In the page 601 of 'HOMOMORPHISMS OF BANACH ALGEBRAS'(Bade and Curtis) https://www.jstor.org/stable/pdf/2372972.pdf, we get an inequality
\begin{align*} ||\nu(x_m - x_n)|| \geq \rho_{\mu(C(\Omega))}(\mu (x_m - x_n)) \end{align*}
where
- $\nu \colon C(\Omega) \to \mathfrak{B}$ is an arbitrary homomorphism of a Bananach algebra,
- $\mu \colon C(\Omega) \to \mathfrak{B}$ is a continuous homomorphism with a closed range which depends on $\nu$,
- $\rho_{\mathfrak{X}}(x)$ is the spectral radius of $x \in \mathfrak{X}$, and
- $(x_n)$ is a sequence such that $\nu (x_n)$ is convergent.
I think that $\rho_{\mu(C(\Omega))}(\mu (x_m - x_n)) = || x_m - x_n ||$ holds because the spectrum is defined by algebraic condition and the spectrum of $x \in C(\Omega)$ is equal to the range of $x$.
So we have $\nu (x_m - x_n) \geq ||x_m - x_n||$, however, I think that this is a contradiction.
$\nu $ is an arbitrary homomorphism, so we can assume that $\nu(x) = \nu (y)$ for some $x \neq y$. Define a sequence $(x_n)$ by $x_{2n} = x$ and $x_{2n + 1} = y$. Then, $\nu(x_n)$ is convergent but $x_n$ is not convergent.
What went wrong?
I think the faulty assumption is that $\rho_{\mu(C(\Omega))}(\mu(x_m - x_n)) = \|x_m - x_n\|$.
The dependence of $\mu$ on $\nu$ is important: $\mu$ is the continuous part of $\nu$. Note that therefore $\mu$ can have a non-trivial kernel (I'm tempted to guess that $\ker \nu \subset \ker \mu$?). So we might have $$0 = \rho(\mu(x_m - x_n)) < \|x_m - x_n\|,$$ which would avoid your contradiction.