Spring 2011 - Problem 1

Solution.

1. $f_n \to f$ weakly in $L^2\p{\br{0,1}}$ if for every $g \in L^2\p{\br{0,1}}$, we have

   $$\int f_ng \,\diff{t} \xrightarrow{n\to\infty} \int fg \,\diff{t}.$$

2. Notice that by weak convergence, $\sup_n \abs{\inner{f_n, g}} < \infty$ for any $g \in L^2\p{\br{0,1}}$. $L^2\p{\br{0,1}}$ is its own dual, we may view $\set{f_n}_n \subseteq \p{L^2\p{\br{0,1}}}^*$ and so by the uniform boundedness principle,

   $$\sup_n\, \abs{\inner{f_n, g}} < \infty \quad \forall g \in L^2\p{\br{0,1}} \implies M \coloneqq \sup_n\,\norm{f_n}_{L^2} < \infty.$$

   Thus for $x_0 \in \br{0, 1}$, Cauchy-Schwarz gives

   $$\abs{F_n\p{x} - F_n\p{x_0}} \leq \abs{\int_x^{x_0} f_n\p{t} \,\diff{t}} \leq \norm{\chi_{\br{x,x_0}}}_{L^2}\norm{f_n}_{L^2} \leq M\abs{x - x_0}^{1/2} \xrightarrow{x\to x_0} 0.$$

   Notice that the upper bound is uniform over $n \geq 1$, so $\set{F_n}_n$ is actually equicontinuous. Similarly,

   $$\abs{F_n\p{x}} \leq \int_0^1 \abs{f_n\p{t}} \,\diff{t} \leq \norm{f_n}_{L^2} \leq M,$$

   so $\set{F_n}_n$ is uniformly bounded. By Arzelà-Ascoli, $\set{F_n}_n$ admits a uniformly convergent subsequence. Moreover, since $\chi_{\br{0,x}} \in L^2\p{\br{0,1}}$, weak convergence also implies

   $$F_n\p{x} = \int f_n\p{t} \chi_{\br{0,x}}\p{t} \,\diff{t} \xrightarrow{n\to\infty} \int f\p{t} \chi_{\br{0,x}}\p{t} \,\diff{t} = F\p{x}.$$

   Thus, any convergent subsequence of $\set{F_n}_n$ must converge to $F$. This implies that $F_n \to F$ uniformly itself. Otherwise, there exists $\epsilon > 0$ such that there exists $x_{n_1} \in \br{0, 1}$ with $\abs{F_{n_1}\p{x_{n_1}} - F\p{x_{n_1}}} \geq \epsilon$. By induction, we get a subsequence $n_k$ such that $\abs{F_{n_k}\p{x_{n_k}} - F\p{x_{n_k}}} \geq \epsilon$ for all $k \geq 1$, so no subsequence of $\set{F_{n_k}}_k$ may converge uniformly to $F$. But by Arzelà-Ascoli, $\set{F_{n_k}}_k$ must admit a uniformly convergent subsequence to $F$, a contradiction. Hence, $F_n \to F$ uniformly to begin with, which completes the proof.