Let $g_{\mu}(x) = \mu^{d/2}\exp(-\pi\mu|x|^2)$ for every $\mu > 0$. Prove that $$ \int_{\mathbb R^{d}}\left|(-\Delta)^{\frac{s}{2}} u\right|^{2} \geq \int_{\mathbb R^{d}}\left|(-\Delta)^{\frac{s}{2}} \sqrt{|u|^{2}*g_{\mu}}\right|^{2} $$ for every $u \in H^s(\mathbb R^d)$ with $0<s<1$.
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1$\begingroup$ Can I ask how you know that this is true? (Also, $\mathbb R^4$ in the second integral seems a typo.) $\endgroup$– Christian RemlingCommented Feb 12 at 14:41
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2$\begingroup$ One can show, by Fourier transform that $$\int_{\mathbb R^{d}}\left|(-\Delta)^{\frac{s}{2}} u\right|^{2} \geq \int_{\mathbb R^{d}}\left|(-\Delta)^{\frac{s}{2}} (u*g_{\mu})\right|^{2}.$$ The inequality that I asked is an improvement but it might not be true. Then I hope there is an estimate between the two terms with an error depending on $\mu$. $\endgroup$– MuniainCommented Feb 12 at 14:47
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