Frechet differentiability with a Hölder continuous map

Question

A bit of context: I try to verify that the following map is Frechet differentiable everywhere: Let $\Omega\subset\mathbb{R}^n$ be convex, bounded and open. Furthermore let $0<\alpha<\beta\leq 1$ and $\varphi\in C^{1,\beta}(\mathbb{R})$. Then the map in question is $f: C^{2,\alpha}(\Omega)\rightarrow C^{0,\alpha}(\Omega)$ with $$f(u):= -\Delta u + \varphi(u).$$ Here is what I already achieved: First of all I calculated a directional derivative at $u$ in direction $v$: $$\lim_{t\searrow 0}\frac{f(u+tv) - f(u)}{t} = \lim_{t\searrow 0}\frac{-\Delta(u+tv) + \Delta(u) + \varphi(u+tv)-\varphi(u)}{t} = -\Delta v + v\varphi'(u).$$ Now I like to verify that this directional derivative is also the Frechet derivative. Since differentiability is linear, it is enough to examine $u\mapsto -\Delta u$ and $u\mapsto \varphi(u)$ seperately. The map $u\mapsto -\Delta u$ did not pose a challenge for me, because the Laplace operator is linear. On the other hand the map $u\mapsto \varphi(u)$ is not so easy for me. Hence what I still need to show is $$\|\varphi(u+v) - \varphi(v) - v\varphi'(u)\|_{C^{0,\alpha}(\Omega)} = o(\|v\|_{C^{2,\alpha}(\Omega)})$$ Since the norm is given by $$\|u \|_{C^{0,\alpha}(\Omega)} = \|u \|_{C^{0}(\Omega)} + \sup_{x\neq y}\frac{|u(x)-u(y)|}{|x-y|^\alpha}$$ I can examine the two summands seperately. Using the mean value theorem on $\varphi$ I managed to show the result for the supremum norm. The Hölder-semi norm on the other hand still eludes me. I thought of applying the mean value theorem to $x\mapsto \varphi(u(x))$, because $\Omega$ is convex. This leads me to (for $x,y\in \Omega$) \begin{align*} &|\varphi(u(x)+v(x)) - \varphi(u(x)) - v(x)\varphi'(u(x)) - \varphi(u(y)+v(y)) + \varphi(u(y)) + v(y)\varphi'(u(y))|\\ =&|\nabla(u(\xi)+v(\xi))(x-y)\varphi'(u(\xi)+v(\xi)) - \nabla(u(\xi))(x-y)\varphi'(u(\xi)) - v(x)\varphi'(u(x)) +v(y)\varphi'(u(y))| \end{align*} Now I would have to estimate this expression to something like $$|x-y|^\alpha o(\|v\|_{C^{2,\alpha}(\Omega)}).$$ Since $\varphi$ only possesses one derivative I cannot apply the mean value theorem again or with respect to $\varphi'$. I also do not know how I can rearrange the resulting terms (using the Hölder continuity of $\varphi$) in a good manner to proceed. Hopefully one of you can give me a hint in this regard.

Answer 1

I think I got it. I will leave a hint for future references. The idea is to use the following inequality for $h\in C^{0,\beta}(\Omega)$, which is true for all $\delta>0$. $$h\ddot{o}l_{\Omega,\alpha}h \leq \delta^{\beta-\alpha}h\ddot{o}l_{\Omega,\beta}h + C(\delta)\|h\|_{L^\infty(\Omega)}$$ Applying this inequality for $$h= \varphi(u+v)-\varphi(u) - \nabla v\varphi'(u)$$ should yield the desired result. The inequality itself can be proven by distuingishing two cases: $|x-y|\leq \delta$ and $|x-y|\geq \delta$.