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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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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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    @JackD'Aurizio Well I guess I tacitly assumed that M_0<\infty, in which case $M_2=0$ forces $M_1=0$ as the argument shows (and if M_2>0 we could just take $h=\sqrt{M_0/M_2}$ as you say, avoiding the algebraic step at the end). But I agree that if we allow $M_0=\infty$ then the inequality won't make sense as stated for $M_2=0$. – 2015-01-12