How much can you improve a Hölder function by composing it with another?

Question

Let $f: [0, 1] \to \mathbb R$ be a continuous function. Define the local Hölder exponent $H(f, x)$ of $f$ at $x \in [0, 1]$ by

$$H(f, x) := \sup\left\{0 \leq \alpha \leq 1\mid\lim_{\delta \to 0_+} \sup_{y, z \in B_\delta (x)} \frac{|f(y) - f(z)|}{|y-z|^\alpha}\ \, \text{is finite}\right\}.$$

Let $F: \mathbb R \to \mathbb R$ be an another continuous function. We define the local Hölder exponents for $F$ in the same way. Is it true that for all $x \in [0, 1]$, we have

$$H(F \circ f, x) \leq 1 + H(f, x) - H(F, f(x))?$$

Answer 1

For any $\alpha>0$, put $$f(x)=\begin{cases}-e^{-\frac{1}{x^2}},&x<0\\x^\alpha,&x>0\end{cases}\quad \text{and}\quad F(x)=\begin{cases}x,&x<0\\e^{-\frac{1}{x^2}},&x>0\end{cases},$$

the composition $F\circ f$ is smooth and so your inequality becomes $$ 1=H(F\circ f,0)\leq 1+H(f,0)-H(F,0)=\alpha. $$