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Consider the map from $\mathbb{C}$ to a unital Banach algebra $B$ given by $x \mapsto \exp(xb)$ for $b\in B$. I proved that this map is continuous by using the definition of $\exp(xb)$ as a contour integral around the spectrum of $b$.

However, I am wondering if there is an easier way to prove this without using the integral expression, for example by using the fact that the exponential map is continuous as a map $\mathbb{C}\rightarrow \mathbb{C}$, and/or by using some properties of the algebra of holomorphic functions on A? I have tried using the composition $x \mapsto xb \mapsto \exp(xb)$, but without succes...

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  • $\begingroup$ You can just use the power series. Would you consider that easier? $\endgroup$ – Daniel Fischer Nov 25 '13 at 20:11
  • $\begingroup$ @DanielFischer I was wondering if I could exploit the isomorphism between holomorphic functions on $\mathbb{C}$ and holomorphic functions on $A$, I didn't figure out how though. As for the power series: multiplication is continuous on Banach algebras, but how to deal with the limit? $\endgroup$ – ScroogeMcDuck Nov 25 '13 at 20:19
  • $\begingroup$ Just a moment, I think I have something, I will edit this post in a few minutes $\endgroup$ – ScroogeMcDuck Nov 25 '13 at 20:20
  • $\begingroup$ @DanielFischer: (I can't edit my comment anymore) I was thinking of using that $\sum_{i=1}^N \frac{x^i}{i!}\rightarrow exp(x)$ uniformly as $N\rightarrow \infty$, since then $\exp(b)$ is such a limit as well (by the algebra isomorphism), but it didn't help me as I am considering a map from $\mathbb{C}\rightarrow A$ rather than a map between algebras of functions... I could work out the norm of the power series case but I don't think it's that much easier. $\endgroup$ – ScroogeMcDuck Nov 25 '13 at 20:33
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    $\begingroup$ $$x \mapsto \sum_{k=0}^\infty \frac{x^kb^k}{k!}$$ It's immediate that that's locally uniformly convergent, hence the sum is continuous. $\endgroup$ – Daniel Fischer Nov 25 '13 at 20:37
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Answered by Daniel Fischer: $$ \sum_{k=0}^\infty \frac{x^kb^k}{k!}$$ is a locally uniformly convergent series, hence the sum is continuous. Indeed, $$\lVert x^kb^k\rVert \leqslant \lvert x\rvert^k\lVert b\rVert^k$$ for every $k$. Since the exponential series converges everywhere in $\mathbb C$, the convergence is uniform in $|x|\le R$ for all $R<\infty$.

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