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This question already has an answer here:

In reading a book All the mathematics you missed I came across this line:

(The Cauchy Riemann equations coupled with the condition that the partial derivatives are continuous are sufficient to prove a function is analytic.) This extra assumption regarding the continuity of the various partials is not needed, but without it the proof is quite a bit harder.

Now I am pretty sure that there is some mistake here because the Cauchy Riemann equations alone cannot guarantee analyticity. Can someone clarify whether the book is wrong? It says the proof is quite a bit harder without this extra requirement. What proof is the author alluding to?

Here a function from an open subset of $\mathbb{C}$ to $\mathbb{C}$ is being considered. No other assumption is given.

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marked as duplicate by Jack, hardmath, Arnaud D., Daniel W. Farlow, suomynonA Jan 8 '17 at 17:48

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He is talking about the following theorem:

Looman–Menchoff theorem, if f is continuous, u and v have first partial derivatives (but not necessarily continuous), and they satisfy the Cauchy–Riemann equations, then f is holomorphic.

I haven't heard about this, but simply get it from wiki.

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  • $\begingroup$ Okay. Although he omits saying that $f$ is continuous. So I assume that is a mistake in the book? $\endgroup$ – Shahab Feb 21 '14 at 2:11
  • $\begingroup$ @Shahab: Yes, one needs $f$ to be continuous. Or, at least, one needs more than just the Cauchy-Riemann equations alone. For an example and a thorough discussion, see here: en.wikipedia.org/wiki/… $\endgroup$ – Jesse Madnick Feb 21 '14 at 4:58
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Remember, if a complex function satisfies the CR equations, it is harmonic. Harmonic functions possess derivatives of all orders.

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  • $\begingroup$ Thanks, But I am not sure that this helps. I am talking about complex analyticity. $\endgroup$ – Shahab Feb 21 '14 at 2:12
  • $\begingroup$ @ncmathsadist: "if a complex function satisfies the CR equations, it is harmonic." This is true if the function is $C^2$, but I'm not sure if it's true in general... $\endgroup$ – Jesse Madnick Feb 21 '14 at 2:32

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