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

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.

• An intuitive idea, for you to make into a proof: If $f$ got extremely big somewhere between $a$ and $b$, then, by the intermediate value theorem and the assumption that $f(a)=f(b)=0$, $f'$ would have to have a big positive value somewhere and a big negative value somewhere else between $a$ and $b$. Then, again by the intermediate value theorem, $f''$ would be big somewhere, contrary to hypothesis. Your task is to make all of this ("extremely big", "big") quantitative. – Andreas Blass Mar 18 at 14:39

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
$$\frac M2 (x-a)(b-x) \pm f(x) \ge 0 \, .$$

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.