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 $
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 $
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.