TITLE: Inequality involving the sup of a function and its first and second derivatives
QUESTION [10 upvotes]: Let $f:\mathbb{R} \to \mathbb{R}$ be a twice differentiable function. And let 
$$\eqalign{   & M_0  = \sup \left|f(x)\right|  \cr 
              & M_1  = \sup \left|\frac{d}{dx} f(x) \right|  \cr 
              & M_2  = \sup \left|\frac{d^2}{dx^2} f(x) \right| \cr }$$
Prove that 
$$M_1 ^2  \leqslant 4M_0 M_2$$
I can not think how I can relate these values ​​in some inequality )=

REPLY [11 votes]: This is an exercise in Rudin's Principles of Mathematical Analysis, if I remember correctly, where it appears with a generous hint. I think the solution alluded to runs as follows.
Fix $x\in\mathbb R$. For $h>0$, you can expand $f(x+2h)$ via Taylor's theorem:
$$
f(x+2h)
=
f(x)+f'(x)2h+\frac{f''(\xi)}{2!}(2h)^2
$$
where $\xi\in(x,x+2h)$. So
$$
f'(x)=\frac{1}{2h}\left[f(x+2h)-f(x)\right]-f''(\xi)h
$$
From here you can bound $|f'(x)|$:
$$
|f'(x)|\leq\frac{1}{2h}|f(x+2h)-f(x)|+|f''(\xi)|h\leq\frac{M_0}{h}+M_2h.
$$
Since $x$ is arbitrary, you can replace $|f'(x)|$ with $M_1$. If $M_2=0$, let $h\to\infty$ to get $M_1=0$. Otherwise, rearrange to get
$$
0\leq M_2h^2-M_1h+M_0=M_2\left(h-\frac{M_1}{2M_2}\right)^2+\frac{4M_2M_0-M_1^2}{4M_2}.
$$
Take $h=\frac{M_1}{2M_2}$ to get the inequality.