# About Sobolev embedding theorem of the case $W^{s,2}$.

Let $s>n/2, \; f \in W^{s,2}(\Bbb R^n)$ . Then how can I show that there is an embedding into the space of uniformly bounded, continuous functions, that is, $$|f(x)| \leqslant C\| f \|_{W^{s,2}}$$ for almost all $x \in \Bbb R^n$ ?

I think the general Sobolev embedding theorem cannot be applied in this case because the domain is $\Bbb R^n$ which is open.

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You should try to do it yourself. It's a good homework exercise. – Deane Yang Oct 13 '12 at 23:19
can you please explain what you call a "uniformly bounded continuous function"? or do you mean bounded uniformly continuous functions? How is the norm you consider different from that of $C_b(\mathbb R^n)$? – Delio Mugnolo Oct 13 '12 at 23:24
@DelioMugnolo It means that uniformly bounded and continuous functions, $f \in C_b \cap C^0$ I think. – Andrew Oct 13 '12 at 23:27
A global $W^{s,2}$ bound implies more than the Sobolev theorem for a bounded domain, because it not only implies continuity but also implies that the function has to decay fast enough at infinity. – Deane Yang Oct 13 '12 at 23:38

The Fourier transform $\hat{f}$ of $f$ is in $L^2$ for the weight $(1+|x|^2)^s$. Since $s>n/2$, the constant function $1$ is in that weighted $L^2$ space, so by Cauchy-Schwarz-Bunyakowsky, $\int_{\mathbb R^n} |\hat{f}| = \int |\hat{f}|\cdot 1\le |\hat{f}|_s \cdot |1|_t$, where $t=-s$ [to avoid TeX bug?] and the subscript denotes the weight. Thus, $\hat{f}$ is in $L^1$. Thus, by Fourier inversion and Riemann-Lebesgue, $f=\hat{\hat{f}}$ is uniformly continuous and goes to $0$ at infinity.
Dear Paul: presumably $1$ has the weight $-s$ instead of the weight $s$? – Willie Wong Nov 5 '12 at 10:29