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Fall 2017 - Problem 2

Lp spaces

Let $\set{f_n}_n$ denote a bounded sequence in $L^2\p{\br{0,1}}$ . Suppose the sequence $\set{f_n}_n$ also converges almost everywhere. Show that then $\set{f_n}_n$ converges in the weak topology on $L^2\p{\br{0,1}}$ .

Solution.

Let $g \in L^2\p{\br{0,1}}$ . We will show that

\abs{\int \p{f_n - f}g \,\diff{x}} \leq \int \abs{f_n - f}\abs{g} \,\diff{x} \xrightarrow{n\to\infty} 0.

Let $M = \sup_n \norm{f_n}_{L^2}$ . By Fatou's lemma, if $f$ denotes the pointwise limit of $\set{f_n}_n$ , then

\int \abs{f}^2 \,\diff{x} \leq \liminf_{n\to\infty} \int \abs{f_n}^2 \,\diff{x} \leq M,

so $f \in L^2$ .

Let $\epsilon > 0$ . Since $g \in L^2\p{\br{0,1}}$ , there exists $R > 0$ such that $\norm{g\chi_{B\p{0,R}^\comp}}_{L^2} < \epsilon$ . Also, the map $E \mapsto \int_E \abs{g}^2 \,\diff{x}$ defines a measure absolutely continuous with respect to the Lebesgue measure, so there exists $\delta > 0$ such that if $m\p{E} < \delta$ , then $\int_E \abs{g}^2 \,\diff{x} < \epsilon^2$ . For $B\p{0, R}$ , let $A_n = \set{x \in B\p{0, R} \mid \abs{f_n\p{x} - f\p{x}} \geq \epsilon}$ . Since $f_n \to f$ almost everywhere, $f_n \to f$ in measure, so there exists $N \in \N$ so that if $n \geq N$ , then $m\p{A_n} \leq \delta$ . Thus, by Cauchy-Schwarz,

\begin{aligned} \int \abs{f_n - f}\abs{g} \,\diff{x} &= \int_{B\p{0,R}^\comp} \abs{f_n - f}\abs{g} \,\diff{x} + \int_{A_n} \abs{f_n - f}\abs{g} \,\diff{x} + \int_{B\p{0,R} \setminus A_n} \abs{f_n - f}\abs{g} \,\diff{x} \\ &\leq \norm{f_n - f}_{L^2} \norm{g\chi_{B\p{0,R}^\comp}}_{L^2} + \norm{f_n - f}_{L^2} \p{\int_{A_n} \abs{g}^2 \,\diff{x}}^{1/2} + \epsilon \int_{B\p{0,R} \setminus A_n} \abs{g}^2 \,\diff{x} \\ &\leq 2M\epsilon + 2M\epsilon + \epsilon \norm{g}_{L^2}. \end{aligned}

Sending $\epsilon \to 0$ , we get the claim.