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?

Actually something stronger is true. We dont need $h$ twice differentiable on an interval, just that $h''(0)$ exists. Using the MVT:
$$h(x) = h(x) - h(0) = h'(c_x)x = (h'(c_x)-h'(0))x = [(h'(c_x)-h'(0))/c_x]c_x \cdot x.$$
Because $h''(0)$ exists, $[(h'(c_x)-h'(0))/c_x]$ is bounded in absolute value for small $x,$ say by $M.$ For such $x,$ $|h(x)| \le M|c_x\cdot x| \le Mx^2.$
