Analysis Problem of the Day 28

Today’s quite intimidating problem appeared as Problem 5 on the UCLA Spring 2023 Analysis Qual:

Problem 5. Let $e_1,e_2,e_3 \in \mathbb{R}^3$ be the standard basis for $\mathbb{R}^3,$ and let $w \in \mathbb{R}^3.$

a) Show that there is no $f \in L^2(\mathbb{R}^3)$ satisfying

$f(x) = \frac16 \sum_{j=1}^3 (f(x+e_j)+f(x-e_j)) +e^{iw \cdot x - |x|^2/2}$

a.e.

b) Show that for every $\epsilon > 0$ there exists $f \in L^2(\mathbb{R}^3)$ such that

$\left\|f(x)-\left(\frac16 \sum_{j=1}^3 (f(x+e_j)+f(x-e_j)) +e^{iw \cdot x - |x|^2/2}\right)\right\|_2^2 < \epsilon.$

Solution: a) The presence of a complex exponential, a Gaussian, and the fact that we are working over $L^2$ highly suggests that this problem is related to the Fourier transform. Thus, let’s take the Fourier transform on both sides to get

$\widehat{f}(\xi) = \frac16 \sum_{j=1}^3 \widehat{f}(\xi) (e^{i e_j \cdot \xi}+e^{-i e_j \cdot \xi}) + e^{-|\xi -w|^2/2},$

where we use the fact that the Fourier transform of a Gaussian is a Gaussian and translation corresponds to a phase shift on the frequency side. Simplifying and solving for $\widehat{f}$ yields

$\widehat{f}(\xi) = \frac{e^{-|\xi-w|^2/2}}{1-\frac13\sum_{j=1}^3 \cos (\xi_i)}.$

Notice that the right hand side blows up whenever $\xi_i = 2\pi n.$ To understand how fast this blow up occurs, we Taylor expand by using $\cos x \approx 1 -\frac{x^2}{2}.$ Then, the denominator simplifies as $\frac16 (\xi_1^2+\xi_2^2+\xi_3^2) = \frac16 |\xi|^2.$ On $\mathbb{R}^3,$ integrating in spherical coordinates yields a Jacobian factor proportional to $r^2,$ so since the Gaussian is bounded near the origin, the $L^2$ norm of $\widehat{f}$ is bounded below by $\int \frac{1}{r^4} r^2 dr = \infty,$ i.e. $\widehat{f} \not \in L^2(\mathbb{R}^3).$ By Plancherel’s theorem, this implies that $f \not \in L^2(\mathbb{R}^3),$ so no $f$ satisfying such conditions can exist.

b) From the above discussion, the issue with $L^2$ integrability clearly occurs precisely at the lattice points $2\pi \mathbb{Z}^3.$ To remedy this, we thus may modify $\widehat{f}$ in a small neighborhood of those lattice points. For small enough $\delta>0,$ pick balls of radii $B_j$ of radii $\frac{\delta}{\sqrt[3]{\frac43\pi}2^{\frac{j}{3}}}$ centered at the lattice points so that $\int_{\bigcup_j B_j} |\widehat{f}|^2 dx < \epsilon$ (this can be done by absolute continuity). Then, set $\widehat{f}(\xi)$ to be

$\widehat{f}(\xi) := \frac{e^{-|\xi-w|^2/2}}{1-\frac13\sum_{j=1}^3 \cos (\xi_i)}$

outside of those balls and some fixed $\epsilon>0$ on $\bigcup_j B_j.$ Note that $\widehat{f}$ has finite $L^2$ norm on $\bigcup_j B_j$ rougly bounded by $\epsilon \delta.$ Moreover, $\widehat{f}$ has finite $L^2$ norm on $\left(\bigcup_j B_j\right)^c,$ since the Gaussian is in $L^2$ and near the lattice points, $\sum_j e^{-|w-\xi|^2} \frac{2^j \pi}{\delta} = O(1)$ (since $e^{-x^2}$ decays faster than the growth of $2^x.$ ) Thus, $\widehat{f} \in L^2(\mathbb{R}^3),$ so by Plancherel’s, $f \in L^2(\mathbb{R}^3).$ Finally, by shrinking $\delta>0$ if necessary,

$\widehat{f}(\xi) - \left(\frac16 \sum_{j=1}^3 \widehat{f}(\xi) (e^{i e_j \cdot \xi}+e^{-i e_j \cdot \xi}) + e^{-|\xi -w|^2/2}\right)$

has $L^2$ norm less than $\epsilon$ since the function vanishes on $\left(\bigcup_j B_j\right)^c,$ while on all of the balls, $\widehat{f}$ contributes at most a constant multiple of $\epsilon,$ and so does the Gaussian. Thus, the inverse Fourier transform $f$ of $\widehat{f}$ as constructed satisfies the statement of the problem.

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