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I have difficulty understanding an inequality. Let $\phi\in D(\mathbb{R}^d)$ with value 1 near the unit ball, where $D(\mathbb{R}^d)$ is the space of smooth compactly supported functions. Let $g=\mathcal{F}^{-1}\phi$, $r\geq 1$, $k\geq 1$. Then why we have the inequality $$||\partial^{\alpha}g||_{L^r}\leq ||\partial^{\alpha}g||_{L^{\infty}}+||\partial^{\alpha}g||_{L^1}\leq C||(1+|\cdot|^2)^d\partial^{\alpha}g||_{L^{\infty}}$$ where $|\alpha|=k$. Thanks for any hint!

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Let $f=\partial^\alpha g$. Since $g\in\mathcal{D}(\mathbb{R}^n)$, $f$ is in the Schwarz class. For the first inequality $$ \|f\|_r=\Bigl(\int_{\mathbb{R}^n}|f|^r\,dx\Bigr)^{1/r}\le\Bigl(\|f\|_\infty^{r-1}\,\|f\|_1\Bigr)^{1/r}=\|f\|_\infty^{1-1/r}\,\|f\|_1^{1/r}\le\|f\|_\infty+\|f\|_1. $$ The last inequality follows prom Young's inequality: if $a,b\ge0$, and $p,q\ge1$ such that $1/p+1/q=1$, then $$ a\,b\le\frac1p\,a^p+\frac1q\,b^q\le a^p+b^q. $$ For the second inequality use $$ \|f\|_1=\int_{\mathbb{R}^n}(1+|x|^2)^d\,|f(x)|\,(1+|x|^2)^{-d}\,dx\le\|(1+|\cdot|^2)^df\|_\infty\int_{\mathbb{R}^n}(1+|x|^2)^{-d}\,dx. $$ The last integral is finite.

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  • $\begingroup$ Thank you! Just one question, why g is compactly supported? The Fourier transform of g is but g itself may not be I think. I guess if we don't know it your proof still works. Thanks again! $\endgroup$ – cali Sep 18 '15 at 2:59
  • $\begingroup$ The support of $g$ is not compact. $g$ is in the Schwarz class: $C^\infty$ and rapidly decaying. $\endgroup$ – Julián Aguirre Sep 18 '15 at 9:37
  • $\begingroup$ Okay. I see. Thank you! $\endgroup$ – cali Sep 18 '15 at 13:19
  • $\begingroup$ Yes! Definitely! $\endgroup$ – cali Sep 18 '15 at 15:11

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