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I am stuck with something standard...

Let $f_n \in L^\infty(\mathbb{R}^d)\cap L^1(\mathbb{R}^d)$, $n\geq1$, be such that $$ \sup_{n\geq1} \|f_n\|_{L^\infty(\mathbb{R}^d)}<\infty $$ and $$ \lim_{n\to\infty} \int_{\mathbb{R}^d}|f_n(x)-f(x)| g(x)\,dx=0, \qquad \text{for all } 0\leq g\in L^1(\mathbb{R}^d). $$ Is it correct that $f_n(x)\to f(x)$ for a.a. $x\in\mathbb{R}^d$ ?

It can be reformulated as follows: if a sequence from $L^\infty$ converges to 0 in the weak-* topology, does it converge to 0 almost everywhere?

(Update: in the reformulation, I omitted that $f_n\in L^1$ itself, not sure whether this is relevant.)

Dmitri
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  • I might be wrong here, but isn't your final claim weaker than what you state before? After all, the dual of $L^\infty$ is larger than $L^1$. – Leo163 May 03 '20 at 17:21
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    Weak-star convergence $(f_n)$ on $X^$ means, I believe, that $\langle f_n,x\rangle\to \langle f,x\rangle$ for all $x\in X$. – Dmitri May 03 '20 at 17:34

2 Answers2

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The best we can say is that a subsequence of $(f_n)_{n\geqslant 1}$ converges to $f$ almost everywhere. Indeed, for all $k\geqslant 1$, choosing $g$ as the indicator function of $[-k,k]$, one can see that $f_n\to f$ in $\mathbb L^1[-k,k]$. Therefore, for each $k\geqslant 1$ and all $N$, we can find an integer $n_k\geqslant N$ such that $\lVert f_{n_k}-f\rVert_{[-k,k]}\leqslant 4^{-k}$. We can thus find an increasing sequence of integers (that we can constructed inductively) $(n_k)$ such that for all $k\geqslant 1$, $\lVert f_{n_k}-f\rVert_{[-k,k]}\leqslant 4^{-k}$ annd we get that $f_{n_k}\to f$ almost everywhere since $\sum_{k\geqslant 1} \lvert f_{n_k}-f\rvert$ is integrable on any interval.

But it is possible that $(f_n)$ does not converge almost everywhere to $f$. If $n$ has the form $2^N+k$ for $N\geqslant 1$ and $ 1\leqslant k\leqslant 2^{N}$, let $f_n$ be the indicator function of $[(k-1)2^{-N},k2^{-N})$. One can see that $\lim_{n\to\infty} \int_{\mathbb{R}^d}|f_n(x)| g(x)\,dx=0$ holds for all integrable functions by absolute continuity of the integral (we integrate on sets whose measure goes to zero) but the almost everywhere convergence does not hold. Note that here, the subsequence $(f_{2^N+1})_{N\geqslant 1}$ converges to $0$ almost everywhere.

Davide Giraudo
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I think it is not true in general. Further more you cannot control all elements of $L^{\infty,*}$ using $L^1$ because $L^1$ it is not reflexive.

Finally you can state that if $(f_n)_n$ converges weak* only if you allow to take a subsequence of it because the unitary ball is weak* compact. ($L^{\infty,*}=L^1$).

Filippo Giovagnini
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    The fact that $L^1$ is not reflexive means that dual to $L^\infty$ is not $L^1$ (much larger), but I consider not weak convergence on $L^\infty$, but weak. Next, Banach-Alaoglu gives information about a weak convergence subsequence, here it is just assumed; not sure that this theorem can help/disprove something here. – Dmitri May 03 '20 at 17:37