Suppose that H = L²(0,1) with respect to the Lebesgue measure. Then (Vf)(x) = ò₀^xf(t)dt defines an operator V Î B(H) which is called Volterra operator. Clearly the closure A of {p(V) ; p  is  a  polynomial  in  z } in B(H) is the commutative Banach subalgebra of B(H) generated by V and the identity operator I.
An straightforward computation shows that (Vⁿf)(x) = ò₀^x[( (x-t)^n-1)/( (n-1)!)] f(t)dt and so ||Vⁿ || £ [1/((n-1)!)] and r(V) = lim_n ||Vⁿ ||^[1/n] = 0. Hence sp(B(H),V) = {0}. But by [Con, VII.Theorem 5.4], sp(A,V) is equal to the polynomially convex hull of sp(B(H),V), hence sp(A,V) = {0}. But the maximal ideal space of A is homeomorphic to sp(A,V) = {0}. So the only character on A is f(l) = l and f(x) = 0 for x Î A - C. Since f is continuous, the unique maximal ideal space is Rad(A) = Ker(f) = the closure of {p(V) ; p is a polynomial in z and p(0) = 0}.

Ref.

[Con]J.B. Conway, A course in functional analysis, New York, Springer-Verlag, 1990.