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I need help proving that the set $B \cap L^2\big( (0,T) \times (0,1)\big)$ is a closed subset of $L^2\big( (0,T) \times (0,1)\big)$, where $B$ is defined as:

$$B=\Big\{x \in L^{\infty}\big(0,T;L^1(0,1)\big) \ : \|x(t)\|_{L^1(0,1)} \le 1 , \ \ \text{a.a in } \ (0,T) \Big\} $$

Any suggestions or proofs would be greatly appreciated!

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$\newcommand{\R}{\mathbb R}\newcommand{\C}{\mathbb C} $To attach a meaning to the intersection $C:=B\cap L^2\big((0,T)\times(0,1)\big)$ of the subset $B$ of the space $L^{\infty}\big(0,T;L^1(0,1)\big)$ with the space $L^2\big((0,T)\times(0,1)\big)$, we need to identify functions $(0,T)\times(0,1)\ni(t,s)\mapsto x(t,s)\in\C$ in $L^2\big((0,T)\times(0,1)\big)$ with functions $(0,T)\ni t\mapsto x_t\in L^{\infty}\big(0,T;L^1(0,1)\big)$, which we will do by the standard formula $x_t(s):=x(t,s)$.

Let $(x^{(n)})$ be a sequence in $C$ converging to some $x\in L^2\big((0,T)\times(0,1)\big)$ in $L^2\big((0,T)\times(0,1)\big)$: \begin{equation} d_n:=\int_0^T dt\,\int_0^1 ds\, |x^{(n)}(t,s)-x(t,s)|^2\to0 \end{equation} (as $n\to\infty$). We have to show that then $x\in B$, so that $x$ is actually in $C$.

Using the Cauchy--Schwarz inequality, we have \begin{equation} \begin{aligned} &\int_0^T dt\,\int_0^1 ds\, |x^{(n)}(t,s)-x(t,s)| \\ &\le\sqrt{T\times1}\Big(\int_0^T dt\,\int_0^1 ds\, |x^{(n)}(t,s)-x(t,s)|^2\Big)^{1/2} =\sqrt T\sqrt{d_n}\to0. \end{aligned} \end{equation} That is, $\int_0^T dt\, g^{(n)}(t)\to0$, where $g^{(n)}(t):=\|x^{(n)}_t-x_t\|_{L^1(0,1)}$. That is, the sequence $(g^{(n)})$ of nonnegative functions converges to $0$ in $L^1(0,T)$ and hence in measure over the interval $(0,T)$. So, passing to a subsequence, without loss of generality we may assume that the sequence $(g^{(n)})$ converges to $0$ almost everywhere on $(0,T)$. That is, $\|x^{(n)}_t-x_t\|_{L^1(0,1)}\to0$ for a.a. $t\in(0,T)$. Recall that for each $n$ we have $x^{(n)}\in C\subseteq B$, whence $\|x^{(n)}_t\|_{L^1(0,1)}\le1$ for a.a. $t\in(0,T)$. So, \begin{equation} \|x_t\|_{L^1(0,1)}=\lim_n\|x^{(n)}_t\|_{L^1(0,1)}\le1 \end{equation} for a.a. $t\in(0,T)$. Thus, $x\in B$. Thus, $x\in B$. $\quad\Box$

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  • $\begingroup$ Could you please clarify whether the transition from convergence in measure of $(g^{n})$ to almost everywhere convergence is established by employing the Riesz-Fischer theorem? $\endgroup$
    – elmas
    Commented Jun 14, 2023 at 13:13
  • $\begingroup$ @elmas : No, the Riesz–Fischer theorem is not used here. That the convergence in measure implies the almost everywhere convergence for a subsequence is a simple theorem, found I think in almost any textbook on measure theory -- see e.g. Theorem 2.30 on p. 61 of Folland, Real Analysis, Second Edition. (You can also find this here. More immediately here, you can use Corollary 2.32 on p. 62 of Folland's book, stating that the convergence in $L^1$ implies the almost everywhere convergence for a subsequence. $\endgroup$
    – Iosif Pinelis
    Commented Jun 14, 2023 at 16:14

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