Is $f \mapsto (x\mapsto \int_0^x f\,\mathrm d\lambda)$ weak-strong continuous from $L^1([0,1])$ to $C([0,1])$?

A mapping $T : L^1([0,1]) \to C([0,1])$ is called weak-strong continuous at a point $f\in L^1([0,1])$, if for every sequence $(f_n)_n$ of functions in $L^1([0,1])$ that converges weakly to $f$ $\left(\text{i.e.}~\forall g\in L^\infty([0,1]): \lim_{n\to\infty} \left| \int_0^1 f_n\cdot g\,\mathrm d\lambda - \int_0^1 f\cdot g\,\mathrm d\lambda \right| = 0\right)$, we have $$ \lim_{n\to\infty}\left\lVert T(f) - T(f_n) \right\rVert_\infty := \lim_{n\to\infty} \max_{x\in[0,1]} T(f)(x) - T(f_n)(x) = 0 .$$

I am interested in the Volterra operator $T:L^1([0,1]) \to C([0,1]), f\mapsto (x\mapsto \int_0^x f\,\mathrm d\lambda)$. Is it weak-strong continuous in the sense above?

Edit: I have tried the standard "template" for a counterexample: $f_n = n \cdot \mathbf 1_{[0,1/n]}$ and $f \equiv 1$, but $f_n$ does not converge weakly to $f$ as the witness $g= \mathbf 1_{[0,1/2]}$ demonstrates.

Solution 1:

The following proof uses the Dunford-Pettis theorem. The OP indicated in the comments that he doesn't know this result, so a proof that avoids it would still be desirable.

If $f_n\to f$ weakly in $L^1$, then $$ T f_n(x)=\int_0^1 1_{[0,x]}(z)f_n(z)\,dz\to \int_0^1 1_{[0,x]}(z)f(z)\,dz=Tf(x). $$ Hence it suffices to show that $(T f_n)$ has a uniformly convergent subsequence. For this purpose we can apply the Arzelà-Ascoli theorem: We need to show that $(Tf_n)$ is bounded in $C([0,1])$ and (uniformly) equicontinuous, that is, for every $\epsilon>0$ there exists $\delta>0$ such that $|Tf_n(x)-Tf_n(y)|\leq \epsilon$ whenever $|x-y|<\delta$. Boundedness follows immediately from the boundedness of $(f_n)$ and the boundedness of $T$

For the equicontinuity the Dunford-Pettis theorem comes in handy. As $(f_n)$ converges weakly, the set $\{f_n\mid n\in\mathbb N\}$ is weakly compact, hence equi-integrable, that is, for every $\epsilon>0$ there exists $\delta>0$ such that $$ \int_A |f_n(z)|\,dz<\epsilon $$ whenever $A\subset [0,1]$ is a measurable set with $|A|<\delta$. In particular this implies $$ |Tf_n(x)-T f_n(y)|\leq\int_x^y |f_n(z)|\,dz<\epsilon $$ whenever $|x-y|<\delta$, as desired.

Solution 2:

To complete MaoWao's answer, here's how you can do without the Dunford-Pettis theorem: suppose that $f_n$ converges weakly to $f$ in $L^1$. Let $\epsilon >0$, then there exists $\delta >0$ such that $$\tag{1}\left|\int_A f(z)\, dz\right|\leq\int_A |f(z)|\, dz < \epsilon $$ whenever $|A|< \delta$. Now weak convergence implies that there exists $N$ such that $$\left|\int_x^y f_n(z)\, dz \right| < 2\epsilon$$ for every $n >N$ and every $x,y\in [0,1]$ with $|x-y|<\delta$. It remains to control the first $N$ terms and for this we can repeat the argument used for $(1)$: for every $n\leq N$, there exists $\delta_n>0$ such that $$\left|\int_x^y f_n(z)\, dz \right| <2\epsilon$$ for every $x,y \in [0,1]$ such that $|x-y| < \delta_n$. Now take $\delta' = \min(\delta, \min_{n\leq N} \delta_n)$. This implies that $$|Tf_n(x) - Tf_n(y)| =\left|\int_x^y f_n(z)\, dz \right|< 2\epsilon$$ for every $n$ whenever $|x-y|< \delta'$.