If $fg_n$ converges in $L^p$ for all $f\in L^p$, then $g_n$ converges in $L^\infty$. Let $(X,\mathcal{A},\mu)$ be a measure space with $\mu(X)<\infty$. I know that if $g:X\rightarrow\mathbb{R}$ satisfies that $fg\in L^p$ for all $f\in L^p$, then $g\in L^\infty$.
My question is whether there is a "sequential version" of this result: if $fg_n$ converges in $L^p$ for all $f\in L^p$, then $g_n$ converges in $L^\infty$. I would like a reference of this result (if it is true).
 A: The claim is not true.
Consider $X=(0,1)$ with the borel measure.
and consider the functions
$$
g_n(x) = \chi_{(0,1/n)}.
$$
Note that $g_n(x)\to 0$ a.e. in $\Omega$.
Then for all $f\in L^p$, it can be shown that
we habe
$$
\| f g_n \|_{L^p}^p = \int_{(0,1/n)} |f(x)|^p \mathrm dx \to 0,
$$
so $f g_n$ converges for all $f\in L^P$ to $0$ in the $L^p$-norm.
However, $g_n$ does not converge to $0$ in $L^\infty$-norm, because
$
\|g_n\|_{L^\infty} = 1
$
and using the pointwise convergence, only $g=0$ would be possible as a limit.
Alternatively, it is easy to see that
$
\|g_n - g_m\|_{L^\infty} = 1
$
for $n\neq m$, and therefore $\{g_n\}_{n\in\mathbb N}$ is not a Cauchy sequence in $L^\infty$,
and thus not convergent.
A: It seems that the best we can deduce is boundedness in $\mathbb L^\infty $ of $(g_n)$.
Let 
$$
F_N:=\bigcap_{n\geqslant N}\left\{f\in\mathbb L^p\mid \left\lVert fg_n\right\rVert_p\leqslant N   \right\}.
$$
The set $F_N$ is closed as an intersection of closed sets. To see that $\left\{f\in\mathbb L^p\mid \left\lVert fg_n\right\rVert_p\leqslant N   \right\}$ is closed, extract from $\left(\left\lvert fg_n\right\rvert\right)_{n\geqslant 1}$ an almost everywhere convergent subsequence and use Fatou's lemma.
Since the union of the sets $F_N$ is $\mathbb L^p$, we know from Baire theorem that there is a $N_0$ such that $F_{N_0}$ has an empty interior. In particular, it contains a ball of radius $r$ centered at some $f_0$. Since for all $f\in\mathbb L^p$, the function 
$$
 A(f):=  f_0+\frac r2\frac{1}{\left\lVert f\right\rVert_p+1}f
$$
is such that $\left\lVert A(f)-f_0\right\rVert_p\leqslant r/2$, we have 
for all $n\geqslant N_0$ that $$\left\lVert A(f)g_n\right\rVert_p\leqslant N\mbox{ and }\left\lVert f_0g_n\right\rVert_p\leqslant N.$$The combination of these estimates gives $$\sup_{n\geqslant N_0}\left\lVert fg_n\right\rVert_p\leqslant C\left\lVert f\right\rVert_p  $$
where $C$ is independent of $f$.
