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 $c$.



How to prove that the L-infinity norm is smaller than the Besov norm?
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
$$
\Vert u\Vert\_{L^\infty}\le \sum\_{\nu \ge 0}2^{\nu \alpha}\Vert\phi\_\nu(D)u\Vert\_{L^\infty}2^{-\nu \alpha}
\le \Vert u\Vert\_{B^\alpha\_{\infty, \infty}}
\underbrace{\sum\_{\nu \ge 0}2^{-\nu \alpha}}\_{c\_\alpha}.
$$


