Let $H_1$ and $H_2$ be Hilbert spaces and $K(w):H_1 \rightarrow H_2$ be an operator for given $w\in\Omega\subset\mathbb{C}$.(like a resolvent. It would not be resolvent.)
In complex analysis, a function $f:\Omega \rightarrow \mathbb{C}$ is holomorphic if
$$f'(w_0):=\lim_{w\rightarrow w_0} \frac{f(w)-f(w_0)}{w-w_0}$$
exists for every $w_0\in\Omega$. And its limit takes modulus.
However, since an operator valued function $w \mapsto K(w) $ is not complex value, $\left|\frac{K(w)-K(w_0)}{w-w_0}\right|$ cannot be defined.
So, I can GUESS its limit is defined in operator norm, i.e. if $K'(w_0)$ is defined such that
$$\left\|\frac{K(w)-K(w_0)}{w-w_0} – K'(w_0)\right\|_{H_1 \rightarrow H_2} \rightarrow 0 \quad as \quad w \rightarrow w_0,$$ $K(w_0)$ is holomorphic for $w_0 \in \Omega$. In my book, there is no mention about this detail.
By the way, consider the Neumann series,
$$(I-K(w))^{-1} = \sum_k^\infty (K(w))^k.$$
In order to show $(I-K(w))^{-1}$ is holomorphic, I can think about 'analytic' first (since analytic and holomorphic is equivalent in $\mathbb{C}$) and Neumann series say '$K(w)$ is analytic for each $w$'. But, again, it is not complex number.
Questions
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What is definition of 'Operator valued function is holomorphic' ? Is my guess right ?
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Does it suffice to show existence of its Neumann series ? If then, why ?
Best Answer
Let $F : \mathcal{O}\subset\mathbb{C}\rightarrow \mathcal{L}(X,Y)$ where $\mathcal{O}$ is an open subset of $\mathbb{C}$, and $\mathcal{L}(X,Y)$ denotes the bounded linear operators from a complex Banach space $X$ to another complex Banach space $Y$. Then $F$ is holomorphic iff the following limits exist for all $z\in\mathcal{O}$: $$ \lim_{w\rightarrow z}\frac{1}{w-z}(F(w)-F(z)). $$ In that case, the limit is denoted by $F'(z)$.
It can be proved that $F$ is holomorphic iff $z\mapsto F(z)x$ is a holomorphic vector function for all $x\in X$. And $F(z)x$ is holomorphic iff $z\mapsto y^*(F(z)x)$ is a complex holomorphic function for all $y^*\in Y^*$. The proof follows from the uniform boundedness principle, and a power series argument. This remarkable bootstrapping is due to the Uniform Boundedness Principle and a power series argument, where the existence of a complex derivative of a complex function $f : \mathcal{O}\rightarrow\mathbb{C}$ amounts to boundedness of the following for all $w$ in a punctured neighborhood of $z$: $$ \frac{1}{w-z}\left[\frac{f(w)-f(z)}{w-z}-f'(z)\right],\;\;\;w\in B_r(z)\setminus \{z\}. $$