The classical Sobolev embedding theorem asserts that, under suitable conditions on the exponents $s,p$ and $n$, the Sobolev space $W^{s,p}(\mathbb{R}^n)$ embeds into an Holder space $C^{r,\alpha}(\mathbb{R}^n)$. Suppose now to work with the more general SobolevLorentz space $$W^{s,(p,q)}(\mathbb{R}^n):=\{f\in L^{p,q}(\mathbb{R}^n)\,s.t.\nabla^sf\in L^{p,q}(\mathbb{R}^n)\}$$ I wonder if is it possible to have (when $q<p$) some local logarithmic refinement of the Holder estimate, namely $$f\in W^{s,(p,q)}(\mathbb{R}^n)\Rightarrow \big(D^{\mu}f(x)D^{\mu}f(y)\big)\frac{\ln^{\beta}{xy}}{xy^{\alpha}}\lesssim 1$$ for every multiindex $\mu=r$, for $xy\leq C<1$ and for some positive parameter $\beta$.
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$\begingroup$ That implication is off. What is $D^\mu(x)$ supposed to mean? And where does $f$ come into play on the RHS? $\endgroup$– Johannes HahnSep 19, 2017 at 19:00

1$\begingroup$ I add f in the RHS. $D^{\mu}f$ is the derivative of $f$ with respect to the multiindex $\mu$. $\endgroup$– CapublancaSep 23, 2017 at 22:33
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Perhaps the answer is not. As far as I know, the estimates in Sobolev embedding are sharp, as long as the indexes satisfy the scaling relationship. And Lorentz spaces come from interpolation of normal Lebesgue spaces, which means the SobolevLorentz spaces have similar embeddings as normal Sobolev spaces.


$\begingroup$ Sorry for ambiguity. I mean the answer is not. $\endgroup$– Jacob LuSep 26, 2017 at 5:22

1$\begingroup$ I really don't get your argument. I totally agree that the Sobolev embedding are sharp, in the sense that one cannot improve the Holder exponent. But i'm asking for logarithmic improvement, which are not sensitive of such exponent. $\endgroup$ Nov 10, 2017 at 1:19