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Suppose that $X\subset \mathbb C^n$ is a subvariety (locally given by holomorphic equations) and $f: X\to \mathbb C$ is a function. Suppose that $f$ is

1) continuous,

2) holomorphic on the smooth part of $X$.

Is it true that for any $x\in X$ there exists an open subset $x\in U\subset \mathbb C^n$ and a holomorphic function $F: U\to \mathbb C$ such that $f|_{U\cap X} = F|_{U\cap X}$?

If not, will it be true if in addition $X$ is locally irreducible?

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  • $\begingroup$ I think X would be called an analytic subspace rather than a subvariety. $\endgroup$
    – Avi Steiner
    Commented Oct 30, 2017 at 18:49
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    $\begingroup$ Consider the variety X; y^2=x^3 in C^2 and the function y/x restricted to X . If X has normal singularities what you hope for is true. $\endgroup$
    – Mohan Ramachandran
    Commented Oct 30, 2017 at 18:58
  • $\begingroup$ Thanks for your comment! Can you explain (or give a reference) why what I hope is true if $X$ has normal singularities? $\endgroup$
    – Mikhail
    Commented Oct 30, 2017 at 20:28
  • $\begingroup$ @Mikhail I believe Mohan’s comment is an exercise in Hartshorne. $\endgroup$
    – Avi Steiner
    Commented Oct 31, 2017 at 16:03
  • $\begingroup$ @Mikhail : One needs to talk of X with a sheaf of holomorphic functions. The pair (X,O_X )s normal then f is automatically holomorphic and therefore has local extension . See Grauert Remmert Coherent analytic sheaves page 144. For f to be holomorphic we only need the pair (X,O_X) to be seminormal see Grauert Several Complex Variables vol 7 page 92 . $\endgroup$
    – Mohan Ramachandran
    Commented Oct 31, 2017 at 20:38

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