How to prove this integral inequality $\int_{0}^{1}[f''(x)]^2dx\ge 12$?

Question

Let $f\in C^{2}[0,1]$ such that $f(0)=0,f(1/2)=f(1)=1$. Show that

$$\int_{0}^{1}[f''(x)]^2dx\ge 12.$$

My idea: try to find a function $f(x)$ such that the equality holds and use Cauchy-Schwarz. So I tried some "good" functions to achieve the minimum value.

1. Assume $f(x)=ax^3+bx^2+cx$ is a polynomial. Then I find that when $f(x)=-2x^2+3x$, the integral $\int_{0}^{1}[f''(x)]^2dx$ has minimum value $16$. It seems it doesn't help to solve the problem.

2. Since $f''(x)=(f(x)-ax-b)''$ and replace $f(x)$ by $f(x)-x$, we can change the condition to $f(0)=f(1)=0, f(1/2)=1/2$. I find that $f(x)=\frac{1}{2}\sin(\pi x)$ satisfies the condition and $\int_{0}^{1}[f''(x)]^2dx=\frac{\pi^2}{8}=12.1761...$ which is very closed to $12$. But it still doesn't help.

It seems that it isn't easy to find such $f(x)$ such that the equality holds. So how to prove this inequality? And what is the minumum value of this integral? May be not 12? Thank you!

Answer 1

I'll work with the condition $f(0)=f(1)=0, f(1/2)=1/2$.

The idea of the following calculation is to minimize $\int_0^1 f''(x)^2 \, dx $ on the intervals $[0, 1/2]$ and $[1/2, 1]$ separately. Without further restrictions that would give a minimum of zero, attained by a piecewise linear function, which is not (twice) differentiable on $[0, 1]$.

Therefore we introduce an additional parameter $a \in \Bbb R$ and minimize the integral over the left interval subject to the conditions $f(0) = 0, f(1/2) = 1/2, f'(1/2) = a$, and over the right interval subject to the conditions $f(1/2) = 1/2, f'(1/2) = a, f(1) = 0$.

Doing integration by parts, $$ \frac 12 = f\left( \frac 12\right) - f(0) = \int_0^{1/2} f'(x) \, dx = \left .x f'(x)\right]_{x=0}^{x=1/2} - \int_0^{1/2} x f''(x) \, dx \\ = \frac a2 - \int_0^{1/2} x f''(x) \, dx \, , $$ so that $$ \left( \frac{a-1}{2}\right)^2 = \left( \int_0^{1/2} x f''(x) \, dx\right)^2 \le \int_0^{1/2} x^2 \, dx \cdot \int_0^{1/2} f''(x)^2 \, dx \\ = \frac{1}{24} \int_0^{1/2} f''(x)^2 \, dx \, . $$ In the same way one can derive $$ \left( \frac{a+1}{2}\right)^2 = \left( \int_{1/2}^1 (1-x) f''(x) \, dx\right)^2 \le \frac{1}{24} \int_{1/2}^1 f''(x)^2 \, dx \, . $$ Combining these estimates one gets $$ \int_0^1 f''(x)^2 \, dx \ge 24 \left( \frac{a-1}{2}\right)^2+ 24 \left( \frac{a+1}{2}\right)^2 = 12 (a^2 + 1) \ge 12 \, . $$

Equality holds if $a=f'(1/2) = 0$ and $f''(x) = \text{const} \cdot x$ on $[0, 1/2]$ and $f''(x) = \text{const} \cdot (1-x)$ on $[1/2, 1]$, that leads to $$ f(x) = \begin{cases} -2 x^3 + \frac 3 2 x & \text{ for } 0 \le x \le \frac 12 \, , \\ 2 x^3 -6 x^2 + \frac 92 x -1/2 & \text{ for } \frac 12 \le x \le 1 \, . \end{cases} $$

The following graph (created with wxMaxima) shows that function (blue) and, for comparison, the function $\sin(\pi x)/2$ (red).

Answer 2

You can split the integral in to two separate ones $$ \int_0^{\frac12} (f''(x))^2 \, \mathrm{d}x+\int_{\frac12}^{1} (f''(x))^2 \, \mathrm{d}x $$ and find the smallest function for both using calculus of variations and combine them piecewise. The reason that you have to do this is because calculus of variations only makes statements about the boundary. The solution is $$ f(x)= \begin{cases} -2x^3+\frac52 x& x \in [0,\frac12]\\ 2x^3-6x^2+\frac{11}2x -\frac12& x \in (\frac12,1]. \end{cases} $$ This achieves the lower bound of 12. From there on it is indeed Cauchy-Schwarz, similar to this How prove this $\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}$ .

Answer 3

Thanks for the answer. Now I can write down the remaining steps. Let $p(x)$ be the piecewise polynomial in the follwing answer. Integration by parts, we can get

$$\int_0^{\frac12} (f''^2- p''^2)dx=\int_0^{\frac12}(f''-p'')^2dx-2p''(1/2)f'(1/2).$$

Similarly, we have

$$\int_{\frac12}^{1} (f''^2- p''^2)dx=\int_{\frac12}^{1}(f''-p'')^2dx+2p''(1/2)f'(1/2).$$

Then we have $\int_0^{1} f''^2dx\ge\int_0^{1} p''^2dx=12$.

Answer 4

Given $\phi(x) = f''(x)$, one recovers $f(x)$ up to two constants

$$f(x) = a+b x + \int_{0}^x(x-t)\phi(t)\,d t$$

Given the conditions $f(0)=0$, $f(1)=1$ and $f(\frac{1}{2})=1$, one gets $a=0$ and

$$b + \int_0^1 (1-t)\phi(t)\,d t = 1\\ \frac{b}{2} + \int_{0}^{\frac{1}{2}}(\frac{1}{2}-t) \phi(t) \,d t = 1$$

so

$$1-\int_0^1(1-t)\phi(t) d t = 2\left(1-\int_{0}^{\frac{1}{2}}(\frac{1}{2}-t) \phi(t) \,d t\right)$$ or $$\int_{0}^1 u(t) \phi(t) \, d t = 1$$

where $u(t) = 2\max(\frac{1}{2}-t,0)-(1-t)$. Now one uses the C-B-S inequality, noticing that $\int_0^1u^2(t) d t = \frac{1}{12}$.

Idea: express the solution of a Cauchy problem, and get extra conditions, then it's an $L^2$ optimization