Consider $V:L^2([0,1],\mathbb{C}) \rightarrow L^2([0,1],\mathbb{C})$ defined by $$(Vu)(x)=\int_0^xu(t)dt$$
(i) Calculate the adjoint $V^*$.
$$V^*=\int_0^x \bar{u(t)}\ \ dt$$
(ii) Suppose $u \in C([0,1];\mathbb{C})$. Show that $$(V^*Vu)(x)(y)=1-y\int_0^yu(t)dt+\int_y^1(1-t)u(t)dt.$$
(iii) Find all non-zero eigenvalues of $V^*V$
(v) Compute $||V||$
I am really getting stuck with those questions, I would appreciate any help or hint.
It must be $<Vf,g>=<f,V^*g>$ for all $f,g\in L^2([0,1])$. We have $<Vf,g>=\displaystyle{\int_0^1\int_0^xf(y)\overline{g(x)}dydx=\int_0^1\int_y^1f(y)\overline{g(x)}dxdy}$, by Fubini's theorem (I believe you can justify this). Define $U:L^2[0,1]\to L^2[0,1]$ by $Uh(y)=\displaystyle{\int_y^1h(x)dx}$. This is a well-defined and bounded linear operator (check this too). Now we have that $<Vf,g>=\displaystyle{\int_0^1\int_y^1f(y)\overline{g(x)}dxdy=\int_0^1\bigg{(}f(y)\int_y^1\overline{g(x)}dx\bigg{)}dy=\int_0^1\bigg{(}f(y)\overline{\int_y^1g(x)dx}\bigg{)}dy=<f,Ug>}$
Now it is $<f,V^*g>=<f,Ug>$ for all $f,g\in L^2[0,1]$, hence $<f, (V^*-U)g>=0$ for all $f,g$. That implies $U=V^*$.
For (ii), we have $V^*Vu(y)=\displaystyle{\int_y^1\int_0^xu(t)dtdx}$. Draw the region of integration, it breaks into a square and a triangle. Integrate separately on each and you get $V^*Vu(y)=\displaystyle{\int_0^y\int_y^1u(t)dtdx}+\int_y^1\int_t^1u(t)dxdt=y\int_y^1u(t)dt+\int_y^1(1-t)u(t)dt$.
