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Assume $ f \in C^2 ( \mathbb R) \cap L^2 ( \mathbb R) , \; f'' \in L^2 ( \mathbb R)$. Assume the situation, $(b-a)^2 \int_{a}^b | f''|^2 \leqslant (b-a)^{-2} \int_a^b |f|^2 $. I want to prove that there exists $ b_2 \geqslant b$ such that $$ (b_2 -a)^2 \int_a^{b_2} |f''|^2 = (b_2 - a)^{-2} \int_{a}^{b_2} |f|^2 .$$

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Define $G(x)=(x-a)^4\int_a^x|f''|^2-\int_a^x|f|^2$. This is a continuous map and $G(b)\leq 0$. If we can show that there is $c\geq b$ such that $G(c)\geq 0$, we are done, by the intermediate value theorem. If for all $c\geq b$ we have $$(c-a)^4\int_a^c|f''|^2<\int_a^c|f|^2,$$ then, because $f\in L^2$, $$\int_a^c|f''|^2\leq\frac 1{(c-a)^4}\int_a^{+\infty}|f|^2$$ hence $f''(x)=0$ if $x\geq a$. So $f$ has the form $Ax+B$, but since $f$ is square integrable, $f=0$, and in this case $G$ is identically $0$.

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Thank you. You assumed that for all $ c \geqslant b$, $(c-a)^4\int_a^c|f''|^2<\int_a^c|f|^2 $ to get a contradiction. But I cannot understand here what the contradiction is. Would you tell me more about this? :) –  Misaj Jun 17 '12 at 11:07
    
It's not really a contradiction, in fact I deal with two cases. Both yield to the wanted conclusion. –  Davide Giraudo Jun 17 '12 at 12:36
    
oh, you mean that the case $f'' \neq 0, f \neq 0$ and the other case? –  Misaj Jun 17 '12 at 13:00
    
No, I mean the case "there is $c\geq b$ such that $G(c)\geq 0$" and the other one. –  Davide Giraudo Jun 17 '12 at 13:01
    
Oh then finally we get $b_2$ such that $ b \leqslant b_2 \leqslant c ,\; G(b_2) = 0 $ by IVT. Is this right? I'm sorry for many comments.. –  Misaj Jun 17 '12 at 13:15

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