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.


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