# Why is $f(t) = e^{ta}$ differentiable in a unital Banach algebra?

Let $$A$$ be a unital Banach algebra. For $$a\in A$$, we define $$\exp(a):= \sum_{n=0}^\infty \frac{a^n}{n!}$$

Consider the function $$f: \Bbb{R} \to A: t \mapsto \exp(ta) = \sum_{n=0}^\infty \frac{t^n a^n}{n!}$$

In the book I'm reading, it is claimed that $$f'(t) = af(t)$$ by differentiating term by term. How can we justify the differentiation term by term? Or what is another way to show that $$f$$ is differentiable with $$f'(t) = af(t)$$. Maybe some argument with functionals?

Here is an elementary proof inspired on the classical proof for power series.

For $$t \in \Bbb{R}$$, put $$g(t):= \sum_{k=1}^\infty k\frac{t^{k-1}a^{k}}{k!}$$ $$S_n(t) := \sum_{k=0}^n \frac{t^ka^k}{k!}$$ $$R_n(t) := \sum_{k=n+1}^\infty \frac{t^ka^k}{k!}$$

All these series converge since $$A$$ is a Banach space.

Fix $$t \in \Bbb{R}$$ and let $$\epsilon > 0$$.

Note first that $$\lim_n S_n'(t) = g(t)$$, so there is $$N_1$$ such that $$n \geq N_1 \implies \Vert S_n'(t)-g(t)\Vert < \epsilon/3$$

Also, choose $$N_2$$ such that $$n \geq N_2 \implies \sum_{k=n+1}^\infty\frac{\Vert a \Vert^k}{k!} k (|t|+1)^{k-1} < \epsilon/3$$

Put $$n:= \max \{N_1, N_2\}$$. Choose $$\delta> 0$$ such that $$0 < |s-t| < \delta \implies \left\Vert \frac{S_n(s)-S_n(t)}{s-t}- S_n'(t)\right\Vert< \epsilon/3$$

Then for any $$s \neq t$$ with $$|s-t| < \delta \land 1$$, we have $$\left \Vert\frac{f(s)-f(t)}{s-t}- g(t)\right\Vert$$ $$\leq \left\Vert\frac{S_n(s)-S_n(t)}{s-t}-S_n'(t)\right\Vert+\Vert S_n'(t)-g(t)\Vert + \frac{\Vert R_n(s)-R_n(t)\Vert}{|s-t|}$$

But $$\left|\frac{s^k-t^k}{s-t}\right|= |t^{k-1}+ t^{k-2}s + \dots + ts^{k-2} + s^{k-1}| \leq k (|t|+1)^{k-1}$$ Hence $$\frac{\Vert R_n(s)-R_n(t)\Vert}{|s-t|}=\frac{\Vert \sum_{k=n+1}^\infty \frac{s^k-t^k}{k!} a^k\Vert}{|s-t|}\leq \sum_{k=n+1}^\infty \left|\frac{s^k-t^k}{s-t}\right|\Vert a\Vert^k/k! < \epsilon/3$$ and we conclude $$\left \Vert\frac{f(s)-f(t)}{s-t}- g(t)\right\Vert < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon$$

We thus have shown that $$f'(t) = g(t) = a f(t)$$ and the proof is done.

Reference: Conway's "Functions of one complex variable I" (I modified the proof I saw there).

• Very nice, it is definitely a +1 from me. Apparently the idea of power series theory being true for Banach Algebras (at least when the coefficients commute) is not that crazy! – JustDroppedIn Jul 9 at 18:00
• Yes, I did some googling around and I found some papers about the theory of analytic functions developed in Banach algebras! Cool stuff! – MathQED Jul 9 at 18:01
• @user745578 If I were you I would be accepting this answer! – JustDroppedIn Jul 9 at 18:02
• It is quite cool. I had an older professor that had a lot of experience in FA who still used Banach spaces of equivalence classes of Cauchy sequences of polynomials to develop distribution theory for students – Duncan W Jul 10 at 9:11

Your intuition is right, we need functionals. This proof uses arguments like those used in the proof of the spectrum being non-empty in Banach algebras. Have a look:

Let $$\tau\in A^*$$. Then $$\tau\circ f:\mathbb{R}\to\mathbb{C}$$ is a continuous function and we have that $$\tau\circ f(t)=\tau(e^{ta})=\tau\bigg(\sum_{n=0}^\infty (ta)^n/n!\bigg)=\sum_{n=0}^\infty \frac{t^n\tau(a^n)}{n!}.$$ (we use continuity and linearity of $$\tau$$).

