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Fall 2014 - Problem 3

Lp spaces

Let $\set{f_n}_n$ be a bounded sequence in $L^2\p{\R}$ and suppose that $f_n \to 0$ Lebesgue almost everywhere. Show that $f_n \to 0$ in the weak topology on $L^2\p{\R}$ .

Solution.

Let $g \in L^2\p{\R}$ and $\epsilon > 0$ . Then there exists $R > 0$ such that

\p{\int_{\abs{x}>R} \abs{g}^2 \,\diff{x}}^{1/2} < \epsilon

since $\abs{g}^2$ is integrable. Also, $E \mapsto \int_E \abs{g}^2 \,\diff{x}$ defines a measure absolutely continuous with respect to the Lebesgue measure, so there exists $\delta > 0$ such that if $m\p{E} < \delta$ , then $\int_E \abs{g}^2 \,\diff{x} < \epsilon^2$ .

On $\br{-R, R}$ , which is a finite measure set, we may apply Egorov's theorem to get a measurable set $E \subseteq \br{-R, R}$ such that $m\p{E} < \delta$ and so that $f_n \to 0$ uniformly on $\br{-R, R} \setminus E$ . Thus, if $M = \sup_n \norm{f_n}_{L^2}$ and $n$ is large enough, we have $\abs{f_n} \leq \epsilon$ on $\br{-R, R} \setminus E$ and so by Cauchy-Schwarz,

\begin{aligned} \abs{\int_\R f_ng \,\diff{x}} &\leq \int_{\br{-R,R} \setminus E} \abs{f_ng} \,\diff{x} + \int_E \abs{f_ng} \,\diff{x} + \int_{\set{\abs{x}>R}} \abs{f_ng} \,\diff{x} \\ &\leq \p{\int_{\br{-R,R} \setminus E} \abs{f_n}^2 \,\diff{x}}^{1/2} \norm{g}_{L^2} + \norm{f_n}_{L^2}\p{\int_E \abs{g}^2 \,\diff{x}}^{1/2} + \norm{f_n}_{L^2}\p{\int_{\set{\abs{x}>R}} \abs{g}^2 \,\diff{x}}^{1/2} \\ &\leq \norm{g}_{L^2}\epsilon + M\epsilon + M\epsilon. \end{aligned}

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