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