Is composition of analytic functions itself analytic?

Question

Is there a proof that, say,

$$f(x)=e^{\frac{x^2+1}{x^2-1}}$$

analytic?

@Giuseppe Negro and what? these problems do not make it non-analytic and non-holomorphic. — Anixx, Jun 14 '14 at 12:34
Yes, it is still analytic on $(-\infty, -1)\cup (-1, 1)\cup (1, \infty)$. — Giuseppe Negro, Jun 14 '14 at 12:35
See https://math.stackexchange.com/questions/1197737, https://math.stackexchange.com/questions/1761148/ — Watson, Jun 05 '21 at 19:30

score 3 · Answer 1 · answered Jun 14 '14 at 12:37

3

The composition of analytic functions indeed is analytic. The fastest proof surely relies on complex analysis: every analytic function of one real variable is the restriction of a holomorphic function of one complex variable, so the statement is a consequence of the (complex variable) chain rule.

answered Jun 14 '14 at 12:37

Giuseppe Negro

34,508

This requires first the proof that any differentiable complex function is analytic, which I think is a non-trivial theorem. Anyway, what about real analysis? Can this be proven without complex numbers? – Anixx Jun 14 '14 at 12:40
I don't know. I guess that it can, but probably this requires quite a bit of algebra. – Giuseppe Negro Jun 14 '14 at 20:01
What do u think about this? http://math.stackexchange.com/questions/834257/if-two-functions-are-equal-to-their-newton-series-is-their-composition-also-equ – Anixx Jun 14 '14 at 20:09
"every analytic function of one real variable is the restriction of a holomorphic function" - this is not true. Take logarithm for example. – Anixx Oct 20 '21 at 16:34
I don't see why this is a counterexample. There are branches of the complex logarithm that coincide with the real logarithm on $(0, \infty)$. Those are holomorphic. – Giuseppe Negro Oct 20 '21 at 17:13
@Anixx - For real analysis, see my answer to the linked question. – mr_e_man Jun 16 '22 at 18:49
@mr_e_man which one? – Anixx Jun 16 '22 at 19:18
Composition of real-analytic functions is real-analytic – mr_e_man Jun 16 '22 at 19:18
@mr_e_man: you forgot the link: https://math.stackexchange.com/a/4474145/8157 I could find your answer anyway, thank you for the notification. – Giuseppe Negro Jun 16 '22 at 22:00

score -2 · Answer 2 · answered Jun 14 '14 at 12:37

Your example is a composition/combination of several functions $$ u(x) = e^x \\ v(x) = x^2 + 1\\ h(x) = x^2 - 1 \\ m(x, y) = x / y \\ f(x) = u( m( v(x), h(x) ) ) $$ The function $m$ is not very nice when $y = 0$, but $y = 0$ happens to be in the image of $h$.

What are you assuming is the domain of $f$? And are you asking "Is it analytic on that domain?"?

Is composition of analytic functions itself analytic?

2 Answers2