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

Question

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?

Answer 1

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).

Answer 2

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.

Answer 3

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 $$f'(t) = \sum_{n=0}^\infty \lim_{t\to t_0}\frac{t^n-t_0^n}{t-t_0} \frac{a^n}{n!} = \sum_{n=0}^\infty nt_0^{n-1} \frac{a^n}{n!} = a\exp(t_0a) = af(t).$$

Answer 4

An additon to @JustDroppedIn's answer (and the comments below that answer) in the case that $A$ is over the complex numbers:

Let's call a map $f: U \to A$ an open $U \subseteq \mathbb{C}$ strongly holomorphic if it is differentiable in the usual sense, i.e. the limit

$$ \lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0} $$

exists for any $z_0 \in U$. This is the same as Fréchet-differentiability if one identifies $\mathcal{L}(\mathbb{C},A) \cong A$.

Then there is also a notion of weak holomorphicity in the sense that for any $l \in A^*$ the map $l \circ f \in \mathbb{C}^U$ is holomorphic. The surprising result is now

(Dunford) For any map from an open subset of $\mathbb{C}$ to a complex unital Banach algebra weak and strong holomorphicity are equivalent.

The proof essentially uses the Cauchy integral formula. Using this the answer of @JustDroppedIn not only proves the formula for the derivative, but also the differentiability itself. But for all of this magic to hold we of course have to work over the complex numbers (as always).