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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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    This is a great observation!2012-06-14

1 Answers 1

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(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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    Dear @Joshua, but I didn't mention Spivak ?!2013-07-24