I'm studying Murphy's book: C*-Algebras and Operator Theory, and got stuck on exercise 8 from chapter 1:
"Show that if $B$ is a maximal abelian subalgebra of a unital Banach algebra $A$, then $B$ is closed and contains the unit. Show that $\sigma_A(b)=\sigma_B(b)$ for all $b\in B$." (where $\sigma_A(b)=\left\{\lambda\in\mathbb{C}:\lambda 1-b\text{ is not invertible in }A\right\}$, and $\sigma_B(b)$ is defined analogously)
I tried to argue by contradiction: suppose that $B$ is not closed. Then $\overline{B}$ is a closed commutative subalgebra of $A$ that contains $B$ strictly. By maximality, we have $\overline{B}=A$, hence $B$ is a dense maximal subalgebra of a unital commutative Banach algebra. But I don't see how this leads to a contradiction, although it smells like Gelfand Transform…
EDIT: Actually, I tried to show that $B$ must contain the unit, and maybe the problem is wrong: Let $B$ be a commutative non-unital Banach algebra, for example, $B=C_0(\Omega)$, where $\Omega$ is a non-compact, locally compact, Hausdorff topological space, say $\mathbb{R}$. Let $A=B\oplus\mathbb{C}$ be its unitization, that is $A=B\times\mathbb{C}$ as a set and the operations on $A$ are defined as
$$(b,\alpha)+\gamma(c,\beta)=(b+\gamma c,\alpha+\gamma\beta),\quad (b,\alpha)(c,\beta)=(bc+\alpha c+\beta b,\alpha\beta)$$
and $A$ has the norm $\Vert(b,\alpha)\Vert=\Vert b\Vert+|\alpha|$. It is easily checked that $A$ is a commutative Banach algebra with unit $(0,1)$.
If we identify $B$ with the subalgebra $B\oplus\left\{ 0\right\}$ of $A$, then it's clear that $B$ is a maximal commutative subalgebra of $A$ which does not contain the unit.
If my arguments were correct, then it is not true in general that $B$ contains the unit. But maybe we can still assure that $B$ is closed in $A$ and that, if $B$ contains the unit, then $\sigma_A(b)=\sigma_B(b)$ for every $b\in B$.
Best Answer
If $A$ is not abelian, then clearly $B\not=A$ for any abelian subalgebra $B\subseteq A$. If $A$ is abelian, then $B=A$ is the only maximal abelian subalgebra of $A$ (and all the claims are trivially true).
If $A$ itself is a unital Banach algebra (as stated in Murphy's exercise), then any maximal abelian subalgebra $B\subseteq A$ is closed and unital (because $B_1=\{\lambda 1 + b;\lambda\in C, b\in B\}$ ($C$ the field of complex numbers) and $\overline{B}$ are abelian subalgebras of $A$ and they contain $B$, so by maximality of $B$ we get $B=\overline{B}=B_1$).
We have $\sigma_A(b)\subset\sigma_B(b)$ for all $b\in B$, because if $\lambda 1-b$ has an inverse in $B$, then it also has an inverse in $A$ (so if it doesn't have an inverse in $A$, then it can't have an inverse in $B$, either).
But note that the converse is true also: if for an element $b\in B$, $\lambda 1-b$ has an inverse in $A$, say $c$, then $c$ belongs to $B$. This is because the relation $(\lambda 1-b)b'=b'(\lambda 1-b)$ (satisfied for all $b,b'\in B$ because $B$ is abelian) implies $cb'=b'c$ for all $b'\in B$. Then $B_c=\{p(c)b';p(c)$ a polynomial in $c, b'\in B\}$ is abelian, contains $B$, so since $B$ is maximal abelian we have $B_c=B$ and $c\in B$, i.e. $\lambda 1-b$ has an inverse in $B$. Therefore $\sigma_B(b)\subset\sigma_A(b)$.