Let $f(x) \in C^2[0,1]$, with $f(0)=f(1)=0$, and $f(x)\neq 0$ when $x \in(0,1)$. Show that $$\int_0^1 \left| \frac{f''(x)}{f(x)} \right| dx \geq 4.$$
Prove that $\int_0^1 \left| \frac{f^{''}(x)}{f(x)} \right| dx \geq4$.
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3If $f$ 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 $\epsilon$. – 2012-03-09
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5@yohBS Except if $n=1$. Consider $f(x)=\sin(\pi\,x)$; then $|f''/f|=\pi^2$. – 2012-03-09
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0And, 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. – 2012-03-09
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0@Norbert: Correct me if I am wrong. But I don't think $f_n(x)$ is twice differentiable in $(0,1)$ – 2012-03-09
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0@Norbert: $\left[1 + \frac1{2n},1 \right]$? and $f'(x)$ does-not exist at $x = 1 - \frac1{2n}$? – 2012-03-09
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0@SivaramAmbikasaran, sorry that was a typo. The sequence of functions $$f_n(x)=\begin{cases}2x\quad x\in[0,\frac{1}{2}-\frac{1}{2n}]\\\frac{n^3 (2x-1)^4}{8}-\frac{3n (2x-1)^2}{4}-\frac{3}{8n}+1\quad x\in[\frac{1}{2}-\frac{1}{2n},\frac{1}{2}+\frac{1}{2n}]\\2-2x\quad x\in[\frac{1}{2}+\frac{1}{2n},1]\end{cases}$$ will minimize lhs of inequality – 2012-03-09
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0$4$ can be approached by approximating the function $f(x)=1/2-|1/2-x|$. At $x=1/2$, $f(x)\approx1/2$ and the integral of $f''(x)$ is $-2$ ($f'$ changes from $1$ to $-1$). Thus, $\pi^2$ is not optimal. – 2012-03-10
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0I'm wondering whether one could successfully approach it by first approximating $f''(x) \approx \sum_n c_n\chi_{[a_n, a_{n+1})}(x)$ for some $0=a_0< a_1 < \dots < a_N = 1$ and then integrating this twice to get $f(x) \approx \sum_n c_n \chi_{[a_n, a_{n+1})}(x) (x^2/2 + d_nx + e_n)$. So at least one would have an easy expression on each interval $[a_n, a_{n+1})$ to integrate: $$\int_0^1 \left|\frac{f''(x)}{f(x)}\right| \; dx = \sum_n \int_{a_n}^{a_{n+1}} \left|\frac{f''(x)}{f(x)}\right| \; dx \approx \sum_n \int_{a_n}^{a_{n+1}} \frac{dx}{\frac 12 x^2 + d_n x + e_n}$$ – 2012-03-10
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0And, of course, trying to prove it is a lot easier than actually proving it:) – 2012-03-10
4 Answers
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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0Can I know the motivation for this idea? – 2012-03-11
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0@Sivaram Thanks! Actually this problem is very similar to a Putnam problem I had solved before ([2007, B2](http://amc.maa.org/a-activities/a7-problems/putnam/-pdf/2007.pdf)), 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. – 2012-03-11
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0Thanks! Wish I could up-vote this answer couple of more times. – 2012-03-11
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0@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? – 2012-03-11
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0@Gingerjin The negation of that statement is $\int_0^c |f''(t)|\,dt > (1-c)I$ and $\int_c^1|f''(t)|\,dt > cI$. This leads to $I = \int_0^c |f''(t)|\,dt + \int_c^1 |f''(t)|\,dt > (1 - c)I + cI = I$, which is contradictory. – 2012-03-11
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0@NickStrehlke you are my hero. – 2012-03-19
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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1I'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? – 2012-03-10
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0@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? – 2012-03-10
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0How did you calculate the integral in (3)? – 2012-03-10
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0@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. – 2012-03-10
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0@Sam: I have just posted the more rigorous exposition that I mentioned I was working on. – 2012-03-10
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0This looks much more convincing, but it's not quite watertight yet. If you pass from $f$ to $g$, then $g$ is not $C^2$ anymore: For example take $f(x) = \sin^2(2\pi x)$. Then $g$ is constant on $[1/4,3/4]$, so $g''(x) = 0$ there. But approaching $x= 1/4$ from the left, we have $g''(x-0)=f''(x-0) = 8\pi^2 ( \cos^2(2\pi x)- \sin^2(2\pi x)) = -8\pi^2 \ne 0$. Can this flaw be fixed? – 2012-03-10
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0Note also that for piecewise-$C^2$ functions (I'm guessing a generic $g$ will be that) the bound fails (as a simple triangle shows). So this problem of the passage $f \to g$ has to be addressed. – 2012-03-10
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0@Sam: $g$ is $C^1$ and piecewise $C^2$, and this proof works for that class of functions, which includes $C^2$. If we consider a sawtooth (or triangle) function as the limit of such functions, then the integral of $\frac{f''}{f}$ around a vertex is the jump in $f'$ divided by $f$ at that vertex. Thus, if we include that in $\int_0^1\left|\frac{f''(x)}{f(x)}\right|\,\mathrm{d}x$ the result holds for those functions, too. – 2012-03-10
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)
Let $f(c)=\max_{x\in[0,1]}f(x)$. Then By Mean Value Theorem, $f(c)-f(0)=f'(a)c, \quad f(1)-f(c)=f'(b)(1-c)$ so we have $f'(a)=\frac{f(c)}{c},\quad f'(b)=-\frac{f(c)}{1-c}$. Then $\int^1_0\frac{|f"(x)|}{|f(x)|}dx\ge\frac{1}{|f(c)|}\int^1_0|f''(x)|dx\ge\frac{1}{|f(c)|}\int^b_a|f''(x)|dx\ge\frac{1}{|f(c)|}|\int^b_af''(x)dx|=\frac{1}{|f(c)|}|f'(b)-f'(a)|=\frac{1}{|f(c)|}|\frac{f(c)}{c}+\frac{f(c)}{1-c}|=\frac{1}{c(1-c)}\ge\frac{1}{4}$