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From what I understand, we normally define the operator $\exp$ only on unital Banach algebras, because the triangle inequality ensures that the infinite series converges:

$$\lim_{n\to\infty}\left\|\sum^{\infty}_{i=n}\frac{x^i}{i!}\right\|\leq\lim_{n\to\infty}\sum^{\infty}_{i=n}\left\|\frac{x^i}{i!}\right\|\leq\lim_{n\to\infty}\sum^{\infty}_{i=n}\frac{\|x\|^i}{n^i}$$

And then we use known, low-level results from real analysis to show this limit converges to $0$.

This is evidence that exponentiation will work for unital Banach algebras, but it doesn't show they can't work outside of them.

As a result, I'd like to ask if there are any examples of topological power associative non-metric unital algebras where exponentiation is defined everywhere, or if such algebras do not exist.

Kyan Cheung
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1 Answers1

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If $(A_i)_{i \in I}$ is a family of topological algebras in which the exponential series converges, then the same is true for the topological product algebra $\prod_{i \in I} A_i$. We simply have $\exp(a) = (\exp(a_i))_{i \in I}$ for $a \in \prod_{i \in I} A_i$.

If $I$ is uncountable and the $A_i \neq 0$ are metrizable, then $\prod_{i \in I} A_i$ is not metrizable (SE/4089222). This provides an example for what you are looking for.

Several subalgebras of the product qualify as well. For example, when $X$ is an uncountable Tychonoff space, then $C(X) = \{f : X \to \mathbb{C} \text{ continuous}\}$ with pointwise convergence, that is, as a subspace of $\prod_{x \in X} \mathbb{C}$, is not metrizable (SE/179800). But for $f \in C(X)$ also $\exp(f) = \exp \circ f \in C(X)$.

Martin Brandenburg
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