Analysis Problem of the Day 14

Today’s problem comes as Problem 5 on Stanford’s Spring 2019 Analysis Qual:

Problem 5: Recall that we define the Sobolev space $H^s(\mathbb{R})$ for $s \geq 0$ as the subspace of $L^2(\mathbb{R})$ with norm $\|u\|_{H^s} = \|\langle \xi \rangle^s \widehat{u}(\xi)\|_{L^2}$ , where $\langle \xi \rangle := (1+|\xi|^2)^{\frac12}$ is the Japanese bracket and $\widehat{u}$ is the Fourier transform of $u \in L^2.$

a. Show that $H^s(\mathbb{R}) \hookrightarrow L^\infty(\mathbb{R})$ for $s>\frac12$ and that the inclusion map is continuous.

b. Show that for $u,v \in H^1(\mathbb{R}),$ $uv \in H^1(\mathbb{R})$ and

$\|uv\|_{H^1} \leq C\|u\|_{H^1}\|v\|_{H^1}$

for some constant $C>0.$ (Hint: show that $\|u\|_{H^1} = (\|u\|_2^2+\|u'\|_2^2)^{\frac12}.$ )

Solution: Throughout, we will use $\|u\|_A \lesssim \|u\|_B$ to denote $\|u\|_A \leq C \|u\|_B$ for some constant $C>0.$ Additionally, we will use the fact that the Fourier transform is an isometry on $L^2,$ i.e. $\|u\|_2 = \|\widehat{u}\|_2$ for $u \in L^2.$

a. For $s>\frac12,$

$\|u\|_{L^\infty} \lesssim \|\widehat{u}\|_{L^1} \lesssim \|\widehat{u} \langle \xi \rangle^s \langle \xi \rangle^{-s}\|_{L^1} \lesssim \|\widehat{u} \langle \xi \rangle^s\|_{L^2} \|\langle \xi \rangle^{-s}\|_{L^2} \lesssim \|u\|_{H^s},$

where the first inequality follows from the fact that the Fourier transform (and its inverse) are continuous as maps $\mathcal{F},\mathcal{F}^{-1}: L^1 \to L^\infty,$ the middle inequality is by Holder, and the last inequality follows from the definition of the $H^s$ norm and the fact that $\langle \xi \rangle^{-2s}$ is integrable on $\mathbb{R}$ for $s>\frac12.$

Remark: Note that $\langle \xi \rangle^{-2s}$ is integrable on $\mathbb{R}^n$ for $s>\frac{n}{2}.$ Thus, the exact same argument shows that for $s>\frac{n}{2},$ $H^s(\mathbb{R}^n) \hookrightarrow L^\infty(\mathbb{R}^n).$ This is a reflection of a much more general set of theorems called the Sobolev embedding theorems, which state that $H^{s}(\mathbb{R}^n) \hookrightarrow L^p(\mathbb{R}^n)$ for $\frac{1}{p} = \frac12 - \frac{s}{n}.$

b. We first show the hint by noting that

$\|u\|_{H^1}^2 = \int \langle \xi \rangle^2 \widehat{u}^2 d\xi = \int \widehat{u}^2 d\xi + \int |-i\xi \widehat{u}|^2 d\xi = \|u\|_{L^2}^2+\|u'\|_{L^2}^2,$

where $u' := \mathcal{F}^{-1}(-i\xi \widehat{u}) \in L^2.$ Thus, the problem reduces to showing that

$\|uv\|^2_2+\|(uv)'\|_2^2 \lesssim (\|u\|_2^2+\|u'\|_2^2)(\|v\|_2^2+\|v'\|_2^2).$

The first term is easily bounded by $\|u\|_2^2 \|v\|_2^2$ as a consequence of Holder’s inequality, and for the second term, we use the product rule $(uv)'=u'v + uv'$ to bound $\|(uv)'\|_2^2 \leq \|u'\|_2^2\|v\|_2^2 + \|u\|_2^2\|v'\|_2^2.$ Thus, $\|uv\|_{H^1} \lesssim \|u\|_{H^1} \|v\|_{H^1}.$

Remark: The Leibniz rule is valid for $H^1$ functions since it is valid on $C_c^\infty,$ which is dense in $H^1$ in the $H^1$ norm (by standard use of mollifiers).

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