# Integral Inequality $|f''(x)/f(x)|$

Let $f$ be a $C^2$ function in $[0,1]$ such that $f(0)=f(1)=0$ and $f(x)\neq 0\,\forall x\in(0,1).$ Prove that

$$\int_0^1 \left|\frac{f{''}(x)}{f(x)}\right|dx\ge4$$

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I think you mean $f(0) = f(1) = 0$, not $f(x) = f(1) = 0$. –  Michael Albanese Oct 14 '12 at 5:03
Yep. f(0)=f(1)=0. Sorry about the typo. –  Christmas Bunny Oct 14 '12 at 5:21
I can show that, if $f\in C^2$0,1$$ and $f'(0)=f'(1)=0$ then exist a point $t\in $0,1$$ such that, $f''(t)>4|f(1)-f(0)|$ –  Salech Alhasov Oct 14 '12 at 5:34
This is a duplicate of a question that has been asked before but I am unable to find the other question. –  user17762 Oct 14 '12 at 5:56

Just some considerations. Without loss of generality, we can assume that $f(x)$ is positive on $(0,1)$. Let $\operatorname{graph}(g)$ be the convex hull of $\operatorname{graph}(f)$. We have $g(x)\in C^2([0,1])$ and

$$\int_{0}^{1}\frac{|f''(x)|}{f(x)}dx\geq \int_{0}^{1}\frac{|g''(x)|}{g(x)}dx,$$

so we can also assume that $f$ is a concave function over $[0,1]$, with $f'(0)=\alpha>0,f'(1)=-\beta<0$ and a unique maximum in $x_0\in(0,1)$, for which $f'(x_0)=0$ and $f(x_0)=1$. Let now $\operatorname{graph}(h)$ be the envelope of the tangent lines to $\operatorname{graph}(f)$ for $x\in\left\{0,x_0,1\right\}$. For any $\epsilon>0$ there is a function $u\in C^2([0,1])$ such that $|u-h|<\epsilon$ and:

$$\int_{0}^{1}\frac{|f''(x)|}{f(x)}dx\geq \int_{0}^{1}\frac{|u''(x)|}{u(x)}dx=\alpha+\beta-O(\epsilon),$$

but $\alpha+\beta$ must be greater than $4$, since $\max u$ is below the $y$-coordinate of the intersection of the tangent lines in $x=0$ and $x=1$, so

$$1 = \max u \leq \frac{\alpha\beta}{\alpha+\beta} \leq_{AM-GM} \frac{\alpha+\beta}{4}.$$

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