How to show that the Volterra operator is not normal How to show that the Volterra operator:
$$V:L_2(0,1)\rightarrow L_2(0,1): x\mapsto \int^t_0 x(s) \, ds$$
is not normal.  $t\in (0,1)$
Could you please help with this question.
 A: From this answer we know that
$$
V^*(x)(t)=\int_t^1 x(s)\,ds
$$
Using Fubini theorem you can chek that
$$
VV^*(x)(t)
=\int_0^t\int_s^1x(\tau)\,d\tau \,ds
=\int_0^t\int_0^\tau x(\tau) \,ds\,d\tau
=\int_0^t x(\tau)\int_0^\tau \, ds\,d\tau
=\int_0^t \tau x(\tau)\,d\tau
$$
$$
V^*V(x)(t)
=\int_t^1\int_0^s x(\tau)\,d\tau \,ds
=\int_0^t\int_t^1 x(\tau) \,ds \,d\tau + \int_t^1\int_\tau^1 x(\tau) \,ds \,d\tau\\
=(1-t)\int_0^t x(\tau) \,d\tau + \int_t^1(1-\tau) x(\tau) \,d\tau
$$
So $VV^*\neq V^*V$.
A: From Norbert's post, you have $Vf = \int_{0}^{t}f(s)\,ds$ and $V^{\star}f = \int_{t}^{1}f(s)\,ds$. Consider the simple special case
$$
             V^{\star}V1 = V^{\star} t = \frac{1}{2}(1-t^{2}),\\
             VV^{\star}1 = V(1-t)=t-t^{2}/2.
$$
You can almost guess $V^{\star}V \ne VV^{\star}$ because $Vf$ is continuous and always vanishes at $0$, but not necessarily at $1$, while $V^{\star}g$ is continuous and always vanishes at $1$, but not necessarily at $0$.
Another way: $\frac{d}{dt}V=I$ and $\frac{d}{dt}V^{\star}=-I$. So, if $V^{\star}V-VV^{\star}=0$, then $V+V^{\star}=0$, which is not the case because $(V+V^{\star})f=(f,1)1$. A counter example comes by choosing $f=1$.
