# 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?

$$\|Af\|^2=\int\limits_0^1\left|\int\limits_0^1\frac{f(t)}{\sqrt{|x-t|}}dt\right|^2dx=\int\limits_0^1\left|\int\limits_0^1\frac{1}{|x-t|^\frac{1}{4}}\frac{f(t)}{|x-t|^\frac{1}{4}}dt\right|^2dx\leq$$$$\leq\int\limits_0^1\int\limits_0^1\frac{dt}{\sqrt{|x-t|}}\int\limits_0^1\frac{|f(t)|^2dt}{\sqrt{|x-t|}}dx.$$$$\int\limits_0^1\frac{dt}{\sqrt{|x-t|}}=\int\limits_0^x\frac{dt}{\sqrt{x-t}}+\int\limits_x^1\frac{dt}{\sqrt{t-x}}=2\sqrt{x}+2\sqrt{1-x}\leq4.$$$$\|Af\|^2\leq4\int\limits_0^1\int\limits_0^1\frac{|f(t)|^2dt}{\sqrt{|x-t|}}dx=4\int\limits_0^1|f(t)|^2\int\limits_0^1\frac{dx}{\sqrt{|x-t|}}dt\leq16\|f\|^2.$$
$$|(Af)(x)|=\left|\int_0^1 (f(t) (x-t)^{0.25} )(x-t)^{-0.75} dt\right| \leq\sqrt{ \int_0^1 (|f(t)|^2 |x-t|^{0.5} )dt}\sqrt{\int_0^1 |x-t|^{-1.5}dt} \leq M\sqrt{\int_0^1 |x-t|^{-1.5}dt} \cdot||f||$$ Hence $$||Af||\leq M\sqrt{\int_0^1 \left(\int_0^1 |x-t|^{-1.5}dt\right)dx} \cdot ||f||$$