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 $c$.
With $\sum_{\nu \ge 0}\phi_\nu(\xi)=1$ be a LittlewoodPaley 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 $$ \Vert u\Vert_{L^\infty}\le \sum_{\nu \ge 0}2^{\nu \alpha}\Vert\phi_\nu(D)u\Vert_{L^\infty}2^{\nu \alpha} \le \Vert u\Vert_{B^\alpha_{\infty, \infty}} \underbrace{\sum_{\nu \ge 0}2^{\nu \alpha}}_{c_\alpha}. $$

$\begingroup$ 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. $\endgroup$– InuyashaDec 2 '21 at 16:38

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 $L^\infty$.) $\endgroup$ Dec 2 '21 at 17:47

$\begingroup$ @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 $u$. $\endgroup$– InuyashaDec 2 '21 at 21:07

2$\begingroup$ Let $K = \u\_{B^\alpha_{\infty,\infty}}$, then by definition $\ \phi_\mu(D)u\_{\infty} \leq 2^{\mu\alpha} K$. So $\u_M  u_N\_\infty \leq 2^{(1\min(M,N))\alpha} K$. $\endgroup$ Dec 2 '21 at 21:08

$\begingroup$ @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 $\alpha \in (0,1)$) $\endgroup$ Dec 2 '21 at 21:18