# 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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