Prove $|h(x)|\leq (Mx^2)/2$ I repeated this question because I would really appreciate a hint.
Let $h:[0,a]\rightarrow \mathbb{R}$ be twice differentiable, $h'(0)=h(0)=0$, and $|h''(x)|\leq M$ for all $x\in [0,a]$.
I proceed with the mean value theorem, but I do not have the factor of $1/2$ in the result above (I get$|g(x)|\leq (Mx^2)$). Does anyone have any hints? Let me know if more detail is wanted. Is there a solution only using MVT?

Show that $h(x) = \int_0^x (x-t) h''(t) dt$, then proceed by using the esgtimate for $h''$. 
