence, $h_n^{(\varepsilon )}\rightarrow 0$ almost everywhere. Since by the above inequality,

$$\begin{aligned} \left| |\pi _n+\rho _n|^q - |\pi _n|^q -|\rho _n|^q\right|\le & {} \left| |\pi _n+\rho _n|^q - |\rho _n|^q \right| + |\pi _n|^q\\\le & {} \varepsilon |\rho _n|^q + (1+C_\varepsilon ) |\pi _n|^q , \end{aligned}$$

we have $h_n^{(\varepsilon )} \le (1+C_\varepsilon )|\Pi |^q.$ Thus, by the analogue of the dominated convergence theorem recalled above,

$$\begin{aligned} \int _X \left( \int _Y h_n^{(\varepsilon )}\,dy \right) ^{p/q} dx \rightarrow 0 . \end{aligned}$$

(A.3)

By definition of $h_n^{(\varepsilon )}$ we have

$$\begin{aligned} \left| |\pi _n+\rho _n|^q-|\pi _n|^q -|\rho _n|^q \right| \le \varepsilon |\rho _n|^q + h_n^{(\varepsilon )} \end{aligned}$$

and therefore

$$\begin{aligned}&\int _X \left( \int _Y \left| |\pi _n+\rho _n|^q-|\pi _n|^q -|\rho _n|^q \right| dy \right) ^{p/q} \,dx\\&\quad \le \int _X \left( \int _Y \left( \varepsilon |\rho _n|^q + h_n^{(\varepsilon )}\right) dy \right) ^{p/q} \,dx. \end{aligned}$$

In this inequality we first take the limsup as $n\rightarrow \infty$ and then we let $\varepsilon \rightarrow 0.$ Again by Lemma A.1 and (A.3), we have

$$\begin{aligned} \int _X \left( \int _Y \left( \varepsilon |\rho _n|^q + h_n^{(\varepsilon )}\right) dy \right) ^{p/q} \,dx=\varepsilon ^p\int _X \left( \int _Y |\rho _n|^q dy \right) ^{p/q} \,dx+o_{n\rightarrow {\infty }}(1). \end{aligned}$$

Since the $L^p_xL^q_y$-norm of $f_n,\pi _n,$ and $\sigma _n$ are uniformly bounded in n,  the $L^p_xL^q_y$-norm of $\rho _n$ is uniformly bounded in n by Lemma A.1. This proves (A.2). $\square$

Appendix B: A homogenization result in mixed Lebesgue spaces

The following is an extension of [1, Lem. 5.2] to mixed Lebesgue spaces. We use the convention for the torus

$$\begin{aligned} {{\mathbb {T}} }^k:=({{\mathbb {R}} }/(2\pi {{\mathbb {Z}} }))^k \end{aligned}$$

and denote by $d\theta$ normalized Lebesgue measure on ${{\mathbb {T}} }^k.$

Lemma B.1

Let $r>0,M,N\in {{\mathbb {N}} }^*,$ and

$$\begin{aligned} \psi :{{\mathbb {R}} }^M\times {{\mathbb {R}} }^N\times {{\mathbb {T}} }^M\times {{\mathbb {T}} }^N\rightarrow {{\mathbb {R}} }_+ \end{aligned}$$

a function satisfying the following assumptions :  there exists a zero measure set $E\subset {{\mathbb {R}} }^M\times {{\mathbb {R}} }^N$ such that

• (1) For any $(x_1,x_2)\notin E, (\theta _1,\theta _2)\mapsto \psi (x_1,x_2,\theta _1,\theta _2)$ is continuous on ${{\mathbb {T}} }^M\times {{\mathbb {T}} }^N;$

• (2) For any $(\theta _1,\theta _2)\in {{\mathbb {T}} }^M\times {{\mathbb {T}} }^N, (x_1,x_2)\mapsto \psi (x_1,x_2,\theta _1,\theta _2)$ is measurable on ${{\mathbb {R}} }^M\times {{\mathbb {R}} }^N;$

• (3)
  $$\begin{aligned} \int _{{{\mathbb {R}} }^M}\left( \int _{{{\mathbb {R}} }^N}\sup _{(\theta _1,\theta _2)\in {{\mathbb {T}} }^M\times {{\mathbb {T}} }^N}\psi (x_1,x_2,\theta _1,\theta _2)\,dx_2\right) ^r\,dx_1<{\infty }. \end{aligned}$$

Then,  we have

$$\begin{aligned}&\lim _{\varepsilon \rightarrow 0}\int _{{{\mathbb {R}} }^M}\left( \int _{{{\mathbb {R}} }^N} \psi (x_1,x_2,x_1/\varepsilon ^2,x_2/\varepsilon )\,dx_2\right) ^r\,dx_1\\&\quad =\int _{{{\mathbb {T}} }^M}\int _{{{\mathbb {R}} }^M}\left( \int _{{{\mathbb {T}} }^N}\int _{{{\mathbb {R}} }^N}\psi (x_1,x_2,\theta _1,\theta _2)\,dx_2\,d\theta _2\right) ^r\,dx_1\,d\theta _1. \end{aligned}$$

Remark B.2

We state the lemma with the scale $\varepsilon ^2$ for $x_1$ only for our application; it can be replaced by any scale of the form $f(\varepsilon )$ with $f(\varepsilon )\rightarrow 0$ as $\varepsilon \rightarrow 0.$

