# Inequality involving the sup of a function and its first and second derivatives

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
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 7 '15 at 17:47
@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

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.
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
@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