Let $\Omega \subset \mathbb{R}^n$ be a bounded set with smooth boundary and let $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ be Lipschitz continuous, nondecreasing and $f(0)=0$. Consider the following boundary value problem $$\begin{aligned}u_t-\Delta u &= f(u), & &\text{in }\Omega\times(0,\infty) \\ u&= 0, &&\text{on } \partial\Omega\times(0,\infty) \\ u &= u_0 \geq 0, & &\text{on }\Omega\times\{t=0\}.\end{aligned}$$
In a paper that I'm reading the author uses the unique global classical solution to this problem, however, it is not clear to me why such a solution exists. Does someone know how to prove this or does someone know where I can find a proof for this?
So far I tried to construct a fixed point argument. For $0<M<\infty$ and $<\alpha<1$ let $D_M$ be the space of all functions $v$ on $\Omega\times(0,\infty)$ that are Hölder continuous with Hölder exponent $\alpha$, Hölder norm smaller than $M$, and $v(x,0)=u_0(x)$ on $\Omega$. We define a operator $T$ on $D_M$ such that $w=Tv$ is the solution to the linear problem
$$\begin{aligned}w_t-\Delta w &= f(v), & &\text{in }\Omega\times(0,\infty) \\ w&= 0, &&\text{on } \partial\Omega\times(0,\infty) \\ w &= u_0 \geq 0, & &\text{on }\Omega\times\{t=0\}.\end{aligned}$$
Parabolic Schauder results give us the existence of such a $w$ that is Hölder continuous with exponent $\alpha$. Now I'm stuck with proving that the Hölder norm of $w$ is also bounded by $M$.
Maybe you should have a look at Chapter 6: Some Nonlinear Evolution Equations, in the book: