Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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

share|cite|improve this answer
+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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.