The Exponential Group in a Banach Algebra: a proof of Rudin.

Question

Define $G$ to be the group of invertible elements in a Banach algebra $A$ and $G_1$ the component of $G$ that contains $e\in G.$

Now, $\exp x=\sum_{n=0}^\infty \frac{x^n}{n!}$ and if $\Gamma$ is the subgroup generated by $\exp (A),$ a straightforward argument shows that $\Gamma$ is open: if we define $y=\sum_{n=0}^\infty \frac{-1}{n}(1-x)^n$ then the series converges absolutely whenever $\|1-x\|<1$ and a (at least for me) somewhat messy calculation shows that $\exp y=x$ so $\exp (A)$ contains a neighborhood of $e$, and since $\Gamma$ is a group, it follows that $\Gamma$ is open.

Unfortunately, Rudin does not do it this way. He cites the following theorem:

Theorem Suppose $A$ is a Banach algebra, $x\in A$, and the spectrum $\sigma(x)$ of $x$ does not separate $0$ from $\infty$. Then

(a) $x$ has roots of all orders in $A$,

(b) $x$ has a logarithm in $A$, and

(c) if $\varepsilon>0$, there is a polynomial $P$ such that $\|x^{-1}-P(x)\|<\varepsilon$.

to claim that $\exp(A)$ has nonempty interior in $G$ and uses this fact to prove that $\Gamma$ is open. I tried to do this as follows: take $y\in\exp(A)$ and note that as soon as $\|x-y\|<\|y^{-1}\|^{-1}$ then $\|e-y^{-1}x\|<1.$ It follows that $\sigma(y^{-1}x)\subseteq \{|\lambda-1|<1|\}$ and now the theorem applies to show that $y^{-1}x=\exp z$ for some $z\in A$ and so $x=y\exp z.$ Of course, unless $A$ is commutative, we can't conclude.

Since Rudin claims that $\exp(A)$ is open, as a passing remark, I guess it's easy, but I have not been able to prove it. I would like a hint to get me going in the right direction.

Answer 1

I don't think Rudin claims that $\exp(A)$ is open, but rather that it has a nonempty interior. Moreover the proof I believe he has in mind is not much different from what you have above: in a neighborhood of the identity all elements have spectrum close to 1, which therefore does not separate 0 from $\infty$. By Theorem 10.30.(b) these elements have a logarithm, which is to say that they lie in $\exp(A)$.

Answer 2

$e=1$ has a neighborhood of invertible elements and, crucially, they are in $\exp(A)$, since $\log(1+x)$ is defined (as a series for example) for $x$ small enough.
(This is the extra bit required in your argument: $1-y^{-1}x\in\exp(A)$, not just $y^{-1}x$.)

Let $y\in\exp(A)$, then $y+x=y(1+y^{-1}x)\in\Gamma$ for $x$ small enough, by the above. Thus there is a neighborhood of $y$ inside $\exp(A)$, so $\exp(A)$ is an open subset.

(Note that $y^{-1}\in\exp(A)$.) You still have to figure out how small is small enough to make the argument rigorous.