A question to show that the integral operator is a bounded operator This question was asked in my assignment of Functional analysis and I was not able to solve this particular problem.

Question:(a) Show that the formula $Af(x) = \frac{1} {x} \int_{0}^x f(y) dy$ defines a bounded operator $A: L^2 (( 0,\infty)) \to L^2 ((0,\infty))$ with $||A||\leq 2$.

(b)Show that in fact $||A||=2.$
(c) Determine the adjoint operator of A.
(d) Show that $A^*A= A+A^* =AA^* .$  Deduce that the operator $U= I-A $ is unitary.
Attempt: (a) $ ||A|| \leq || \frac{1} {X} \int^x_0 f(y) dy ||\leq \frac{1} {X} \int_{0}^{x}||f(y)|| dy$ but I am not able to show that $||A||\leq 2$ because I don't understand how to find the minimum value of  integral of $f(y)$.
(b)  Due to this I am also not able to show how ||A||=2.
(c) Using the definition , I got that $A^* =  {x} ({\int^{x}_{0} f(y) dy})^{-1}$ but using this definition I am not able to show that $AA^* = A+ A^*$.
Can you please tell is my $A^*$ wrong and how to rightly calculate it?
 A: The bound of the Hardy operator $Hf(x)=\frac1x\int^x_0f(t)\,dt$ has been discussed in several posting in MSE, for example posting 1 and posting 2
As for the adjoint: for $f,g\in L_2(0,\infty)$, an application of Fubini's theorem yields
\begin{align}
\int^\infty_0\Big(\frac1x\int^x_0f(y)\,dy\Big)\overline{g}(x)\,dx&=\int^\infty_0\Big(\frac1x\int^\infty_0\mathbb{1}_{(0,x]}(y)f(y)\,dy\Big)\overline{g}(x)\,dx\\
&=\int^\infty_0f(y)\int^\infty_0\mathbb{1}_{[y,\infty)}(x)\frac{\overline{g}(x)}{x}\,dx\Big)\,dy\\
&=\int^\infty_0f(y)\Big(\int^\infty_y\frac{\overline{g}(x)}{x}\,dx\Big)\,dy
\end{align}
Hence,
$$A^*g(x)=\int^\infty_x\frac{g(y)}{y}\,dy$$

For part (c), another application of Funibi's theorem yields
\begin{align}
AA^*f(x)&=\frac1x\int^x_0A^*f(y)\,dy=\frac1x\int^x_0\Big(\int^\infty_y\frac{f(t)}{t}\,dt\Big)\,dy\\
&=\frac1x\left(\int^x_0\frac{f(t)}{t}\Big(\int^t_0\, dy\Big)\, dt +\int^\infty_x\frac{f(t)}{t}\Big(\int^x_0\, dy\Big)\,dt\right)\\
&=Af(x) + A^*f(x)
\end{align}
Hr=er th region of integration $R:=\{(y, t):0<y\leq x, y\leq t\}$ is expressed as $R:=\{(y, t): 0<t\leq y, 0<y\leq t\}\cup\{(y, t): x<t,0<y\leq x\}$.
Similarly
\begin{align}
A^*Af(x)&=\int^\infty_x\frac{Af(y)}{y}\,dy=\int^\infty_x\frac1{y^2}\Big(\int^y_0 f(t)\,dt\Big)\,dy\\
&=\int^x_0f(t)\Big(\int^\infty_x \frac{dy}{y^2}\big)\,dt+\int^\infty_xf(t)\Big(\int^\infty_t\frac{dt}{y^2}\Big)\,dt\\
&=Af(x)+A^*f(x)
\end{align}
Here, the region of integration $Q:=\{(y, t): x<y, 0<t\leq y\}$ is split into $\{(y, t): x<y, 0<t\leq x\}\cup\{(y, t): x<t, t<y\}$.

Part (d) of the OP follows from (c).
