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Let $a<b$ and $f(a)=f(b)=0$. I want to prove that $\int_a^b f(x)f''(x)\,\mathrm{d}x\leq0$. I used integration by part to rewrite the indefinite integral to $f'(x)f(x)-\int f'(x)^2 dx$. Now when we do do the definite integral $f'(x)f(x)$ will vanish so we are left with $G(a)-G(b)$ where $G(x)=\int f'(x)^2 dx$. How do I see that $G(a)<G(b)$?

user30523
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    After integration by parts I believe you are left with $-\int_{a}^{b}\left(f'(x)\right)^{2}dx$. Note that the integrand is non-negative. – user71352 Dec 13 '16 at 22:09

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You have shown that $$\int_a^b f(x)f''(x)dx=f(b)f'(b)-f(a)f'(a)-\int_a^b(f'(x))^2dx.$$ Now, since $f(a)=f(b)=0$ you have

$$\int_a^b f(x)f''(x)dx=-\int_a^b(f'(x))^2dx.$$ Finally, note that $(f')^2\ge 0.$ Thus $\int_a^b (f')^2\ge 0$ and $-\int_a^b (f')^2\le 0.$ So, you are done.

mfl
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