Derivative of norm on Banach algebra

Question

Let $\mathcal A$ be a unital Banach algebra. I want to consider $f(z):= \vert \vert e^{-zA}Be^{zA} \vert \vert, z\in \mathbb C$ and $A,B \in \mathcal A$.

How can I properly define the derivative of $f$ w.r.t. $z$?

Answer 1

Let $$ B=\begin{bmatrix}0&1\\ 1&0\end{bmatrix}, A=\begin{bmatrix}1&0\\0&2\end{bmatrix}. $$ Then $$ e^{-zA}Be^{zA}=\begin{bmatrix}e^{-z}&0\\0&e^{-2z}\end{bmatrix}\,\begin{bmatrix}0&1\\ 1&0\end{bmatrix}\,\begin{bmatrix}e^{z}&0\\0&e^{2z}\end{bmatrix}=\begin{bmatrix}0&e^z\\ e^{-z} & 0 \end{bmatrix}. $$ So $$ f(z)=\|e^{-zA}Be^{zA}\|^2=\left\|\begin{bmatrix}0&e^z\\ e^{-z} & 0\end{bmatrix}\,\begin{bmatrix}0&e^{-\bar z}\\ e^{\bar z} & 0 \end{bmatrix}\right\|=\left\| \begin{bmatrix} e^{2\mbox{Re}z}&0\\ 0&e^{-2\mbox{Re }z}\end{bmatrix}\right\|\\=\max\{e^{2\mbox{Re}\,z},e^{-2\mbox{Re}\,z}\}=e^{2|\mbox{Re}\,z|}. $$ This function is not analytic. Consider for example $f(0)=1$. You want $$ f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}h=\lim_{h\to0}\frac{e^{2\mbox{Re}\,h}-1}h=\lim_{(x,y)\to(0,0)}\frac{e^{2x}-1}{x+iy} =\lim_{(x,y)\to(0,0)}\frac{(e^{2x}-1)(x-iy)}{x^2+y^2} $$ But it is easy to see that this limit doesn't exist (for instance, approach zero through the imaginary axis and you go to $0$; approach through the real axis and the limit is $2$).

I would say that in most cases where $AB\ne BA$ your function is not analytic (i.e. your limit doesn't exist).