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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With $\sum_{\nu \ge 0}\phi_\nu(\xi)=1$ be a Littlewood-Paley partition of unity we find that $u=\sum_{\nu \ge 0}\phi_\nu(D)u$ and thus since $$ \Vert u\Vert_{B^\alpha_{\infty, \infty}}=\sup_{\nu\in \mathbb N} 2^{\nu \alpha}\Vert\phi_\nu(D)u\Vert_{L^\infty}, $$ we get $$ \Vert u\Vert_{L^\infty}\le \sum_{\nu \ge 0}2^{\nu \alpha}\Vert\phi_\nu(D)u\Vert_{L^\infty}2^{-\nu \alpha} \le \Vert u\Vert_{B^\alpha_{\infty, \infty}} \underbrace{\sum_{\nu \ge 0}2^{-\nu \alpha}}_{c_\alpha}. $$
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$\begingroup$ But how to show the inequality $\|u\|_{L^\infty} \leqslant \sum_{i\geqslant0}\|\phi_i(D)u\|_{L^\infty}$? Actually I got stuck here, since convergence of $\sum_{i\geqslant0}\phi_i(D)u$ is in the space of tempered distribution, not $L^\infty$. So I do not think we can get $\|u\|_{L^\infty} =\|\sum_{i\geqslant0}\phi_i(D)u\|_{L^\infty} $ and then use triangular inequality. $\endgroup$– InuyashaCommented Dec 2, 2021 at 16:38
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2$\begingroup$ Each $\phi_i(D) u$ is a smooth $L^\infty$ function. So the partial sums $u_M := \sum_{\nu = 0}^M \phi_\nu(D) u$ is a Cauchy sequence in $L^\infty$. (In other words, the series actually converges absolutely in $L^\infty$.) $\endgroup$ Commented Dec 2, 2021 at 17:47
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$\begingroup$ @WillieWong I still do not understand it, sure each $\phi_i(D)u$ is smooth and in $L^\infty$, but how do you show $u_M$ is a Cauchy sequence? Even that is right, then it is strange since a Cauchy sequence of continuous functions in $L^\infty$ should converges to a continuous function, so clearly for general $u\in L^\infty$ the sequence $u_M$ does not converges to $u$. $\endgroup$– InuyashaCommented Dec 2, 2021 at 21:07
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2$\begingroup$ Let $K = \|u\|_{B^\alpha_{\infty,\infty}}$, then by definition $\| \phi_\mu(D)u\|_{\infty} \leq 2^{-\mu\alpha} K$. So $\|u_M - u_N\|_\infty \leq 2^{(1-\min(M,N))\alpha} K$. $\endgroup$ Commented Dec 2, 2021 at 21:08
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$\begingroup$ @Inuyasha Your final sentence seems to be confused. The inequality shows that $B^{\alpha}_{\infty,\infty}$ embeds in $L^\infty$, and not the other way around. Indeed functions that are $B^{\alpha}_{\infty,\infty}$ are continuous (in fact uniformly continuous, since $B^{\alpha}_{\infty,\infty} = C^\alpha$ [Holder space] when $\alpha \in (0,1)$) $\endgroup$ Commented Dec 2, 2021 at 21:18