A criterion for commutative Banach algebra

Question

Suppose $A$ is a Banach algebra and there exists $C>0$ such that $\|xy\| \leqslant C\|yx \|$ for all $x,y \in A$. I am trying to show that in this case $A$ is commutative. It is easy to show that given $x,y \in A$, the following inequality holds: $$ \|\exp(-l x) \cdot y \cdot \exp(l x)\| \leqslant C\|y\|, $$ for all $x,y \in A$, $l\in \mathbb{C}$, where $\exp(x) = \sum_{n=0}^{\infty} \frac{x^n}{n}$. Now I am trying to consider the function $f(l) = \varphi(\exp(-l x) \cdot y \cdot \exp(l x))$ for $\varphi \in A^*$. Then it seems that if $f$ is a bounded entire function, then (by Liouville theorem) it must be a constant, which gives the result, but I am stuck at this point. Can you please help me?

Answer 1

Seems to me it goes like so. Say $V$ is an open subset of $\Bbb C$ and $F:V\to A$. Define $F\in H(V,A)$ if $$F'(z)=\lim_{h\to0}\frac{F(z+h)-F(z)}{h}$$exists for every $z\in V$. We could take various sorts of limits, in fact we can suppose that fraction converges in norm.

Exactly as in ordinary complex analysis you show that the sum of a convergent power series is holomorphic. So if $x\in A$ then the function $l\mapsto\exp(lx)$ is holomorphic in $\Bbb C$.

And exactly as in ordinary complex analysis, or for that matter exactly as in Calc I, you show that the product of two holomorphic functions is holomorphic. So if you define $$F(l)=\exp(-lx)y\exp(lx)$$then $F\in H(\Bbb C,A)$. The inequality in your post shows that $||F(l)||\le C||y||$ for every $l$.

So the question is to show Liouville works for $A$-valued holomorphic functions. It's easy to show from the definition that $F$ holomorphic implies that $\phi\circ F$ is holomorphic for $\phi\in A^*$. So Liouville implies $\phi\circ F$ is constant. Now Hahn-Banach implies that $F$ is constant.

(Hence $\exp(x)y=y\exp(x)$ for every $x$. Now if $z\in A$ there exists $a>0$ so $||az||<1$. The power series for $\log(1-t)$ in the unit disk shows that there exists $x$ with $1-az=\exp(x)$. Hence $(\Bbb 1-az)y=y(\Bbb 1-az)$, and hence $yz=zy$.)