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# If $u \in H^s(\mathbb{R}^n)$ for $s > n/2$, then $u \in L^\infty(\mathbb{R}^n)$?

How do I use the Fourier transform to see that if $u \in H^s(\mathbb{R}^n)$ for $s > n/2$, then $u \in L^\infty(\mathbb{R}^n)$, with the bound$$\|u\|_{L^\infty(\mathbb{R}^n)} \le C\|u\|_{H^s(\mathbb{R}^n)},$$the constant $C$ depending only on $s$ and $n$?

#### Solutions Collecting From Web of "If $u \in H^s(\mathbb{R}^n)$ for $s > n/2$, then $u \in L^\infty(\mathbb{R}^n)$?"

This is a part of the proof continuous inclusions Sobolev theorem. That is, if $s-k > n/2$, than $H^s(\mathbb{R}^n) \hookrightarrow C^k(\mathbb{R}^n)$.

If $|\xi| \geq 1$, let $r=|\xi|$, we have $d\xi=r^{n-1}dr$ with $r \in [1,\infty)$ and

$\displaystyle \int_{\mathbb{R}^n} (1+|\xi|^2)^{k-s} d\xi < \infty$ $\Longleftrightarrow$ $\displaystyle \int_{1}^{\infty} (1+r^2)^{k-s}r^{n-1} dr \cong \int_{1}^{\infty} r^{2k-2s}r^{n-1} dr < \infty$

if $2k-2s+n-1 < -1$, i.e. if $s-k > n/2$. Than, if $\varphi \in \mathcal{S}(\mathbb{R}^n)$, multiply and divide by $\omega_{s-k}(\xi)=(1+|\xi|^2)^{(s-k)/2}$ in

$\displaystyle D^\alpha \varphi(x)= \mathcal{F}^{-1}(\widehat{D^\alpha \varphi})(x) = \int_{\mathbb{R}^n} \widehat{D^\alpha \varphi}(\xi) e^{2\pi i x \cdot \xi} d\xi = \int_{\mathbb{R}^n} (2\pi i \xi)^\alpha \widehat{\varphi}(\xi) e^{2\pi i x \cdot \xi} d \xi$

and by Schwarz inequality, follows that

$\displaystyle |D^\alpha \varphi(x)| \leq \int_{\mathbb{R}^n} |(2\pi \xi)|^\alpha \omega_{s-k}(\xi)|\widehat{\varphi}(\xi)| \omega_{k-s}(\xi) d\xi \leq \left \| D^\alpha \varphi \right \|_{H^{s-k}} \left( \int_{\mathbb{R}^n} (1+|\xi|^2)^{k-s} d\xi \right)^{1/2}$

and then (i) $\left \| D^\alpha \varphi \right \|_{L^{\infty}} \leq C \left \| D^\alpha \varphi \right \|_{H^{s-k}} < \infty$.

Now, if $u \in H^s(\mathbb{R}^n)$, since $\mathcal{S}(\mathbb{R}^n)$ is dense in $H^s(\mathbb{R}^n)$, $\exists \lbrace \varphi_m \rbrace \subset \mathcal{S}(\mathbb{R}^n)$ such that $\varphi_m \rightarrow u$ in $H^s(\mathbb{R}^n)$, but for (i) we have that $\lbrace D^\alpha \varphi_m \rbrace_{m=1}^\infty \subset \mathcal{S}(\mathbb{R}^n)$ is uniformly of Cauchy, and follows that $D^\alpha \varphi_m \rightarrow D^\alpha u$ uniformly $\forall |\alpha| \leq k$, therefore $u \in \mathcal{C}^k(\mathbb{R}^n) \cap L^{\infty}(\mathbb{R}^n)$, and in your case you can take $k=0$.

Note that $\left \| \cdot \right \|_{H^{s-k}}$ and $\left \| \cdot \right \|_{H^{s}}$ are two equivalent norms.

I studied recently this theorem, but it should be correct.

I think it is worth mentioning a proof using the Fourier analytic definition of $H^s$, if only for its succinctness.

We have
$$\| u \|_{L^\infty} \leq \| \hat{u} \|_1 \leq \| \langle \xi \rangle^{-s} \|_2 \| \langle \xi \rangle^s \hat{u} \|_2 \leq C \| u \|_{H^s}.$$
Here $\langle \xi \rangle = \sqrt{1 + \lvert \xi \rvert^2}$. Interpolation with $L^2$ implies
$$\| u \|_p \leq C(s) \| u \|_{H^s} \quad \forall 2 \leq p \leq \infty.$$