Is discontinuity along a line equivalent to branch cut?

Question

Suppose I claim the analytic function $f(z)$ has a branch cut along the positive real line, how would one go on to prove this?

Is it sufficient to prove that $f(z)$ is discontinuous across this line?

Thanks

Answer 1

Here is a related result: Using Morera's theorem, if $f$ is continuous across some portion of the cut, then it can be made analytic across the cut, and thus $f$ can be extended across this portion of the cut. So, I suppose that if $f$ is not continuous across any portion of the cut, then the cut behaves like a branch cut... I'm not sure I would call it a branch cut though, as Jonathan points out, there is a specific meaning for that.

Still interesting, I suppose another question could be, is there always a Riemann surface for which $f$ can be continued analytically?

Answer 2

I suppose it depends on exactly what you mean by branch cut, but I would interpret it in the way @Evan does: that $f$ can be extended to some Riemann surface. In that case, the answer is no. Being discontinuous along the line is not enough. In fact, the function at least has to be continuous from both sides, but even that isn't enough.

A fairly simple example is to construct a holomorphic function $f$ on $\mathbb{C} \setminus L$ with zeros accumulating along one (or both) sides of $L$. By the identity theorem, this $f$ can't be extended across $L$, so there is no hope of a Riemann surface extension.