Are there any $t\in(0,1)$, $p\in[1,\infty)$ such that $W^{t,p}(\mathbb{R})$ is continuously embedded into $L^\infty(\mathbb{R})$? I have been looking several literatures, but I have not yet found this out. Also, I am not familiar with proofs for Sobolev spaces. Can anyone give a reference on whether it can or cannot be done? Thank you.

  • $\begingroup$ It's pretty sure that the embedding holds if and only if $tp>1$. Typically, the fractional exponents are treated for Besov spaces, of which Sobolev spaces are a special case. Try to fish for information in this article and references. $\endgroup$ – you can call me Al Jan 30 '14 at 5:48

The case of $\,p=2\,$ with $\,s>\frac{1}{2}\,$ is rather trivial: $$\max_{\mathbb{R}}|u(x)|\leqslant\frac{1}{2\pi}\!\cdot\!\!\int\limits_{-\infty}^{+\infty}|\hat{u}(\xi)|\,d\xi \leqslant\frac{1}{2\pi}\!\cdot\!\!\biggl(\int\limits_{-\infty}^{+\infty}\frac{d\xi}{(1+{\xi}^2)^s}\,\biggr)^{1/2}\!\!\!\cdot\!\biggl(\int\limits_{-\infty}^{+\infty}(1+{\xi}^2)^s|\hat{u}(\xi)|^2 d\xi\biggr)^{1/2},$$ where $\,\hat{u}\,$ denotes the Fourier transform of $u$. Hence follows the required estimate $$ \max_{x\in\mathbb{R}}|u(x)|\leqslant C\|u\|_{H^s(\mathbb{R})}\,.$$

The case of $\,p\in(1,2)\,$ with $\,s>\frac{1}{p}\,$ is slightly less trivial: $$\max_{\mathbb{R}}|u(x)|\leqslant\frac{1}{2\pi}\!\cdot\!\!\int\limits_{-\infty}^{+\infty}|\hat{u}(\xi)|\,d\xi \leqslant\frac{1}{2\pi}\!\cdot\!\!\biggl(\int \limits_{-\infty}^{+\infty}\frac{d\xi}{(1+{\xi}^2)^{\frac{sp}{2}}}\,\biggr)^{1/2}\!\!\!\cdot\!\biggl(\int\limits_{-\infty}^{+\infty}\bigl|(1+{\xi}^2)^{\frac{s}{2}}\hat{u}(\xi)\bigr|^{p'} d\xi\biggr)^{1/2}$$ with the Hölder conjugate $p'=p/(p-1)$.  Applying the Hausdorff-Young inequality $$\bigl\|(1+{\xi}^2)^{\frac{s}{2}}\hat{u}(\xi)\bigr\|_{L^{p'}(\mathbb{R})} \leqslant C\bigl\|F^{-1}\bigl[(1+{\xi}^2)^{\frac{s}{2}}\hat{u}(\xi)\bigr]\bigr\|_{L^{p} (\mathbb{R})}\,,$$ notice that for a function $u$, the inverse Fourier transform $F^{-1}\bigl[(1+{\xi}^2)^{\frac{s}{2}}\hat{u}(\xi)\bigr]$ coincides with its Bessel potential of order $-s$. Hence follows the required estimate $$ \max_{x\in\mathbb{R}}|u(x)|\leqslant C\|u\|_{W^{s,p}(\mathbb{R})}\,.$$

