# holomorphy in infinite dimensions (holomorphic families of operators)

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?

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  and .

References:

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

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

• The following paper of Bonet, Frerick, and Jordá Extension of vector-valued holomorphic and harmonic functions, Studia Math. 183 (2007), no. 3, 225–248 could also interesting for you. – Jochen Wengenroth Oct 10 at 15:13

It has long been known that the result is true in the Banach space situation---even without local boundedness and under the condition of holomorphicity for the weak operator topology. In the more general context, you will have problems associated with non completeness of the operator space but it will be true with fairly weak conditions on the underlying lcs. A good reference is Grothendieck's masterpiece on analyticity for vector space valed functions.

• (1) Excuse my ignorance: what is the exact title of the Grothendieck reference? (2) Concerning "it will be true with fairly weak conditions on the lcs", may I ask for the precise conditions you have in mind? – André Henriques Oct 9 at 14:08
• (1): Presumably Sur certains espaces de fonctions holomorphes. I and II? – André Henriques Oct 9 at 14:17