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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?

thanks

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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
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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

1 Answer 1

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 $\mathbb{C}$.'. –  sos440 Apr 17 '11 at 13:05
    
Thank you all for the help. –  A.A Apr 17 '11 at 13:51

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