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Added an explicit title, typo fixed and minor Math Jaxing (used $\|\cdot\|$ instead of $||\cdot||$)
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Daniele Tampieri
Daniele Tampieri
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How to prove that the L-infinity norm is smaller than the Besov norm?

Suppose we have a distribution $u\in B_{\infty,\infty}^\alpha$, the Besov space with regularity coefficient $\alpha>0$. How to proofprove the folowing inequality?$$||u||_{L^\infty}\leqslant c||u||_{B_{\infty,\infty}^\alpha}$$ $$ \|u\|_{L^\infty}\leqslant c\|u\|_{B_{\infty,\infty}^\alpha} $$ for some constant $c$.

L-infinity norm is smaller than Besov norm

Suppose we have a distribution $u\in B_{\infty,\infty}^\alpha$, the Besov space with regularity coefficient $\alpha>0$. How to proof the folowing inequality?$$||u||_{L^\infty}\leqslant c||u||_{B_{\infty,\infty}^\alpha}$$for some constant $c$.

How to prove that the L-infinity norm is smaller than the Besov norm?

Suppose we have a distribution $u\in B_{\infty,\infty}^\alpha$, the Besov space with regularity coefficient $\alpha>0$. How to prove the folowing inequality? $$ \|u\|_{L^\infty}\leqslant c\|u\|_{B_{\infty,\infty}^\alpha} $$ for some constant $c$.

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Inuyasha
Inuyasha
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L-infinity norm is smaller than Besov norm

Suppose we have a distribution $u\in B_{\infty,\infty}^\alpha$, the Besov space with regularity coefficient $\alpha>0$. How to proof the folowing inequality?$$||u||_{L^\infty}\leqslant c||u||_{B_{\infty,\infty}^\alpha}$$for some constant $c$.