Existence of a `Partial Weak Limit' in $L^1$ 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 $g\in S$?
 A: 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.