Proof

We mimic the proof of [1, Lem. 5.2], adapting it to the context of mixed Lebesgue spaces. Notice that our assumptions imply that $\psi$ is of Carathéodory type [15, Def. VIII.1.2] so that with the help of Fubini’s theorem, all the integrals that we consider are well-defined (the measurability is the hard part; however by [15, Prop. VIII.1.1] the function $\psi$ coincides with a measurable function on ${{\mathbb {R}} }^M\times {{\mathbb {R}} }^N\times {{\mathbb {T}} }^M\times {{\mathbb {T}} }^N$ a.e. in ${{\mathbb {R}} }^M\times {{\mathbb {R}} }^N,$ which imply that all the functions we consider are measurable on the appropriate space). Let $(Y_i)$ a paving of ${{\mathbb {T}} }^M$ by disjoint cubes of side length 1 / n. We first prove the result for the function

$$\begin{aligned} \psi _n(x_1,x_2,\theta _1,\theta _2):=\sum _i\psi (x_1,x_2,y_i,\theta _2){\mathbb 1 }_{Y_i}(\theta _1), \end{aligned}$$

where $(y_i)$ are arbitrary points in $Y_i.$ We have

$$\begin{aligned}&\int _{{{\mathbb {R}} }^M}\left( \int _{{{\mathbb {R}} }^N} \psi _n(x_1,x_2,x_1/\varepsilon ^2,x_2/\varepsilon )\,dx_2\right) ^r\,dx_1\\&\quad =\sum _i\int _{{{\mathbb {R}} }^M}{\mathbb 1 }_{Y_i}(x_1/\varepsilon ^2)\left( \int _{{{\mathbb {R}} }^N}\psi (x_1,x_2,y_i,x_2/\varepsilon )\,dx_2\right) ^r\,dx_1. \end{aligned}$$

Using Fubini’s theorem and [1, Lem. 5.2], we have for a.e. $x_1\in {{\mathbb {R}} }^M$ that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\int _{{{\mathbb {R}} }^N}\psi (x_1,x_2,y_i,x_2/\varepsilon )\,dx_2=\int _{{{\mathbb {T}} }^N}\int _{{{\mathbb {R}} }^N}\psi (x_1,x_2,y_i,\theta _2)\,dx_2\,d\theta _2. \end{aligned}$$

Furthermore, we have the uniform bound

$$\begin{aligned}&\left| \left( \int _{{{\mathbb {R}} }^N}\psi (x_1,x_2,y_i,x_2/\varepsilon )\,dx_2\right) ^r-\left( \int _{{{\mathbb {T}} }^N}\int _{{{\mathbb {R}} }^N}\psi (x_1,x_2,y_i,\theta _2)\,dx_2\,d\theta _2\right) ^r\right| \\&\quad \leqslant 2\left( \int _{{{\mathbb {R}} }^N}\sup _{(\theta _1,\theta _2)\in {{\mathbb {T}} }^M\times {{\mathbb {T}} }^N}\psi (x_1,x_2,\theta _1,\theta _2)\,dx_2\right) ^r \end{aligned}$$

which is integrable on ${{\mathbb {R}} }^M$ by assumption. By Lebesgue’s dominated convergence theorem, we deduce that

$$\begin{aligned}&\int _{{{\mathbb {R}} }^M}\left( \int _{{{\mathbb {R}} }^N} \psi _n(x_1,x_2,x_1/\varepsilon ^2,x_2/\varepsilon )\,dx_2\right) ^r\,dx_1\\&\quad =\sum _i\int _{{{\mathbb {R}} }^M}{\mathbb 1 }_{Y_i}(x_1/\varepsilon ^2)\left( \int _{{{\mathbb {T}} }^N}\int _{{{\mathbb {R}} }^N}\psi (x_1,x_2,y_i,\theta _2)\,dx_2\,d\theta _2\right) ^r\,dx_1+o_{\varepsilon \rightarrow 0}(1). \end{aligned}$$

Applying [1, Lem. 5.2] to the function

$$\begin{aligned} (x_1,\theta _1) \mapsto {\mathbb 1 }_{Y_i}(\theta _1) \left( \int _{{{\mathbb {T}} }^N}\int _{{{\mathbb {R}} }^N}\psi (x_1,x_2,y_i,\theta _2)\,dx_2\,d\theta _2\right) ^r \end{aligned}$$

we obtain

$$\begin{aligned}&\int _{{{\mathbb {R}} }^M}\left( \int _{{{\mathbb {R}} }^N} \psi _n(x_1,x_2,x_1/\varepsilon ^2,x_2/\varepsilon )\,dx_2\right) ^r\,dx_1\\&\quad =\sum _i \int _{{{\mathbb {T}} }^M} \int _{{{\mathbb {R}} }^M}{\mathbb 1 }_{Y_i}(\theta _1)\left( \int _{{{\mathbb {T}} }^N}\int _{{{\mathbb {R}} }^N}\psi (x_1,x_2,y_i,\theta _2)\,dx_2\,d\theta _2\right) ^r\,dx_1\,d\theta _1 +o_{\varepsilon \rightarrow 0}(1) \\&\quad = \int _{{{\mathbb {T}} }^M} \int _{{{\mathbb {R}} }^M} \left( \int _{{{\mathbb {T}} }^N}\int _{{{\mathbb {R}} }^N}\psi _n(x_1,x_2,\theta _1,\theta _2)\,dx_2\,d\theta _2\right) ^r\,dx_1\,d\theta _1 +o_{\varepsilon \rightarrow 0}(1) , \end{aligned}$$

which is the claimed formula for $\psi _n$ instead of $\psi .$

In the remainder of the proof we derive the formula for $\psi$ by showing that $\psi _n$ approximates $\psi$ in a suitable topology. Indeed, as in [1, Lem. 5.2], we know that the function

$$\begin{aligned} \delta _n(x_1,x_2):=\sup _{(\theta _1,\theta _2)\in {{\mathbb {T}} }^M\times {{\mathbb {T}} }^N}|\psi _n(x_1,x_2,\theta _1,\theta _2)-\psi (x_1,x_2,\theta _1,\theta _2)| \end{aligned}$$

satisfies $\delta _n\rightarrow 0$ a.e. in $(x_1,x_2)$ and that $0\leqslant \delta _n(x_1,x_2)\leqslant g(x_1,x_2)$ with

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle g(x_1,x_2):=2\sup _{(\theta _1,\theta _2)\in {{\mathbb {T}} }^M\times {{\mathbb {T}} }^N}\psi (x_1,x_2,\theta _1,\theta _2),\\ \displaystyle \int _{{{\mathbb {R}} }^M}\left( \int _{{{\mathbb {R}} }^N}g(x_1,x_2)\,dx_2\right) ^r\,dx_1<{\infty }. \end{array}\right. } \end{aligned}$$

