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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 )=

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Oh sorry , I did it <.< !! –  August Oct 9 '11 at 15:21
That's great August! When that happens, it is perfectly acceptable to post your own answer below, so that others may learn how to do the problem! –  Zev Chonoles Oct 9 '11 at 15:28
Oh no , sorry again , im the more stupid guy in the world <.< , what i did is wrong, sorry <.<.<<<.< –  August Oct 9 '11 at 16:25

1 Answer 1

up vote 3 down vote accepted

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.

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