What is 'Operator valued function is holomorphic' ? Does Neumann series imply holomorphic? 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

*

*What is definition of 'Operator valued function is holomorphic' ? Is my guess right ?


*Does it suffice to show existence of its Neumann series ? If then, why ?
 A: I think it is wise to be uneasy about whether or not Cauchy (-Goursat) complex function theory works for operator-valued functions the same as it works for complex-valued functions. Happily (some details below) it does, and in a fashion which is basically idiot-proof! As often happens, the proofs that everything works amazingly well are more sophisticated than the assertions themselves.
(To carefully distinguish, I'd prefer to use complex_differentiable for existence of a complex derivative, and complex_analytic for existence of a (convergent) power series expression.)
First, yes, we can talk about the operator norm topology on operators, making the space a Banach space. But, also, the same discussion applies to the strong operator topology (given by seminorms $\mu_x(T)=\sup_{|x|\le 1}|Tx|$), and other topologies on operators. These other topologies are fancier than Banach spaces, but still do have necessary completeness properties to make vector-valued Cauchy-Goursat theory succeed. Maybe this is not the main point for you, but it does illustrate the robustness of the truth of the analogues.
As in Abel's and others' proofs about convergent power series giving complex differentiable functions... the same argument (with obvious adjustments) proves the same for Banach-space valued power series. (The fact that they're Banach spaces of operators doesn't enter.)
The converse, that complex differentiability of Banach-space valued functions implies a Cauchy theorem, Cauchy formula, etc., is less trivial, but holds. A very general case of vector-valued holomorphic functions was treated by Grothendieck in the early 1950s. Again, some of the robustness of the situation is illustrated by the theorem that a weakly holomorphic $V$-valued function $f$ is holomorphic: that is, if for all continuous linear functionals $\lambda$ on $V$ the composite $\lambda\circ f$ is a holomorphic scaler-valued function, then $f$ is holomorphic $V$-valued. This holds for quasi-complete, locally convex spaces $V$, so applies to Frechet spaces, spaces of test functions, weak duals of these, strong duals, and nearly anything that arises naturally with any reasonable completeness.
Many functional analysis books at least treat the Banach-space-valued case.
A: 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 $z$:
$$
              \frac{1}{w-z}\left[\frac{f(w)-f(z)}{w-z}-f'(z)\right],\;\;\;w\in B_r(z)\setminus \{z\}. 
$$
