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?
 A: Define
$$
g:[0,a]\rightarrow \mathbb{R}, g(x) = h(x) - M\frac{x^2}{2} \, .
$$
Then $g(0) = g'(0) = 0$ and $g''(x) \le 0$ in $[0, a]$.
Apply the MVT twice to obtain $g(x) \le 0$ in $[0, a]$.
For the other direction, use $\tilde g(x) = h(x) + M\frac{x^2}{2}$  .
A: Hint: Use Taylor theorem with Lagrange remainder to expand $h$ at $0$:
$$h(x) = h(0) + h'(0)x + \frac{1}{2}h''(\theta x) x^2.$$
where $\theta \in (0, 1)$.

For any $x_0 \in (0, a]$, let $A$ be defined such that $h(x_0) = h(0) + h'(0)x_0 + Ax_0^2 = Ax_0^2$. And define 
$$g(x) = h(x) - (h(0) - h'(0)x) - Ax^2 = h(x) - Ax^2, x \in [0, x_0].$$
By condition, $g(0) = g(x_0) = 0$, hence by MVT, there exists $\xi_1 \in (0, x_0)$ such that $g'(\xi) = h'(\xi) - 2A\xi = 0$. On the other hand, $g'(0) = h'(0) - 2A\times 0 = h'(0) = 0$. So $g'(\xi) = g'(0) = 0$ and another application of MVT then gives $g''(\eta) = 0$ for some $\eta \in (0, \xi)$. Since $g''(\eta) = h''(\eta) + 2A$, we deduce that 
$$|A| = \frac{1}{2}|h''(\eta)| \leq \frac{1}{2}M,$$
which implies that 
$$|h(x_0)| = |Ax_0^2| \leq \frac{1}{2}Mx_0^2.$$
A: Show that $h(x) = \int_0^x (x-t) h''(t) dt$, then proceed by using the esgtimate for $h''$. 
A: 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 $x,$ $|h(x)| \le M|c_x\cdot x| \le Mx^2.$