Again by Fubini’s theorem and Lebesgue’s dominated convergence theorem, we deduce that

$$\begin{aligned} \lim _{n\rightarrow {\infty }}\int _{{{\mathbb {R}} }^M}\left( \int _{{{\mathbb {R}} }^N}\delta _n(x_1,x_2)\,dx_2\right) ^r\,dx_1=0. \end{aligned}$$

For shortness, let us introduce the notations

$$\begin{aligned} I_{1,2,\varepsilon }[\psi ]:= & {} \int _{{{\mathbb {R}} }^M}\left( \int _{{{\mathbb {R}} }^N} \psi (x_1,x_2,x_1/\varepsilon ^2,x_2/\varepsilon )\,dx_2\right) ^r\,dx_1,\\ \overline{I_{1,2}}[\psi ]:= & {} \int _{{{\mathbb {T}} }^M}\int _{{{\mathbb {R}} }^M}\left( \int _{{{\mathbb {T}} }^N}\int _{{{\mathbb {R}} }^N}\psi (x_1,x_2,\theta _1,\theta _2)\,dx_2\,d\theta _2\right) ^r\,dx_1\,d\theta _1. \end{aligned}$$

We need to show that $I_{1,2,\varepsilon }[\psi ]\rightarrow \overline{I_{1,2}}[\psi ]$ as $\varepsilon \rightarrow 0$ and, to do so, we distinguish whether $r\leqslant 1$ or $r>1.$

If $r\leqslant 1,$ we split

$$\begin{aligned} I_{1,2,\varepsilon }[\psi ]-\overline{I_{1,2}}[\psi ]= & {} I_{1,2,\varepsilon }[\psi ]-I_{1,2,\varepsilon } [\psi _n]+I_{1,2,\varepsilon }[\psi _n]-\overline{I_{1,2}}[\psi _n]\\&+\,\overline{I_{1,2}}[\psi _n] -\overline{I_{1,2}}[\psi ]. \end{aligned}$$

Using $|a^r-b^r|\leqslant |a-b|^r,$ we deduce that

$$\begin{aligned} |I_{1,2,\varepsilon }[\psi ]-I_{1,2,\varepsilon }[\psi _n]|\leqslant \left| \! \left| \delta _n \right| \! \right| _{L^rL^1}^r ,\quad |\overline{I_{1,2}}[\psi _n]-\overline{I_{1,2}}[\psi ]|\leqslant \left| \! \left| \delta _n \right| \! \right| _{L^rL^1}^r, \end{aligned}$$

so that for all $\alpha >0,$ there is n large enough so that for all $\varepsilon >0,$

$$\begin{aligned} |I_{1,2,\varepsilon }[\psi ]-\overline{I_{1,2}}[\psi ]|\leqslant |I_{1,2,\varepsilon }[\psi _n] -\overline{I_{1,2}}[\psi _n]|+\alpha . \end{aligned}$$

Taking the limit $\varepsilon \rightarrow 0,$ we find the desired result.

If $r>1,$ we introduce the notation

$$\begin{aligned} I_{2,\varepsilon }[\psi ](x_1):= & {} \int _{{{\mathbb {R}} }^N} \psi (x_1,x_2,x_1/\varepsilon ^2,x_2/\varepsilon )\,dx_2,\\ \overline{I_2}[\psi ](\theta _1,x_1):= & {} \int _{{{\mathbb {T}} }^N}\int _{{{\mathbb {R}} }^N} \psi (x_1,x_2,\theta _1,\theta _2)\,dx_2\,d\theta _2, \end{aligned}$$

so that $I_{1,2,\varepsilon }[\psi ]^{1/r}= \left| \! \left| I_{2,\varepsilon }[\psi ] \right| \! \right| _{L^r_{x_1}}$ and $\overline{I_{1,2}}[\psi ]^{1/r}= \left| \! \left| \overline{I_2}[\psi ] \right| \! \right| _{L^r_{\theta _1,x_1}}.$ We now split

$$\begin{aligned} I_{1,2,\varepsilon }[\psi ]^{1/r}-\overline{I_{1,2}}[\psi ]^{1/r}= & {} \left| \! \left| I_{2,\varepsilon }[\psi ] \right| \! \right| _{L^r_{x_1}}- \left| \! \left| I_{2,\varepsilon }[\psi _n] \right| \! \right| _{L^r_{x_1}}\\&+\, \left| \! \left| I_{2,\varepsilon }[\psi _n] \right| \! \right| _{L^r_{x_1}}- \left| \! \left| \overline{I_2}[\psi _n] \right| \! \right| _{L^r_{\theta _1,x_1}}\\&+\, \left| \! \left| \overline{I_2}[\psi _n] \right| \! \right| _{L^r_{\theta _1,x_1}}- \left| \! \left| \overline{I_2}[\psi ] \right| \! \right| _{L^r_{\theta _1,x_1}}. \end{aligned}$$

We now use the estimates

$$\begin{aligned}&\left| \left| \! \left| I_{2,\varepsilon }[\psi ] \right| \! \right| _{L^r_{x_1}}- \left| \! \left| I_{2,\varepsilon }[\psi _n] \right| \! \right| _{L^r_{x_1}}\right| \leqslant \left| \! \left| I_{2,\varepsilon }[\psi -\psi _n] \right| \! \right| _{L^r_{x_1}}\leqslant \left| \! \left| \delta _n \right| \! \right| _{L^rL^1},\\&\left| \left| \! \left| \overline{I_2}[\psi _n] \right| \! \right| _{L^r_{\theta _1,x_1}}- \left| \! \left| \overline{I_2}[\psi ] \right| \! \right| _{L^r_{\theta _1,x_1}}\right| \leqslant \left| \! \left| \overline{I_2}[\psi _n-\psi ] \right| \! \right| _{L^r_{\theta _1,x_1}}\leqslant \left| \! \left| \delta _n \right| \! \right| _{L^rL^1}, \end{aligned}$$

to deduce similarly as for $r\le 1$ that $I_{1,2,\varepsilon }[\psi ]^{1/r} \rightarrow \overline{I_{1,2}}[\psi ]^{1/r}$ as $\varepsilon \rightarrow 0.$ This completes the proof of the lemma. $\square$

