DefinitionLet be a banach space, be the space of bounded linear operator from . If , then say a complex number is in the resolvent set of if has a bounded inverse. Then we define the resolvent of at to be . If , then is said to be in the spectrum of T.

We will learn some properties of the resolvent and the spectrum.

Lemma 1is a banach algebra under composition.

*Proof:* Let , then

Lemma 2Let be a banach algebra with identity . If with , then exists.

*Proof:* Since is a banach algebra, we have . Then is absolutely convergent in . Note that

Let we will have . Therefore .

Theorem (First resolvent identity)If , then and commute. Furthermore,

*Proof:*

Since we can interchange and , and commute.

Theoremis an open subset of . Fixed , is an analytic – valued function.

*Proof:* If , first define . From the previous lemma, we know that if , the series defines the inverse of . Therefore,

if . So is open. Since can be expressed as infinite series as , it is analytic.

Now we are going to show the spectrum of is non-empty (actually works for any complex banach algebra), which gives a nice corollary by Gelfand and Mazur. Before proving that, we need a banach-valued version of Liouville theorem:

Theorem (Liouville theorem for banach-valued function)Let be a banach space. If is a function from , which is entire and bounded, then is constant.

*Proof:* First note that if , then the function is a complex-valued analytic function.

Pick a and let . For any , we have is complex-valued analytic function and bounded in . Then by Liouville theorem is a constant, and , therefore for all .

Assume , by Hahn-Banach theorem there exists a nonzero linear bounded functional such that . By contraposition this implies , and hence is a constant.

**Remark**: There might be a proof without using axiom of choice. I just don’t know it yet.

Theoremis a non-empty compact subset in .

*Proof:* First define . By the previous lemma again, if , then defines the inverse of . Therefore is a bounded subset. Since is open, is closed. Hence is compact.

To see is non-empty. Assume it is, then is entire by the previous theorem. For , , and goes to 0 as goes to . Then by Liouville theorem, , which is not invertible. This gives a contradiction.

**Remark**: As mentioned before, this argument works for any complex banach algebra.

Corollary (Gelfand-Mazur)Every complex banach division algebra is an isometric isomorphic to .

*Proof:* Since is a division algebra, the only non-invertible element is 0. By the previous theorem, we know that for any , there exists a such that . Then . So the map gives an isometric isomorphism from to .