# A inequality of calculus [duplicate]

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Let $f \in C^2[a,b]$ and $f(a) = f(b) = 0$, $f'(a) = 1$,$f'(b) = 0$, prove that $$\int_a^b|f''(x)|^2\,dx \geq \frac{4}{b-a}$$ Remark:

1. This question is in the book functional analysis of Peking University;
2. We have$$u(x) = \int_a^xu'(t)\,dt$$so $|u(x)|^2 \leq (b-a)\int_a^bu'(x)\,dx$ by applying the Cauthy-Schwartz inequality. but I cannot get the number 4
3. I have construct a function of which satisfies the condition using quadratic function，and the infimum is attained, and $4$ is got from differentiating and squaring.

## marked as duplicate by Davide Giraudo, user63181, Shuchang, abiessu, DanMar 6 '14 at 17:59

We have local information on $a$ and $b$ for the behaviour of $f$, so we use the Taylor formula: $$f(x)= f(a) + (x-a)f'(a) + \int_a^x (t-a) f''(t)dt = (x-a) + \int_a^x (t-a) f''(t)dt \\= f(b) - (b-x)f'(b) + \int_x^b (b-t) f''(t)dt = \int_x^b (b-t) f''(t)dt$$ so $$x-a = \int_a^x (a-t) f''(t)dt + \int_x^b (b-t) f''(t)dt\le \sqrt{\int_a^b f''(t)^2 dt} \sqrt{\int_a^x (a-t)^2 dt + \int_x^b (b-t)^2 dt}\\= \sqrt{\int_a^b f''(t)^2 dt}\sqrt{ \frac{(x-a)^3}3 + \frac{(b-x)^3}3 }$$ now take $x = \frac 13(a+2b)$ gives the optimal result.
• this is just the Taylor formula applied twice, and optimization of a real function. the idea is to use information you have locally in $a$ and $b$, and the Taylor furmula is just the right tool for that. – mookid Mar 8 '14 at 7:51