Appendix C: A complex interpolation result

Proposition C.1

Let $1<p_0,p_1,q_0,q_1<{\infty },\alpha _0,\alpha _1>0$ and $\theta \in (0,1).$ Define

$$\begin{aligned} \frac{1}{p_\theta }=\frac{\theta }{p_1}+\frac{1-\theta }{p_0},\quad \frac{1}{q_\theta }=\frac{\theta }{q_1}+\frac{1-\theta }{q_0},\quad \alpha _\theta =\theta \alpha _1+(1-\theta )\alpha _0. \end{aligned}$$

Then,  there exists $C>0$ such that for all $f:{{\mathbb {R}} }_t\times {{\mathbb {R}} }_x\rightarrow {{\mathbb {C}} }$ such that the right side is well-defined,  we have

$$\begin{aligned} \left| \! \left| |D_x|^{\alpha _\theta }f \right| \! \right| _{L^{p_\theta }_tL^{q_\theta }_x}\leqslant C \left| \! \left| |D_x|^{\alpha _1}f \right| \! \right| _{L^{p_1}_tL^{q_1}_x}^\theta \left| \! \left| |D_x|^{\alpha _0}f \right| \! \right| _{L^{p_0}_tL^{q_0}_x}^{1-\theta }. \end{aligned}$$

Proof

By density, it suffices to prove the inequality for any f such that $\widehat{f}\in C^{\infty }_0({{\mathbb {R}} }^2\setminus \{(t,0),\,t\in {{\mathbb {R}} }\}),$ where $\widehat{f}$ is the x-Fourier transform of f. By duality, it is enough to prove that there exists $C>0$ such that for all $g\in L^{p_\theta '}L^{q_\theta '}$ we have

$$\begin{aligned} \langle g,|D_x|^{\alpha _\theta }f\rangle \leqslant C \left| \! \left| g \right| \! \right| _{L^{p_\theta '}L^{q_\theta '}} \left| \! \left| |D_x|^{\alpha _1}f \right| \! \right| _{L^{p_1}_tL^{q_1}_x}^\theta \left| \! \left| |D_x|^{\alpha _0}f \right| \! \right| _{L^{p_0}_tL^{q_0}_x}^{1-\theta }. \end{aligned}$$

(C.1)

Hence, let $g\in L^{p_\theta '}L^{q_\theta '}.$ We write g as $g=|g|h$ with |g| and h measurable, $|h|\leqslant 1.$ For $z\in {{\mathbb {C}} },$ consider the function

$$\begin{aligned}&\varphi (z)=(1+z)^{-1}\langle g_z,|D_x|^{z\alpha _1 + (1-z)\alpha _0}f\rangle ,\\&\quad g_z(t,x):= \left| \! \left| g(t,\cdot ) \right| \! \right| _{L^{q_\theta '}}^{cz+d}|g(t,x)|^{az+b}h(t,x), \end{aligned}$$

with the convention $0^z:=0,$ and where the parameters a, b, c, d are chosen so that

$$\begin{aligned} a=\frac{q_0'-q_1'}{\theta q_0'+(1-\theta )q_1'},\quad b=\frac{q_\theta '}{q_0'}=\frac{q_1'}{\theta q_0'+(1-\theta )q_1'},\quad c=-\frac{d}{\theta },\quad d=\frac{p_\theta '}{p_0'}-\frac{q_\theta '}{q_0'}. \end{aligned}$$

These assumptions imply the relations

$$\begin{aligned} a+b=\frac{q_\theta '}{q_1'},\quad b+d=\frac{p_\theta '}{p_0'},\quad a+b+c+d=\frac{p_\theta '}{p_1'}. \end{aligned}$$

Let $S=\{\lambda +is,\,\lambda \in (0,1),\,s\in {{\mathbb {R}} }\}$ a strip in the complex plane, and let us show that $\varphi$ is analytic on S,  continuous on $\overline{S}.$ For a.e. (t, x),  the function $z\mapsto \overline{g_z(t,x)}(|D_x|^{z\alpha _1+(1-z)\alpha _0}f)(t,x)$ is analytic on S,  continuous on $\overline{S}$ by the support assumptions made on $\widehat{f}.$ They also imply that there exists a $T>0$ such that for any $N\in {{\mathbb {N}} },$ there exists $C_{N,f}$ such that for any $z=\lambda +is\in \overline{S}$ and for any $(t,x)\in {{\mathbb {R}} }^2$ we have

$$\begin{aligned} |(|D_x|^{z\alpha _1+(1-z)\alpha _0}f)(t,x)|\leqslant C_{N,f}e^{|s|}{\mathbb 1 }_{[-T,T]}(t)(1+|x|)^{-N}. \end{aligned}$$

This can be done by integration by parts in the x-Fourier variables, the factor $e^{|s|}$ coming from the derivatives of $|\xi |^{z\alpha _1+(1-z)\alpha _2}$ which can be bounded by $|s|^M$ for some power M depending on N,  which we choose to bound independently by $e^{|s|}.$ Furthermore, for any $z\in \overline{S},$ the extremal values of $a\lambda +b$ are $b=q_\theta '/q_0'$ and $a+b=q_\theta '/q_1'$ so that

$$\begin{aligned} |g(t,x)|^{a\lambda +b}\leqslant |g(t,x)|^{q_\theta '/q_0'}+|g(t,x)|^{q_\theta '/q_1'}\in L^{q_0'}_x+L^{q_1'}_x. \end{aligned}$$

As a consequence, we infer that for a.e. $t\in {{\mathbb {R}} },$

$$\begin{aligned} z\mapsto \int _{{\mathbb {R}} }\overline{g_z(t,x)}(|D_x|^{z\alpha _1+(1-z)\alpha _0}f)(t,x)\,dx \end{aligned}$$

is analytic on S and continuous on $\overline{S}.$ It satisfies the bound for any $z\in \overline{S}$ and a.e. $t\in {{\mathbb {R}} }$

$$\begin{aligned} \left| \int _{{\mathbb {R}} }\overline{g_z(t,x)}(|D_x|^{z\alpha _1+(1-z)\alpha _0}f)(t,x)\,dx\right| \leqslant C_{N,f}{\mathbb 1 }_{[-T,T]}(t)e^{|s|} \left| \! \left| g(t,\cdot ) \right| \! \right| _{L^{q_\theta '}}^{(a+c)\lambda +b+d}. \end{aligned}$$

