7
$\begingroup$

Let $f$ be a function analytic on an open subset $D\subset \mathbb{C}$, and let $\gamma:[0,1] \to D$ be a line segment. $g = f\circ\gamma$ is another curve in the complex plane; is it possible to for $g$ to cross a straight line infinitely often, where the crossing points accumulate towards a point? That is, does there exist a point $\alpha$ and a ray $R$ emanating from the point $\alpha$ such that for all $\epsilon > 0$, $g$ crosses the ray $R$ infinitely many times in the $\epsilon$-ball around $\alpha$?

We're trying to show something about analytic continuation, but we cannot rule out pathological beasts like these.

Thanks!

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Oh dear, this question looks as though it has the potential to obsess. I thought I had an answer but it collapsed as I tried to write it down. $\endgroup$
    – gowers
    Commented Apr 22, 2010 at 7:02
  • $\begingroup$ Please define "cross the ray". Maybe I am missing something, but if I take $f(z) = z$ and $R = \gamma$ then $g$ is valued in the ray. Does that count? $\endgroup$
    – BCnrd
    Commented Apr 22, 2010 at 7:19
  • 1
    $\begingroup$ Are you thinking of the case when the inverse image of these crossings converges to a point outside D? $\endgroup$
    – Zsbán Ambrus
    Commented Apr 22, 2010 at 8:57

2 Answers 2

Reset to default
9
$\begingroup$

The image of $[0,1]$ is compact and so must contain the purported accumulation point. It makes no loss to assume that $\gamma(t)=t$, $f(0)=0$ is the accumulation point, and the line in question is the real axis. Then $f(z)=a_n z^n+a_{n+1}z^{n+1}+\cdots$ where $a_n$ is nonzero and $n$ is a positive integer.

At this stage I'll assume there is a sequence of reals $t_1>t_2>\cdots$ tending to zero with each $f(t_j)$ real. we want to show that all the $a_k$ are real. Then considering $f(t_j)/t_j^n$ we get $a_n$ real. Now consider $f(z)-a_n z^n$ in place of $f(z)$. We get $a_{n+1}$ real etc. So $f$ takes reals to reals so there's no "crossing" of the real axis.

In general, there must be a sequence of distinct points $(t_j)$ in $[0,1]$ where the curve crosses the line and which tends (by Bolzano-Weierstrass) to a point $t\in[0,1]$. Replace $[0,1]$ by $[0,t]$ or $[t,1]$ (one of these has infinitely many $t_j$) and rescale the interval to $[0,1]$.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Very nice. My mistake was to try to use the no-isolated-zeros theorem rather than imitating its proof. $\endgroup$
    – gowers
    Commented Apr 22, 2010 at 12:12
  • 1
    $\begingroup$ I follow everything, except for the base step to establish $a_n\in\mathbb{R}$. I presume that it's because $f(z)/z^n$ has an accumulation of real values on values from the $[0,1]$ interval, and since the image of $[0,1]$ contains the accumulation point, $f(0)/0^n = a_n$ must be real valued as well. $\endgroup$
    – Henry Yuen
    Commented Apr 22, 2010 at 17:18
  • 1
    $\begingroup$ Each $f(t_j)/t_j^n$ is real. As $j\to\infty$, $t_j\to0$ and so $f(t_j)/t_j^n\to a_n$. As the limit of a sequence of real numbers, $a_n$ is real. $\endgroup$
    – Robin Chapman
    Commented Apr 22, 2010 at 17:39
4
$\begingroup$

I am little worried by Gowers' comment as this seems fairly straightforward to me.

Let $L:{\mathbb C} \to {\mathbb R}$ be a nonconstant affine map vanishing on your ray. I assume that by "line segment" you really mean that $\gamma$ is linear (or affine). In any case, as long as $\gamma$ is real analytic (by which I mean that it extends to a real analytic function on some open neighborhood of $[0,1]$), so is $h := L \circ f \circ \gamma$. A real analytic function on a compact interval is either always zero or has only finitely many zeros. If $h$ is identically zero, then your curve will stay on the line containing your ray and can enter or leave the ray only finitely many times. If $h$ is not identically zero, then your curve meets the ray only finitely many times.

Improve this answer
$\endgroup$
2
  • $\begingroup$ My comment does not imply that the question isn't straightforward. I am struggling to understand your argument, however. Are your "real analytic" functions complex-valued? I should add that the fact that I am struggling to understand your argument is compatible with its being clear and correct. $\endgroup$
    – gowers
    Commented Apr 22, 2010 at 12:09
  • $\begingroup$ No, my functions are real valued. The function L has the form L(z) = a Re(z) + b Im(z) + c for some real numbers a, b and c. I am using the fact that if F(z) is complex analytic, then via the usual identification of C with R^2, the induced function G:R^2 -> R^2 given by (a,b) -> (Re(F(a+ib)),Im(F(a+ib))) is real analytic, a result one can see by expanding the Taylor series for F(a+ib). $\endgroup$
    – Thomas Scanlon
    Commented Apr 22, 2010 at 16:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .