I have to prove the following remark, but I have no idea how. I searched everywhere but didn't find anything.
Remark 1: If $\nabla{F}$ is only Hölder-continous with exponent $\gamma$ (instead of Lipschitz-continous) with $0<\gamma<1$ and $L>0$, than Lemma 1 (below) is true for all $x\in\mathcal{B}(z,\eta)$ with $0<\eta<\left(\dfrac{c}{L\beta}\right)^{1/\gamma}$ with fixed $0<c<1$.
Lemma 1: $F:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ is continuous differentiable in the open set $D\subset\mathbb{R}^n$ and $\nabla{F}$ is Lipschitz-continous in $D$ (with constant $L>0$). $\nabla{F}(z)^{-1}$ exists with fixed $z\in{D}$. Futher assume that there exists $\beta>0$ such that $\|\nabla{F}(z)^{-1}\|\le\beta$. Then for all $x\in\mathcal{B}(z,\eta)$ with $0<\eta<(c/(L\beta))$ and $0<c<1$ fixed, $\nabla{F}(x)$ is nonsingular and satisfies $$\|\nabla{F}(x)^{-1}\|\le\dfrac{\beta}{1-c}.$$
Does anyone know how I can proof the remark or where I can find it? Thanks
I believe everything rests on the following expansion $$ (I - A)^{-1} = I + A + A^2 + A^3 + \cdots $$ which converges whenever $\|A\| < 1$. Writing $(B-A)^{-1} = (B(I-B^{-1}A)^{-1} = (I-B^{-1}A)^{-1}B^{-1},$ this also implies $$ (B - A)^{-1} = (I-B^{-1}A)^{-1}B^{-1} = B^{-1} + B^{-1}AB^{-1} + (B^{-1}A)^2 B^{-1} + \cdots. $$ when $\|B^{-1}A\| < 1$. Using the triangle inequality, the fact that $\|CD\|\leq \|C\|\|D\|$, and summing the geometric series, we arrive at the bound $$ \|(B-A)^{-1}\| \leq \frac{\|B^{-1}\|}{1-\|B^{-1}A\|}. $$
We can now use this to prove Lemma 1, substituting $B = \nabla F(z)$ and $A = \nabla F(z) - \nabla F(x)$. This gives us $$ \|\nabla F(x)^{-1}\| \leq \frac{\|\nabla F(z)\|}{1-\|\nabla F(z)^{-1}(\nabla F(z) - \nabla F(x)\|} \leq \frac{\beta}{1 - \beta\|\nabla F(z) -\nabla F(x)\|}. $$ Now suppose that $|x-z| \leq \eta < \left(\frac{c}{L\beta}\right)^{1/\gamma}$. Then by the Holder continuity assumption on $\nabla F$, $$ \|\nabla F(z) - \nabla F(x)\| \leq \frac{c}{\beta}. $$ Plugging this into the above inequality yields $$ \|\nabla F(x)^{-1}\| \leq \frac{\beta}{1 - c}. $$