# To give an estimate for the maximal function associated to the Schrödinger group by using a measurable selector function

I am consulting some papers (references below) about the Carleson's problem for the pointwise convergence of the Schrödinger group \begin{equation} S_t=e^{i t \Delta}. \end{equation} In this context it is important to obtain $L^p$-estimates for the maximal function \begin{equation} S^\star f(x)=\sup_{t\in(0,1)}\lvert \left(S_tf\right)(x)\rvert, \end{equation} where $f\in H^s(\mathbb{R})$. Precisely, we want to find conditions on $s\ge 0$ such that the following inequality holds for some $p\in [1, \infty)$ and all $f\in H^s(\mathbb{R})$: \begin{equation}\tag{1} \lVert S^\star f\rVert_{L^p(\left[-1, 1\right])}\le C \lVert f\rVert_{H^s(\mathbb{R})}. \end{equation}

Question. Both referenced papers claim without further explanation that (1) follows from \begin{equation}\tag{2} \left\lVert \left(S_{t(x)}f\right)(x)\right\rVert_{L^p([-1,1], dx)}\le C \lVert f\rVert_{H^s(\mathbb{R})}, \qquad \forall t\colon \mathbb{R}\to\mathbb{R}\ \text{measurable}.\end{equation} Can you prove the implication $(2)\Rightarrow (1)$ in full detail?

I guess that we should begin by assuming that $f$ is Schwartz, so that $S_tf(x)$ is a continuous function of $(x, t)$. In this case we can use selection theorems (cfr. this answer by Ilya on MathSE) to write $$S^\star f(x)=\left\lvert S_{t(x)}f(x)\right \rvert,$$ for a Borel-measurable function $t=t(x)$, and so obtain (1) from (2). This answers the question when $f$ is Schwartz.

I am not sure on how to eliminate this restriction. Can you help me?

References:

• B. Dahlberg, C. Kenig, "A note on the almost everywhere behaviour of solutions to the Schrödinger equation", Harmonic Analysis (Minneapolis, Minn. 1981), Lecture Notes in Math., vol. 908, Springer, Berlin, 1982, pp.205-209;
• K. Rogers, A. Vargas, L. Vega, "Pointwise convergence of solutions to the nonelliptic Schrödinger equation", Indiana University Mathematics Journal, Vol. 55 No. 6 (2006) pp. 1893-1906.

I have had the opportunity to have a chat with some of the authors, so that I am now able to answer this question myself.

For the sake of clarity let us formulate the answer as a theorem.

Theorem Notation for $S_t$ and $S^\star$ being as above, and $\mathscr{S}$ denoting the Schwartz class, the following propositions are equivalent:

1. There exists an absolute constant $C>0$ such that $$\lVert S^\star f\rVert_{L^p([-1, 1])}\le C \lVert f\rVert_{H^s},\qquad \forall f\in H^s(\mathbb{R}).$$
2. The following inequality holds true (where $C>0$ is the same constant as in 1.): $$\lVert S^\star f \rVert_{L^p([-1,1])}\le C \lVert f \rVert_{H^s},\qquad \forall f \in \mathscr{S}(\mathbb{R}).$$
3. The following inequality holds true (where $C>0$ is the same constant as in 1 and 2): $$\left\lVert \left(S_{t(x)} f\right)(x) \right\rVert_{L^p([-1,1], dx)}\le C \lVert f \rVert_{H^s},\qquad \forall f \in \mathscr{S}(\mathbb{R}),\ \forall t\colon \mathbb{R}\to (0,1)\ \text{Borel-meas.}$$

Proof (sketch) We need to prove the implications $3.\Rightarrow 2. \Rightarrow 1.$

$3.\Rightarrow 2.$ This is a consequence of the existence of Borel-measurable selectors (cfr. this aforementioned link). Actually, it is enough to prove the existence of approximate selectors. That is, for fixed $f\in \mathscr{S}(\mathbb{R})$ we can prove that for every $\varepsilon>0$ there exists a Borel measurable function $t_\varepsilon\colon \mathbb{R}\to (0,1)$ such that $$\left\lVert \left(S_{t(x)}f\right)- S^\star f\right\rVert_{L^p([-1, 1], dx)}\le \varepsilon.$$

$2.\Rightarrow 1.$ Let $f_j\to f$ in $H^s$ and all $f_j$ are Schwartz. Since \begin{equation} \left\lvert S^\star f_j -S^\star f_l \right\rvert \le S^\star(f_j-f_l), \end{equation} the sequence $S^\star f_j$ is $L^p$-Cauchy and so, exchanging $f_j$ for a subsequence if necessary, \begin{equation} \lim_{j\to \infty} S^\star f_j(x)\ \text{exists a.e.} \end{equation} Now fix $t\in (0,1)$. Since $S_t$ is $L^2$-continuous, there exists a subsequence $f_{j_k}$ (with $j_k$ depending on $t$) such that \begin{equation} S_tf(x)=\lim_{k \to \infty} S_t f_{j_k}(x),\qquad x\text{-a.e.} \end{equation} So \begin{equation} \begin{split} \left \lvert S_tf (x)\right\rvert &= \lim_{k\to\infty} \left\lvert S_t f_{j_k}(x)\right\rvert \\ &\le \limsup_{k\to\infty} S^\star f_{j_k}(x) \\ &= \lim_{j\to \infty} S^\star f_j(x). \end{split} \end{equation} Taking the $\sup$ for $t\in (0,1)$ we get
$$S^\star f(x)\le \lim_{j\to \infty} S^\star f_j(x)$$ and applying Fatou's lemma \begin{equation} \left\lVert S^\star f\right\rVert_{L^p([-1, 1])}\le \lim_{j \to \infty} \left\lVert S^\star f_j\right\rVert_{L^p([-1, 1])}\le C\lVert f \rVert_{H^s}. \end{equation}