Analysis Problem of the Day 77

Today’s problem appeared as Problem 2 on the UCLA Fall 2009 Analysis Qual:

Problem 2: Consider the unbounded densely-defined operator $I-\Delta: P \subset L^\infty(\mathbb{T}^2) \to L^2(\mathbb{T}^2),$ where $\Delta$ is the Laplacian, $P$ is the subspace of trigonometric polynomials, and $\mathbb{T}$ is the unit torus. Show that $I-\Delta$ is bounded below, i.e. $\|v\|_\infty \leq C\|(I-\Delta)v\|_2$ for all $v \in P.$

Solution: We rely on the properties of the Fourier transform. Indeed, recall that the inverse Fourier transform maps $l^1$ boundedly to $L^\infty,$ i.e. $\|v\|_{L^\infty} \lesssim \|\widehat{v}\|_{l^1}.$ Additionally, $I-\Delta$ is an operator that corresponds to multiplication by $1+n^2+m^2$ on the Fourier side. Finally, by Plancherel’s theorem, one has $\|u\|_{L^2} = \|\widehat{u}\|_{l^2}.$ Thus, to show the claim, it suffices to prove that $\|\widehat{v}\|_{l^1} \lesssim \|(1+n^2+m^2)\widehat{v}\|_{l^2}.$ But this follows by Cauchy-Schwarz since

$\sum_{n,m}|a_{n,m}| \leq \|((n^2+m^2) a_{n,m})\|_{l^2} \|(n^{2}+m^2)^{-1}\|_{l^2} \lesssim \|(1+n^2+m^2)a_{n,m}\|_{l^2}$

for $a_{n,m} := \widehat{v}(n,m).$ We thus conclude the claim.

Remark: From a functional-analytic perspective, this shows that $I-\Delta,$ which in fact can be extended to a closed operator on the maximal domain that is the Sobolev space $H^2(\mathbb{T}^2),$ is injective and has closed range. In fact, it is invertible on $H^2,$ since the Fourier multiplier $1+n^2+m^2$ is invertible as a function.

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