Proposition 6

Let A be a unital Banach algebra and B a closed unital subalgebra of A. For every $x\in B$, one has Sp$_B(x)\supset$ Sp$_A(x)$, and the boundary of Sp$_A(x)$ in $\mathbf{C}$ contains the boundary of Sp$_B(x)$ in $\mathbf{C}$. In particular, if Sp$_B(x)\subset \mathbf{R}$, then one has Sp$_B(x) =$ Sp$_A(x)$.

One has Sp$_B(x)\supset$ Sp$_A(x)$ (remark 6 of I, p. 3). If $\lambda$ is a point of the boundary of Sp$_B(x)$ in $\mathbf{C}$, there exists a sequence $(\lambda_n)$ of points exterior to Sp$_B(x)$ tending to $\lambda$. Then $x-\lambda_n$ is invertible in B and tends to $x-\lambda$, which is not invertible in B; hence $x-\lambda$ is a left or right topological zero divisor in B (Proposition 4 of I, p. 24), hence in A. Thus, $\lambda \in$ Sp$_A(x)$. But since Sp$_A(x)\subset$ Sp$_B(x)$, the complex number $\lambda \in$ Fr$_{\mathbf{C}}$(Sp$_B(x)$) cannot be interior to Sp$_A(x)$, and therefore belongs to its boundary.