Let $f:C:=[0,1]^{d}\longrightarrow \mathbb{R}$ be continuous, and define the (usual) modulus of continuity of $f$ of order $\epsilon$ as
$\omega(f;\epsilon):=\sup\{|f(x)-f(y)|:x,y\in C, \|x-y\|\leq \epsilon\}$ ,
where $\|\cdot\|$ denotes the Euclidean norm. Consider an orientation-preserving homeomorphisms, or simplily an homeomorphism, $\phi:C\longrightarrow C$. What is the relation between $\omega(f;\epsilon)$ and $\omega(f\circ \phi;\epsilon)$? Can be stated, for a such $\phi$, an inequality of the form
$\omega(f\circ \phi;\epsilon) \leq k \omega(f;\epsilon)$
$k$ being a constat independent of $\phi$?
It is seems clear, from the definition, that $\omega(f\circ \phi;\epsilon) =\omega(f;\omega(\phi;\epsilon))$, but I do not see clear if an inequality like above is possible. Or, in other words, if there is an upper bound for $\omega(\phi;\epsilon)$, independent of $\phi$.
Many thanks in advances for your comments.
In fact for just one homeomorphism this can fail. On $[0,1],$ let $f(x)=x,\phi(x) = \sqrt x.$ Suppose we have $\omega(f\circ \phi,\epsilon)\le k \omega(f,\epsilon)$ for all $\epsilon.$ Then
$$f\circ \phi(\epsilon) - f\circ \phi(0) = \sqrt \epsilon \le \omega(f\circ \phi,\epsilon) \le k\omega(f,\epsilon) =k\epsilon$$
for all $\epsilon>0.$ Dividing by $\sqrt \epsilon,$ we obtain $1\le k\sqrt \epsilon$ for all $\epsilon>0.$ Letting $\epsilon\to 0$ then shows $1\le 0,$ contradiction.