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Let $i : X \hookrightarrow Y$ be a dense embedding of complex Hilbert spaces.

Let $f : \mathbb{D} \to X$ be a function, such that $i \circ f$ is holomorphic ($\mathbb{D}$ is the open unit disk). Is $f$ necessarily holomorphic?

Here is what I know so far:

  • By a standard Baire argument, $f$ is holomorphic on a dense open subset $U \subset \mathbb{D}$.
  • There are counterexamples for $X$ Banach.
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No. This is stated in Arendt and Nikolski [Vector-valued holomorphic functions revisited. Math. Z. 234 (2000), no. 4, 777–805] Theorem 1.6:

Let $X$ be a Banach space which is continuously embedded into another Banach space $Z$. If $X$ is not closed in $Z$, then there exists a function $f : \mathbb D \to X$ which not holomorphic, such that $f:\mathbb D \to Z$ is holomorphic.

They write that this is due I. Globevnik for the case $X = \ell_p , Z = \ell_q$ and to Wrobel in the general case, however with a more complicated proof.

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