Is that linear operator in $L^{2}(0,1)$ bounded?

Question

The question is "is the follow linear operator in $L^{2}(0,1)$ bounded?" The linear operator is the following

$(Af)(x)=\frac{1}{\sqrt{x}}\int_{0}^{x}f(t)dt , 0< x \leq 1$

i know that

$\left \| A \right \|=\frac{\left \| Af \right \|}{\left \|f \right \|} $

And i got stucked here

$\left \| A \right \|=\frac{\int_{0}^{1}\frac{1}{x}(\left | \int_{0}^{x}f(t)dt \right|)^{2} dx }{\int_{0}^{1}f(x)dx}$

my discipline is poorly formulated, the professor is teaching about linear operators in hillbert space without even commenting on the necessary mathematical background

Answer 1

Well, applying Cauchy-Schwarz to $\int_0^xf(t)\,dt=\int_0^x1\cdot f(t)\,dt$ shows that $$\begin{align}\int_0^1|Af(x)|^2\,dx&=\int_0^1\frac1{ x}\left|\int_0^xf(t)\,dt\right|^2\,dx \\&\le\int_0^1\int_0^x|f(t)|^2\,dt\,dx \\&=\int_0^1\int_0^t|f(t)|^2\,dx\,dt \\&=\int_0^1t|f(t)|^2\,dt \\&\le\int_0^1|f(t)|^2\,dt,\end{align}$$so...