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.



For Hilbert spaces, does weak analyticity with respect to a dense subspace of functionals imply analyticity?
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*.


