Let $F\subset \mathbb{R}^n$ be an $F_\sigma$-set and $\Phi:F\to \mathbb{R}^d$ be an $\alpha$-Holder continuous map, that is $$|\Phi(x)-\Phi(y)|_\infty \le L|x-y|_\infty^\alpha\; \forall x,y\in F$$ with constant $L$ and exponent $\alpha \in (0,1]$.
Now assume that $J$ is a square of side-length $s<1$ and centre $c\in \mathbb{R}^n$. Then because of the above condition, we have $$\Phi(J)=\Phi\Big(\prod_{k=1}^n[c_k-\frac{1}{2}s,c_k+\frac{1}{2}s)\Big)\subset\prod_{k=1}^d[\Phi_k (c)-\frac{L}{2}s^{1/\alpha},\Phi_k(c)+\frac{L}{2}s^{1/\alpha})$$
$|\cdot|_\infty$ is the maximum norm of the vector in the Euclidean space.
I tried showing this by putting $y\in J$, then $|y_k-c_k|\le \frac{1}{2}s$, so $|\Phi(y_k)-\Phi(c_k)|\le |\Phi(y)-\Phi(c)|_\infty \le L|y-c|_\infty^\alpha \le L (\frac{s}{2})^\alpha$. But I don't see how to get the bound $\frac{L}{2}s^{1/\alpha}$ as required. I would greatly appreciate it if anyone could help me.
The statement seems to be wrong. Take $\Phi(x)=\sqrt{|x|}$.
For $0\le x<y$, $$\sqrt{y}-\sqrt{x} \\ =(y-x)/(\sqrt{y}+\sqrt{x}) \\ \le(y-x)^{1/2}(y+x)^{1/2}/(\sqrt{y}+\sqrt{x}) \\ \le(y-x)^{1/2}(y^{1/2}+x^{1/2})/(\sqrt{y}+\sqrt{x}) \\ =(y-x)^{1/2},$$ similarly for $y<x\le 0$, and for $x<0<y$, $$|\sqrt{y}-\sqrt{-x}| \\ \le |y-(-x)|^{1/2} \\ \le (|y|+|x|)^{1/2} \\ =(y-x)^{1/2}.$$
So, it is $\tfrac{1}{2}$-Holder with $L=1$. Now, $[-1/4,1/4)$ is mapped onto $[0,1/2]$ which is not a subset of $[-1/8,1/8)$.
Unless I am missing something here.