The case $p\in(2,\infty)$ is a little tricky, but just a little. The latter case generally requires some suitable integral representation for a function $u\in W^{s,p} (\mathbb{R})$, thus making it possible to merge all cases into one: $p\in(1,\infty)$ with $\,s>\frac{1}{p}\,$. Under the circumstances, the most appropriate will be a representation in terms of the Bessel potentials. This is due to the definition of $u\in W^{s,p}(\mathbb{R})$ which can be formulated to cover all real values of $s\in\mathbb{R}$. Generally, most helpful is to define $u\in W^{s,p}(\mathbb{R})$ as a subspace in the Schwartz space of tempered distributions $$W^{s,p}(\mathbb{R})\overset{def}{=}\bigl\{u\in S'(\mathbb{R})\,\colon\, \hat{u}\in L^1_{loc}(\mathbb{R}),\;F^{-1}\bigl[(1+|\xi|^2)^{\frac{s}{2}}\hat{u}(\xi)\bigr]\in L^p(\mathbb{R})\bigr\},$$ where $F^{-1}$ denotes the inverse of the Fourier transform $F\colon\, S'(\mathbb{R})\to S'(\mathbb{R})$. The Bessel potential $B_{\mu}\colon\, S'(\mathbb{R})\to S'(\mathbb{R})$ is generally defined as a linear operator $$B_{\mu}(v)=F^{-1}\bigl[(1+|\xi|^2)^{-\frac{\mu}{2}}\hat{v}(\xi)\bigr], \quad \mu\in \mathbb{R}. $$ The space $W^{s,p}(\mathbb{R})$ is often called a space of the Bessel potentials with densities in $L^p(\mathbb{R})$. To find the desired suitable representation for a function $u\in W^{s,p}(\mathbb{R})$, notice that $$ \hat{u}(\xi)=\frac{\hat{u}(\xi)}{1+|\xi|^2}+\frac{\xi^2}{(1+|\xi|^2)^{1+\frac{s}{2}}}\!\cdot\!{(1+|\xi|^2)^{\frac{s}{2}}}\hat{u}(\xi),\quad s\in \mathbb{R}, $$ which implies that $$u(x)=B_2(u)-\frac{d^2\,}{dx^2}B_{2+s}\Bigl(F^{-1}\bigl[(1+|\xi|^2)^{\frac{s}{2}}\hat{u}(\xi)\bigr]\Bigr),\quad s\in \mathbb{R}. $$ Most important is the fact that for $\mu>0$, the Bessel potential turns into an integral operator $$ B_{\mu}(v)=\!\int\limits_{-\infty}^{+\infty}\!G_{\mu}(x-y)v(y)\,dy $$ with very nice kernel function $$G_{\mu}(x)\overset{def}{=}F^{-1}\bigl[(1+|\xi|^2)^{-\frac{\mu}{2}}\bigr], $$ a simple representation of which is step by step explicitly constructed on pages 131-132 of Singular Integrals and Differentiability Properties of Functions by E.M. Stein (Princeton, 1970). Using this representation one can readily verify that $\, G_{\mu}\in L^r(\mathbb{R}\,) $ if $\,\mu>\frac{1}{r'}\,$ with $\,r'=r/(r-1)\,$ being the Hölder conjugate to $\,r\in [1,\infty)$, while $\, G''_{\mu}\in L^r(\mathbb{R})\,$ if $\,\mu>2+\frac{1}{r'}\,$ with the same Hölder conjugate to $\,r\in [1,\infty)$ . Hence, by Hölder's inequality follows the estimate $$\max_{x\in\mathbb{R}}|u(x)| \leqslant \|G_2\|_{L^{p'}(\mathbb{R})}\!\!\times\!\|u\|_{L^{p}(\mathbb{R})}+\|G''_{2+s}\|_{L^{p'}(\mathbb{R})}\!\!\times\!\|u\|_{W^{s,p}(\mathbb{R})}\leqslant C\|u\|_{W^{s,p}(\mathbb{R})}$$ whenever $p\in (1,\infty)$ with $s>\frac{1}{p}\,$.

  • $\begingroup$ I don't understand the first inequality in the first line: how can you get rid of the absolute value of $e^{ix\cdot\xi}$ if you don't know that $\xi$ is real? For a general function, the F-transform may be defined for complex values of $\xi$. Am I missing something? $\endgroup$ – bartgol Oct 28 '14 at 15:08

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