Fix $\Omega \subseteq \mathbb{R}^N$ a bounded domain (of whatever smoothness you end up needing, let's say $C^1$ domain for definiteness) and fix some $0 <T < \infty$. In considering evolution equations in PDE, we may concern ourselves with spaces of functions $[0,T] \to X$ where $X$ is a Banach space of functions $\Omega \to \mathbb{R}$ ($X = L^p(\Omega), W^{k,p}(\Omega)$, etc.). For functions of this form, we can introduce notions of measurability and weak (time) derivatives in connection with the Bochner integral (e.g. as in Evans PDE).

But also to any function $\textbf{f} : [0,T] \to X$ we can associate a function $f : [0,T] \times \Omega \to \mathbb{R}$ in the obvious way: $f(t,x) = \textbf{f}(t)(x)$. Then $f$ is a function defined on a subset of $\mathbb{R}^{N+1}$ and so has natural notions of measurability and weak time partial derivatives. This suggests the fairly natural questions: is $f$ measurable iff $\textbf{f}$ is (note that if $X$ isn't separable, there are a few different notions of measurability to consider for $\textbf{f}$)? And is $\textbf{f}$ weakly differentiable in time iff $\partial_t f$ exists in the weak sense?

More generically, the question is how can the highly related notions defined separately for function-space-valued functions and for functions on $\mathbb{R}^{N+1}$ be connected? It seems like we'd really want them to agree, at least in nice cases, but I'm not quite convinced they will.

The book I've looked at (Evans, Wloka) haven't touched on these seemingly natural questions, so any pointers to resources that cover this sort of thing would also be useful.

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    $\begingroup$ What about looking at what is dense in such spaces, e.g.smooth functions, and conclude if the norms are algebraically the same? $\endgroup$
    – username
    Commented Jun 20, 2021 at 16:48
  • $\begingroup$ @username Not sure what you mean by checking if the norms are the same. I think the first concern one should have is measurability. Does the fairly strong notion of measurability of, say, functions $[0,T] \to L^p(\Omega)$ make the associated function $[0,T] \times \Omega \to \mathbb{R}$ measurable in the usual way? If this is the case, it's fairly easy to see that if we consider the space $L^p([0,T]; L^p(\Omega))$ (defined in the obvious e.g. in Evans 5.9.2), then the norms are the same (i.e. this norm is equal to $L^p$ norm on $[0,T] \times \Omega$. $\endgroup$ Commented Jun 20, 2021 at 16:54
  • $\begingroup$ Yes, and then defining both spaces as the completion for that norm does it, doesn't it? $\endgroup$
    – username
    Commented Jun 20, 2021 at 17:01
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    $\begingroup$ @username Yeah that makes sense. I think that'll work for showing that elements of $L^p([0,T]; L^p(\Omega))$ are measurable. The general measurability question requires something more though. In this related post, Gerald Edgar suggests that it's obvious as measurable functions taking values in e.g. $L^1(\Omega)$ are the pointwise a.e. limit of simple functions, but the limit at each time is $L^1$ not a.e., so I don't see how we get measurability out of that fact. $\endgroup$ Commented Jun 20, 2021 at 17:45

1 Answer 1


First note that the vectors of the Lebesgue spaces $L^p(\Omega)$ are not functions $\Omega\to\mathbb R$ or ($\mathbb C$, depending on whether real or complex valued functions are considered) but a.e. equivalence classes of them. In usual PDE considerations of evolution equations, there are made lots of implicit "identifications" that make understanding the correct state of matters difficult to the beginner. Obviously from this the OP question arises. In the classical book

Dunford, Nelson; Schwartz, Jacob T., Linear operators. I. General theory. Pure and Applied Mathematics. Vol. 7. New York and London: Interscience Publishers. (1958)

there is an explicit theorem (probably Theorem 17 on Page 198, although I have not had the possibility to check this) that for $1\le p<+\infty$ justifies the identification that $\iota:E=L^p(I,L^p(\Omega))\simeq F=L^p(I\times\Omega)$ holds in the sense that $\iota$ is a topological linear homeomorphism $E\to F$ when defined as follows. Given a vector $\vec e$ in $E$ that is an a.e. class of functions $\bar e:I\to L^p(\Omega)$, let $\bar e\in\vec e$ be arbitrary. Then $\bar e$ is a function $I\to L^p(\Omega)$ and there is a measurable function $f:I\times\Omega\to\mathbb K$ ($=\mathbb R$ or $\mathbb C$) such that the function $\bar f:I\to L^p(\Omega)$ defined by $t\mapsto[\,f(t,\cdot\,)\,]$ is a.e. equivalent to $\bar e$ and hence is in $\vec e$. If $g$ is a.e. equivalent to $f$, also $\bar g\in\vec e$ holds, and conversely any representative $f$ of a vector $\vec f$ in $F$ determines in this manner a unique vector in $E$. Then we put $\iota:\vec e\mapsto\vec f$. Note, however, that there are also nonmeasurable functions $f:I\times\Omega\to\mathbb K$ such that the function $t\mapsto[\,f(t,\cdot\,)\,]$ defines a vector in $E$, provided that ZF is suitably complemented. Put in more detail, ZFC complemented with the continuum hypothesis allowes one to prove that every representative $\bar c$ of every vector of $E$ has uncountably many choice functions $c$ such that the function $\hat c$ defined by $(t,s)\mapsto c\,t\,s$ is nonmeasurable.

The above result is the basis for the implicit "exponential" identifications made in PDE theory.

  • $\begingroup$ I can't find the corresponding result in Dunford and Schwartz. This only really resolves the aspect of the question for which a solution was already suggested in the comments by username. Can we say anything about the measurability of generic strongly measurable functions $I \to L^1(\Omega)$ (as suggested here) and can we say anything about weak derivatives? $\endgroup$ Commented Jun 21, 2021 at 14:54
  • $\begingroup$ According to the Hitchhiker's Guide, the theorem that TaQ is referring to is Theorem 17 on Page 198 of Dunford and Schwartz. (I also don't have a copy of D&S to check.) $\endgroup$ Commented Jun 21, 2021 at 16:20
  • $\begingroup$ @WillieWong Thanks. That reference is correct. I'll go through the proof presented there and see what insight it has to offer. The remaining aspect of the original question still stand. $\endgroup$ Commented Jun 21, 2021 at 16:34
  • $\begingroup$ @KeeferRowan: "can we say anything about ..." Of course, but this kind of questions are too vague to be answered properly in reasonable space. You should be more specific about what precisely you want to know. $\endgroup$
    – TaQ
    Commented Jun 22, 2021 at 18:11
  • $\begingroup$ Yeah I see that. That's why I'm also soliciting resources. The linked post be GEdgar suggests that strong measurability of functions $I \to L^1(\Omega)$ is equivalent to measurability of the associated function $I \times \Omega \to \mathbb{R}$, which would be something like the final word on the matter. If that's not the case, I see that it would be a somewhat messier problem. I'm somewhat satisfied with the measurability concerns by consulting Dunford and Schwartz. I still have no clue about the derivative stuff. $\endgroup$ Commented Jun 22, 2021 at 18:28

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