A problem on Sobolev Hilbert space using Fourier transform, from PDE book by Evans.

Question

Let $u \in H^s(\mathbb{R}^n)$.Given that $s \in \mathbb{N}$ and $s>\frac{n}{2}$.

I need to prove that $$u\in L^{\infty}(\mathbb{R}^n)$$ and $$\|u\|_{L^{\infty}(\mathbb{R}^n)} \le C\|u\|_{H^s(\mathbb{R}^n)}$$ $C$ depending only on $s$ and $n$.

Appreciate your help in getting some hints on proving this using Fourier transform. I know Plancheral Theorem is the one to use, but don't know how to link it with $L^{\infty}$-norm.

PS : This is Problem 18 in Chapter 5 from PDE by Lawrence C. Evans

PS2 : I can assure you that I am not a student, and hence this isn't a homework problem.

Answer 1

Since the $H^s$ norm is defined via the Fourier transform, it is natural to express $u$ in terms of $\hat u$ via the inverse transform. The main trick is to multiply by 1 in disguise, with $\langle y\rangle := \sqrt{1+|y|^2}$, $$u(x) = \int_{\mathbb R^n} \hat u(y)e^{2\pi i y\cdot x} dy = \int_{\mathbb R^n} \hat u(y) \langle y\rangle^s \langle y\rangle^{-s} e^{2\pi i y\cdot x} dy, $$ so that by Holder, $$ |u(x)| \le \int_{\mathbb R^n} \left |\hat u(y)\langle y\rangle^s \right| \langle y\rangle^{-s} dy \le \|\hat u(y)\langle y\rangle^{s}\|_{L^2} \|\langle y\rangle^{-s}\|_{L^2}.$$

$\|\hat u(y)\langle y\rangle^{s}\|_{L^2} = \|u\|_{H^{s}}$, and the constant depending on $s,n$ is $\|\langle y\rangle^{-s}\|_{L^2}$. It's finite since $s > n/2$ by assumption.

Answer 2

To get pointwise information about $u$ from its Fourier transform $\widehat u$, it makes sense to use the Fourier inversion formula: $$ u(x)=\int e^{2\pi i x\cdot \xi}\widehat u(\xi)d\xi. $$ Applying $\|.\|_\infty$ (Lebesgue norm not Sobolev) to both sides gives $$ \|u\|_\infty\le\|\widehat u\|_1. $$ In order to extract the decay from $\widehat u$, we let $\lambda(\xi)=(1+|\xi|^2)^{1/2}$; writing $\widehat u=(\lambda^{n+\delta/2}\widehat u)\frac{1}{\lambda^{n+\delta/2}}$ and applying Cauchy-Schwarz gives $$ \|u\|_\infty\le \left\|\frac{1}{\lambda^{n+\delta}}\right\|_1\|\lambda^{n+\delta}|\widehat u|^2\|_1. $$ This is why the inequality $s>n/2$ is strict, so that we can put $\delta=2(s-n/2)$. Choosing $\delta$ this way bounds the $\widehat u$ term by $\|\widehat u\|_{H^s} $, and the first term is finite since $\delta>0$ prevents the logarithmic divergence.