Let $V$ denote the Volterra integral operator on $L^2[0, 1]$ defined by $$ Vf(s)=\int_0^s f(t) dt $$ and let $A$ be the closed subalgebra of $\mathcal{B}(L^2[0,1])$ generated by $V.$ Show that $A$ has precisely one maximal ideal.
My attempt:
I first showed that spectral radius of $V, r_A(V)=0.$
Now, $A$ is the norm closure in $\mathcal{B}(L^2[0,1])$ of all polynomials in $V$ of the form $$\sum_{i=1}^n a_i V^i, a_i \in \mathbb C, n \in \mathbb N.$$
Since $r_A$ is subadditive, submultiplicative and continuous. Therefore it follows that $r_A(S) = 0$ for all $S \in A.$ Let $\Delta(A)$ denote the set of all nonzero multiplicative linear functionals on $A.$ We know that if $\phi \in \Delta(A),$ then $|\phi(S)| \leq r_A(S)$ for all $S \in A.$ Thus $\phi \equiv 0,$ which is not possible. Hence, $\Delta(A)$ must be empty.
Since there exists a bijection between $\Delta(A)$ and maximal modular ideals of $A,$ we conclude that $A$ has no maximal modular ideals.
How do I proceed from here?