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Let $\Omega\subseteq\Bbb C^2$ be open bounded (and connected), $f:\Omega\to\Bbb C$ separately holomorphic (i.e. $f$ is holomorphic in each variable when the other is fixed).

Hartogs theorem is not allowed, so we can't say $f$ is holomorphic on $\Omega$.

Now I think that in this case $f$ is continous in $\bar\Omega$ iff $f$ is integrable in $\Omega$. In other words $f\in\mathcal{C}^0(\bar\Omega)\Leftrightarrow f\in L^1(\Omega)$.

Clearly one implication ($\Rightarrow$) is always true. But the other?

At the moment I can't get no proof nor counterexamples. Can someone help me? Many thanks

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  • $\begingroup$ you are asking integrable implies and is implied by continuous. which is a good question, depends which field you are looking @. $\endgroup$
    – JMP
    Apr 19, 2015 at 13:50
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    $\begingroup$ What do you mean by "Hartog's theorem is not allowed"? $\endgroup$
    – Malik Younsi
    Apr 19, 2015 at 14:20
  • $\begingroup$ I mean that we can't use it; we can't say "the Hartogs thm hyp are ok so we can say $f$ is holomorphic". However, I don't think this is a restriction. Maybe I can delete it. $\endgroup$
    – Joe
    Apr 19, 2015 at 15:27
  • $\begingroup$ @JonMarkPerry: I'm looking for $\Bbb C$; the functions we're dealing with are $f:\Omega\to\Bbb C$, where $\Omega\subseteq\Bbb C^2$, and it is open bounded and connected. $\endgroup$
    – Joe
    Apr 19, 2015 at 15:28
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    $\begingroup$ $f(w,z)=1/z$ on $|w|,|z|<1$, $z\not= 0$ is a counterexample. $\endgroup$
    – Christian Remling
    Apr 19, 2015 at 16:15

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This will not work if $L^1$ refers to area measure: the function $f(w,z)=1/z$ on (let's say) $|w|,|z|<1$, $z\not= 0$ is a counterexample.

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