Let V be the Volterra operator on $\mathscr{L^2(0,1)}$.$V(f)(x)=\int_{0}^{x}{f(y)dy}$.
Show that $C^*(V)$, the smallest C* algebra generated with V, is $\mathbb{C}+\mathscr{B_0(L^2(0,1))}$ where $\mathbb{C}$ denotes the complex field and $ \mathscr{B_0(L^2(0,1))}$ denotes the algebra of compact operators on $\mathscr{L^2(0,1)}$.
I know how to prove $ \mathscr{B_0(L^2(0,1))}\subset C^*(V)$ and the quotient algebra $ C^*(V)/\mathscr{B_0(L^2(0,1))}$ is abelian. In order to prove $C^*(V)=\mathbb{C}+\mathscr{B_0(L^2(0,1))}$, I want to prove that $ C^*(V)/\mathscr{B_0(L^2(0,1))}$ is a division banach algebra, hence $ C^*(V)/\mathscr{B_0(L^2(0,1))}=\mathbb{C}$.
But i have no idea on proving nonzero element of $ C^*(V)/\mathscr{B_0(L^2(0,1))}$ is invertible.
Any help would be appreciated.
Since $V$ is compact, $C^*(V)$ is the set of norm limits of polynomials in $V$ and $V^*$. All these polynomials are of the form $\lambda I+K$, with $K$ compact.
Now, if $T\in C^*(V)$, then $T=\lim \lambda_n I+K_n$. If $T$ is compact, we have nothing to prove; so we assume that $T$ is not compact. This implies that the sequence $\lambda_n$ does not converge to $0$ (that would make $T$ compact). It cannot be unbounded, because that would make $T$ unbounded (when $K$ is compact, $\|\lambda I+K\|\geq|\lambda|$). So we may assume that $\lambda_n\to c\ne0$ (by replacing $\{\lambda_n\}$ and $\{K_n\}$ with appropriate subsequences). We now have $$ \|K_n-K_m\|\leq\|K_n+\lambda_nI-(K_m+\lambda_mI)\|+|\lambda_n-\lambda_m|\to0. $$ It follows that the sequence $\{K_n\}$ is convergent (to a compact operator $K$). Thus, $T=cI+K$. This shows that $C^*(V)\subseteq\mathbb C +\mathcal B_0 (\mathcal L^2 (0,1)) $. The other inclusion follows directly from $\mathcal B_0 (\mathcal L^2 (0,1))\subseteq C^*(V) $.