Let be a Banach space over and be a continuous linear map from to itself. The usual norm on is , also denoted by . Setting , we want to show:

**Theorem:** .

The proof is interesting because it uses both functional analysis and complex analysis. The note finishes with some corollaries about complex Banach algebra and matrices.

**Proof.** For convenience, let .

**Claim 1:** is compact and included in the disk .

The map is continuous and where is the set of isomorphisms. So we can show that is closed just by noticing that is open in . Let and such that . Then is absolutely convergent, and a fortiori convergent since is complete; moreover, it is the inverse of so is invertible and is finally closed.

Then, if then is absolutely convergent and it can be verifyed that . Therefore, . In particulier, is compact.

**Claim 2:** is nonempty.

First, introduce and notice that it is holomorphic. Indeed:

for . By continuity, is holomorphic with .

For all , let . If , then ; in particular, so and is bounded. According to Liouville’s theorem, if then is constant (in fact identically zero because of the preceding limit). However, according to Hahn-Banach theorem, there exists such that is non identically zero. Therefore, ie. .

Thus, using claims 1 and 2, is well defined.

**Claim 3:** For all , .

For any , being holomorphic on , according uniqueness of power series expansion, for every such that . In particuler, for every . Therefore, using the lemma following the proof, there exists such that for all . Therefore, .

To conclude our proof, notice that the equality implies that if then ; using claim 1, we get . Finally, .

**Lemma:** Let be a Banach space over and a sequence in . If for all , , then .

**Proof.** For any , there exists a linear functional such that for all . Moreover so . According to Hahn-Banach theorem, there exists such that and , so . Finally, . By assumption, for any , ; using Banach-Steinhaus theorem, we conclude that .

A simple corollary is the following (related to Frobenius theorem and Hurwitz theorem):

**Theorem:** (Gelfand-Mazur) Let be a complex Banach algebra. If is a division algebra then is isometrically isomorphic to .

**Proof.** Let such that . Suppose that there exists . By assumption, is invertible for all hence , contradicting claim 2. Therefore, . Now, it is easy to verify that from to is an isometric isomorphism.

Notice that if is a subspace of a complex Banach space then the spectral radius formula still holds in (but with the spectral radius computed in ); in particular, it can be useful to extend the formula to some real Banach spaces. For instance:

**Property:** Let be a matrix with or . Then .

**Corollary:** For all , .

In your proof of Claim 3 (the very first line), I’m struggling to see how the series for is convergent for every ; you quote “uniqueness of power series expansion”. I don’t see how does it help.

In fact, I don’t see why will have an expansion centred at which includes the points .

I agree that there is an expansion in the smaller disc though.

First, you probably noticed that there is a typo: the equality is in fact , as in claim 2.

Then, I agree that the justification is not well-written, as several points in this note, I have to update it.

A possibility is to notice from the proof of the analyticity of holomorphic functions that, if a function is holomorphic on the disk , then it can be written as a power series on for every . Now, let ; we know that is holomorphic on , and on the other hand, we know that on . But if you write as a power series on , with , because it agrees with the previous one on , they are in fact identical. Therefore, the series converges for all , where ; equivalently, the series converges for all .