The extremal values of $(a+c)\lambda +b+d$ are $b+d=p_\theta '/p_0'$ and $a+b+c+d=p_\theta '/p_1'$ so that

$$\begin{aligned} \left| \! \left| g(t,\cdot ) \right| \! \right| _{L^{q_\theta '}}^{(a+c)\lambda +b+d}\leqslant \left| \! \left| g(t,\cdot ) \right| \! \right| _{L^{q_\theta '}}^{p_\theta '/p_0'}+ \left| \! \left| g(t,\cdot ) \right| \! \right| _{L^{q_\theta '}}^{p_\theta '/p_1'}\in L^{p_0'}_t+L^{p_1'}_t. \end{aligned}$$

This implies that $\varphi$ is analytic on S and continuous on $\overline{S},$ with the bound valid for any $z=\lambda +is\in \overline{S},$

$$\begin{aligned} |\varphi (z)|\leqslant C_{N,f}e^{|s|} \left| \! \left| g \right| \! \right| _{L^{p_\theta '}_tL^{q_\theta '}_x}. \end{aligned}$$

On the boundary of S,  let us show more precise bounds. For any $s\in {{\mathbb {R}} },$ we have

$$\begin{aligned} \left| \int _{{\mathbb {R}} }\overline{g_{is}(t,x)}(|D_x|^{\alpha _0+is(\alpha _1-\alpha _0)}f)(t,x)\,dx\right| \leqslant \left| \! \left| g_{is}(t,\cdot ) \right| \! \right| _{L^{q_0'}_x} \left| \! \left| |D_x|^{is(\alpha _1-\alpha _0)}|D_x|^{\alpha _0}f \right| \! \right| _{L^{q_0}_x}. \end{aligned}$$

For any $\eta \in {{\mathbb {R}} },$ the Fourier multiplier by $m_\eta (\xi ):=|\xi |^{i\eta }$ satisfies the bounds

$$\begin{aligned} |m_\eta (\xi )|\leqslant 1,\quad |\xi ||m_\eta '(\xi )|\leqslant |\eta |. \end{aligned}$$

By the Marcinkiewicz multiplier theorem [24, Thm. 5.2.2], this implies the $L^p$-bound for all $p>1$:

$$\begin{aligned} \left| \! \left| |D_x|^{i\eta }g \right| \! \right| _{L^p}\leqslant C(1+|\eta |) \left| \! \left| g \right| \! \right| _{L^p}, \end{aligned}$$

which in our case gives

$$\begin{aligned} \left| \! \left| |D_x|^{is(\alpha _1-\alpha _0)}|D_x|^{\alpha _0}f \right| \! \right| _{L^{q_0}_x}\leqslant C(1+|s|) \left| \! \left| |D_x|^{\alpha _0}f \right| \! \right| _{L^{q_0}_x}. \end{aligned}$$

Using the bound

$$\begin{aligned} \left| \! \left| g_{is}(t,\cdot ) \right| \! \right| _{L^{q_0'}_x}\leqslant \left| \! \left| g(t,\cdot ) \right| \! \right| _{L^{q_\theta '}}^{b+d}, \end{aligned}$$

together with the relation $b+d=p_\theta '/p_0',$ we deduce that

$$\begin{aligned} |\varphi (is)|\leqslant C \left| \! \left| g \right| \! \right| _{L^{p_\theta '}_tL^{q_\theta '}_x} \left| \! \left| |D_x|^{\alpha _0}f \right| \! \right| _{L^{p_0}_tL^{q_0}_x}. \end{aligned}$$

Here, we see the role of the prefactor $(1+z)^{-1}$ in front of $\varphi (z)$ to compensate the growth of the $L^p$-multiplier norm of $m_\eta .$ By the same method, we have the estimate for all $s\in {{\mathbb {R}} }$

$$\begin{aligned} |\varphi (is)|\leqslant C \left| \! \left| g \right| \! \right| _{L^{p_\theta '}_tL^{q_\theta '}_x} \left| \! \left| |D_x|^{\alpha _1}f \right| \! \right| _{L^{p_1}_tL^{q_1}_x} \end{aligned}$$

using the relations $a+b=q_\theta '/q_1'$ and $a+b+c+d=p_\theta '/p_1'.$ Using Hadamard’s three line lemma [47, Thm. 5.2.1], we deduce (C.1), which ends the proof. $\square$

Appendix D: Weak compactness of the Airy–Strichartz map

Lemma D.1

Let $\alpha \in (-1/2,1)$ and $(u_n)\subset L^2({{\mathbb {R}} })$ a sequence converging weakly to zero in $L^2({{\mathbb {R}} }).$ Then,  up to a subsequence,  $|D_x|^\alpha e^{-t\partial _x^3}u_n\rightarrow 0$ a.e. in ${{\mathbb {R}} }^2.$

The proof follows from some local smoothing properties of the Airy kernel:

Lemma D.2

Let $a\in L^1({{\mathbb {R}} })$ a non-negative function. Then,  for all $u\in L^2({{\mathbb {R}} })$ we have

$$\begin{aligned} \int _{{{\mathbb {R}} }^2}a(x)\left| |D_x|e^{-t\partial _x^3}u(x)\right| ^2dx\,dt\leqslant \frac{1}{3} \left| \! \left| a \right| \! \right| _{L^1} \left| \! \left| u \right| \! \right| _{L^2}^2. \end{aligned}$$

Proof

By the Plancherel identity, we have

$$\begin{aligned}&\int _{{{\mathbb {R}} }^2}a(x)\left| |D_x|e^{-t\partial _x^3}u(x)\right| ^2dx\,dt\\&\quad =\sqrt{2\pi }\int _{{{\mathbb {R}} }^2}\widehat{a}(\xi '-\xi )|\xi ||\xi '|\widehat{u}(\xi )\overline{\widehat{u}(\xi ')}\delta (\xi ^3-\xi '^3)\,d\xi \,d\xi '. \end{aligned}$$

