# Prove $\int_0^1 \left| \frac{f^{''}(x)}{f(x)} \right| dx \geq4$

$f''(x)$ is continuous in $[0,1]$, $f(0)=f(1)=0$, $f(x)\neq 0$ when $x \in(0,1)$, try to prove: $$\int_0^1 \left| \frac{f''(x)}{f(x)} \right| dx \geq 4.$$

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If $f$ is analytic then close to $0$, $f(x)=x^n+O(x^{n+1})$. Therefore, $$\left|\frac{f''}{f}\right|\approx \frac{n(n-1)}{x^2}+\mbox{something}$$ and the integral over $[0,\epsilon]$ diverges for all $\epsilon$. –  yohBS Mar 9 '12 at 8:45
@yohBS Except if $n=1$. Consider $f(x)=\sin(\pi\,x)$; then $|f''/f|=\pi^2$. –  Julián Aguirre Mar 9 '12 at 10:08
And, if $f(x) = \sum_{k=1}^{\infty} a_k x^k$ with $a_1 \neq 0$ then we need $a_2 = 0$ or else the integral diverges. It looks like your example @JuliánAguirre is very revealing. –  Antonio Vargas Mar 9 '12 at 22:29
@Norbert: Correct me if I am wrong. But I don't think $f_n(x)$ is twice differentiable in $(0,1)$ –  user17762 Mar 9 '12 at 23:44
@Norbert: $\left[1 + \frac1{2n},1 \right]$? and $f'(x)$ does-not exist at $x = 1 - \frac1{2n}$? –  user17762 Mar 9 '12 at 23:50

Here's another answer, which avoids reducing to the case in which $f$ is concave. It is plainly enough to prove that $$\sup_{x \in [0,1]}{|f(x)|} \leq \frac{1}{4}I,\quad \text{where} \quad I := \int_0^1|f''(x)|\,dx. \tag{1}$$ First of all, since $f(0) = f(1) = 0$ and $f$ is nonzero in $(0,1)$, we know that the supremum on the left-hand side is attained at some $c \in (0,1)$, and moreover, that $f'(c) = 0$. By using Taylor's theorem with remainder (which is really just repeated integration by parts) to expand $f$ around the point $c$, we have $$f(x) = f(c) + f'(c)(x-c) + \int_c^x (x - t)f''(t)\,dt = f(c) + \int_0^c (x - t)f''(t)\,dt,$$ for any $x \in [0,1]$. Successively taking $x = 0$ and $x = 1$ gives $$f(c) = -\int_0^c tf''(t)\,dt = -\int_c^1 (1-t)f''(t)\,dt,$$ because $f(0) = f(1) = 0$. This means that $$|f(c)| \leq c\int_0^c |f''(t)|\,dt \quad \text{and} \quad |f(c)| \leq (1-c) \int_c^1 |f''(t)|\,dt. \tag{2}$$ Since $$\int_0^c|f''(t)|\,dt + \int_c^1|f''(t)|\,dt = I = (1-c)I + cI,$$ we must have either $\int_0^c|f''(t)|\,dt \leq (1-c) I$ or $\int_c^1|f''(t)|\,dt \leq c I$. Either way $(2)$ shows that $$|f(c)| \leq c(1-c)I = \frac{1}{4}I - \left(\frac{1}{2} - c\right)^2I \leq \frac{1}{4}I,$$ and $(1)$ is therefore proved.

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+1. Wow! Nice! The argument is really elegant. –  user17762 Mar 11 '12 at 5:08
Can I know the motivation for this idea? –  user17762 Mar 11 '12 at 5:11
@Sivaram Thanks! Actually this problem is very similar to a Putnam problem I had solved before (2007, B2), and it was a little more natural to consider Taylor's theorem in that problem. I just adapted the solution to get a more precise estimate. –  Nick Strehlke Mar 11 '12 at 5:25
Thanks! Wish I could up-vote this answer couple of more times. –  user17762 Mar 11 '12 at 6:03
@NickStrehlke I don't understand either $\int_0^c|f''(t)|\,dt \leq (1-c) I$ or $\int_c^1|f''(t)|\,dt \leq c I$. Can you show me that? –  89085731 Mar 11 '12 at 8:33

Without loss of generality, assume that $f(x)\ge0$ on $[0,1]$. Furthermore, we can assume that $f''(x)\le0$. If not, we can replace $f$ by $g$ where the graph of $g$ is the convex hull of the graph of $f$. Note that where $g(x)\not=f(x)$, $g''(x)=0$, therefore, $\int_0^1\left|\frac{g''(x)}{g(x)}\right|\,\mathrm{d}x\le\int_0^1\left|\frac{f''(x)}{f(x)}\right|\,\mathrm{d}x$.

