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 09 '11 at 15:21
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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 09 '11 at 15:28
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2Oh no , sorry again , im the more stupid guy in the world <.< , what i did is wrong, sorry <.<.<<<.< – August Oct 09 '11 at 16:25
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One should maybe remark here that the question implicitly assumes $M_0$ to be finite (or that $\infty\cdot 0=\infty$), because otherwise we can take $f (x)=x$, for which the inequality would read $1=M_0 \leq 4\cdot M_0 M_2= 4\cdot \infty \cdot 0$. – PhoemueX Jan 07 '15 at 17:47
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@PhoemueX: not only, we also have to assume that $f''$ is not identically zero, otherwise for $f(x)=ax+b$ on $(0,1)$ the inequality states $a^2\leq 0$. – Jack D'Aurizio Jan 12 '15 at 20:46
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@JackD'Aurizio Sorry to leave a comment here, but it's about the question http://math.stackexchange.com/questions/1305324 that you marked as duplicate of the present; this is not true; look carefully at the question you marked as duplicate: it is not the Kolmogorov Inequality (please read my comment to math110: the other question is about the max of $|f'|^2/f$, not about the max of $|f'|^2$ divided by the max of $f$). – gniourf_gniourf May 30 '15 at 16:57
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Possible duplicate of Proving that $(\sup_{x\in R}|f'(x)|)^2\leq 4\sup_{x\in R}|f(x)|\cdot\sup_{x\in R}|f''(x)|$. – Peter_Pan Apr 11 '19 at 04:32
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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.
Nick Strehlke
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Remark: we must assume $M_2>0$, since the inequality does not hold for linear functions. Given that, we can just take $h=\sqrt{\frac{M_0}{M_2}}$ in the bound for $|f'(x)|$. – Jack D'Aurizio Jan 12 '15 at 21:01
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2@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 $M_2=0$. – Nick Strehlke Jan 12 '15 at 21:17