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Suppose that I have a sequence of functions $f_n\in L^1(\mathbb{R}^d)$ for which $\lim_{n\rightarrow \infty} \int_{\mathbb{R}^d} f_n(x)g(x)dx$ exists for all $g\in S\subset L^\infty(\mathbb{R}^d)$, where $S$ is a closed subspace of $L^\infty(\mathbb{R}^d)$.

Is it possible (and if not, under which conditions) to conclude the existence of a `partial weak limit' $f$ which satisfies $\|f\|_{L^1}\leq \lim\inf_{n\rightarrow \infty} \|f_n\|_{L^1}$ and \begin{equation} \int_{\mathbb{R}^d} f(x)g(x)dx = \lim_{n\rightarrow \infty} \int_{\mathbb{R}^d} f_n(x)g(x)dx, \end{equation} for all $g\in S$?

jwsiegel
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1 Answers1

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Note: this answer currently only suggests an idea, maybe these are helpful for others.

Maybe there is a way if $S$ is weakly-$*$ closed, and I will provide some ideas on how this could be done below. Of course, the resulting partial weak limit does not need to be unique.

I will also use a more abstract notation, with a separable Banach space $X$ (here $L^1(\Bbb R^d)$) and $S\subset X'$ a closed subspace of the dual space of $X$. The arguments should work in the abstract setting.

Let $f_n\in X$ be a sequence such that $g(f_n)$ converges for all $g\in S$. First, the set $S^\perp:=\{ f\in X : s(f)=0 \forall s\in S\}\subset X$ is a closed subspace of $X$. Note that the quotient $X/(S^\perp)$ is again a Banach space and that its dual space can be identified with $S$ (this requires that $S$ is weakly-$*$ closed and makes use of a theorem that states $S^{\perp\perp}=S$).

Then we have that $f_n+S^\perp$ converges weakly in $X/(S^\perp)$. Thus a weak limit $f+S^\perp$ should exist.

However, I am not sure yet how to get the desired norm inequality.

supinf
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  • I just realized that the statement is false, unfortunately. For instance, take $S=C_0(\mathbb{R}^d)$, and let take $f_n = n\chi_{0,n^{-1}}$. Then the $f_n$ converge to $\delta_0$ in the dual of $S$, which is not in $L^1(\mathbb{R}^d)$. Thanks for the suggestion, though. – jwsiegel Oct 01 '20 at 15:32
  • @jwsiegel that is a good observation. My suggestion was only for the case that $S$ is weakly-$$ closed anyways, and the set $S=C_0(\Bbb R^d)$ is not weakly-$$ closed (if i am not mistaken). – supinf Oct 01 '20 at 15:40
  • I see, you're right. In the case where $S$ is weakly-* closed it still may be correct. – jwsiegel Oct 01 '20 at 19:58