Suppose that $f'(0)=a$ and $f'(1)=-b$. Since $f$ is concave, $f(x)\le ax$ and $f(x)\le b(1-x)$. Therefore, $$\max_{[0,1]}f(x)\le\frac{ab}{a+b}\tag{1}$$ Furthermore, \begin{align} \int_0^1|f''(x)|\,\mathrm{d}x &\ge\left|\int_0^1f''(x)\,\mathrm{d}x\right|\\[6pt] &=|f'(1)-f'(0)|\\[6pt] &=a+b\tag{2} \end{align} Therefore, since $\min\limits_{\mathbb{R^+}}\frac{(1+t)^2}{t}=4$, \begin{align} \int_0^1\left|\frac{f''(x)}{f(x)}\right|\,\mathrm{d}x &\ge\frac{1}{\max\limits_{[0,1]}f(x)}\int_0^1|f''(x)|\,\mathrm{d}x\\ &\ge\frac{(a+b)^2}{ab}\\ &=\frac{(1+b/a)^2}{b/a}\\ &\ge4\tag{3} \end{align}

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I've been thinking along these lines too, but the problem I have is that $f''$ needs to be continuous, and maybe in that very small interval where the turn takes place, you have imposed a very high value for $f''$. So that briefly extreme $f''$ is responsible for the lower bound. Why can't we afford larger regions with a slightly nonzero $f''$ in the hopes to greatly reduce the extremely large value for $f''$ at your turn? Why would it be impossible for that to yield a smaller integral? –  alex.jordan Mar 10 '12 at 4:05
@alex: The whole point is that $\int_0^1f''(x)\mathrm{d}x$ is fixed by $f'(1)-f'(0)$. No matter how small or large an interval through which we take that turn, the total integral of $f''$ will be the same. Thus, we want to take that turn at the very maximum of $f$ that we can to minimize $\int_0^1\left|\frac{f''(x)}{f(x)}\right|\mathrm{d}x$. Does that make sense? –  robjohn Mar 10 '12 at 4:19
How did you calculate the integral in (3)? –  Antonio Vargas Mar 10 '12 at 6:38
@Antonio: $f''(x)=0$ except near $x=\frac{b}{a+b}$ where $f(x)$ is near $\frac{ab}{a+b}$ and $\int|f''(x)|\mathrm{d}x=a+b$. Thus, in the region where $f''(x)\not=0$, $f(x)$ is near $\frac{ab}{a+b}$, so $\int_0^1\left|\frac{f''(x)}{f(x)}\right|\mathrm{d}x=\frac{(a+b)^2}{ab}$. This is not as rigorous as I would like, and I am working on a more rigorous exposition. –  robjohn Mar 10 '12 at 10:01
@Sam: I have just posted the more rigorous exposition that I mentioned I was working on. –  robjohn Mar 10 '12 at 12:30

There is an elementary proof for this inequation, by mean-value theorem.

CASE I: $f(x)\geq 0$ on $[0,1]$

$f(0)=f(1)=0$ There is a point $x_0\in (0,1)$ s.t $\max_{[0,1]}f(x)=f(x_0)$

By mean-value theorem ,there are $\lambda_1 \in (0,x_0),\lambda_2 \in (x_0,1)$

s.t \begin{align} f(x_0)-f(0)=\int_0^{x_0} f'(x) dx = (x_0 - 0)f'(\lambda_1)\Rightarrow f'(\lambda_1)=\frac{f(x_0)}{x_0}\end{align}

\begin{align} f(1)-f(x_0)=\int_{x_0}^{1} f'(x) dx = (1-x_0)f'(\lambda_2)\Rightarrow f'(\lambda_2)=\frac{-f(x_0)}{1-x_0}\end{align}

\begin{align} \int_0^1 \left| \frac{f''(x)}{f(x)} \right| dx \ge \int_{\lambda_1}^{\lambda_2} \left| \frac{f''(x)}{f(x)} \right| dx \ge \int_{\lambda_1}^{\lambda_2} \frac{|f''(x)|}{f(x_0)} dx \\= \frac {1}{f(x_0)} \int_{\lambda_1}^{\lambda_2} |f''(x)| \ge \frac {1}{f(x_0)} |\int_{\lambda_1}^{\lambda_2} f''(x) |\end{align}

And \begin{align} \frac {1}{f(x_0)}| \int_{\lambda_1}^{\lambda_2} f''(x)| =\frac {1}{f(x_0)}|f'(\lambda_2)-f'(\lambda_1)| =\frac {1}{f(x_0)}|\frac{-f(x_0)}{1-x_0}-\frac{f(x_0)}{x_0}|=\frac {1}{(1-x_0)(x_0)}\ge 4\end{align}

CASE II: $f(x)$ may be nagetive on $[0,1]$.

Since $\forall x\in [0,1],|f(x)|\le M$ for some M,let $g(x) = f(x) + 2M$.

Then $\forall x\in [0,1],|g(x)|\ge M \ge |f(x)|$ and $g''(x)=f''(x),g(x)\ge 0$ on $[0,1]$

\begin{align} \int_0^1 \left| \frac{f''(x)}{f(x)} \right| dx =\int_0^1 \left| \frac{g''(x)}{f(x)} \right| dx\ge \int_0^1 \left| \frac{g''(x)}{g(x)} \right| dx \ge 4\end{align} (By CASE I)

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