4
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This question already has an answer here:

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.
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marked as duplicate by Davide Giraudo, user63181, Shuchang, abiessu, Dan Mar 6 '14 at 17:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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.

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  • $\begingroup$ I think your method is too technical and hard to think, can you tell me how you think this problem? And I've read the the same problem, and the solution is just as well, find a special function,and do the calculus, and then came the answer. For us, how to think it? I want a way to do this. $\endgroup$ – Chen Jie Mar 8 '14 at 7:48
  • $\begingroup$ 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. $\endgroup$ – mookid Mar 8 '14 at 7:51

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