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

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

• 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 Rubenstein 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

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

Proving the Lower Bound

Without loss of generality, assume that $f(x)\gt0$ for $x\in(0,1)$.

Suppose that $f(x_0)=y_0=\max\limits_{x\in[0,1]}f(x)$. Then, $f'(x_0)=0$.

By the Mean Value Theorem, for some $x_1\in(0,x_0)$, $f'(x_1)=\frac{y_0}{x_0}$. Therefore, $$\int_0^{x_0}|f''(x)|\,\mathrm{d}x\ge\frac{y_0}{x_0}$$ Furthermore, for some $x_2\in(x_0,1)$, $f'(x_2)=-\frac{y_0}{1-x_0}$. Therefore, $$\int_{x_0}^1|f''(x)|\,\mathrm{d}x\ge\frac{y_0}{1-x_0}$$ Since $f(x)\le y_0$, \begin{align} \int_0^1\left|\,\frac{f''(x)}{f(x)}\,\right|\,\mathrm{d}x &\ge\frac{\frac{y_0}{x_0}+\frac{y_0}{1-x_0}}{y_0}\\ &=\frac1{x_0}+\frac1{1-x_0}\\ &=\frac1{\frac14-\left(x_0-\frac12\right)^2}\\[3pt] &\ge4 \end{align}

The Lower Bound is Sharp

Let $$f_a(x)=\sin^{-1}\left(\frac{\sin(\pi x)}{1+a^2\sin^2(\pi x)}\right)$$ then $$\lim_{a\to0}\int_0^1\left|\,\frac{f_a''(x)}{f_a(x)}\,\right|\,\mathrm{d}x=4$$ since $f_a''(x)$ is tends to $0$ except near $x=\frac12$, and $\int_0^1f_a''(x)\,\mathrm{d}x$ tends to $-2\pi$, whereas $f_a\!\left(\frac12\right)$ tends to $\frac\pi2$.