If $u\in H^s(\mathbb R^d),\,s>d/2$, then $\|u\|_{L^\infty(\mathbb R^d)}\leq C\|u\|_{H^s(\mathbb R^d)}$ via Fourier transform

Question

Problem: Use the Fourier transform to prove that if $u\in H^s(\mathbb R^d)$ for $s>d/2$, then $u\in L^\infty(\mathbb R^d)$, with the bound $$\|u\|_{L^\infty(\mathbb R^d)}\leq C\|u\|_{H^s(\mathbb R^d)}$$ for some constant $C$ depending only on $s$ and $d$.

My Thoughts: We will use the Fourier characterization of the Sobolev space $H^s(\mathbb R^d)$, where we define the Fourier tranform by $$\widehat{u}(\xi)=\frac{1}{(2\pi)^{d/2}}\int_{\mathbb R^d}u(x)e^{-ix\cdot\xi}\,dx.$$ Let $x\in\mathbb R^d$ be arbitrary but fixed. Then using the Fourier inversion theorem and the Cauchy-Schwarz inequality we get that \begin{align*} \vert u(x)\vert&=\frac{1}{(2\pi)^{d/2}}\left\vert\int_{\mathbb R^d}e^{ix\cdot \xi}\widehat{u}(\xi)\,d\xi\right\vert\\ &\leq\frac{1}{(2\pi)^{d/2}}\int_{\mathbb R^d}\vert \widehat{u}(\xi)\vert\,d\xi\\ &=\frac{1}{(2\pi)^{d/2}}\int_{\mathbb R^d}(1+\vert\xi\vert^2)^{s/2}\vert \widehat{u}(\xi)\vert(1+\vert\xi\vert^2)^{-s/2}\,d\xi\\ &\leq\frac{1}{(2\pi)^{d/2}}\left[\int_{\mathbb R^d}(1+\vert\xi\vert^2)^s\vert\widehat{u}(\xi)\vert^2\,d\xi\right]^{1/2}\left[\int_{\mathbb R^d}(1+\vert\xi\vert^2)^{-s}\,d\xi\right]^{1/2}\\ &= C\|u\|_{H^s(\mathbb R^d)}, \end{align*} where $$C=\frac{1}{(2\pi)^{d/2}}\int_{\mathbb R^d}(1+\vert\xi\vert^2)^{-s}\,d\xi,$$ where the latter integral, using polar integration, can be shown to be finite if and only if $s>d/2,$ hence the dependance of $C$ on $d$ and $s$.
Since $x\in\mathbb R^d$ above was arbitrary, it follows that $$\|u\|_{L^\infty(\mathbb R^d)}\leq C\|u\|_{H^s(\mathbb R^d)}$$ and the proof is complete.

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Do you agree with my proof above? Any feedback is most welcomed and appreciated.
Thank you for your time.

Answer 1

Yes, that's exactly it. Technically this proof establishes the estimate on a dense subspace of $H^s$ such as Schwartz space, because you are assuming that you can write $u$ as the inverse Fourier transform of an $L^1$ function, whereas the Fourier transform on $L^2$ (which the Fourier characterization of $H^s$ depends on) is defined through a limiting process. So you might also have to provide the standard density argument showing how this estimate extends to arbitrary $H^s$ functions, depending on what sort of details you're willing to sweep under the rug.