Let $X$ be a Banach space (over $\mathbb C$), and let $\mathcal L(X)$ be its algebra of bounded linear operators.


Let $U\subset \mathbb C^N$ be an open subset, and $f:U\to \mathcal L(X)$ a function that is locally bounded (with respect to the operator norm on $\mathcal L(X)$), and holomorphic when $\mathcal L(X)$ is equipped with the topology of pointwise convergence (the strong operator topology).



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> Does it then automatically follow that $f$ is holomorphic when one equips $\mathcal L(X)$ with respect to the topology of uniform convergence on bounded sets (the norm topology)?
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Let $X$ be a locally convex topological vector space, and let $\mathcal L(X)$ be its algebra of continuous linear operators.


Let $U\subset \mathbb C^N$ be an open subset, and $f:U\to \mathcal L(X)$ a function that is locally bounded, and holomorphic when $\mathcal L(X)$ is equipped with the topology of pointwise convergence. Here, $f$ being locally bounded means that for every compact $K\subset U$ and every bounded $B\subset X$, the set $\{f(z)(x): z\in U, x\in B\}$ is again bounded in $X$.



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> Does it then automatically follow that $f$ is holomorphic when one equips $\mathcal L(X)$ with respect to the topology of uniform convergence on bounded sets?
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holomorphy in infinite dimensions (holomorphic families of operators)
In addition to the information given by user bathalf15320, I think that a bit more information on the Banach space case could be useful:


Here is a very general theorem about vector valued functions:


**Theorem 1.** Let $Y$ be a complex Banach space and let $f: U \to Y$ be locallly bounded. Let $W \subseteq Y'$ be a subset which is norming for $Y$ (or more generally, *almost norming* as defined in [1, p. 779]). If $z \mapsto \langle y', f(z) \rangle$ is holomorphic for each $y' \in W$, then $f$ is holomorphic with respect to the norm on $Y$.


*Reference:* [1, Theorem 1.3].


**Corollary 2.** Let $X$ be a complex Banach space and let $f: U \to \mathcal{L}(X)$ be such that $z \mapsto \langle x', f(z) x\rangle$ is holomorphic for each $x \in X$ and each $x' \in X'$. Then $f$ is holomorphic with respect to the operator norm.


*Proof:* (a) Note that $f$ is automatically locally bounded as a consequence of the uniform boundedness theorem.


(b) Now apply Theorem 1 to $Y = \mathcal{L}(X)$, where $W$ is to the linear span of the set of all functionals on $\mathcal{L}(X)$ of the form
$$
 \mathcal{L}(X) \ni T \mapsto \langle x', Tx \rangle \in \mathbb{C},
$$
where $x \in X$ and $x' \in X'$. qed


What is, however, probably more surprising is the fact that, in Theorem 1, we can replace the assumption that $W$ be (almost) norming with the assumption that $W$ merely separates the points of $X$. This result can be found in [1, Theorem 3.1].


Further information on the topic can for instance be found in [1] and [2].


**References:**


[1] W. Arendt, N. Nikolski: Vector-valued holomorphic functions revisited (Math. Z., 2000)


[2] W. Arendt, N. Nikolski: Addendum to 'Vector-valued holomorphic functions revisited' (Math. Z., 2006)


