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).

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)?

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 $X$.

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?