Self-Adjoint of Volterra Operator I have the Volterra operator defined as following:
$$
A : L_{2}([0,1]) \to L_{2}([0,1])
$$
$$
A f(x) = \int_{0}^{x} f(t) dt
$$
I was asked to show that:
$$
A^{*}A f(x) = \int_{0}^{1} (1- \max(x,t)) f(t) dt 
$$
This is my attempt:
Using the Fubini's theorem and changes of variables:
I can show that:
$$
<f, Ag> = \int_{0}^{1} f(x) \int_{0}^{x} g(y) dy dx = \int_{0}^{1} g(y) \int_{y}^{1} f(x) dx dy = <A^{*}f, g > 
$$
So, I get the adjoint operator $ A^{*} f(x) = \int_{x}^{1} f(t) dt$. But I do not see how to get the term $\max(x,t)$. From this, I only see that:
$$
A^{*}A f(x) = \int_{x}^{1} \int_{0}^{y} f(z) dz dy
$$
How can I show that $A^{*}A f(x)$ as in the question that asks me to do?

$ \langle A^{*}f,g \rangle= \langle f, Ag \rangle=\int_0^{1} g(y)\int_y^{1} f(x)dxdy$. We can write this as $ \langle A^{*}f,g \rangle= \langle h, g \rangle$ where $h(y)=\int_y^{1} f(x)dx$. Since this is true for all $g$ it follows that $A^{*}f=h$.
Now $A^{*}Af(x)=\int_x^{1} Af(t)dt=\int_x^{1} \int_0^{t} f(s)ds dt$. Switch the order of integration to finish.
