# $L^\infty((0,T)\times\Omega)$ is not equal to $L^\infty(0,T;L^\infty(\Omega))$.

Let $\Omega$ be bounded domain in $\mathbb{R}^n$. We know that $L^\infty((0,T)\times\Omega)$ is not equal to $L^\infty(0,T;L^\infty(\Omega))$.

Are there any circumstances in which we can say that one is a subset of the other though? Is it really only measurability which is a problem for identifying the two spaces together?

For example, suppose that $|f(t,\cdot)| \leq C$ a.e. in $\Omega$ for every $t$. Doesn't this imply $$\sup_{t \in [0,T]}\lVert f(t,\cdot)\rVert_{L^\infty(\Omega)} \leq C?$$ So that $f \in L^\infty(0,T;L^\infty(\Omega))$, provided $f$ is measurable?

• can you explain your notation? Is $\Omega \subset \mathbb{R}^n$? And what is $L^\infty(0,T;L^\infty(\Omega))$? Do you mean bounded functions from $(0,T)$ to $L^\infty(\Omega)$? – user56914 Mar 18 '14 at 14:15
• You are right about $\Omega$. $L^\infty(0,T;L^\infty(\Omega))$ is the Bochner space of measurable functions $u:(0,T) \to L^\infty(\Omega)$ with $\lVert u(t) \rVert_{L^\infty(\Omega)}$ in $L^\infty(0,T)$. – student Mar 18 '14 at 14:16

Yes, only the measurability is a problem. You have $L^\infty(0,T;L^\infty(\Omega)) \subset L^\infty((0,T)\times\Omega)$. The problem for the other way round is only the measurability. That is, if $f \in L^\infty((0,T)\times\Omega)$ is $L^\infty(\Omega)$-measurable, it belongs to $L^\infty(0,T;L^\infty(\Omega))$.
• Is there a good example of $f$ in $L^\infty((0,T)\times\Omega)$ but not in $L^\infty(0,T;L^\infty(\Omega))$, say for one-dimensional $\Omega$? – user127096 Mar 19 '14 at 1:15
• It depends :) It is not hard to show that both norms are actually the same on $L^\infty(L^\infty)$. – gerw Jul 18 '14 at 8:05
For completeness, I post an example of a function in $L^{\infty}([0,1]\times [0,1])$ which is not in $L^\infty([0,1];L^\infty([0,1]))$. The example must be standard, but my source was Nonlinear Partial Differential Equations with Applications by Tomás Roubicek, as suggested by student. $$f(x,y) = \begin{cases} 1 , &x\le y \\ 0,\quad &x>y \end{cases}$$ Clearly, this is in $L^{\infty}([0,1]\times [0,1])$. On the other hand, the induced map $F:[0,1]\to L^\infty([0,1])$, namely $F(t) = \chi_{[0,t]}$, is not Bochner measurable. Indeed, a Bochner measurable function must be essentially separably valued, but the range of $F$ consists of an uncountable set of points at distance $1$ from one another.