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Suppose that $$f\in C^\infty (\mathbb{R})$$ and $f$ is an odd function. ($f(x)=-f(-x)$) What can we say about the zero at zero? Does $f$ have to be of the form $x g(x)$ for some $g\in C^\infty (\mathbb{R})$? I know this is true for complex analytic functions, and I think it is true here, but I don't know how to prove/disprove it.

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You have described an even function, not an odd one. – lhf Apr 20 '11 at 19:51
@lhf ...Typo... – Math Student Apr 20 '11 at 20:00
up vote 8 down vote accepted

Set $g(x) = \begin{cases} f(x)/x \; (x \ne 0) \\ f'(0) \; (x = 0) \end{cases}$. Then check that $g$ has all derivatives at 0, using Taylor polynomials.

Analyticity of $f$ is not needed. If $f$ is real analytic, then so is $g$.

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I don't see how you can check using Taylor polynomials - suppose $f(x) = e^{-\frac{1}{x^2}}$ for $x > 0$ and $-e^{-\frac{1}{x^2}}$ for $x < 0$. Then $f$ is smooth, odd, and all its derivatives are zero - thus its Taylor polynomials are all zero as well. Am I missing something? – MartianInvader Apr 20 '11 at 21:20
@MartianInvader - If $f$ has the (n+1)-th order Taylor polynomial $P(x)$ at $x=0$, then $g$ has the n-th order Taylor polynomial $Q(x) = P(x)/x$. If $Q(x)$ has a term $b_kx^k$, then $g^{(k)}(0) = k!b_k$. Also, $|g(x) - Q(x)| = o(|x|^n)$ as $x \to 0$. This implies that $g \in C^n$, for all $n$. Your example leads to the special case where $P(x) = 0$ for all $n$, but it's not an exception. – Hans Engler Apr 21 '11 at 13:49
Ah, I see. I was thinking the real analyticity would be implicitly used in the proof, but now I see why it's not required. Thanks! – MartianInvader Apr 23 '11 at 12:41

It is actually always true that for any smooth function $f(x)$ in the reals, there exist smooth functions $g,h$ in the reals s.t. $f(x) = g(x^2) + xh(x^2)$. For a proof, I direct you to: L. Hörmander, The Analysis of Linear Partial Differential Operators I, (Distribution theory and Fourier Analysis), 2nd ed, Springer-Verlag, 1990. Exercise 1.2.

I apologize for not proving it in full - I'm a bit rusty. But I happen to have come across this somewhat recently.

However, I can give you a gist of a proof. The idea is very similar to the proof that every function in the reals is the sum of an odd function and an even function. Algebraically, odd and even functions each form a vector space over the real numbers. Then consider a basis argument. Ultimately, because the sum of an odd and an even function is neither, we can eliminate the even function. And so for an odd function $f$, $f = xg(x^2)$.

Note: I use $g(x^2)$ just to emphasize that g is itself even.

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