# Existence of a Hölder homeomorphism satisfying prescribed norm constraints

Let $$\Omega$$ be a convex body$$^{\boldsymbol{1}}$$ in $$\mathbb{R}^n$$ where $$n$$ is a positive integer. Fix a positive integer $$k$$ and some $$0<\alpha\leq 1$$. Let $$k_1> k_2>0$$. Does there necessarily exist a diffeomorphism $$\phi^{k,\alpha}\in C(\Omega,\Omega)$$ satisfying: $$\lVert\phi-1_{\Omega}\rVert_{k,\alpha}= k_1 \text{ and } \lVert\phi-1_{\Omega}\rVert_{\infty}\leq k_2,$$ where $$\lVert\cdot\rVert_{k,\alpha}$$ is the usual norm on the Hölder space $$C^{k,\alpha}(\Omega,\mathbb{R}^n)$$ and $$\lVert\cdot\rVert_{\infty}$$ is the familiar sup-norm on $$C(\Omega,\mathbb{R}^n)$$.

Intuitively, I imagine this can be constructed by starting with some “small homeomorphism” $$\tilde{\phi}:\Omega\rightarrow \Omega$$ and then smoothing it out/mollifying it. But I don't know how to formalize this idea, or if it is even true.

Edit$$^{\boldsymbol{1}}$$: Following @Pietro Majer's point; I should mention that I also assume that $$\Omega$$ is a convex body in $$\mathbb{R}^n$$ (so non-empty interior) and that $$n\in \mathbb{Z}^+$$ (so $$\Omega$$ cannot be a point).

• There is something missing in the question or in my understanding: if $1_\Omega$ is the identity, then $\phi = 1_\Omega$ would do. Aug 10, 2021 at 14:34
• @BenoîtKloeckner You're right. I meant to have an equality $\|\phi-1_{\Omega}\|_{k,\alpha}=k_1$ and not an inequality (or else, the problem becomes trivial as you noted). Aug 10, 2021 at 14:39

As it is the answer is no, by the following counter-example $$.$$
• Oh hahah! It took me a second, ok.. But what is $\Omega$ is say a convex body in $\mathbb{R}^n$ and $n>0$. Is it still obvious Aug 10, 2021 at 15:45
• For a convex body, I would take the flow of a sufficiently large(1) smooth field with sufficiently small(2) support in the interior of $\Omega$. The flow $\eta^t$ at time $t$ is a one parameter family of diffeos $\Omega\to\Omega$, that vary continuously wrto $t$, from $\eta^0=Id$ to a distance (in the chosen norm) larger than $k_1$ by (1), yet less than $k_2$ in uniform norm by (2). By continuity some $t$ should do. Aug 10, 2021 at 16:46