It is well-known 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?
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$\begingroup$ This works, you can just estimate $\|\widehat{u}\|_1$ with Cauchy-Schwarz. (I assume that $|D|$ is multiplication by $|\xi|$ on the Fourier transform side.) $\endgroup$– Christian RemlingCommented May 23, 2021 at 20:43
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$\begingroup$ You may look at the Brezis-Gallouet inequality. $\endgroup$– RaffaeleScandoneCommented May 23, 2021 at 22:44
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$\begingroup$ @RaffaeleScandone, it might be helpful to state the inequality and provide a reference. $\endgroup$– Deane YangCommented May 23, 2021 at 22:45
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2 Answers
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A logarithmic correction to critical Sobolev embedding is given by the Brezis-Gallouet-Weinger inequality: [Brezis-Gallouet] [Brezis-Weinger]:
\begin{equation} \|f\|_{L^{\infty}(\mathbb{R}^d)}\lesssim 1+\|f\|_{W^{d/p,p}(\mathbb{R}^d)}\ln^{(p-1)/p}(e+\|f\|_{W^{d/q+\alpha,q}(\mathbb{R}^d)}),\qquad p,q\in(1,\infty),\;\alpha>0. \end{equation}
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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 $$ \int_1^\infty\frac{dr}{L(r)^2 r}<+\infty. $$