# Embedding theorem for anisotropic Sobolev spaces

Let $d=d_1+d_2$, $s_1,s_2>0$, $p>1$ and $(x_1,x_2)\in \mathbb{R}^{d_1}\times \mathbb{R}^{d_2}$, $(\xi_1,\xi_2)\in \mathbb{R}^{d_1}\times \mathbb{R}^{d_2}$. Define $$W^{s_1,s_2}_{p}:=\left\{f: f=\mathcal{F}^{-1}\left(\frac{\mathcal{F}g(\xi)}{1+|\xi_1|^{s_1}+|\xi_2|^{s_2}}\right);\forall g\in L^p(\mathbb{R}^{d}) \right\}.$$

Is there any reference talk about the Sobolev embedding theorem for this kind of space? say, under which condition one can get $$W^{s_1,s_2}_{p}\hookrightarrow L^\infty.$$

• There are p1 and p2 in the notation but p in the definition – almaz Nov 10 '17 at 1:13
• @almaz Thank you for your comment. I have modified the definition. – Guohuan Zhao Nov 10 '17 at 6:58
• Isn't $min(s_1,s_2) p>n$ enough to ensure the embedding you want? – almaz Nov 10 '17 at 23:53
• @almaz of course $\min\{s_1, s_2\}p>d$ can ensure the embedding. I think the condition should be $\frac{d_1}{s_1p}+\frac{d_2}{s_2p}<1$, but I can't prove it. – Guohuan Zhao Nov 12 '17 at 9:24

There are several ways to proceed. Maybe the most elegant (if you are functional-analysis minded) is to use the fact that in $\mathbb{R}^d$ $(-\Delta)^{\alpha}$, for $0 < \alpha \leq 1$ is the generator of a Markovian semigroup satisfying Sobolev inequalities with ''dimensional constant'' given by $\alpha^{-1} \, d$. This can be seen by subordination, see the Chapter II of [1].
With that in mind, assume $s_1 < s_2$ and fix $$\alpha = \frac{s_1}{s_2}.$$ Now, the operator $A = (-\Delta_{d_1})^{\alpha} + (-\Delta_{d_2})$ generates a Markovian semigroup over $\mathbb{R}^{d_1} \times \mathbb{R}^{d_2}$ with dimensional constant given by $d' = d_1 \alpha^{-1} + d_2$. Your anisotropic Sobolev space will be $W^{s_2,p}_{A}(\mathbb{R}^{d_1} \times \mathbb{R}^{d_2})$ and writing the Sobolev embedding for the new dimension $d'$ gives the result.
Alternatively, you could prove directly the embedding for $p=2$ using the Plancherel inequality to see that any $f \in W^{p,s_1, s_2}$ has integrable Fourier transform for the right set of parameters. Indeed $$\int_{\mathbb{R}^d} |\widehat{f}(\xi)| d\xi = \int_{\mathbb{R}^d} (1 + |\xi_1|^{s_1} + |x_2|^s_2)^{-1} (1 + |\xi_1|^{s_1} + |x_2|^s_2) |\widehat{f}(\xi)| d\xi,$$ and applying the Cauchy-Schwartz inequality for $s_1$ and $s_2$ such that $(1 + |\xi_1|^{s_1} + |x_2|^s_2)^{-1} \in L^2(\mathbb{R}^d)$ gives the desired embedding. A complex interpolation argument will then give the result for $2 < p$ and duality for the Fourier multiplier of symbol $(1 + |\xi_1|^{s_1} + |x_2|^s_2)^{-1}$ will give its $L^1 \to L^2$ boundedness.