Spring 2018 - Problem 2

Solution.

1. By Cauchy-Schwarz and translation invariance,

   $$\abs{\int_\R f\p{x}f\p{x + h} \,\diff{x}} \leq \norm{f}_{L^2}^2 \quad\text{and}\quad \abs{\int_\R f\p{x}f\p{x - h} \,\diff{x}} \leq \norm{f}_{L^2}^2.$$

   Thus,

   $$\begin{aligned} Q\p{f, h} &= \frac{1}{h^2} \int_\R 2\abs{f\p{x}}^2 - f\p{x}f\p{x + h} - f\p{x}f\p{x - h} \,\diff{x} \\ &= \frac{1}{h^2}\br{\p{\norm{f}_{L^2}^2 - \int_\R f\p{x}f\p{x + h} \,\diff{x}} + \p{\norm{f}_{L^2}^2 - \int_\R f\p{x}f\p{x - h} \,\diff{x}}} \\ &\geq 0. \end{aligned}$$

2. By Plancherel, for any $f \in L^2\p{\R}$, we have

   $$\begin{aligned} Q\p{f, h} &= \int_\R \frac{2\hat{f}\p{x} - e^{ihx}\hat{f}\p{x} - e^{-ihx}\hat{f}\p{x}}{h^2} \conj{\hat{f}\p{x}} \,\diff{x} \\ &= \int_\R \frac{2 - 2\cos\p{hx}}{h^2} \abs{\hat{f}\p{x}}^2 \,\diff{x}. \end{aligned}$$

   Let $\set{f_n}_n \subseteq E$ such that $f_n \to f$ in $L^2$. Passing to a subsequence if necessary, we may assume without loss of generality that $f_n \to f$ almost everywhere. By applying Fatou's lemma in $h$,

   $$\begin{aligned} \int_\R x^2 \abs{\hat{f_n}\p{x}}^2 \,\diff{x} &= \int_\R \liminf_{h\to0} \frac{2 - 2\cos\p{hx}}{h^2} \abs{\hat{f_n}\p{x}}^2 \,\diff{x} \\ &\leq \liminf_{h\to0} Q\p{f_n, h} \\ &\leq 1, \end{aligned}$$

   so $x\abs{\hat{f_n}\p{x}} \in L^2\p{\R}$ for each $n$. Now applying Fatou's lemma in $n$,

   $$\int_\R x^2\abs{\hat{f}\p{x}} \,\diff{x} \leq 1.$$

   To complete the proof, observe that

   $$1 - \cos\p{\theta} = \int_0^\theta \sin{t} \,\diff{t} \leq \int_0^\theta t \,\diff{t} = \frac{\theta^2}{2}.$$

   Hence,

   $$\frac{2 - 2\cos\p{hx}}{h^2} \leq x^2.$$

   Thus, by dominated convergence,

   $$\begin{aligned} \lim_{h\to0} Q\p{f, h} &= \int_\R \lim_{h\to0} \frac{2 - 2\cos\p{hx}}{h^2} \abs{\hat{f}\p{x}}^2 \,\diff{x} \\ &= \int_\R x^2 \abs{\hat{f}\p{x}}^2 \,\diff{x} \\ &\leq 1, \end{aligned}$$

   so $f \in E$, which was what we wanted to show.