It is wellknown that $H^{\frac d2}(\mathbb R^d)=W^{\frac d2, 2}(\mathbb R^d)$ is not included in $L^\infty(\mathbb R^d)$, but it seems that there are some logarithmic substitutes. Is it true for instance that $$ \Vert u\Vert_{L^\infty(\mathbb R^d)}\lesssim \big\Vert \vert D \vert^{d/2}\ln(1+\vert D\vert) u\big\Vert_{L^2(\mathbb R^d)}\,, $$ or is there some analogous statement? Is there a clear connection with Gross' logarithmic Sobolev inequalities?

$\begingroup$ This works, you can just estimate $\\widehat{u}\_1$ with CauchySchwarz. (I assume that $D$ is multiplication by $\xi$ on the Fourier transform side.) $\endgroup$– Christian RemlingCommented May 23, 2021 at 20:43

$\begingroup$ You may look at the BrezisGallouet inequality. $\endgroup$– RaffaeleScandoneCommented May 23, 2021 at 22:44

$\begingroup$ @RaffaeleScandone, it might be helpful to state the inequality and provide a reference. $\endgroup$– Deane YangCommented May 23, 2021 at 22:45
2 Answers
A logarithmic correction to critical Sobolev embedding is given by the BrezisGallouetWeinger inequality: [BrezisGallouet] [BrezisWeinger]:
\begin{equation} \f\_{L^{\infty}(\mathbb{R}^d)}\lesssim 1+\f\_{W^{d/p,p}(\mathbb{R}^d)}\ln^{(p1)/p}(e+\f\_{W^{d/q+\alpha,q}(\mathbb{R}^d)}),\qquad p,q\in(1,\infty),\;\alpha>0. \end{equation}
Following Christian Remling suggestion, it seems that $$ \Vert u\Vert_{L^\infty(\mathbb R^d)}\le \gamma(d)\bigl\{ \Vert u\Vert_{L^2(\mathbb R^d)}+ \Vert \vert D\vert^{d/2} L(\vert D\vert) u\Vert_{L^2(\mathbb R^d)} \bigr\} $$ with $$ \int_1^\infty\frac{dr}{L(r)^2 r}<+\infty. $$