If $fg_n$ converges in $L^p$ for all $f\in L^p$, then $g_n$ converges in $L^\infty$.

Question

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).

Answer 1

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.

Answer 2

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 $f$.