About Sobolev embedding theorem of the case $W^{s,2}$. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T07:12:05Z http://mathoverflow.net/feeds/question/109562 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109562/about-sobolev-embedding-theorem-of-the-case-ws-2 About Sobolev embedding theorem of the case $W^{s,2}$. Andrew 2012-10-13T22:34:04Z 2012-11-05T13:42:12Z <p>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 $x \in \Bbb R^n$ ? </p> <p>I think the general Sobolev embedding theorem cannot be applied in this case because the domain is $\Bbb R^n$ which is open. </p> http://mathoverflow.net/questions/109562/about-sobolev-embedding-theorem-of-the-case-ws-2/109572#109572 Answer by paul garrett for About Sobolev embedding theorem of the case $W^{s,2}$. paul garrett 2012-10-14T00:46:32Z 2012-11-05T13:42:12Z <p>This is very standard, but perhaps buried in fancier things:</p> <p>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.</p>