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Let $A$ be a real unital Banach algebra. For $x\in A, r> 0$, denote by $B(x,r)\subset A$ the open ball of radius $r$ centred at $x$. Define $$\log(1+\cdot):B(0,1)\to A,\quad\log(1+x):=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}x^k.$$ Is it always true that $\log(1+\cdot)$ is a homeomorphism onto its image with inverse $\exp(x):=\sum_0^\infty\frac{x^k}{k!}$? If not, can we perhaps say that $\exists\varepsilon>0$ so that $\log(1+\cdot)\vert_{B(0,\varepsilon)}$ is a homeomorphism onto its image?

In the case where $A=\mathbb R$, this is of course true and is proven here, for example. However, I can't seem to generalise that proof as it uses techniques such as differentiating power series term-by-term and the fact that $(x^k)'=kx^{k-1}$, which doesn't hold in noncommutative Banach algebras.

Bowditch
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  • For $\exp(x+y)=\exp x \exp y$ (and also for $\log$ to be a homomorphism) to hold you need $x$ and $y$ to commute. Is that ever a concern here? – Jyrki Lahtonen Dec 06 '17 at 13:39

1 Answers1

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Let $A,\|.\|$ be a complete normed unital algebra where $\|\frac{x}{n}\|=\frac{\|x\|}{n}$. $$\exp(x)-1 \overset{def}= \lim_{N \to \infty}\sum_{n=1}^N \frac{x^n}{n!}, \qquad \log(1+x) \overset{def}= \lim_{K \to \infty}\sum_{k=1}^K \frac{(-1)^{k+1} x^k}{k} $$ For any $x \in A, \|x\| < 1$ and any $y \in A$ $$\|\exp(y)-1\| \le \sum_{n=1}^\infty \frac{\|y\|^n}{n!} = \exp(\|y\|)-1 < \infty, \\ \|\log(1+x)\| \le \sum_{k=1}^\infty \frac{\|x\|^k}{k} = -\log(1-\|x\|) < \infty \\ \|\exp(\log(1+x))-1\| \le \sum_{n=1}^\infty \frac{\|\sum_{k=1}^\infty \frac{\|x\|^k}{k}\|^n}{n!}=\exp(-\log(1-\|x\|))-1 < \infty$$ Thus everything converges absolutely and everything is continuous. Thus we can change the order of summation to write $$(\log(1+x))^n = \sum_{k=1}^\infty a_{n,k} x^k, \qquad \exp(\log(1+x)) -1= \sum_{n=1}^\infty \frac{1}{n!}\sum_{k=1}^\infty a_{n,k} x^k$$ Changing the order of summation again $$\exp(\log(1+x))-1 = \sum_{k=1}^\infty x^k \sum_n \frac{a_{n,k}}{n!}= x$$ where $\sum_n \frac{a_{n,k}}{n!} = 0$ for $k > 1$ is a consequence of our knowledge in real/complex analysis.

YuiTo Cheng
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reuns
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  • Thanks for this, reuns! I just can't quite understand the final line! I can see why $\sum_{n=1}^\infty\frac{a_{n,1}}{n!}=1$, but what is the knowledge from analysis which tells us $\sum_n\frac{a_{n,k}}{n!}=0$ for $k>1$? – Bowditch Dec 06 '17 at 14:52
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    @Bowditch Write $\exp(\log(1+z))-1 = \sum_{k=1}^\infty c_k z^k$, note this is an analytic function. In real/complex analysis you know that $\exp(\log(1+z))-1 = z$ for $|z| < 1$, which means $c_k = 0$ for $k > 1$. Those coefficients are $c_k = \sum_n \frac{a_{n,k}}{n!}$, they stay the same in your Banach algebra. – reuns Dec 06 '17 at 15:08