I am given $f: \mathbb{R} \to \mathbb{R}$, which is convex, and know that it's derivative in 0 exists and that $f'(0) = 1$. Now I have to prove/disprove if $f(x) \geq x ~\forall x \in \mathbb{R}$.
My attempt to find a counterexample goes as follows.
I know that $f$ strictly convex $\iff f'$ monotone. So I took $f(x) := - \ln(1-x)$, for which f'(0) = 1 holds and which's derivative is monotone because $$ f'(x_1) < f'(x_2) \iff \frac{1}{1-x_1} < \frac{1}{1-x_2} \iff 1 - x_2 < 1 - x_1 \iff x_1 < x_2 $$ Am I correct?
Since both the convexity and the derivative of a function are unaffected by adding a constant, it is possible to take a function satisfying all of the above properties and simply add a large enough negative constant to ensure $f(x)<x$ for some $x$.
For instance: $$f(x)=x^2+x-100000000$$ works.