Let $f \in L^1 ([0,1]).$ Prove that for each $0<\alpha<\frac{1}{2}$, $$\int_0^1 \left( \frac{\left|f(x)\right|}{x} \right)^\alpha dx$$ is finite.
I attempted to use Jensen's Inequality, but there is, of course, no guarantee that $\frac{\left|f(x)\right|}{x}$ is integrable. I would appreciate some input on how to properly proceed.
Let $p=1/a$ and $q=(1-a)^{-1}$. Clearly, $p,q>1$ and $1/p+1/q=1$. Then applying Hölder's inequality we obtain \begin{align} \int_0^1 \left(\frac{|f(x)|}{x}\right)^a dx &=\int_0^1 |f(x)|^a x^{-a}dx \le \left(\int_0^1 \big(|f(x)|^a\big)^{1/a}\right)^a\left(\int_0^1 (x^{-a})^\frac{1}{1-a}\right)^{1-a} \\ &\le \left(\int_0^1 |f(x)|\,dx\right)^a \left(\int_0^1 (x^{-a})^\frac{1}{1-a}\right)^{1-a} =\|f\|_{L^1}^a \left(\int_0^1 x^{-\frac{a}{1-a}}dx\right)^{1-a}. \end{align} But $a<1-a$, as $a\in(0,1/2)$, and hence $b=\frac{a}{1-a}<1$, which implies that $$ \int_0^1 x^{-\frac{a}{1-a}}dx=\int_0^1 x^{-b}dx=\frac{1}{1-b}=\frac{1-a}{1-2a}<\infty. $$ Altogether $$ \int_0^1 \left(\frac{|f(x)|}{x}\right)^a dx\le \left(\frac{1-a}{1-2a}\right)^{1-a}\|f\|_{L^1}^a. $$