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Let $\Omega \subset \mathbb{C}$ an open set and $K \subset \mathbb{R}^n$ a compact set. Let $v: \Omega \times K \to \mathbb{C}$ be a continuous function with continuous partial derivatives $v_x, v_y$. Suppose that for all $y \in K$we have $v(\cdot, y) \in H(\Omega)$.

Consider $$u(z) = \int_K v(z,y)dy$$ for $z \in \Omega$.

  1. How do I prove that $u$ is holomorphic?
  2. Can we relax the assumptions (on the sets or on $v$) and obtain the same result?

1 Answers 1

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Assume that $v : \Omega \times K \to \Bbb{C}$ is a jointly measurable function such that

  1. $v$ is locally bounded, or equivalently, for each compact $L \subset \Omega$ we have $\sup_{L\times K}|v| < \infty$,
  2. $v(\cdot, y)$ is holomorphic on $\Omega$ for each $y \in K$.

(Notice that your assumption implies all these conditions.) Then

$$ u(z) = \int_{K} v(z, y) \, dy $$

is holomorphic. This is essentially because $u(z)$ is a linear combination of holomorphic functions. And all the conditions above are cooked up so that the integral is regular enough to be interpreted as linear combination of $v(\cdot, y)$'s.

Indeed, let $\gamma : [0,1] \to \Omega$ be any piecewise $C^1$-loop on $\Omega$. Then the image $\gamma([0,1])$ is a compact subset of $\Omega$ so that the function $v$ is bounded on $\gamma([0,1]) \times K$. Since $\gamma'(t)$ is integrable, we can apply the Fubini's theorem to get:

\begin{align*} \int_{\gamma} u(z) \, dz &= \int_{0}^{1} u(\gamma(t))\gamma'(t) \, dt \\ &= \int_{0}^{1} \left( \int_{K} v(\gamma(t), y) \gamma'(t) \, dy \right) \, dt \\ &= \int_{K} \left( \int_{0}^{1} v(\gamma(t), y) \gamma'(t) \, dt \right) \, dy \\ &= \int_{K} \left( \int_{\gamma} v(z, y) \, dz \right) \, dy = 0 \end{align*}

since $\int_{\gamma} v(z, y) \, dz = 0$ by analyticity of each $v(\cdot, y)$. Therefore by the Morera's theorem, $u$ is holomorphic on $\Omega$.