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!



Is $B \cap L^2\big( (0,T) \times (0,1)\big)$ closed in $L^2\big( (0,T) \times (0,1)\big)$?
$\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, $x\in B$. $\quad\Box$


