Let $a>0$ and consider the operator $$Tf(t,x)= \int_{\mathbb{R}^{n}}e^{ i x\cdot \xi} e^{i t \lvert\xi\rvert^{a}} \widehat{f}(\xi) \, d\xi.$$

When $a=2$, the function $Tf$ solves the Cauchy problem associated with the homogeneous Schrödinger equation: $$i \, \partial_t u+i \, \Delta u=0,\qquad u(0)=f.$$

A classical problem is to study the pointwise convergence of the solution $u(t,x)$ to $f(x)$ for almost every $x$ as $t\to 0$. To this end, it is useful to study the estimate $$\lVert Sf\rVert_q \leq C \lVert f\rVert_{H^s}, \tag1\label{1}$$ where $$Sf(x)=\sup_{0<t<1}\lvert Tf(t,x)\rvert$$ and $H^s$ is the usual Sobolev space of functions $f\in \mathcal{S}^{\prime}$ such that $\lVert(1+\lvert\xi\rvert^s)\widehat{f}\rVert_2<\infty$.

My question:

I have seen this step in many papers on this problem and I can not make it rigorous: To prove \eqref{1}, it suffices to prove that $$\lVert Rf\rVert_q\leq C \lVert f\rVert_{H^s}, \tag2\label{2}$$ where $$Rf(x)= \int_{\mathbb{R}^n}e^{ i x\cdot \xi} e^{i t(x) \lvert \xi\rvert^a} \widehat{f}(\xi) \, d\xi.$$ and $t:\mathbb{R}^n\to (0,1)$ is a measurable function. Is the pointwise relation $\lvert Sf(x)\rvert\leq C \lvert Rf(x)\rvert$ true and why ?

Remark: I have just found an answer to a very closely related question To give an estimate for the maximal function associated to the Schrödinger group by using a measurable selector function. It is very interesting. It treats the estimate \eqref{1} with the $L^p(\mathbb{R}^{n})$ norm on the left side replaced by a an $L^p(B)$ where $B$ is a ball. I will try to adapt the solution to the global estimate.

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    $\begingroup$ There exists a choice of $t$ such that $|Rf(x)| \geq \frac{1}{2} Sf(x)$ (say) for all $x$: one basically chooses $t(x)$ to nearly maximize $|T f(t,x)|$, conceding a factor of 2 so that we don't have to worry about whether the supremum is attained. [There is a small technicality in ensuring that $t$ can be chosen measurably.] This technique is often referred to as "linearizing a maximal operator". I believe it is discussed in Stein's "Harmonic analysis" in the section on maximal operators, though I do not have it handy at present; maybe others can supply more precise references. $\endgroup$
    – Terry Tao
    Jul 19 at 0:42
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    $\begingroup$ If you can't type the ö in Schrödinger, it is acceptable to write Schroedinger. If it were Schrodinger, it would be pronounced differently. $\endgroup$ Jul 19 at 10:49
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    $\begingroup$ Your link to an answer went to a question, so I assumed you meant the sole answer, and edited accordingly. I hope that was correct. \\ @MichaelHardy, thanks for adding \tags in your edit! If you combine those with \labels, then one can \eqref them. I edited accordingly In fact, they are global labels per page, so one can \eqref even in an answer or comment (as to \eqref{1} and \eqref{2}). $\endgroup$
    – LSpice
    Jul 20 at 22:35

1 Answer 1


$\newcommand\dotcup{\mathbin{\dot\cup}}$I shared your confusion until a year ago, when Markus Haase explained to me that \eqref{2} can actually be rewritten as assertion (ii) in the following theorem, which I find much clearer.

Theorem. Let E be a Banach space, let $(\Omega,\mu)$ be a measure space, let $p \in [1,\infty)$ and let $I$ be a non-empty index set. For each $i \in I$ let $T_i: E \to L^p := L^p(\Omega,\mu)$ be a bounded linear operator. The following are equivalent:

(i) The operator family $(T_i)_{i \in I}$ satisfies a maximal inequality, i.e., there is a constant $C \ge 0$ with the following property: for every $f \in E$ there exists $0 \le h \in L^p$ of norm $\lVert h\rVert_{L^p} \le C \lVert f\rVert_{E}$ such that $\lvert T_if\rvert \le h$ for all $i \in I$ (where the inequality is to be understood pointwise almost everywhere).

(ii) There exists a constant $C$ with the following property: for every finite measurable partition $\Omega = A_1 \dotcup \dotsb \dotcup A_n$, all indices $i_1, \dotsc, i_n \in I$, and every $f \in E$ one has $\Bigl\lVert \sum_{k=1}^n 1_{A_k} T_{i_k} f \Bigr\rVert_{L^p} \le C \lVert f\rVert_{E}$.

Proof. "(i) $\Rightarrow$ (ii)" This implication is straighforward to check.

"(ii) $\Rightarrow$ (i)" Let $\mathcal{F}$ denote the set of all non-empty finite subsets of $I$, which is directed with respect to set inclusion. Fix $f \in E$.

