If B is an open subset of the complex numbers which contains the closed disc {z $\in$ C: $\vert z \vert \leq \Vert x_o \Vert$} and $f \in H^{\infty}(B)$ (bounded holomorphic functions). Then I need to show that
$ f(x_0)= f \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} f(0) & f'(0) & 1/2 f''(0) \\ 0 & f(0) & f'(0) \\ 0 & 0 & f(0) \end{pmatrix} $
If $x_0 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ is in a unital commutative Banach Algebra. I have some trouble getting started though. How would one approach this problem?
Note that $$f(z) = \sum_{n=0}^{+\infty}\frac{f^{(n)}(0)}{n!}z^n$$ around zero. Next if $A$ is a matrix with sufficiently small norm, then $$f(A) = \sum_{n=0}^{+\infty}\frac{f^{(n)}(0)}{n!}A^n$$ The matrix you are considering is nilpotent - its cube is zero. So just use the formula above.