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Suppose I have a real-valued function f defined on a complex open connected set D. Am I right in saying that: f is analytic on D if and only if f is n times continuously differentiable on D?

Or does this only apply if f is complex valued?


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What do you mean by $n$? – wildildildlife Apr 17 '11 at 11:54
Sorry, n is just some natural number > 0 – A.A Apr 17 '11 at 12:39
What do you mean by analytic? What do you mean by differentiable? There is a big difference between considering complex differentiability versus real differentiability; the former is much more rigid than the latter. (I seem to remember recently a thread about this, but I can't find it at the moment) – Willie Wong Apr 17 '11 at 12:39
Hmm. Well, I have no control over D; it is either a real open interval or a complex open connected set. I require f(c), ..., f^(n)(c) to exist for certain values of c which are inside D and for the derivatives to be continuous inside D. So instead of saying if D was real that f should be n-times continuously differentiable on D, and saying if D was complex that f should be analytic on D, I thought I'd just say f should be continuously differentiable on D. So I guess by analytic, cts. and differentiable? For differentiabilty, I just need it to exist.. I am not too sure about the differences :| – A.A Apr 17 '11 at 13:54
Complex differentiable functions are much different from real differentiable functions. Whether $D$ is an open interval in the real numbers or it is an open subset of $\mathbb{C}$ makes a huge difference. On the interval, the only notion of differentiability that makes sense is the real one, and it is much much weaker than analyticity. In fact even by having infinitely many derivatives you cannot guarantee analyticity. On a domain in the complex plane, if you consider it as a domain in $\mathbb{R}^2$, you have the – Willie Wong Apr 17 '11 at 14:51

Let $D$ be a region in $\mathbb{C}$. Then $f : D \to \mathbb{R}$ is analytic on $D$ if and only if $f$ is constant. This is easily verified if you consider the Cauchy-Riemann equation.

If we consider a complex-valued function $f : D \to \mathbb{C}$ instead, we obtain the theorem that epitomizes the true nature of complex analysis: $f$ is analytic if and only if $f$ is complex differentiable on $D$. No further assumption on the differentiability of $f$, including $C^1$ condition, is needed. Notice that this drasically constrasts with the situation in the real case, where $C^0 (\Omega) \supsetneq C^1 (\Omega) \supsetneq C^2 (\Omega) \supsetneq \cdots \supsetneq C^{\infty}(\Omega) \supsetneq C^{\omega}(\Omega)$.

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Your first statement is not true (strictly speaking). What you meant to write is "$f:D\to\mathbb{C}$ is a complex analytic function that happens to be purely real if and only if..." If you consider real analytic functions instead, there are plenty of examples. – Willie Wong Apr 17 '11 at 12:28
@Willie Wong - You're right, if we consider $f : D \subset \mathbb{R}^2 \to \mathbb{R}$. A polynomial function with two variables will be a good counterexample. But this interpretation seems unnatural for me in this case, since A.A is considering 'analycitity on a subset of $\mathbb{C}$.'. – Sangchul Lee Apr 17 '11 at 13:05
Thank you all for the help. – A.A Apr 17 '11 at 13:51

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