let $ f \in L^{2} [ 0 , 1 ] $.
define: $ V( f ( t ) ) = \int_{0}^{t} f ( s) ds $.(Volterra operator)
we know that $ V( f ( t ) ) \in C [ 0 , 1 ] $ for all $ f \in L^{2} [ 0 , 1 ] $
can we find $ \sigma ( V ) = \{ 0 \} $ and $ r ( V ) = 0 $?
we have $ V^{*} : L^{2} [ 0 , 1 ] \longrightarrow L^{2} [ 0 , 1 ] $,
so $ < f , V f > = < V^{*} f , f > $
and $ V^{*} ( f ( t )) = \int_{t}^{1} f ( s) ds $
is it right to say $ V^{*} V $ is compact and normal ?
You can construct the resolvent $(V-\lambda I)^{-1}$ by solving for $g$, given $f$. This is a first order ODE that can be solved with an integrating factor. Assuming $\lambda \ne 0$, $$ Vg -\lambda g = f \\ Vg -\lambda \frac{d}{dx}Vg = f $$ The solution $g=R(\lambda)f$ is defined for all $\lambda\ne 0$ by $$ R(\lambda)f = -\frac{1}{\lambda}f-\frac{1}{\lambda^2}\int_{0}^{x}e^{(x-t)/\lambda}f(t)dt. $$ This operator is easily shown to be bounded on $L^2$. So $\sigma(V)=\{0\}$ because the spectrum is contained in $\{0\}$ and cannot be non-empty because $V$ is a bounded operator. $V$ cannot be normal because the norm and spectral radius are the same for a normal operator, and the spectral radius of $V$ is $0$ because $\sigma(V)=\{0\}$.
$V$ is compact. So $V^*V$ is compact. And $V^*V$ is selfadjoint and, hence, normal.