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I have a basic question in my mind and wish to consult your ideas:

Suppose $\Omega_1$ and $\Omega_2$ are regions, $f$ and $g$ are nonconstant functions defined in $\Omega_1$ and $\Omega_2$, respectively, and $f(\Omega_1) \subset \Omega_2$. Define $h=g \circ f$. What can we say about the third function if

(a) both $g$ and $f$ are analytic;

(b) both $g$ and $h$ are analytic;

(c) both $h$ and $f$ are analytic.

Here I consider all possible cases.

I think in part (a) $h$ is analytic being the composition of two differentiable functions.

Actually to my mind, analyticity of $g$ implies analyticity of $h$, am I correct ? Otherwise, I can't find counterexamples on each cases. What is your suggestion?

Thank you.

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Regarding your penultimate paragraph, suppose $g$ is the identity function. –  Rahul Jun 14 '12 at 9:47
1  
This has been crossposted to MO –  mixedmath Jun 14 '12 at 10:36
    
This is a great observation! –  беркай Jun 14 '12 at 11:26

1 Answer 1

up vote 3 down vote accepted

(a) Function $g \circ f$ is analytic : standard.

(b) Cannot deduce $f$ analytic: $g=17$, $f$ non-analytic.

(c) Cannot deduce $g$ analytic$: f=17$, $g$ non-analytic.

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+1 for the Spivak reference. –  Joshua Ciappara Jul 24 '13 at 5:21
    
Dear @Joshua, but I didn't mention Spivak ?! –  Georges Elencwajg Jul 24 '13 at 6:49

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