Analysis Problem of the Day 85

Today’s problem appeared as Problem 2 on the UCLA Spring 2018 Analysis Qual:

Problem 2. Let $f \in L^2(\mathbb{R})$ and for $h>0$ define

$Q(f,h):= \int_\mathbb{R} \frac{2f(x)-f(x+h)-f(x-h)}{h^2} f(x) dx.$

a) Show that $Q(f,h) \geq 0$ for all $f \in L^2(\mathbb{R})$ and all $h>0.$

b) Show that $E:=\{f \in L^2(\mathbb{R}): \limsup_{h \to 0} Q(f,h) \leq 1\}$ is closed in $L^2(\mathbb{R}).$

Solution: a) Note that by Cauchy-Schwarz, $\left|\int_\mathbb{R} f(x)f(x+h)dx \right| \leq \|f\|_2\|f\|_2 = \int |f(x)|^2 dx,$ and similarly $\left|\int_\mathbb{R} f(x)f(x-h)dx\right| \leq \int |f(x)|^2 dx.$ Factoring $\frac{1}{h^2}$ out of $Q(f,h)$ and using the above estimates yields

$\frac{1}{h^2} \int_\mathbb{R} f(x+h)f(x)+f(x-h)f(x) dx \leq \frac{1}{h^2} \int_\mathbb{R} 2f(x)^2 dx,$

and rearranging yields the desired inequality $Q(f,h) \geq 0.$

b) We transform the integrand of $Q(f,h)$ to the Fourier side, motivated by the fact that we are working over $L^2(\mathbb{R}).$ On the Fourier side, since $\langle \widehat{f},\widehat{g}\rangle = \langle f,g\rangle,$ the integral becomes

$Q(f,h)=\int_{\mathbb{R}} \frac{2-e^{i\xi h}-e^{-i\xi h}}{h^2} \widehat{f}^2(\xi) d\xi= \int_\mathbb{R} \frac{2-2\cos(\xi h)}{h^2} \widehat{f}^2(\xi)d\xi.$

Now, one can easily check that $|\frac{2-2\cos(\xi h)}{(\xi h)^2}| \leq 1,$ so that $|\frac{2-2\cos(\xi h)}{h^2}| \leq |\xi|^2$ is uniformly bounded as $h \to 0.$ Thus, if $\|g-f\|_2<\epsilon,$ by Plancherel’s theorem, $\|\widehat{g}-\widehat{f}\|_2<\epsilon,$ so that

$|Q(g,h)-Q(f,h)| =\left|\int_\mathbb{R} \frac{2-2\cos(\xi h)}{h^2} (\widehat{g}^2-\widehat{f}^2)d\xi\right| \leq \int_{|\xi|<\epsilon} |\xi|^2 |\widehat{g}^2-\widehat{f}^2|d\xi$

$+ \left|\int_{|\xi| \geq \epsilon} \frac{2-2\cos(\xi h)}{h^2} (\widehat{g}^2-\widehat{f}^2)d\xi\right| < 2C \epsilon$

uniformly in $h$ by applying Cauchy-Schwarz with $\widehat{g}-\widehat{f}$ in both terms. Thus, if $\|g-f\|_2<\epsilon$ and $Q(f,h_n)\geq 1+5C\epsilon$ for $h_n \to 0,$ then $Q(g,h_n) \geq 1+C\epsilon$ for all $n$ for $h_n \to 0,$ implying that $E^c$ is open and therefore showing that $E$ is closed.

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