$f:[0,1] \to \mathbb R$ is a continuously differentiable function, and $f(\frac{1}{2})=0$. I'm trying to show $$( \int_0^1{f(x)} \, dx)^2 \le \frac{1}{4} \int_0^1 (f'(x))^2 \, dx $$ I tried using the mean value theorem but got nowhere. Are there any hints? Thank you!
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2You may find this problem useful, as it is similar in nature. – Decaf-Math Nov 23 '19 at 20:01
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@cmk: Certainly similar, but not a duplicate. – copper.hat Nov 23 '19 at 20:38
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Note that $f(x) = \int_{1 \over 2}^x f'(t)dt$ and Cauchy Schwartz gives $|f(x)| \le \|f'\| \sqrt{|x-{1 \over 2}|}$.
Since $\int_0^1 \sqrt{|x-{1 \over 2}|}dx = {\sqrt{2} \over 3}$, we get $(\int_0^1 f(x) dx)^2 \le (\int_0^1 |f(x)| dx)^2 \le {2 \over 9} \int_0^1 (f'(x))^2 dx$ with a little to spare :-).
copper.hat
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Sorry, so $$f(x)^2 \leq \left( \int_{1/2}^x (f'(t))^2 , {\rm d}t\right) \left( \int_{1/2}^x 1 , {\rm d}t \right) , ?$$ – Diger Nov 23 '19 at 23:36
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Close. Apply Cauchy Schwartz to $\int f' \cdot 1_{[{1 \over 2},x]}$. – copper.hat Nov 23 '19 at 23:39
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? That's what I did, or what is wrong with my notation? Or is the integral $\int_{1/2}^x ...$ no vector space? Do I need to extent it to $\int_0^1 ...$ and introduce the indicator function? – Diger Nov 23 '19 at 23:47
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