Odd $C^\infty$ function 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.
 A: 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 $g$. 
A: 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.
