Inverse Function Theorem on Zygmund Spaces, is the inverse in the same Zygmund Space?

Preliminary Definitions

Let $\Omega \subset \mathbb{R}^n$ be open. We define the Zygmund spaces $C^r_{*}(\Omega)$ with $r>0$, $r \in \mathbb{R}$ in the following way: (all the functions are allowed to have values in $\mathbb{R}^m$, via working with the components)

Let $r=l+\alpha$ with $l \in \mathbb{N}$ and $0\le \alpha <1$.

Case 1) If $\alpha >0$, $C^r_{*}(\Omega)$ is just the Hölder space $C^{l,\alpha}(\Omega)$.

Case 2) If $\alpha =0$, $C^l_{*}(\Omega)$ is the space of functions $u$ whose partial derivatives up to order $l-1$ are continuous and bounded, and if $\beta \in \mathbb{N}^n$ denotes a multiindex of order $l-1$, then the second order difference quotients are bounded, i.e,

$$\sum_{|\beta|=l-1} \displaystyle \sup_{h \ne 0, x,h} \frac{|\partial^{\beta}u(x+h)| - 2\partial^{\beta}u(x) + \partial^{\beta}u(x-h)}{|h|} < \infty$$

where $x \in \Omega$ and $h$ is taken so that $x+h, x-h \in \Omega$.

It is easy to see that $C^{l,0}(\Omega) \subset C^{l-1,1}(\Omega) \subset C^l_{*}(\Omega)$.

The motivation to work with Zygmund spaces is that these spaces are well-behaved for elliptic regularity results. They are useful in non linear elliptic PDE and they appear in the theory of mildly regular pseudodifferential operators.

Remark

An equivalent definition of Zygmund spaces when the domain of definition $\Omega$ of the functions is $\mathbb{R}^n$ can be given with the Fourier transform:

A tempered distribution $u \in \mathcal{S}'(\mathbb{R}^n)$ is said to be in the Zygmund space $C^r_{*}(\mathbb{R}^n)$ if we have that $$\|u\|_{C^r_{*}(\mathbb{R}^n)}:=\sup_{k \in \mathbb{N}} 2^{kr} \|\varphi_k(D)u\|_{L^{\infty}} < \infty$$

Where $\varphi_j$, $j \ge 0$ is a Littlewood–Paley partition of unity. This definition is just the definition of the Besov space $B^s_{\infty,\infty}(\mathbb{R}^n)$, which coincides with the Zygmund space $C^r_{*}(\mathbb{R}^n)$.

Question :

Let $s>2$, and let $\Phi=(\phi_1,\dots,\phi_n):\mathbb{R}^n \to \mathbb{R}^n$ be a $C^{s}_{*}$ mapping whose differential at $0 \in \mathbb{R}^n$ is an isomorphism, so the Inverse Function Theorem applies. My question: is the local inverse of $\Phi$ also $C^{s}_{*}$?

The answer must be affirmative but I don't know how to prove it.

Attempt at solution

Here it goes how I would try to prove it. Suppose for simplicity $\Phi(0)=0$.

Put $s=k+\delta$ for $k\ge 2$ and $0 \le \delta <1$.

By the Inverse Function Theorem we know that $\Phi$ is a $C^k$ diffeomorphism in some neighborhood of $0$, i.e, $\Phi:B_{\eta} \to B_{\varepsilon}$ is a $C^k$ diffeomorphism. In particular $\Phi^{-1}$ is $C^{k} \subset C^{s-1}$. Moreover the IFT says that

$D (\Phi^{-1})=[D \Phi \circ \Phi^{-1}]^{-1} : B_{\varepsilon} \to Gl(n)$

I want to prove that $D (\Phi^{-1})$ is $C^{s-1}_{*}$

As $D \Phi$ is $C^{s-1}_{*}$ and $\Phi^{-1}$ is $C^{s-1}$, one is tempted to think that the composition $D \Phi \circ \Phi^{-1}$ is $C^{s-1}_{*}$. Is this true? I don't know how to prove it. Suppose this is true, and we keep going.

As $C^{s-1}_{*}$ is an algebra under pointwise multiplication we see that the determinant $Det:=Det(D \Phi \circ \Phi^{-1})$ is also $C^{s-1}_{*}$, so I guess this implies that $\frac{1}{Det}$ is also $C^{s-1}_{*}$.

By the formula of the inverse of a matrix (the transpose of the adjoint multiplied by the inverse of the determinant) we conclude that $D (\Phi^{-1})$ is $C^{s-1}_{*}$, and so $\Phi^{-1}$ is $C^{s}_{*}$, and this concludes that $\Phi$ is a $C^{s}_{*}$ diffeomorphism.

Conclusion: for this proof to work, I would need to prove:

1) If $f \in C^{s-1}_{*}$ and $g \in C^{s-1}$, then the composition $f \circ g$ is $C^{s-1}_{*}$. It looks reasonable for me but I don't know whether it is true. This is used to see that $D \Phi \circ \Phi^{-1}$ is $C^{s-1}_{*}$.

2) The space $C^{s-1}_{*}$ is an algebra under pointwise multiplication. This is used to see that $Det \in C^{s-1}_{*}$. This is true for sure, and I am going to search the proof in a reference I've been given. However if anyone knows a short proof or another reference, is welcome. Edit: I already found the reference for this, on Triebel's book Function Spaces

3) If a function $f \in C^{s-1}_{*}$ is such that $|f(x)| \geqslant c >0$ for some constant $c$, its inverse $\frac{1}{f}$ is also $C^{s-1}_{*}$. This is used to see that $\frac{1}{Det} \in C^s_{*}$. It seems true, and I would say the proof should be similar to the proof that $C^{s-1}_{*}$ is an algebra for pointwise multiplication. Edit: I have already proved this The proof for Holder spaces is trivial, so it remains to see the proof for the Besov functions of integer exponent $\ge 2$, and this is easy using the chian rule.

Remark

All properties (1), (2) and (3) above are true and easy to prove if $s$ is not a natural number (i.e., in Hölder spaces). But if $s=k \in \mathbb{N}$ the proofs become more difficult because of the second order difference quotients.

Any help on this would be welcome (references, guesses, proofs). Of course if anyone know a shorter way to prove it than my attempt, it would be great.

Edit

Property (1) remains open. I was given the reference FUNCTIONAL CALCULUS IN HOLDER-ZYGMUND SPACES written by G. BOURDAUD AND MASSIMO LANZA DE CRISTOFORIS

There, the composition of two besov functions of exponent greater that one is proved, but it is done only in the scalar case, and it remains to adapt this proof (if possible) for the vector valued case. I have not had the time to try and adapt the proof yet, and I dont know whether I will do it. Don´t get me wrong, I am a geometer and its hard for me to understand that paper, so I am looking for a reference if possible.

• Please do not cross-post the same question. – Dirk Oct 15 '14 at 9:32
• I just edited the question saying that I have a partial answer, don't know what do you mean. – juan rojo Oct 15 '14 at 9:42
• What I mean is that you should post the same question here and at math.stackexchange simultaneously. This produces unrelated threads and wasted time of the community, see here. – Dirk Oct 15 '14 at 10:19
• Ok, sorry I didn't know this was not OK. Then I could delete the stack exchange question. – juan rojo Oct 16 '14 at 18:45