# If $f\geq0 , f(0)=f(1)=0$,and $\int_0^1|f''|/f dx$ exists,how to prove $\int_0^1f''/f dx\geq \pi^2$

If $$f\in C^2 , f\geq0$$ on $$[0,1]$$ , $$f(0)=f(1)=0$$,and $$\int_0^1f''/f dx$$ exists,how to prove $$\int_0^1\frac{|f''|}{f}dx\geq \pi^2$$

The lower bound $$\pi^2$$ was guessed by myself, if it was not true ,then how to find this lower bound ?

My way is to have an analytic extension with $$T=2$$, and use the Fourier Series to have an evaluation .

But it seems like that I was wrong.

• I'm not guessing anything, just pointing out that the integral is typically going to be negative (e.g. for $f(x) = \sin(\pi x)$ it is $-\pi^2$). Commented Mar 22, 2019 at 12:48
• What if $f$ is chosen to be piecewise linear?
– daw
Commented Mar 22, 2019 at 13:16
• Did your bound come from applying a sine function? How did you arrive at this question? Commented Mar 22, 2019 at 13:18
• @Mefitico It's quite hard to find a function in which $\int_0^1 \frac{|f''(x)|}{f(x)} dx$ exists other than $\sin(x)$. Commented Mar 22, 2019 at 13:42
• Well, the original question is asked to show that integral $\geq 4$ , and the method is by using the mean value theorem , but I think the lower can be raised up to $\pi^2$，I use the Fourier Expansion to evaluate, and by the mistake of $\sum a_n\leq|\sum n^2a_n|$, I "deduced" the inequality , but later I found that inequality is not true , however, I still feel that result is correct , so I post it as a question. Commented Mar 22, 2019 at 13:45

The proposition you're trying to prove is false. This integral can be made arbitrarily close to $$4$$. A family of examples: $$f(x) = \begin{cases}2x&0\le x\le \frac12-\epsilon\\ 1-\frac{3\epsilon}{4} - \frac{3}{2\epsilon}(x-\frac12)^2 + \frac1{4\epsilon^3}(x-\frac12)^4 & \frac12-\epsilon\le x\le \frac12+\epsilon\\ 2-2x&\frac12+\epsilon\le x\le 1\end{cases}$$ Why is this $$C^2$$? We check the values and derivatives where the formulas change. At $$\frac12+\epsilon$$ using the middle formula, \begin{align*}f\left(\frac12+\epsilon\right) &= 1-\frac{3\epsilon}{4} - \frac{3}{2\epsilon}\epsilon^2 + \frac1{4\epsilon^3}\epsilon^4 = 1 +\epsilon\left(-\frac34-\frac32+\frac14\right) = 1-2\epsilon\\ f'\left(\frac12+\epsilon\right) &= -\frac{3}{\epsilon}\epsilon+\frac1{\epsilon^3}\epsilon^3 = -2\\ f''\left(\frac12+\epsilon\right) &= -\frac{3}{\epsilon} + \frac{3}{\epsilon^3}\epsilon^2 = 0\end{align*} Those match the values from the other formula it meets, and the first two derivatives are continuous there. At $$\frac12-\epsilon$$, we appeal to symmetry; $$f(x)=f(1-x)$$, so continuity of the first two derivatives at $$\frac12+\epsilon$$ implies it at $$\frac12-\epsilon$$.

What is $$\int_0^1 \frac{|f''(x)|}{f(x)}\,dx$$? Rather than calculate it exactly, I'll estimate. We have $$f''(x) = \frac3{\epsilon}\left(-1+\left(\frac{x-\frac12}{\epsilon}\right)^2\right) \le 0$$ for $$\frac12-\epsilon\le x\le\frac12+\epsilon$$, so $$\int_{1/2-\epsilon}^{1/2+\epsilon}|f''(x)|\,dx = \left|f'\left(\frac12+\epsilon\right)-f'\left(\frac12-\epsilon\right)\right| = 4$$. Now, we divide by $$f(x)$$ inside the integral; from $$1-2\epsilon\le f(x)\le 1-\frac34\epsilon$$ on the interval, $$\frac{4}{1-\frac34\epsilon}\le\int_{1/2-\epsilon}^{1/2+\epsilon}\frac{|f''(x)|}{f(x)}\,dx \le\frac{4}{1-2\epsilon}$$ What about $$f$$ outside that subinterval? The second derivative out there is identically zero, so that part contributes nothing to the integral, and $$\int_0^1 \frac{|f''(x)|}{f(x)}\,dx$$ lies between those bounds.

As $$\epsilon\to 0$$, both of those bounds tend to $$4$$, and therefore the integral does by the squeeze theorem. The limiting function isn't $$C^2$$, but we can make the integral arbitrarily close to $$4$$ this way.

Let:

$$f(x) = \sin(\pi x)+0.1\sin(2 \pi x)$$

Then $$f'(x) = \pi\cos(\pi x)+0.2\pi\cos(2 \pi x)$$

$$f''(x) =-\pi^2\sin(\pi x)-0.4\pi^2\sin(2 \pi x)$$

So: $$\int_0^1 \frac{|f''(x)|}{f(x)} dx=\int_0^1 \frac{\pi^2\sin(\pi x)+0.4\pi^2\sin(2 \pi x)}{\sin(\pi x)+0.1\sin(2 \pi x)}dx = 9.25905... < \pi^2 = 9.8696$$

Indeed it is hard to come up with a non-sine function that allows the integral to converge, but if you make a composition of such kind, you can likely lower your bound. I would suspect however, that there should be a sequence of functions $$f_n$$ such that the integral converges to zero.

Edit My reasoning to claim that the integral can be made arbitrarily close to zero is that with a large enough sine series, it should be possible to build $$f(x)$$ such that the integral exists due to the characteristics of the sine on the extremes, but outside the extremes and some mid point, the function is "flat", i.e. with arbitrarily small $$f''(x)$$ but with large values for $$f(x)$$, hence decreasing the value of the integral.