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I'm working through the proof of Theorem 2.4 in Chapter 1 of Stein/Shakarchi's Complex Analysis. I'm looking for clarification on where exactly we use the hypothesis that $u$ and $v$ are continuously differentiable. My guess is that it's used right away when writing $$u(x + h_1, y + h_2) - u(x,y) = \frac{\partial u}{\partial x}h_1 + \frac{\partial u}{\partial y}h_2 + |h|\psi(h)$$ and similarly for $v$, since that equation comes from $u$ being differentiable as a map $\mathbb{R}^2 \to \mathbb{R}$, which we can only say provided we know that the partials of $u$ are continuously differentiable. Is this correct?

Matthew Leingang
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Nick A.
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    continuous differentiability is redundant. Holomorphy of $f:\Bbb{C}\to\Bbb{C}$ at a point is equivalent to being differentiable in the sense of $\Bbb{R}^2\to\Bbb{R}^2$ along with Cauchy-Riemann (at that point), which in turn is equivalent to $u,v$ being differentiable in the sense of $\Bbb{R}^2\to\Bbb{R}$ at that point and $u,v$ satisfying Cauchy-Riemann equations (at that point). (The Cauchy RIemann equations are nothing but the condition that an $\Bbb{R}$-linear transformation $\Bbb{R}^2\to\Bbb{R}^2$ actually give rise to a $\Bbb{C}$-linear transformation $\Bbb{C}\to\Bbb{C}$). – peek-a-boo Apr 06 '21 at 18:10
  • @peek-a-boo Really?! Then why is it that David C. Ullrich, who wrote at Complex Analysis textbook, posted this question here? – José Carlos Santos Apr 06 '21 at 18:20
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    @JoséCarlosSantos I didn't say merely that the partials of $u,v$ exist at that point and satisfy Cauchy RIemann. I said $u,v$ are differentiable at that point (in the sense of Frechet as maps $\Bbb{R}^2\to\Bbb{R}$) and satisfy the Cauchy-RIemann equations. Therefore, what I wrote really is an equivalence, and it's different from what's being asked there. The equivalence I talk about can be found in Henri Cartan's textbook on Differential calculus in Banach spaces (in the section relating complex-differentiability and real-differentiability). – peek-a-boo Apr 06 '21 at 18:22
  • @peek-a-boo, in the context of the theorem being referred to, continuous differentiability isn't redundant. Theorem 2.4 in Stein/Shakarchi states that if $f$ is continuously differentiable and satisfies the Cauchy–Riemann equations, then it is holomorphic, which is what you said. – Frank Apr 06 '21 at 18:26
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    @Frank sure, they may have written down continuous differentiability, but they never actually use continuity. This "Taylor expansion" is pretty much true by definition of differentiability of functions $\Bbb{R}^2\to\Bbb{R}$, no where does it use continuity of the partials. – peek-a-boo Apr 06 '21 at 18:28
  • Oh I see what you mean, thanks for clarifying. – Frank Apr 06 '21 at 18:29
  • By the way, if you take a look at theorem 2 of the linked paper in the accepted answer in the question referenced above by @JoséCarlosSantos, you'll see pretty much what I said (except I phrased everything as a pointwise issue). – peek-a-boo Apr 06 '21 at 18:35
  • @peek-a-boo The way I understand it, we need continuity of the partial derivatives to ensure that $u,v$ are differentiable as you require, so indeed it wouldn't be enough to replace "$u,v$ are continuously differentiable" with "the partial derivatives of $u,v$ exist", right? – Nick A. Apr 06 '21 at 20:19
  • right. The condition we actually need is differentiability of $u,v$ at the given point. This is implied by continuity of partials, but is NOT implied by mere existence of partials. – peek-a-boo Apr 06 '21 at 20:20

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