Differentiability in fractional Zygmund spaces I'm trying to understand why for $s$ not an integer, if
$$S_0u\in L^\infty\text{ and }\sup_{j\geq0}2^{js}\|\Delta_ju\|_{L^\infty}<\infty,$$
then $u\in C^s$. Here I write $\Delta_j$ for the Littlewood-Paley block with frequencies in $2^{j-1}\leq|\xi|\leq 2^{j+1}$, and $S_0$ for frequencies $|\xi|\leq1$. We can define these as, eg.
$$\widehat{S_0u}(\xi)=\varphi(\xi)\hat u(\xi)\text{ and }\widehat{\Delta_j u}(\xi)=\psi(\xi/2^j)\hat u(\xi)$$
where $\varphi\in C_0^\infty$ is $1$ if $|\xi|\leq1/2$ and $0$ if $|\xi|\geq1$, and $\psi(\xi)=\varphi(\xi/2)-\varphi(\xi)$.
Proofs I have found of this fact always say it is sufficient to suppose $0<s<1$ (eg. vol. III of Taylor PDE, p. 41). Why is that enough? That is, why do we have

If $S_0u\in L^\infty$ and $\sup_{j\geq0}2^{j(k+\alpha)}\|\Delta_ju\|_{L^\infty}<\infty$ then $u\in C^k$ for $k\in\mathbb N$ and $0<\alpha<1$.

I feel like I'm missing something very obvious since no one bothers to prove this.
 A: Short answer: Bernstein.
First note that since $\widehat{S_0u}$ has compact support $S_0u$ is smooth, in fact real-analytic. So we forget about $S_0u$, at least for now.
Say that dyadic block $2^{j-1}\leq|\xi|\leq 2^{j+1}$ is $A_j$. There exists a $C^\infty_c$ function which equals $i\xi$ on $A_0$, hence there exists a Schwarz function $F$ with $$\hat F(\xi)=i\xi\quad(\xi\in A_0).$$Let $$F_j(t)=4^jF(2^jt).$$So $$||F_j||_1=2^j||F_0||_1$$and $$\hat F_j(\xi)=i\xi\quad(\xi\in A_j).$$If you look at the Fourier transform you see $$\Delta_j(u')=(\Delta_j u)'=F_j*{\Delta_ju};$$hence $$||\Delta_ju'||_\infty\le||F_j||_1||\Delta_ju||_\infty
=c2^j||\Delta_ju||_\infty.$$A similar argument, using a function with Fourier transform $-i/\xi$ on $A_0$, and (exercise) normalizing the dilations differently, shows that $$||\Delta_ju||_\infty\le c2^{-j}||\Delta_ju'||_\infty.$$So if $0<\alpha<1$ we have $$\sup_j2^{j(1+\alpha)}||\Delta_ju||_\infty<\infty$$if and only if
$$\sup_j2^{j\alpha}||\Delta_ju'||_\infty<\infty.$$ Assuming we've proved the result for $0<\alpha<1$, the last display implies $u'\in C^\alpha$, which says $u\in C^{1+\alpha}$ by definition.

We were a little glib about $S_ou$; just saying "smooth" is not really enough, we need uniform smoothness. And of course we never used the hypothesis that $S_0u\in L^\infty$. An argument as above, but simpler, shows that $$||S_0u'||_\infty\le c||S_0u||_\infty;$$in fact $S_ou$ bounded implies every derivative of $S_ou$ is bounded.
