Suppose that $f$ is a holomorphic function on a domain $D$ in $\mathbb{C}^n$, $\partial D$ is smooth, and $f$ is $C^1$ on $\partial D$. Then, the Bochner-Martinelli formula states that
$f(z) = \int\_{\partial D} f(\zeta) \omega(\zeta, z)$,
where $\omega(\zeta, z)$ is the Bochner-Martinelli kernel. I am wondering if a $C^1$ function satisfying this formula is necessarily holomorphic.


This question arose when I was trying to prove that if a sequence of $C^1$ functions $f\_n:D\rightarrow\mathbb{C}$ satisfies $f\_n\rightarrow f$ uniformly and $\partial\bar{f}\_n\rightarrow0$ uniformly, then $f$ is holomorphic. Is this statement correct?



Inverse of Bochner–Martinelli formula
The result you need is a now classical result of Aronov and Kytmanov [see for example [1] chapter 4 §15.1, p. 161, theorem 15.1]: if $D$ is a bounded domain in $\Bbb C^n$ with piecewise smooth boundary and $f\in C^1(\bar D)$ then
$$
f\in \mathscr{O}(D) \iff f(z) = \int\_{\partial D} f(\zeta) \omega(\zeta, z)
$$
Therefore if you have a sequence of $C^1$ functions $\{f\_n\}\_{n\in\Bbb N}$ which are representable as Bochner-Martinelli integrals converging uniformly to a function $f$ representable in the same way, both each single function in the sequence and their limit function are holomorphic in $D$.


**Note**


Aizenberg and Kytmanov proved that the same result is true also for $f\in C^0(\bar{D})$, provided the boundary of $D$ satisfies a higher regularity requirement (namely $\partial D\in C^2$, see [1] chapter 4 §15.2, p. 162, theorem 15.4 for the details).


**Reference**


[1] Alexander M. Kytmanov (1995) [1992], *The Bochner-Martinelli integral and its applications*, Birkhäuser Verlag, pp. xii+305, doi:10.1007/978-3-0348-9094-6, ISBN 978-3-7643-5240-0, [MR1409816](https://mathscinet.ams.org/mathscinet-getitem?mr=1409816), [Zbl 0834.32001](https://zbmath.org/0834.32001).


