How to show Zygmund space is Hölder space? The motivation of this question is to show that Zygmund space is Hölder space, in certain cases.
For simplicity, take $s\in (0,1)$, 
I want to show 
$$\|f\| = \|f\|_\infty + \sup_{x,y\in \mathbb{R}^d,x\neq y}\frac{|f(x)-f(y)|}{|x-y|^s}$$
and 
$$\|f\|_* = \|f\|_\infty + \sup_{x\in\mathbb{R}^d}\sup_{h\in\mathbb{R}^d,h\neq 0}\frac{|f(x+2h)-2f(x+h)+f(x)|}{|h|^s}$$
given all terms are finite.
To show $\|f\|_*\leq C \|f\|_s$ is simple. How could I show the other way?
 A: For further simplicity, let's assume $f(0)=0$ and try to get a bound of the form $|f(x)|\le C|x|^s$. Applying the definition of $\|f\|_*$ with $h-x$ yields
$$|f(2x)-2f(x)|\le \|f\|_* |x|^s \tag1$$
More generally,
$$|f(2^{k+1}x)-2f(2^k x)|\le \|f\|_* 2^{ks}| x|^s , \quad k=0,\dots,n-1 \tag2$$
Multiplying each inequality in $(2)$ by $2^{n-k-1}$ and summing the results, we get
$$
\sum_{k=0}^{n-1}|2^{n-k-1}f(2^{k+1}x)- 2^{n-k} f(2^kx)| \le \|f\|_*| x|^s \sum_{k=0}^{n-1} 2^{ks+n-k-1} \le 2^n C \|f\|_*| x|^s \tag3
$$
where $C$ comes from summing the geometric progression.  The left side of $(3)$ fits into the generalized triangle inequality, which yields
$$
|f(2^nx)-2^n f(x)|\le 2^n C \|f\|_*| x|^s \tag4
$$
Since also $|f(2^nx)|\le \|f\|_*$, we conclude with
$$
|f(x)|\le \tilde C \|f\|_*| x|^s \tag5
$$
as required.
A: One actually does not even need to use the $L^{\infty}$ norm: even the seminorms are equivalent. As a very short proof:
let $\|f\|_{\dot{\mathcal C}^s} := \|f\|_* - \|f\|_\infty$ and $\|f\|_{\dot{C}^{0,s}} := \|f\| - \|f\|_\infty$ be the two suprema defined in the question.
Then
\begin{align*}
 \frac{|f(x+z) - f(x)|}{|z|^s} &= \left|\frac{2\,f(x+z)- f(x)-f(x+2z)}{2\,\left|z\right|^s} + \frac{f(x+2z)-f(x)}{2\,\left|z\right|^s}\right|
 \\
 &\leq \frac{\|f\|_{\dot{\mathcal C}^s}}{2} + \frac{\|f\|_{\dot{C}^{0,s}}}{2^{1-s}}.
\end{align*}
Taking the supremum in $x,z$ we obtain $\|f\|_{\dot{C}^{0,s}} \leq 2^{-1} \|f\|_{\dot{\mathcal C}^s} + 2^{s-1}\,\|f\|_{\dot{C}^{0,s}}$, and so multiplying by $2$ and putting all the $\|f\|_{\dot{C}^{0,s}}$ on the left side leads to
$$
(2-2^s)\,\|f\|_{\dot{C}^{0,s}} \leq \|f\|_{\dot{\mathcal C}^s}.
$$
