Here is a full context and a more detailed question.
There is an operator $T: \mathcal{C}^1[0,1] \rightarrow \mathcal{C}[0,1]$. It is defined as follows:
$$(Tf)(x) = x + (1-\epsilon) f(x) - \epsilon x(1-x) f'(x)$$
where $\epsilon \approx 0$ is a strictly positive constant, $A \in (0,1)$.
I am interested in showing the following: $$ f'(x) \ge 0 \Rightarrow (Tf) \in \mathcal{C}^1.$$
If $(Tf) \in \mathcal{C}^1$, then I can argue that $(1-x)(Tf)'(x)$ goes to $0$ as $x$ goes to $1$. This was the reason why I asked the earlier question, which I think is closely related. The question is, if $f \in \mathcal{C}[0,1]$ such that $f'$ exists and is non-negative $\forall x$, then, can we argue that $\lim_{x \rightarrow 1} (1-x) f'(x) = 0$?
I have tried building counterexamples but they either involve functions like $x^2 \sin(1/x)$ which are not monotonic or, functions like, $\int_0^x e^{-1/(1-y)} dy$ which obviously go to $0$.
For $f'$ being continuous on the compact interval $[0,1]$, then for some $C>0$, $|f'(x)|\leq C$ for all $x\in[0,1]$. So $|(1-x)f'(x)|\leq C|1-x|\rightarrow 0$ as $x\rightarrow 1$, so by Squeeze Theorem we have $(1-x)f'(x)\rightarrow 0$ as $x\rightarrow 1$.
$f''$ need no exist, so the second question is hard to tell.