Twice continuously differentiable bounded functions with non negative second derivative

As you concluded, the only continuous bounded convex functions on $\mathbb R$ are constant functions. One characterization of convex functions is that their graphs lie above their tangent lines: $f(x)\ge f(a)+f'(a)(x-a)$ for all $x$ and $a$. If $f$ is bounded, so if $x\mapsto f(a)+f'(a)(x-a)$, so $f'(a)=0$, and so on.

It is sufficient to prove that $f$ is constant in $]-\infty,+\infty[$.

If it was not constant there would exists $x_0\in \mathbb{R}$ such that $f’(x_0)\ne0 $. There are only two possibilities: $f’(x_0)>0$ or $f’(x_0)<0$.

Without loss of generality we can suppose that $f’(x_0)>0$.

The hypothesis $f’’(x)\ge0$ for all $x\in \mathbb{R}$ implies that $f’(x)$ is a monotone nondecreasing function, so $f’(x)\ge f’(x_0)>0$ for all $x\in ]x_0,+\infty[$.

Moreover for all $x \in ]x_0, +\infty[$ we can apply Langrange Theorem to the interval $[x_0,x]$, so there exists $c \in ]x_0,x[$ such that $f(x)-f(x_0)=f’(c)(x-x_0)\ge f’(x_0)(x-x_0)$.

It means that $f(x)\ge f(x_0)+f’(x_0)(x-x_0)$ for all $x \in ]x_0, +\infty[$. So $f$ is not bounded from above in $]x_0, +\infty[$, but it is absurd because it contradicts one of the hypothesis.

Hence it is impossible that $f$ is not constant in $]-\infty,+\infty[$.

It means $f$ is constant in $]-\infty,+\infty[$, so it is infinitely differentiable.