Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
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
5  
@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

2 Answers 2

up vote 20 down vote accepted

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.

share|improve this answer
    
+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? –  Joe 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 $\min\limits_{\mathbb{R^+}}\frac{(1+t)^2}{t}=4$, $$ \begin{align} \int_0^1\left|\frac{f''(x)}{f(x)}\right|\,\mathrm{d}x &\ge\frac{1}{\max\limits_{[0,1]}f(x)}\int_0^1|f''(x)|\,\mathrm{d}x\\ &\ge\frac{(a+b)^2}{ab}\\ &=\frac{(1+b/a)^2}{b/a}\\ &\ge4\tag{3} \end{align} $$

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.