analysis

Banach Algebra Homomorphism


I’m in no mood to do something challenging after this last ditch effort to learn analysis before my prelim, so I’ll do something nice (functional analytic like I promised) that never ceases to amaze me.

Theorem: If \phi is a complex homomorphism on a Banach algebra A, then the norm of \phi, as a linear functional, is at most 1.

Recall that a Banach algebra is just a Banach space (complete normed linear space) with a multiplication that satisfies \|xy\|\leq\|x\|\|y\|, associativity, distributivity, and (\alpha x)y=x(\alpha y)=\alpha (xy) for any scalar \alpha.

Complex homomorphisms are just linear functionals that preserve multiplication \phi(\alpha x+\beta y)=\alpha\phi(x)+\beta\phi(y) and \phi(xy)=\phi(x)\phi(y).

Assume not, i.e. there exists x_0\in A such that |\phi(x_0)|>\|x_0\|. To simplify notation, let \displaystyle \lambda=\phi(x_0) and let \displaystyle x=\frac{x_0}{\lambda}. Then \displaystyle \|x\|=\frac{\|x_0\|}{\lambda}<1 and \displaystyle\phi(x)=\phi(\frac{x_0}{\lambda})=1.

Now \|x^n\|\leq\|x\|^n so s_n=-x-x^2-\cdots-x^n \in A form a Cauchy sequence. Now A is a Banach space and hence complete, so there exists y\in A such that \|y-s_n\|\to 0. But now factor to see that x+s_n=xs_{n-1} and take the limit to get x+y=xy. Now take the homomorphism of both sides, and we have a contradiction \phi(x)+\phi(y)=\phi(x)\phi(y) (in particular 1+\phi(y)=\phi(y)).

So some reasons why this may not be all that shocking: we require these to be complex, and complex things tend to work out nicer than real. Also, these are pretty stringent conditions on what constitutes a Banach algebra and what constitutes a homomorphism. We should be able to get some nice structure with all the tools available. It isn’t like we got a lot. Really we’re just saying that these things are bounded and hence continuous, which isn’t all that surprising.

OK. I’ll stop down playing it. It does surprise me.

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