Suppose $\Omega\subset \mathbb{C}^n$ is a nonempty open set and $V\subset \Omega$ is a complex analytic (closed) manifold. Let $f$ be a holomorphic function on $\Omega \backslash V$ and for every $p\in V$, there is a open neighborhood $U\subset \Omega$ of $p$ such that $f|_{U\backslash V}$ is bounded. Prove that f can be extended holomorphicly to $\Omega$.
I have just started to learn complex analysis in several variables and got in touch with the Hartogs extension theorem. I wonder how to prove the above proposition. Can we use the Hartogs theorem?
Appreciate any help!
As suggested in the comment below, I have read the Riemann extension theorem in high dimension. It seems we need to find a function $g\not\equiv 0$ such that $V=g^{-1}(0)$. Is that possible?
There are many versions of higher dimensional Riemann extension theorem, what you are trying to prove is one of classical. Let me just give you couple of hints:
You do not need to know, and even cannot, that the set is globally analytic, i.e. given by zeros of one holomorphic function globally - it suffices that around any $p \in V$ you have $V \cap B_r(p) \subset \phi^{-1}(0)$ for a (non-constant zero) holomorphic function in $B_r(p)$. Is it the case in your situation?
It suffices that you extend your function locally around any $p \in V$ separately - why is that, do different extensions have to agree on the intersection of their domains?
Now try to argue according to 2) locally in a setting of 1) by using one dimensional - isolated singular point - Riemann extension theorem.
More precisely since $V$ is of essential codimension you may assume, for example, that $p=0$ in $\mathbb{C}^n$ and $$\phi(0,...,0,\cdot)$$ has isolated zero as function of just one - last - coordinate. From Rouché’s theorem it fallows that also $$\phi(z_1,...,z_{n-1},\cdot)$$ has only isolated zeros. Actually close to $0 \in \mathbb{C}^{n-1}$ [$(z_1,...,z_{n-1})$ in $C^{n-1}$], those functions have at most the order of the zero of $\phi(0,...,0,\cdot)$ zeros - it is only important that they are not constant zero functions since then $f$ restricted to the intersection of this copies of $\mathbb{C}$ with $V$ woul only have singularities - in particular they would not be isolated. Having all of this you can extend $$f_{|(V\cap B_r(p)) \setminus \phi^{-1}(z_1,...,z_{n−1},⋅)(0)}$$ across this discrete set $\phi^{-1}(z_1,...,z_{n−1},\cdot)(0)$ to say $\tilde{f}(z_1,...,z_{n-1},\cdot)$. The last thing is to show that those fibrewise extensions glue to the holomorphic function in $B_r(p)$ which is up to a moment of thinking - Hartogs’ theorem.