# Absolute continuity inquality

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

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

-
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
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.