Let A and B be a Banach algebra with $1\in A\subset B$ (1 is the identity). Let $a\in A $ then show that $$1).Spectrum_a\ B\subset Spectrum_aA$$ $$2). Boundary(Spectrum_aA)\subset Boundary(Spectrum_aB)$$
First, let's take $x\in Spectrum_a\ B$, I'm going to show that is in $Spectrum_aA$.
since $x\in Spectrum_a\ B$
$a-x*1$ is not invertible. Since $a\in A$ , $_\ $ contains x.
This implies $Spectrum_a\ B\subset Spectrum_aA$.
Is this correct? And how we show the second part?
Lemma. Assume that $A$ is a unital Banach algebra and that $\{a_n\}_n\subseteq A$ is a sequence of invertible elements, converging to a non-invertible element $a$. Then $a$ is a topological zero-divisor, meaning that there exists a sequence $\{x_n\}_n$ in $A$, with $\|x_n\|=1$, and $\displaystyle\lim_{n\to \infty }ax_n = 0$.
Proof. We begin by claiming that there is no positive constant $c$, such that $$ \|ax\|\geq c\|x\|,\quad\forall x\in A. $$ In order to prove the claim we assume otherwise and hence $$ c\|a_n^{-1}\|\leq \|aa_n^{-1}\| = \|(a-a_n+a_n)a_n^{-1}\| \leq $$$$ \leq \|a-a_n\|\|a_n^{-1}\| + 1, $$ from where it follows that $$ \|a_n^{-1}\| \leq \frac 1{c-\|a-a_n\|}, $$ as long as the denominator is positive, which it sure is for large enough $n$. This also shows that $\|a_n^{-1}\|$ is bounded. Given any $n$ and $m$ we then have that $$ \Vert a_n^{-1}-a_m^{-1}\Vert = \Vert a_n^{-1}(a_m-a_n)a_m^{-1}\Vert \leq $$$$ \leq \Vert a_n^{-1}\|\|a_m-a_n\|\|a_m^{-1}\Vert \leq \|a_m-a_n\|\big(\sup_k\|a_k^{-1}\Vert \big)^2. $$ This says that $\{a_n^{-1}\}_n$ is a Cauchy sequence. Setting $\displaystyle b= \lim_{n\to \infty }a_n^{-1}$, we then have that $$ ab = \big (\lim_{n\to \infty }a_n\big )\big (\lim_{n\to \infty }a_n^{-1}\big ) = \lim_{n\to \infty }a_na_n^{-1} = 1, $$ and likewise $ba=1$, so $a$ is invertible, a contradiction, hence proving the claim.
Consequently, for every $n>0$, the inequality "$\|ax\|\geq (1/n)\|x\|$" does not always hold, so there exists some $x_n$ such that $$ \|ax_n\|< (1/n)\|x_n\|. $$ The conclusion then follows by replacing the $x_n$ with $x_n/\|x_n\|$. QED
Back to the question, suppose that $\lambda $ is in the boundary of $\sigma _A(a)$. Then $a-\lambda $ is not invertible, but it is clearly a limit of invertible elements of the form $a-\rho _n$, where $\rho _n$ lies in complement of $\sigma (A)$. By the Lemma, we deduce that $a-\lambda $ is a topological zero-divisor, and this obviously prevents $a-\lambda $ from being invertible in $B$, so we see that $\lambda \in \sigma _B(a)$.
This shows that $$ \partial \sigma _A(a)\subseteq \sigma _B(a), $$ and from this one easily shows that in fact $$ \partial \sigma _A(a)\subseteq \partial \sigma _B(a). $$