Using $\delta (\xi ^3-\xi '^3)=\delta (\xi -\xi ')/(3\xi ^2),$ we deduce

$$\begin{aligned} \int _{{{\mathbb {R}} }^2}a(x)\left| |D_x|e^{-t\partial _x^3}u(x)\right| ^2dx\,dt=\frac{\sqrt{2\pi }}{3}\int _{{{\mathbb {R}} }^2}\widehat{a}(0)|\widehat{u}(\xi )|^2\,d\xi \leqslant \frac{1}{3} \left| \! \left| a \right| \! \right| _{L^1} \left| \! \left| u \right| \! \right| _{L^2}^2 \end{aligned}$$

$\square$

Proof of Lemma D.1

We prove that $|D_x|^\alpha e^{-t\partial _x^3}u_n\rightarrow 0$ in $L^2_\text {loc}({{\mathbb {R}} }^2),$ which implies the result. Hence, let $K\subset {{\mathbb {R}} }^2$ a bounded set, and let us show that $\chi _K|D_x|^\alpha e^{-t\partial _x^3}u_n\rightarrow 0$ in $L^2({{\mathbb {R}} }^2).$ To this end, let $\varepsilon >0$ and $\Lambda >0.$ Define $P_\Lambda$ the Fourier multiplier on $L^2_x({{\mathbb {R}} })$ by ${\mathbb 1 }(|\xi |\leqslant \Lambda ),$ and $P_\Lambda ^\perp :=1-P_\Lambda .$ We split $\chi _K|D_x|^\alpha e^{-t\partial _x^3}u_n=\chi _KP_\Lambda |D_x|^\alpha e^{-t\partial _x^3}u_n+\chi _KP_\Lambda ^\perp |D_x|^\alpha e^{-t\partial _x^3}u_n,$ and notice that

$$\begin{aligned} \left| \! \left| \chi _KP_\Lambda ^\perp |D_x|^\alpha e^{-t\partial _x^3}u_n \right| \! \right| _{L^2_{t,x}}\leqslant & {} \left| \! \left| \chi _K e^{x^2} \right| \! \right| _{L^{\infty }_{t,x}} \left| \! \left| e^{-x^2}|D_x|e^{-t\partial _x^3} \right| \! \right| _{L^2_x\rightarrow L^2_{t,x}}\\&\times \left| \! \left| P_\Lambda ^\perp |D_x|^{\alpha -1} \right| \! \right| _{L^2_x\rightarrow L^2_x} \left| \! \left| u_n \right| \! \right| _{L^2}\\\leqslant & {} C_K\Lambda ^{\alpha -1}, \end{aligned}$$

for some constant $C_K>0$ independent of n,  by Lemma D.1 and the boundedness of $(u_n)$ in $L^2({{\mathbb {R}} }).$ Hence, for $\Lambda$ large enough independent of n,  we have

$$\begin{aligned} \left| \! \left| \chi _KP_\Lambda ^\perp |D_x|^\alpha e^{-t\partial _x^3}u_n \right| \! \right| _{L^2_{t,x}}\leqslant \varepsilon . \end{aligned}$$

For any fixed $t\in {{\mathbb {R}} },$ the operator $\chi _K(t,\cdot )P_\Lambda |D_x|^\alpha e^{-t\partial _x^3}$ is compact on $L^2_x({{\mathbb {R}} }),$ hence

$$\begin{aligned} \chi _K(t,\cdot )P_\Lambda |D_x|^\alpha e^{-t\partial _x^3}u_n\rightarrow 0 \end{aligned}$$

strongly in $L^2_x({{\mathbb {R}} })$ as $n\rightarrow {\infty },$ by weak convergence of $(u_n)$ in $L^2({{\mathbb {R}} }).$ Furthermore, we always have

$$\begin{aligned} \left| \! \left| \chi _K(t,\cdot )|D_x|^\alpha P_\Lambda e^{-t\partial _x^3}u_n \right| \! \right| _{L^2_x}^2\leqslant C\Lambda ^{\alpha +\frac{1}{2}} \left| \! \left| \chi _K(t,\cdot ) \right| \! \right| _{L^2_x}^2, \end{aligned}$$

with $C>0$ independent of n. By Lebesgue’s dominated convergence theorem, we deduce that $\chi _KP_\Lambda |D_x|^\alpha e^{-t\partial _x^3}u_n\rightarrow 0$ in $L^2({{\mathbb {R}} }^2)$ as $n\rightarrow {\infty },$ from which the result follows. $\square$

Appendix E: Maximizers in the subcritical case

In the subcritical case $\gamma <1/p,$ the existence of maximizers is simpler and unconditional. Define

$$\begin{aligned} \mathcal {A}_{\gamma ,p}:=\sup _{u\ne 0}\frac{\displaystyle \int _{{\mathbb {R}} }\left( \int _{{\mathbb {R}} }\left| |D_x|^\gamma (e^{-t\partial _x^3}u)(x)\right| ^qdx\right) ^{p/q}dt}{ \left| \! \left| u \right| \! \right| _{L^2}^p} \end{aligned}$$

(E.1)

where q is determined by p and $\gamma$ as in (1.3). Then, we can prove the following result.

Theorem 4

Let $p>4,-1/2<\gamma <1/p,$ and q such that $-\gamma +3/p+1/q=1/2.$ Then,  any maximizing sequence for $\mathcal {A}_{\gamma ,p}$ is precompact up to symmetries and,  in particular,  there exists maximizers for $\mathcal {A}_{\gamma ,p}.$

This result with $p=q=8$ is due to [25].

Remark E.1

The same result holds for real-valued functions, with the same proof.

