convex and differentiable imply twice differentiable

Let $f:\mathbb{R}\to\mathbb{R}$ be convex and differentiable, $a\in \mathbb{R}$ such that for all real $h$, we have : $f(a+h)=f(a)+hf'(a)+o(h^2)$.Show that $f''(a)$ exists and $f''(a)=0$.

I have a proof which does not use the convexity assumption and I cannot find why the proof would be incorrect.

Since $f(a+h-h)=f(a+h)-hf'(a+h)+o(h^2)$ then $f'(a+h)=\frac{f(a+h)-f(a)+o(h^2)}{h}$.Since $f(a+h)=f(a)+hf'(a)+o(h^2)$ then $f'(a)=\frac{f(a+h)-f(a)+o(h^2)}{h}$.

Therefore,$$\frac{f'(a+h)-f'(a)}{h}=\frac{o(h)}{h}=o(1).$$Hence $f''(a)$ exists and is $0$.