Weak convergence in $L^2$.

Question

Let $g_n,g \in L^2[a,b]$ and we define $$G_n(x)=\int_a^x g_n(t) \,dt, \quad\quad G(x) = \int_a^x g(t) \,dt.$$ If $g_n$ converges weakly to $g$, that is $\langle g_n,f\rangle \rightarrow \langle g,f \rangle$ for all $f \in L^2[a,b]$, with $\langle g,f \rangle = \int_a^b g \bar{f}$, and $\sup ||g_n||_2$ is finite, then $G_n$ converges uniformly to $G$.

My initial thought of this problem was naive. Here was my argument, we have $|G_n-G| = |\int_a^x (g_n-g)| < \epsilon$ since we can take $f=1$ and use the weak convergence assumption. But of course, this does not work since this would only prove a pointwise convergence. I know I need to make use of $\sup||g_n||_2$ somewhere but I am stuck. Any help is welcomed.

Answer 1

You proved already pointwise convergence. To show uniform convergence, note by Holder's inequality,

$$ |G_n(x) - G_n(y)| =\left| \int_x^y g_n \right| \le \sup_n \|g_n\|_{L^2}\sqrt{|x-y|}.$$

Together with $G_n(a) = 0$, $\{ G_n\}$ is an equicontinuous family of continuous function on $[a, b]$ and by Arzela-Ascoli theorem, every subsequence of $\{G_n\}$ has a further subsequence which converges uniformly (to $G$, since $G_n\to G$ pointwisely). This shows already that $G_n$ converges uniformly to $G$ (by a contradiction argument).

Remark

• The condition $\sup_n \| g_n\| <\infty$ is implied by the weak convergence; see here

• This argument is just a combination of the fact that (i) $T: L^2([a, b]\to C[a, b]$, $Tf(x) = \int_a^x f(t)dt$ is a compact operator, and (ii) compact operators send weakly convergent sequence to strongly convergent sequence