Proof of Theorem 4

We mimic the proof in Sect. 2. The analogue of Proposition 2.3 is valid, with the same proof using Lemma D.1 and the condition $\gamma >-1/2.$ We now show that $\mathcal {A}_{\gamma ,p}^*=0,$ from which the result follows. To do so, we argue by contradiction and assume $\mathcal {A}_{\gamma ,p}^*>0,$ and let $(u_n)$ a sequence such that $\left| \! \left| u_n \right| \! \right| _{L^2}=1,u_n\rightharpoonup _{\text {sym}}0,$ and

$$\begin{aligned} \limsup _{n\rightarrow {\infty }} \left| \! \left| |D_x|^\gamma e^{-t\partial _x^3}u_n \right| \! \right| _{L^p_tL^q_x}^p\geqslant \frac{1}{2}\mathcal {A}^*_{\gamma ,p}. \end{aligned}$$

In particular, we have $|D_x|^\gamma e^{-t\partial _x^3}u_n\not \rightarrow 0$ in $L^p_t L^q_x.$ In the subcritical case, we have results identical to Corollaries 3.1 and 3.2 by interpolating $(\gamma ,p,q)$ between $(\widetilde{\gamma },p,\widetilde{q})$ and $(1/p,p,2p/(p-4))$ with $-1/2<\widetilde{\gamma }<\gamma$ and using Proposition C.1. Hence, there exists $(g_n)\subset G$ and $(\eta _n)\subset {{\mathbb {R}} }$ with $|\eta _n|\geqslant 1/2$ such that $(\widehat{g_nu_n}(\cdot +\eta _n))$ has a non-zero weak limit v in $L^2$ (here, we do not need to distinguish between positive and negative frequencies), with a lower bound

$$\begin{aligned} \left| \! \left| v \right| \! \right| _{L^2}\geqslant \varepsilon >0, \end{aligned}$$

where $\varepsilon$ only depends on $p,\gamma .$ Again, we must have $|\eta _n|\rightarrow {\infty }.$ Writing again $\delta _n:=1/\eta _n\rightarrow 0$ and $\widehat{g_nu_n}(\cdot +\eta _n)=\widehat{v}+\widehat{r_n},$ we have

$$\begin{aligned} \left| \! \left| |D_x|^\gamma e^{-t\partial _x^3}u_n \right| \! \right| _{L^p_tL^q_x}=|\delta _n|^{\frac{1}{p}-\gamma } \left| \! \left| T_{\gamma ,\delta _n}v+T_{\gamma ,\delta _n}r_n \right| \! \right| _{L^p_tL^q_x}, \end{aligned}$$

where the approximate operator $T_{\gamma ,\delta }$ is

$$\begin{aligned} (T_{\gamma ,\delta }u)(x):=\frac{1}{\sqrt{2\pi }}\int _{{\mathbb {R}} }|1+\delta \xi |^\gamma e^{ix\xi +it(3\xi ^2+\delta \xi ^3)}\widehat{u}(\xi )\,d\xi . \end{aligned}$$

As in Lemma 4.1, we have

$$\begin{aligned} \left| \! \left| T_{\gamma ,\delta } \right| \! \right| _{L^2_x\rightarrow L^p_t L^q_x}=C\delta ^{\gamma -\frac{1}{p}}, \end{aligned}$$

and for all $u\in L^2_x({{\mathbb {R}} })$ and all $\gamma <1/p,$

$$\begin{aligned} \lim _{\delta \rightarrow 0}|\delta |^{\frac{1}{p}-\gamma } \left| \! \left| T_{\gamma ,\delta }u \right| \! \right| _{L^p_tL^q_x}=0. \end{aligned}$$

As a consequence, we find that

$$\begin{aligned} \left| \! \left| |D_x|^\gamma e^{-t\partial _x^3}u_n \right| \! \right| _{L^p_tL^q_x}=|\delta _n|^{\frac{1}{p}-\gamma } \left| \! \left| T_{\gamma ,\delta _n}r_n \right| \! \right| _{L^p_tL^q_x}+o_{n\rightarrow {\infty }}(1), \end{aligned}$$

and undoing the change of variables shows that

$$\begin{aligned} \left| \! \left| |D_x|^\gamma e^{-t\partial _x^3}u_n \right| \! \right| _{L^p_tL^q_x}^p= \left| \! \left| |D_x|^\gamma e^{-t\partial _x^3}w_n \right| \! \right| _{L^p_tL^q_x}^p+o_{n\rightarrow {\infty }}(1), \end{aligned}$$

(E.2)

where $\widehat{w_n}=\widehat{r_n}(\cdot -\eta _n).$ By weak convergence of $r_n$ to zero, we know that $\left| \! \left| r_n \right| \! \right| _{L^2}^2\rightarrow 1- \left| \! \left| v \right| \! \right| _{L^2}^2.$ Hence, as in the proof of Theorem 2, we have $w_n\rightharpoonup _\text {sym}0$ and

$$\begin{aligned} \limsup _{n\rightarrow {\infty }} \left| \! \left| |D_x|^\gamma e^{-t\partial _x^3}w_n \right| \! \right| _{L^p_tL^q_x}^p\leqslant \mathcal {A}_{\gamma ,p}^*(1- \left| \! \left| v \right| \! \right| _{L^2}^2)^{p/2}, \end{aligned}$$

which we insert in (E.2) to obtain

$$\begin{aligned} \limsup _{n\rightarrow {\infty }} \left| \! \left| |D_x|^\gamma e^{-t\partial _x^3}u_n \right| \! \right| _{L^p_tL^q_x}^p\leqslant \mathcal {A}_{\gamma ,p}^*(1- \left| \! \left| v \right| \! \right| _{L^2}^2)^{p/2}\leqslant \mathcal {A}_{\gamma ,p}^*(1-\varepsilon ^2)^{p/2}. \end{aligned}$$

Taking the supremum over all such sequences $(u_n),$ we find

$$\begin{aligned} \mathcal {A}^*_{\gamma ,p}\leqslant \mathcal {A}_{\gamma ,p}^*(1-\varepsilon ^2)^{p/2}<\mathcal {A}^*_{\gamma ,p}, \end{aligned}$$

leading to a contradiction. We thus have $\mathcal {A}_{\gamma ,p}^*=0,$ which finishes the proof. $\square$

Appendix F: Symmetries for extension problems

In this section we show that the argument provided in Lemma 2.8 about real-valuedness of maximizers extends to a more general setting. A similar remark was made independently in [8]. If $N\geqslant 1,S\subset {{\mathbb {R}} }^N, \sigma$ is a Borel measure on S,  and $f\in L^1(S,\sigma ),$ we define its Fourier transform as

$$\begin{aligned} \forall x\in {{\mathbb {R}} }^N,\,\check{f}(x):=\int _S e^{ix\cdot \omega }f(\omega )\,d\sigma (\omega ). \end{aligned}$$

