Spring 2021 - Problem 3

Solution.

1. Notice that if $g \in L^2\p{\R}$, then by Cauchy-Schwarz,

   $$\begin{gathered} f_n\p{x} \leq \int_0^{2\pi} \p{f\p{x + t} - g\p{x + t}} \cos\p{nt} \,\diff{t} + \int_0^{2\pi} g\p{x + t} \cos\p{nt} \,\diff{t} \\ \abs{f_n\p{x}} \leq 2\pi\norm{f - g}_{L^2} + \abs{\int_0^{2\pi} g\p{x + t} \cos\p{nt} \,\diff{t}}. \end{gathered}$$

   Hence, it suffices to show this on a dense subclass of $L^2\p{\R}$, so without loss of generality, we may assume that $f$ is a step function. Moreover, by linearity of the problem, we may assume without loss of generality that $f = \chi_{\br{a,b}}$ for some $0 \leq a < b \leq 2\pi$. Then by direct calculation,

   $$\int_a^b \cos\p{nt} \,\diff{t} = \frac{\sin\p{nb} - \sin\p{na}}{n} \xrightarrow{n\to\infty} 0,$$

   so $f$ converges to $0$ everywhere. For the second claim, observe first that

   $$\begin{aligned} \norm{f_n}_{L^2}^2 &= \int_\R \abs{f_n\p{x}}^2 \,\diff{x} \\ &\leq 2\pi \int_\R \int_0^{2\pi} \abs{f\p{x + t}}^2 \,\diff{t} \,\diff{x} && \p{\text{Cauchy-Schwarz}} \\ &= 2\pi \int_0^{2\pi} \int_\R \abs{f\p{x + t}}^2 \,\diff{x} \,\diff{t} && \p{\text{Fubini-Tonelli}} \\ &= 4\pi^2 \norm{f}_{L^2}^2, &&& \p{*} \end{aligned}$$

   which is essentially an uniform integrability condition. Observe that ($*$) holds for any $f \in L^2\p{\R}$. Let $\epsilon > 0$, and let $R > 0$ be so large so that $\norm{f\chi_{B\p{0,R}^\comp}}_{L^2} < \epsilon$. By absolute continuity, there exists $\delta > 0$ such that if $m\p{E} < \delta$, then $\norm{f\chi_E}_{L^2} < \epsilon$. Since $f_n \to 0$ on $B\p{0, R}$, Egorov's theorem gives $E \subseteq B\p{0, R}$ such that $f_n \to 0$ uniformly on $B\p{0, R} \setminus E$ and so that $m\p{E} < \delta$. In particular, for $n$ large, we have $\abs{f_n} \leq \frac{\epsilon}{\sqrt{2R}}$ on $B\p{0, R} \setminus E$. Hence, applying ($*$) to $f\chi_E$ and $f\chi_{B\p{0,R}^\comp}$, we get

   $$\begin{aligned} \norm{f_n}_{L^2} &\leq \norm{f_n \chi_{B\p{0,R}}}_{L^2} + \norm{f_n \chi_{B\p{0,R}^\comp}}_{L^2} \\ &\leq \norm{f_n \chi_E}_{L^2} + \norm{f_n \chi_{B\p{0,R} \setminus E}}_{L^2} + 2\pi\norm{f \chi_{B\p{0,R}^\comp}}_{L^2} \\ &\leq 2\pi \norm{f \chi_E}_{L^2} + \frac{\epsilon}{\sqrt{2R}}\norm{\chi_{B\p{0,R}}}_{L^2} + 2\pi\epsilon \\ &\leq 4\pi\epsilon + \epsilon, \end{aligned}$$

   and the claim follows from sending $\epsilon \to 0$.