I'm currently working through Jay Cummings' Real Analysis textbook (self study), and have got stuck on a question on differentiation. The question is as follows:
Suppose that $h:[0,5]->\mathbb{R}$ is twice differentiable, that $h'(0)=h(0)=0$, and $|h''(x)|\leq M$. Prove that $\forall x \in [0,5], |h(x)|\leq Mx^2/2$.
So far I've tried using the MVT twice:
Let $x \in [0,5]$. By MVT, $\exists c \in (0,x)$ s.t. $h'(c)=h(x)/x$. Again, by MVT, $\exists d \in (0,c)$ s.t. $h''(d)=h'(c)/c$. This implies $|h(x)|\leq Mxc$, but I can't seem to get much further from here...
I also tried using MVT once, then differentiating:
Let $x \in [0,5]$. By MVT, $\exists c \in (0,x)$ s.t. $h'(c)=h(x)/x. h''(c)=\frac{xh'(x)-h(x)}{x^2}$ giving $|xh'(x)-h(c)|\leq Mx^2$. Again, I can't seem to get much further from here either...
Hints would be appreciated!
