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Show for an absolutely continuous function $u(x)$ on $[0,1]$ that satisfies $u(0) = 0$,

$$ \int_0^1{\frac{u(x)^2}{x^{3/2}}dx} \leq 2\int_0^1 (u'(x))^2 dx $$

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You may have neglected some additional assumptions: after all, for $u(x)=1$ the rhs is zero while the lhs diverges toward $+\infty$. – whuber Dec 27 '12 at 22:23
Oops, thanks! I forgot to mention that u(0) = 0. – Javalina Dec 27 '12 at 22:29
up vote 4 down vote accepted

By Cauchy-Schwarz inequality, $$u(x)^2=\bigg(\int_0^x u'(t)dt\bigg)^2\le x\int_0^xu'(t)^2dt.$$

Therefore, $$\int_0^1\frac{u(x)^2}{x^{3/2}}dx\le\int_0^1\frac{\int_0^xu'(t)^2dt}{x^{1/2}}dx=\int_0^1\bigg(\int_t^1x^{-1/2}dx\bigg)u'(t)^2dt\le2\int_0^1u'(t)^2dt,$$ where the equality in the second step above is due to Fubini's theorem.

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very nice!!!!!!! – Bombyx mori Dec 28 '12 at 17:51
@user32240: Thank you! – 23rd Dec 28 '12 at 17:53
Excellent, thank you so much! – Javalina Dec 28 '12 at 18:19
@Javalina: You are welcome! – 23rd Dec 29 '12 at 16:22

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