Show that $A : L^2 \to L^2$, $(Af)(x) =\int_0^1 \frac{f(t)}{\sqrt{|x-t|}}dt$ is a bounded operator Prove that the operator $A:L^2([0,1])\rightarrow L^2([0,1]) $ where
$$(Af)(x) =\int_0^1 \frac{f(t)}{\sqrt{|x-t|}}dt$$
is bounded. A hint is provided: consider using the fact that $\sqrt{|x-t|}=(|x-t|)^{1/4}(|x-t|)^{1/4}$.
How can I show that this operator is bounded?

$$|(Af)(x)|=\left|\int_0^1 (f(t) (x-t)^{0.25} )(x-t)^{-0.75} dt\right| \leq\sqrt{ \int_0^1 (|f(t)|^2 |x-t|^{0.5} )dt}\sqrt{\int_0^1 |x-t|^{-1.5}dt} \leq M\sqrt{\int_0^1 |x-t|^{-1.5}dt}  \cdot||f||$$
Hence
$$||Af||\leq M\sqrt{\int_0^1 \left(\int_0^1 |x-t|^{-1.5}dt\right)dx} \cdot ||f||$$
