If $f(a)=f(b)=0$ and $|f''(x)|\le M$ prove $|\int_a^bf(x)\mathrm{d}x| \le \frac{M}{12}(b-a)^3$

Question

If $f(a)=f(b)=0$ and $|f''(x)|\le M$. Prove $$|\int_a^bf(x)\mathrm{d}x| \le \frac{M}{12}(b-a)^3$$

I have thought about that since $f(a) = f(b) = 0 $ there is $\xi$ such that $f'(\xi) = 0$. Then when $x \le \xi$, $|f'(x)| \le (\xi - x)M$ and when $x \ge \xi$, $|f'(x)| \le (x-\xi)M$. After that $|f(x)| \le \frac{\xi^2 - (\xi -x)^2}{2}$ when $x \le \xi$ and $|f(x)| \le \frac{(b-\xi)^2 - (x - \xi)^2}{2}$ when $x \ge \xi$. Therefore $$|\int_a^b f(x)\mathrm{d}x| \le \int_a^\xi |f(x)| + \int_\xi^b |f(x)| = \frac{\xi^3}{3} + \frac{(b-\xi)^3}{3}$$ If $x = \frac{a+b}{2}$, we have $$\frac{x^3}{3}+\frac{(b-x)^3}{3} = \frac{(b-a)^3}{12}$$

But in this situation $\frac{x^3}{3}+\frac{(b-x)^3}{3}$ is the minimal value. So I can't go on.

Answer 1

Show that $f(a)=f(b)=0$ and $|f''(x)|\le M$ implies $$ \tag{*} |f(x)| \le \frac M2 (x-a)(b-x) $$ for $a \le x \le b$, i.e. $f$ is bounded by the parabola with constant second derivative $-M$ and zeros at $x=a$ and $x=b$.

Then $$ \left|\int_a^b f(x)\, dx \right| \le \frac M2\int_a^b (x-a)(b-x) \, dx = \frac{M}{12}(b-a)^3 $$ follows.

In order to prove $(*)$, consider for fixed $y \in (a, b)$ the function $$ h(x) = f(x) (y-a)(b-y) - f(y) (x-a)(b-x) \, . $$ $h$ satisfies $h(a) = h(y) = h(b) = 0$, so that a repeated application of Rolle's theorem implies $$ h''(\xi) = f''(\xi)(y-a)(b-y) +2f(y) = 0 $$ for some $\xi \in (a, b)$.

Alternatively note that the two functions $$ \frac M2 (x-a)(b-x) \pm f(x) $$ are concave in $[a, b]$ because their second derivative is $\le 0$. A concave functions attains its minimum on an interval at one of the boundary points, and therefore
$$ \frac M2 (x-a)(b-x) \pm f(x) \ge 0 \, . $$

Answer 2

For simplicity let $a = -1, b = 1$. Let $\varepsilon > 0$. Consider the two functions $g_+(x) = \frac{M + \varepsilon}{2}(1-x^2)$ and $g_-(x) = -g_+(x)$. Then $$ g_\pm(-1) = g_\pm(1) = 0, \; g_+''(x) = -M - \varepsilon, \; g_-''(x) = +M + \varepsilon $$ and $$ \int_{-1}^1 g_+(x) = 2\frac{M+ \varepsilon}{3} = - \int_{-1}^1 g_-(x) dx \, . $$ The claim now is that $g_-(x) \le f(x) \le g_+(x)$ for all $x$.

Indeed, suppose $h_-(x) = f(x) - g_-(x)$ is negative somewhere in $(-1,1)$. Then there is an absolute minimum $c \in (-1,1)$ at which $h_-''(c) \ge 0$ (second derivative test). But $$ h_-''(c) = f''(x) - g_-''(c) = f''(c) - M - \varepsilon \le |f''(c)| - M - \varepsilon \le - \varepsilon < 0 $$ and therefore this is impossible.

Consequently $g_-(x) \le f(x)$ for all $x$. Similarly $f(x) \le g_+(x)$ for all $x$. It follows that $$ -2\frac{M + \varepsilon}{3} \le \int_{-1}^1 f(x) dx \le 2\frac{M + \varepsilon}{3} \, . $$
Since $\varepsilon > 0$ was arbitrary, the desired estimate follows.