Previously, we considered the case $N=2, S=\{(\xi ,\xi ^3),\,\xi \in {{\mathbb {R}} }\}$ the cubic curve and the measure $\sigma$ being the push-forward of the measure $|\xi |^\gamma d\xi$ through the map $\xi \in {{\mathbb {R}} }\mapsto (\xi ,\xi ^3)\in S.$ Notice that in the optimization problem (1.4), the $L^2$-norm is taken with respect to another measure on S than $\sigma .$ As a consequence, let $\sigma '$ be another Borel measure on S,  and $q\geqslant 2.$ Define

$$\begin{aligned} \mathcal {M}(S,\sigma ,\sigma ',q):=\sup \left\{ \frac{ \left| \! \left| \check{f} \right| \! \right| _{L^q({{\mathbb {R}} }^N)}}{ \left| \! \left| f \right| \! \right| _{L^2(S,\sigma ')}}:\ f\in L^1(S,\sigma )\cap L^2(S,\sigma ')\setminus \{0\}\right\} \end{aligned}$$

and, under the symmetry assumption $S=-S,$ also its ‘symmetric’ version

We then have the following statement.

Lemma F.1

Let $N\geqslant 1,S\subset {{\mathbb {R}} }^N$ and $\sigma ,\sigma '$ Borel measures on S. Assume that $(S,\sigma )$ and $(S,\sigma ')$ are symmetric with respect to the origin,  that is,  $S=-S$ and $\sigma (A)=\sigma (-A),\sigma '(A)=\sigma '(-A)$ for all A Borel subset of S. Then,  for any $q\geqslant 2,$ we have

$$\begin{aligned} \mathcal {M}(S,\sigma ,\sigma ',q)=\mathcal {M}_\mathrm{sym}(S,\sigma ,\sigma ',q). \end{aligned}$$

Moreover,  there is an optimizer for $\mathcal {M}(S,\sigma ,\sigma ',q)$ if and only if there is one for $\mathcal {M}_\mathrm{sym}(S,\sigma ,\sigma ',q).$

We emphasize that in the definition of $\mathcal {M}(S,\sigma ,\sigma ',q)$ and $\mathcal {M}_{\text {sym}}(S,\sigma ,\sigma ',q)$ we impose the condition $f\in L^1(S,\sigma )$ only in order to have $\check{f}$ a priori well-defined. Once it is shown that $\mathcal {M}(S,\sigma ,\sigma ',q)<\infty$ it follows that $\check{f}\in L^q({{\mathbb {R}} }^N)$ for any $f\in L^2(S,\sigma ')$ and the condition $f\in L^1(S,\sigma )$ can be dropped. In particular, for an optimizer for $\mathcal {M}(S,\sigma ,\sigma ',q)$ or $\mathcal {M}_{\text {sym}}(S,\sigma ,\sigma ',q)$ we do not require this condition.

Remark F.2

This result applies to $S=\mathbb {S}^{N-1}$ and $\sigma =\sigma '$ is the standard surface measure on $\mathbb {S}^{N-1},$ which is the case of the Stein–Tomas theorem.

Proof

Since the inequality $\geqslant$ is trivial, let us prove the inequality $\leqslant .$ Let $f\in L^1(S,\sigma )\cap L^2(S,\sigma '),f\ne 0.$ We split $f=f_1+if_2$ where

$$\begin{aligned} f_1(\omega ):=\frac{f(\omega )+\overline{f(-\omega )}}{2},\quad f_2(\omega ):=\frac{f(\omega )-\overline{f(-\omega )}}{2i}, \end{aligned}$$

so that $f_1(-\omega )=\overline{f_1(\omega )}$ and $f_2(-\omega )=\overline{f_2(\omega )}$ for a.e. $\omega \in S.$ Using the symmetry of $(S,\sigma ),$ we deduce that $\check{f_1}$ and $\check{f_2}$ are real-valued, so that $|\check{f}|^2=|\check{f_1}|^2+|\check{f_2}|^2$ and hence by the triangle inequality in $L^{q/2}({{\mathbb {R}} }^N)$

$$\begin{aligned} \left| \! \left| \check{f} \right| \! \right| _{L^q({{\mathbb {R}} }^N)}= & {} \left| \! \left| |\check{f_1}|^2+|\check{f_2}|^2 \right| \! \right| _{L^{q/2}({{\mathbb {R}} }^N)}^{1/2}\\\leqslant & {} \left( \left| \! \left| \check{f_1} \right| \! \right| _{L^q({{\mathbb {R}} }^N)}^2+ \left| \! \left| \check{f_2} \right| \! \right| _{L^q({{\mathbb {R}} }^N)}^2\right) ^{1/2}\\\leqslant & {} \mathcal {M}_{\text {sym}}(S,\sigma ,\sigma ',q)\left( \left| \! \left| f_1 \right| \! \right| _{L^2(S,\sigma ')}^2+ \left| \! \left| f_2 \right| \! \right| _{L^2(S,\sigma ')}^2\right) ^{1/2}. \end{aligned}$$

We notice now that $|f_1(\omega )|^2+|f_2(\omega )|^2=(1/2) (|f(\omega )|^2 + |f(-\omega )|^2)$ for all $\omega \in S,$ and therefore, by symmetry of S and $\sigma ',$

$$\begin{aligned} \left| \! \left| f_1 \right| \! \right| _{L^2(S,\sigma ')}^2+ \left| \! \left| f_2 \right| \! \right| _{L^2(S,\sigma ')}^2 = \Vert f\Vert _{L^2(S,\sigma ')}^2 . \end{aligned}$$

By taking the supremum over all f,  we therefore obtain the inequality $\leqslant$ in the lemma. Moreover, if f is an optimizer and the suprema are finite, then tracking the case of equality shows that either $f_1$ or $f_2$ is also a maximizer, showing the desired property. $\square$

Remark F.3

The previous proof clearly extends to mixed Lebesgue spaces (as in the case of the cubic curve), as long as the Lebesgue exponents are greater than 2.