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 $b\in B$.

  • $\begingroup$ I think in the definition of "maximal Abelian subalgebra" of $A$, it is not required to be a proper subset of $A$. $\endgroup$
    – 23rd
    Apr 14 '13 at 18:23
  • $\begingroup$ If we allowed $A$ to be a maximal subalgebra, then $A$ would be the only maximal subalgebra of $A$. Hence, maximal subalgebras are need to be proper by definition, to avoid triviality. One thing that isn't clear at all is what it means for $B$ to be a "maximal abelian subalgebra": It might mean both "a proper algebra which is not contained in any other proper subalgebra, and which is abelian" or "a proper algebra which is abelian and which is not contained in any other proper abelian subalgebra" $\endgroup$ Apr 14 '13 at 18:49
  • $\begingroup$ If you insist that it must be proper, then it should be of the latter meaning. However, the first statement in your question could fail then, as shown in your example. $\endgroup$
    – 23rd
    Apr 14 '13 at 19:05
  • $\begingroup$ I think that when you ask $A$ to be non-abelian, the exercise follows. Maybe Murphy forgot to mention that. $\endgroup$
    – Yul Otani
    Apr 18 '13 at 18:30
  • $\begingroup$ @LuizCordeiro: Where did you get the idea that maximal subalgebras should be proper? I can't find that condition anywhere in Murphy's book. Concerning the unitality of $B$, note that $A$ is supposed to be unital, a condition that is not satisfied in your counter-example. $\endgroup$
    – UwF
    Apr 23 '13 at 7:14

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 $\sigma_B(b)\subset\sigma_A(b)$.

  • $\begingroup$ Why should $B_c$ be a subalgebra ? A prodcut like $(\lambda c + b_1')(\mu c + b_2')$ contains the element $\lambda \mu c^2$. $\endgroup$
    – user42761
    Sep 27 '15 at 12:23

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