How to show Zygmund space is Hölder space?

Question

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?

Answer 1

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.

Answer 2

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 $$ (2-2^s)\,\|f\|_{\dot{C}^{0,s}} \leq \|f\|_{\dot{\mathcal C}^s}. $$