By defining $$e^{A} = \sum_{k=0}^\infty \frac{1}{k!}A^k$$ for any $A \in \{\text{"bounded operators on Hilbert space"}\}$, apparently it is valid to take the derivative of the map $$t\mapsto e^{tA}$$ (where $t\in \mathbb{R}$) via term by term differentiation in the series. I do not see why this is valid.
I am aware of a similar notion where $A$ is instead a matrix, and the validity of term by term differentiation is easier to see in this case as each entry of $e^{tA}$ is just some usual power series in $t$.
—- Update —-
I realize that the explanation of my concern is not precise enough so I will say more in this update. The definition of derivative of any operator valued map $\sigma: \mathbb{R}\rightarrow \mathcal{B(H)}$ (where $\mathcal{B(H)}$ denotes space of bounded operators on a Hilbert space $\mathcal{H}$) is, if exists, given by the familiar limit. Namely,
$$\frac{d}{dt}\sigma(t) := \lim_{h\rightarrow 0} \frac{\sigma(t+h)-\sigma(t)}{h}$$
computed in the operator-norm topology. So in principle one should be able to compute the derivative of $t\mapsto e^{tA}$ and $t \mapsto \frac{t^k}{k!}A^k$ using this definition. What I do not understand is the validity in the following formula\proposition
$$\frac{d}{dt}\left(\sum_{k=0}^\infty \frac{t^k}{k!}A^k\right) = \sum_{k=0}^\infty \left(\frac{d}{dt}\frac{t^k}{k!}A^k\right)$$
How do you show this?
Hope this clarifies.
Best Answer
In the case of power series, most of the results in $\mathbb{R}$ or $\mathbb{C}$ readily extends with due modifications. For example, let
$$ f(z) = \sum_{n=0}^{\infty} A_n z^n $$
be a power series in $z \in \mathbb{C}$, where $A_n$'s lie in a Banach space $(\mathscr{X}, \|\cdot\|)$. If we let
$$ R = \frac{1}{\limsup_{n\to\infty} \|A_n\|^{1/n}} \in [0, \infty], $$
then it is clear that the series converges absolutely if $|z| < R$ and diverges if $|z| > R$. That is, this $R$ is precisely the radius of convergence for $f(z)$. It is also easy to check that
$$ g(z) = \sum_{n=1}^{\infty} n A_n z^{n-1} $$
has the same radius of convergence as $f(z)$.
Now suppose $|z| < R$, and fix $r$ so that $|z| < r < R$. Also, consider any $h$ with $0 < |h| < r-|z|$. Then by using the identity
\begin{align*} \frac{(z+h)^n - z^n}{h} - n z^{n-1} &= n \int_{0}^{1} \bigl[ (z + ht)^{n-1} - z^{n-1} \bigr] \, \mathrm{d}t \\ &= n(n-1)h \int_{0}^{1} \int_{0}^{t} (z + hu)^{n-2} \, \mathrm{d}u\mathrm{d}t, \end{align*}
we find that
\begin{align*} \left| \frac{(z+h)^n - z^n}{h} - n z^{n-1} \right| &\leq n(n-1)h \int_{0}^{1} \int_{0}^{t} (|z| + |h|)^{n-2} \, \mathrm{d}u\mathrm{d}t \\ &= |h| \frac{n(n-1)}{2}(|z| + |h|)^{n-2}. \end{align*}
Now by this estimate and the inequality $|z| + |h| < r$ altogether, it follows that
\begin{align*} \left| \frac{f(z+h) - f(z)}{h} - g(z) \right| &\leq \sum_{n=1}^{\infty} \|A_n\| \left| \frac{(z+h)^n - z^n}{h} - n z^{n-1} \right| \\ &\leq \left( \sum_{n=1}^{\infty} \frac{n(n-1)}{2} \|A_n\| r^{n-2} \right) |h| \end{align*}
and therefore $f'(z)$ exists and is equal to $g(z)$.