Is it true that for $f\in\mathcal L^2$ and $\|f\|_2=1$, $$\int_0^\infty f(x)x^{-1/2}dx<\infty?$$ I'm fairly stuck on this...(and I really hope it is true).
In case it helps seeing a generalization, we could also ask for $f\in\mathcal L^{p^*}$ and $\|f\|_{p^*}=1$, is $$\int_0^\infty f(x)x^{-1/p}dx<\infty,$$ where $\frac{1}{p}+\frac{1}{p^*}=1$.
No, it is not true. A counter-example is given by $$ f(x) = \left\{\begin{array}{cc} 0, & 0 \le x < 1, \\ \frac{1}{\sqrt{x}(1+\ln x)}, & 1 \le x < \infty. \end{array}\right. $$ $f \in L^2$ with $\|f\|=1$ because \begin{align} \int_{0}^{\infty} f(x)^2 dx &= \int_{1}^{\infty}\frac{1}{x(1+\ln x)^2}dx \\ &=\int_{1}^{\infty}\frac{1}{(1+\ln x)^2}\frac{d}{dx}(1+\ln x)dx \\ & = \left.-\frac{1}{1+\ln x}\right|_{x=1}^{\infty}=1. \end{align} However, \begin{align} \int_{0}^{\infty}f(x)\frac{1}{\sqrt{x}}dx & = \int_{1}^{\infty}\frac{1}{x(1+\ln x)}dx \\ & = \int_{1}^{\infty}\frac{1}{1+\ln x}\frac{d}{dx}(1+\ln x)dx \\ & = \ln(1+\ln x)|_{x=1}^{\infty}=\infty. \end{align}