For each $F \in \mathcal{F}$ the vector $0 \le \bigvee_{i \in F} \lvert T_if\rvert \in L^p$ (where $\lvert\cdot\rvert$ denotes the pointwise almost everywhere modulus and $\bigvee$ denotes the pointwise almost everywhere supremum) is norm bounded by $C\lVert f\rVert_{E}$. To see this, enumerate the elements of $F$ as $i_1, \dotsc, i_n$ and choose a measurable partition of $\Omega$ into sets $A_1, \dotsc, A_n$ such that $$ \bigvee_{i \in F} \lvert T_if\rvert = \sum_{k=1}^n 1_{A_k} \lvert T_{i_k} f\rvert = \Bigl\lVert \sum_{k=1}^n 1_{A_k} T_{i_k} f \Bigr\rVert . $$ Since the net $\bigl(\bigvee_{i \in F} |T_if|\bigr)_{F \in \mathcal{F}}$ in $L^p$ is increasing and norm bounded by $C\lVert f\rVert_{E}$, it is norm convergent to a vector $0 \le h \in L^p$ of norm $\le C\lVert f\rVert_{E}$ (here was used that $p \in [1,\infty)$). Clearly, $h$ satisfies the property claimed in (i). $\quad \square$

Remarks. (a) Assertion (i) is a convenient way to write down a maximal inequality if the index set $I$ is not assumed to be countable and if one does not make any regularity assumptions regarding the dependence of $T_i$ on $i$. Writing down the maximal operator before knowing whether a maximal inequality holds leads to measurability problems in this case. (But once the maximal inequality is established, one can define the maximal operator by using the supremum within the ordered space $L^p$ rather than the almost everywhere supremum.)

In cases where the maximal operator can be reasonably defined as a pointwise supremum assertion (i) is equivalent to the $L^p$-boundedness of the maximal operator.

(b) In the situation of the question, assertion (ii) of the theorem is precisely the estimate \eqref{2} for simple functions $t: \mathbb{R}^n \to (0,1)$. So this shows, in particular, that it suffices to consider simple functions in \eqref{2}.

This does not only resolve all measurability issues regarding $t$; rather, in the situation of the theorem such a measurability is not even defined since the index set $I$ is not assumed to carry a $\sigma$-algebra.

In the same vein, note that assertion (ii) makes sense in this general setting, while for an object such as $T_{i(x)}f(x)$ for a non-simple function $i: \Omega \to I$ it is not even clear how to define it if the $T_i$ do not have any particular structure.

(c) The measure space $(\Omega, \mu)$ is not assumed to be $\sigma$-finite. In fact, the proof shows that the theorem does not really rely on the structure of $L^p$-spaces, but stays true on Banach lattices where every increasing norm bounded net is norm convergent (these are the so-called KB-spaces, of which $L^p$ for $p \in [1,\infty)$ are examples). To make sense of (ii) one needs to replace the multiplication with indicator functions with band projections in this abstract setting.

(d) The proof shows that $C$ can be chosen as the same number in (i) and (ii).

(e) Fun fact: The existence of $C$ is actually redundant in (i). Indeed, if one only assumes that for each $f \in E$ there exists $0 \le h \in L^p$ such that $\lvert T_i f\rvert \le h$ for all $i \in I$, then there automatically exists a number $C \ge 0$ such that $h$ can always be chosen to satisfy $\lVert h\rVert_{L^p} \le C \lVert f\rVert_{E}$. This can be shown by means of the closed graph theorem, see for instance (warning: shameless self-promotion ahead) Theorem 4.1 in the preprint Order boundedness and order continuity properties of positive operator semigroups with Michael Kaplin.

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    $\begingroup$ As you noticed, the spacing in $A \dot\cup B$ A \dot\cup B requires correction, which you did. However, you can have TeX do this correction for you, rather than having to eyeball it yourself, using $A \mathbin{\dot\cup} B$ A \mathbin{\dot\cup} B (which, as usual, can be defined as a macro if you're going to do a lot of disjoint-unioning). I edited accordingly. I also edited in \eqrefs to \eqref{2} in the original post; I hope that was what you meant by (2). $\endgroup$
    – LSpice
    Jul 20 at 22:30
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    $\begingroup$ @LSpice: Thanks for the edit! (I know about the spacing via mathbin, but I couldn't remember the precise name of the command and was admittedly too lazy to look it up... ;-) ) $\endgroup$ Jul 20 at 23:26
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    $\begingroup$ Re, completely understood! \mathbin, \mathop, \mathopen, and \mathclose are so often unneeded that, when they're exactly the right tools for the situation, it's easy to forget the exact syntax, and TeXnical terminology is not always Google-friendly. $\endgroup$
    – LSpice
    Jul 20 at 23:28
  • $\begingroup$ @Jochen Glueck Thanks a lot for your answer. The index set $I$ considered in the proof is finite (or countably infinite). I am not sure how to apply the theorem to a family of operators $\{T_{t}\}$ with $t\in (0,1)$, say. $\endgroup$
    – Medo
    Jul 21 at 1:11
  • $\begingroup$ @Medo: There's no assumption on $I$ to be finite or countable; $I$ is a general non-empty set. So in your case you can take $I = (0,1)$. $\endgroup$ Jul 21 at 5:57

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