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 
$$
W^{s\_1,s\_2}\_{p}\hookrightarrow L^\infty. 
$$



Embedding theorem for anisotropic Sobolev spaces
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.


*Varopoulos, N.Th.; Saloff-Coste, L.; Coulhon, T.*, Analysis and geometry on groups, Cambridge Tracts in Mathematics 100. Cambridge: Cambridge University Press (ISBN 978-0-521-08801-5/pbk). xii, 156 p. (2008). [ZBL1179.22009](https://zbmath.org/?q=an:1179.22009).


