Spring 2017 - Problem 2

Solution.

Let $M = \sup_n \norm{f_n}_{L^2}$. For $t \in \pco{0, 1}$, Cauchy-Schwarz gives

Hence, for a fixed $t \in \pco{0, 1}$, $\set{f_n\p{t}}_n$ is a bounded sequence in $\R$, and so it admits a convergent subsequence. By a diagonal argument, we get a subsequence $\set{f_{n_k}}_k$ which converges pointwise for each $q \in \br{0, 1} \cap \Q$.

We will show that this subsequence converges almost everywhere. For $q \in \br{0, 1} \cap \Q$, set $f\p{q} = \lim_{k\to\infty} f_{n_k}\p{q}$. Observe that if $q < r$ are rationals, then

by taking $k \to \infty$. For any $x \in \br{0, 1}$, set $L_x = \sup_{q < x}{f\p{q}}$ and $U_x = \inf_{x < q}{f\p{q}}$. Then if $x < y$, there exists $q \in \p{x, y}$ which implies $U_x \leq f\p{q} \leq L_y$ and so $\p{L_x, U_x} \cap \p{L_y, U_y} = \emptyset$. Since $\br{0, 1}$ is separable, there can only be countably many disjoint non-trivial intervals, and so $L_x = U_x$ for almost every $x \in \br{0, 1}$. For such an $x$, let $q, r$ be rational such that $q \leq x \leq r$ and so

Taking the supremum over $q$ and infimum over $r$, we see

so these are actually all equalities, which means $\lim_{k\to\infty} f_{n_k}\p{x} = L_x = U_x$. Thus, $f_{n_k}$ converges for almost every $x \in \br{0, 1}$.

To complete the proof, we established earlier that $0 \leq f_n\p{t} \leq \frac{M}{\sqrt{1 - t}} \in L^1\p{\br{0,1}}$, so by dominated convergence, $f_{n_k} \to f$ in $L^1$.