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:
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 .
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 , .