I have been trying to solve the following problem using Holder's or Young's Inequality, but I am just not doing the right manipulations I think. It shouldn't require anything fancy.
Let $f \in \mathfrak{L}^2$$(0, 1)$, and let $g(x) = \int_{[0,x]} \frac{f(t)}{\sqrt{1-t^2}}dt$. Prove $g \in \mathfrak{L}^2$$(0, 1)$.
Let $f \in \mathfrak{L}^2$$(0, 1)$, and let $g(x) = \int_{[0,x]} \frac{f(t)}{\sqrt{1-t^2}}dt$. Prove $g \in \mathfrak{L}^2$$(0, 1)$.
$$\int_0^1 g(x)^2 dx = \int_0^1 \left( \int_0^x \frac{f(t)}{\sqrt{1-t^2}}dt \right)^2 dx$$
then, you use Jensen inequality (with a rescaling):
$$\leq \int_0^1 \frac{1}{x}\int_0^x \left( x\frac{f(t)}{\sqrt{1-t^2}} \right)^2 dt dx$$
$$\leq \int_0^1 x\int_0^x \left(\frac{f(t)}{\sqrt{1-t^2}} \right)^2 dt dx$$
You intervert the integrals
$$\leq \int_0^1 \frac{f(t)^2}{1-t^2} \int_t^1 x dx dt $$
$$\leq \int_0^1\frac{f(t)^2}{1-t^2} \frac{1-t^2}{2} dt $$
$$\leq \frac{1}{2} \int_0^1f(t)^2 dt $$