# Estimate for non-differentiable implicit function theorem

Let $$F(x,s)$$ be a continuous function $$F:\mathbb R^m\times\mathbb R^n\to\mathbb R^m$$ such that $$\nabla_xF$$ is a Holder $$C^\alpha$$-function, say $$\frac12<\alpha<1$$.

Suppose $$F(0,0)=0$$ and suppose $$\nabla_xF(x,s)$$ is a invertible matrix for every $$(x,s)$$.

By non-differentiable implicit function theorem mention in Wikipedia, locally there is a unique function $$\phi:U\subset\mathbb R^m\to\mathbb R^n$$ near $$0$$, such that $$F(\phi(t),t)=0$$.

My question is, does $$\phi\in C^\alpha$$? The non-differentiable IFT only guarantee $$\phi$$ to be continuous, but not something further.

And for the distributional derivative $$\nabla\phi(t)=-(F_x^{-1}\cdot F_s)|_{(\phi(t),t)}$$, the product make sense as a $$C^{\alpha-1}$$ distribution when $$\alpha>\frac12$$. But the composition of $$C^{\alpha-1}\circ C^\alpha$$ is not sure to be well-defined

I just assume $$\phi$$ is already continuous due to the non-differentiable IFT. Remain to check the Holder condition.
Use $$F(\phi(t),t)=F(\phi(t'),t')=0$$, we have $$|\phi(t)-\phi(t')|\le\frac1{\inf|(\nabla_x F)^{-1}|}|F(\phi(t),t)-F(\phi(t'),t)|=\frac1{\inf|(\nabla_x F)^{-1}|}|F(\phi(t'),t')-F(\phi(t'),t)|\le\frac{[F(\phi(t'),\cdot)]_\alpha}{\inf|(\nabla_x F)^{-1}|}|t-t'|^\alpha\le\frac{[F]_\alpha}{\inf|(\nabla_x F)^{-1}|}|t-t'|^\alpha$$
The problem is still annoying if we consider the Zygmund-1 function class, namely those functions $$f$$ satisfies $$|f(x)+f(y)-2f(\frac{x+y}2)|\le C|x-y|$$. I believe they are still true. But the estimate may require using mean value theorem, where we should replace the first inequality into equality by choosing suitable coefficients.