Assume that $B$ is a unital Banach algebra ($1$ is a unit), and $A \subset B$ is a closed subalgebra that contains $1$. Is it possible that there exists $a \in A$ such that $\sigma_A(a) \ne \sigma_B(a)$ ($\sigma_A$ is spectrum in the sense of algebra $A$, and the same thing is $\sigma_B$)? The inspiration for this question is that the spectral radius $\rho(a)$ does not depend on algebra ($A$ or $B$) since it is equal to $\lim ||a^n||^{\frac{1}{n}}$.
It is easy to see that $\sigma_B(a) \subset \sigma_A(a)$ for all $a \in A$. If $\lambda \notin \sigma_B(a)$ and $\lambda > \rho(a)$ then $(a - \lambda 1)^{-1}$ belongs to $A$ (since it can be approximated by polynomials of $a$). But for arbitrary $\lambda \notin \sigma_B(a)$ I can't prove that $(a - \lambda 1)^{-1}$ belongs to $A$ (and also I can't create a counterexample).