So $$\tau\circ f$$ is a power series and it converges everywhere, since all the above are well defined. As a power series, this is differentiable and we may differentiate term-by-term, so we have that $$\frac{d}{dt}(\tau\circ f)(t)=\sum_{n=1}^\infty\frac{t^{n-1}\tau(a^n)}{(n-1)!}=\sum_{n=0}^\infty\frac{t^n\tau(a^{n+1})}{n!}$$

Set $$g(t)=af(t):\mathbb{R}\to A$$. Note that for $$\tau\in A^*$$ it is $$\tau\circ g(t)=\tau\bigg(a\sum_{n=0}^\infty\frac{t^na^n}{n!}\bigg)=\tau\bigg(\sum_{n=0}^\infty\frac{t^na^{n+1}}{n!}\bigg)=\sum_{n=0}^\infty\frac{t^n\tau(a^{n+1})}{n!}$$ (we use continuity and linearity of $$\tau$$). Now observe that $$\frac{d}{dt}(\tau\circ f)(t)=\lim_{h\to0}\frac{\tau(f(t+h))-\tau(f(t))}{h}=\lim_{h\to0}\tau\bigg(\frac{f(t+h)-f(t)}{h}\bigg)=\tau(f'(t))$$ by continuity and linearity of $$\tau$$. By the above we get $$\tau(f'(t))=\tau(g(t))$$ for all $$t\in\mathbb{R}$$ and all $$\tau\in A^*$$. By Hahn-Banach we conclude that $$f'(t)=g(t)$$ for all $$t$$ and we are done.

• In your last calculation, didn't you assume that $f'(t)$ exists? – user745578 Jul 9 at 15:15
• I guess we need something like $\tau \circ f$ differentiable for all $\tau \in A^*$ implies $f$ differentiable. The converse is obvious. – user745578 Jul 9 at 15:19
• @user745578 You're right, I did assume that $f'(t)$ exists. At the moment I am not sure on how to show that it is indeed differentiable, let me see if I can deal with this. – JustDroppedIn Jul 9 at 15:30
• @user745578 Here is a crazy idea: what if power-series theory is also valid for functions taking values in a Banach algebra? Skimming through the proofs it seems to me that this is actually the case, so this would also explain what the author (I guess Murphy) means by simply saying "differentiating term-by-term". Maybe if something could go wrong it would have to do with commutativity, but let's assume that we have a power series with commuting coefficients, our case falls in here. – JustDroppedIn Jul 9 at 15:52
• Hmmm, well I will have to look into my complex analysis books and see if the proofs generalise. Thanks for the help/ideas! – user745578 Jul 9 at 16:10

We can also use some integration theory on Banach spaces. We have $$f'(t) = \lim_{t\to t_0} \frac{f(t)-f(t_0)}{t-t_0} = \lim_{t\to t_0} \sum_{n=0}^\infty \frac{t^n-t_0^n}{t-t_0} \frac{a^n}{n!}$$

Now, for every $$t \in [t_0-1, t_0+1]$$ by the mean value theorem we can dominate $$\left\|\frac{t^n-t_0^n}{t-t_0} \frac{a^n}{n!} \right\| \le \left|\frac{t^n-t_0^n}{t-t_0}\right| \frac{\|a\|^n}{n!} \le n(t_0+1)^{n-1} \frac{\|a\|^n}{n!}$$ which is an integrable function since $$\sum_{n=0}^\infty n(t_0+1)^{n-1} \frac{\|a\|^n}{n!} \le \|a\|\exp((t_0+1)\|a\|) < +\infty.$$

Therefore, by the Lebesgue Dominated Convergence Theorem we have $$f'(t) = \sum_{n=0}^\infty \lim_{t\to t_0}\frac{t^n-t_0^n}{t-t_0} \frac{a^n}{n!} = \sum_{n=0}^\infty nt_0^{n-1} \frac{a^n}{n!} = a\exp(t_0a) = af(t).$$