Sobolev estimates $\|\nabla\phi\|_{\infty}\leq C\|\phi\|_{H^2}$

Question

This is in continuation to this question.

What I want to prove is

$\|\nabla\phi\|_{L^{\infty}((a,b))}\leq C\|\phi\|_{H^2((a,b))}$

Answer 1

I will prove it for $I = \mathbb R$ and for all $1 \le p < \infty$ (in your case just take $p = 2$), the general case follows from the extension theorem. The aim is to show that $$\|u\|_{L^\infty(I)} \le C \|u\|_{W^{1, p}(I)}$$ for some $C$ depending only on $I$ (not on $p$). Let $v\in C^\infty_0(I)$ and set $$G(s) = |s|^{p - 1} s.$$ By setting $w = G(v)$ (which is in $C^\infty_0(I)$), we get (exercice) $$w' = G'(v) v' = p|v|^{p-1}v'.$$ Therefore, \begin{align} G(v(x)) &= \int_{-\infty}^x p|v(x)|^{p-1}v'(x)dx \end{align} by Holder we get \begin{align} |v(x)|^p &\le p \|v\|_{L^p(I)}^{p-1}\|v'\|_{L^p(I)} \end{align} and by Young inequality \begin{align} |v(x)| &\le C\|v\|_{W^{1, p}(I)} \end{align} for all $v \in C^\infty_0(I)$. Note that $C$ does not depend on $p$ as $p^{1/p} \le e^{1/e}$ for all $p \ge 1$. We deduce the desired result by density.

Remark. This shows that the inclusion $W^{1,p}(I) \subset L^\infty(I)$ is continuous for all $p \in [1, \infty]$ (the case $p = \infty$ is trivial). Moreover, when $I$ is bounded, we may show that $$W^{1,p}(I) \subset C(\overline{I})$$ is compact for $1 < p \le \infty$.

Answer 2

Another way using the Fourier transform. By the Fourier inversion theorem and the integral formula of the Fourier transform $$ \|\nabla\varphi\|_{L^\infty(\Bbb R)} = \|\mathcal{F}\big(x\,\widehat{\varphi}\big)\|_{L^\infty(\Bbb R)} \leq \|x\,\widehat{\varphi}\|_{L^1(\Bbb R)} $$ and by the Cauchy-Schwarz inequality $$\begin{align*} \|x\,\widehat{\varphi}\|_{L^1(\Bbb R)} &\leq \|(1+x^2)^{-1/2}\|_{L^2(\Bbb R)}\, \|(1+x^2)^{1/2}\,x\,\widehat{\varphi}\|_{L^2(\Bbb R)} \\ &\leq \pi\, \|(1+x^2)\,\widehat{\varphi}\|_{L^2(\Bbb R)} = \pi\, \|\varphi\|_{H^2(\Bbb R)} \end{align*}$$ using the fact that $\|(1+x^2)^{-1/2}\|_{L^2(\Bbb R)} = \int_{\Bbb R} (1+x^2)^{-1}\,\mathrm d x = \pi$ and the fact that $x\leq (1+x^2)^{1/2}$.