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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 Sobolev-Lorentz 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 $$f\in W^{s,(p,q)}(\mathbb{R}^n)\Rightarrow \big(D^{\mu}f(x)-D^{\mu}f(y)\big)\frac{\ln^{\beta}{|x-y|}}{|x-y|^{\alpha}}\lesssim 1$$ for every multi-index $|\mu|=r$, for $|x-y|\leq C<1$ 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? $\endgroup$ Sep 19, 2017 at 19:00
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    $\begingroup$ I add f in the RHS. $D^{\mu}f$ is the derivative of $f$ with respect to the multiindex $\mu$. $\endgroup$
    – Capublanca
    Sep 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 Sobolev-Lorentz spaces have similar embeddings as normal Sobolev spaces.

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  • $\begingroup$ What means "perhaps" here? $\endgroup$
    – Stefan Kohl
    Sep 24, 2017 at 22:29
  • $\begingroup$ Sorry for ambiguity. I mean the answer is not. $\endgroup$
    – Jacob Lu
    Sep 26, 2017 at 5:22
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    $\begingroup$ 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. $\endgroup$
    – Capublanca
    Nov 10, 2017 at 1:19

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