# 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 $n$?

• The proof is in advanced part of real analysis. Google Fourier transform and Sobolev embedding theorem should lead you some references. Commented Feb 22, 2016 at 3:13

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 $$\| u \|_p \leq C(s) \| u \|_{H^s} \quad \forall 2 \leq p \leq \infty.$$

• why $\| u \|_{L^\infty} \leq \| \hat{u} \|_1$? Commented Jun 20, 2019 at 7:41
• @eraldcoil By the Fourier inversion, we have $\lvert u(x) \rvert = \lvert \int \hat{u}(\xi) \exp(2\pi i\xi \cdot x) d\xi \rvert$. Now move the absolute value inside the integral to get $\| \hat{u} \|_1$. Commented Jun 22, 2019 at 19:16
• @EricThoma, I like your answer! Is there any chance to be a bit more precise on the constant $C$ in the last inequality at the first line? At least when $n=1$? Thank you in advance. Commented Nov 27, 2023 at 15:46
• @C.Bishop What do you mean by ‘be a bit more precise’? We have $C(s)=\sqrt{\int_{\Bbb{R}^n}\frac{d\xi}{(1+|\xi|^2)^s}}$, and this is finite as long as $2s>n$, i.e $s>\frac{n}{2}$; but $C(s)\to\infty$ as $s\to \frac{n}{2}$ from above, which suggests that (this proof of) the embedding fails when $s=\frac{n}{2}$, and indeed there are counterexamples here. Commented Nov 28, 2023 at 14:01

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.

• In which book did you see the statement with $s-k>n/2$? Commented Mar 23, 2020 at 20:11
• The proof there is on Functional Analysis by J. Cerda Commented Mar 24, 2020 at 22:51
• thanks. Do you know if there is a similar result in $L^p$? that is to say, if $s-k>n/p$ then $H^{s, p}(\mathbb{R}^n)$ in $C^k$? Commented Mar 24, 2020 at 23:13
• I don't know, probably something yes Commented Mar 25, 2020 at 15:17
• How can make the proof work for $T= \mathbb{R}/ 2 \pi \mathbb{Z}$. Commented Jun 8, 2021 at 5:48