# Extending a smooth odd function

Say I have a smooth odd function $f(z):\mathbb{C} \to \mathbb{C}$. Then is the function $$g = \begin{cases} \frac{f(z)}{z} & \text{ if } z \neq 0 \\ \frac{d f}{dz} (0) & \text{ if } z = 0\\ \end{cases}$$ also smooth?

The limit of $f(z)/z$ as $z\to 0$ is just $f^{\prime}(0)$ by L'Hospital's rule, so that's where I get the $g(0)$. That also shows $g(z)$ is continuous. From this, I can also show that it is differentiable, but how do I show it is smooth? (I tried taking multiple derivatives, but that got messy really quickly) I know that away from zero, $f(z)/z$ is smooth, that $f(z)/z$ is even, that the derivative of an even function is odd (so is $0$ at $z=0$) and the derivative of an odd function is even. I don't see how this wouldn't be smooth but there might be some weird case. And I certainly don't know how to prove it at any rate.

Or, if my piecewise function above isn't the correct one, but there is another way to extend $f(z)/z$ over $0$ I'd be very appreciative.

• "smooth" is a strange term for complex analysis. please define it.
– zhw.
Jul 31, 2017 at 4:11
• @zhw That's a good question. I mean $C^{\infty}$, but what does that mean? I guess it means $C^{\infty}$ thought of as a function $\mathbb{R}^2\to\mathbb{R}^2$. This function is like a section of a complex vector bundle. So not necessarily holomorphic. Jul 31, 2017 at 16:15

If $f$ is only $C^\infty$ and odd, then you're in trouble right at the beginning. That's because there is no reason to think the complex derivative $f'(0)$ exists. For example, let $f(x+iy) = x-iy.$ Then $f$ is smooth and odd, but $f(x)/x = 1,$ $f(iy)/(iy) = -1.$ Thus $f'(0)$ fails to exist. (In fact, $f'(z)$ fails to exist for all $z\in \mathbb C.$)