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This question is related to Holomorphic Parameter Integral, and I want to make sure that we do really have to require that the map $(w,z) \mapsto \partial{f}/\partial{z}(w,z)$ is continuous. Here are the details:

I am looking for a path $\gamma\colon[0,1] \to \mathbb{C}$, some open subset $U\subset \mathbb{C}$, and a continuous function $f\colon \gamma([0,1]) \times U \to \mathbb{C}$ such that for every fixed element $w \in \gamma([0,1])$ the induced map $z \mapsto f(w,z)$ is holomorphic but the map $(w,z) \mapsto \partial{f}/\partial{z}(w,z)$ is not continuous on $\gamma([0,1]) \times U$.

Community
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user363120
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1 Answers1

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Such an example does not exist. Fix $z_0\in U$, and choose $\varepsilon > 0$ so small that $B_{2\varepsilon}(z_0) \subset U$. Then we have

$$\frac{\partial f}{\partial z}(w,z) = \frac{1}{2\pi i} \int_{\lvert \zeta - z_0\rvert = \varepsilon} \frac{f(w,\zeta)}{(\zeta - z)^2}\,d\zeta$$

for $\lvert z-z_0\rvert < \varepsilon$, and the integral is a continuous function on $\gamma([0,1]) \times B_{\varepsilon}(z_0)$ since $f$ is uniformly continuous on the compact set $\gamma([0,1]) \times \overline{B_\varepsilon(z_0)}$.

Daniel Fischer
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  • For $(w_0,z_0)$ and $(w,z)$ with $\vert z-z_0 \vert < \varepsilon$, we have $\vert \frac{\partial{f}}{\partial{z}}(w,z) - \frac{\partial{f}}{\partial{z}}(w_0,z_0) \vert \leq 1/(2\pi) \int 1/(\varepsilon)^2 \vert f(w,\zeta) \frac{(\zeta - z_0)^2}{(\zeta - z)^2} - f(w_0, \zeta) \vert$. How do I get rid of that factor $\frac{(\zeta - z_0)^2}{(\zeta - z)^2}$ so that I get something of the form $\vert f(w,\zeta)-f(w_0,\zeta) \vert$? – user363120 Oct 23 '16 at 16:14
  • Sorry, can't edit that comment: How do I get rid of that factor $\frac{(\zeta - z_0)^2}{(\zeta - z)^2}$ so that I get something of the form $\vert f(w,\zeta) - f(w_0,\zeta) \vert$? – user363120 Oct 23 '16 at 16:21
  • Let's use the subscript $1$, since I used $z_0$ in the answer: Write $$\frac{f(w,\zeta)}{(\zeta - z)^2} - \frac{f(w_1,\zeta)}{(\zeta - z_1)^2} = f(w,\zeta)\biggl(\frac{1}{(\zeta - z)^2} - \frac{1}{(\zeta - z_1)^2}\biggr) + \frac{f(w,\zeta) - f(w_1,\zeta)}{(\zeta - z_1)^2}.$$ For the first term, use that $f$ is bounded on $\gamma([0,1])\times \partial B_{\varepsilon}(z_0)$. When $\lvert z - z_1\rvert < (\varepsilon - \lvert z_1 - z_0\rvert)/2$, we can bound that integrand by $\frac{K\lvert z-z_1\rvert}{(\varepsilon - \lvert z_1 - z_0\rvert)^4}$, and the integral is hence bounded … – Daniel Fischer Oct 23 '16 at 16:29
  • … by $C\lvert z-z_1\rvert$ for a suitable constant $C$. For the second term, use the uniform continuity of $f$ to see that the corresponding integral tends to $0$ as $w \to w_1$. If you know the dominated convergence theorem, applying that is shorter. – Daniel Fischer Oct 23 '16 at 16:30