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Is there a holomorphic nonzero function on a closed bounded connected subset of $\mathbb{C}$ which has infinitely many zeros on the boundary and at most a finitely many zeros in the interior (open connected subset of the closed set)?

If there are infinitely many zeros in the interior, I am not certain whether the function reduces identically to zero by using limit point compactness and constructing a sequence of points with an accumulation point which is certainly in the closure but may not be in the interior.

For clarity: By "holomorphic on a closed bounded connected subset", I mean holomorphic in the interior and continuous on the boundary since holomorphicity is defined only over open sets. For e.g., consider a nonzero function continuous on the closed unit disc and holomorphic on the open unit disc. Can it have only finitely many zeros on the open disc and infinitely many zeros on the boundary?

Thanks.

t.b.
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Herband
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  • If a function is holomorphic on a closed bounded connected subset $B \subset\mathbb{C}$ then it is holomorphic in some neighborhood of $B$, so the boundary points of $B$ are interior for that neighborhood and, as you note, this cannot be. – Andrew Oct 20 '11 at 13:06
  • @Andrew. By "holomorphic on a closed bounded connected subset", I mean holomorpic in the interior and defined on the boundary since holomorphicity is defined only over open sets. For e.g., consider a nonzero function defined on the closed unit disc and holomorphic on the open unit disc. Can it have only finitely many zeros on the open disc and infinitely many zeros on the boundary? – Herband Oct 20 '11 at 13:14
  • It has to be a bit more precise, say holomorphic on the interior and continuous on its closure, otherwise it is possible just to define the function to be zero on the boundary. – Andrew Oct 20 '11 at 13:19
  • @Andrew, edited. Thanks. – Herband Oct 20 '11 at 13:31
  • For an example of a nonzero function where we have an infinite number of zeros inside the domain one could look at the composition $\sin \circ \phi$, where $\phi$ is a Möbius transformation mapping unit circle to the real axis and $1$ to the point at infinity. This is holomorphic everywhere else but at $1$ and has an infinite number of zeros on the unit circle. By restricting to, say, the circle of radius $2$ centered at the origin we get a non-zero function with bounded domain and infinite number of zeros inside. – J. J. Oct 20 '11 at 13:46
  • @J.J., how would that work? Your function would have an essential singularity at 1, so you'd need to exclude that from the function's domain -- and since the OP asks for a closed domain, excluding 1 will require you also to exclude cofinitely many of the zeroes. – hmakholm left over Monica Oct 20 '11 at 13:50
  • @Henning Makholm: I was mainly answering this part of OP's question: "If there are infinitely many zeros in the interior, I am not certain whether the function reduces identically to zero by using limit point compactness and constructing a sequence of points with an accumulation point which is certainly in the closure but may not be in the interior." I might have misinterpreted his question though, since I assumed that he wants to know if this holds for any open (bounded) domain. – J. J. Oct 20 '11 at 13:53
  • @J.J: Oh, I see. – hmakholm left over Monica Oct 20 '11 at 14:08
  • @math101: by "defined on the boundary" do you mean continuous on the closed set? – robjohn Oct 20 '11 at 15:12
  • @robjohn, yes as I have edited. – Herband Oct 22 '11 at 09:56

1 Answers1

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Consider the union, over all rationals $p/q\in [0,1)$, written in lowest terms, of the line segments starting at $0$, having angle $2\pi p/q$ with the positive real axis, and having length $1/q$. Consider the union of this set with the unit circle and call this set $K$. Then $K$ is a compact, locally connected subset of the Riemann sphere (consisting of the unit circle, the interval [0,1] and countably many "spikes" protruding from zero. Let $U$ be the bounded complementary component of $K$, so $U$ is a simply-connected bounded subset of the plane, with locally connected boundary.

Let $\phi:\mathbb{D}\to U$ be a Riemann map, i.e. a conformal isomorphism between the unit disk and the domain $U$. By the Carathéodory-Torhorst theorem, the map $\phi$ extends continuously to the unit circle. This extension will have uncountably many zeros (one for each "access" to zero from the complement of $K$), but of course the map $\phi$ itself has no zeros.

To get an example of a holomorphic function that has infinitely many zeros, extends continuously to the boundary but has only one zero there (the minimum possible due to continuity) is very easy. For example, restrict the function \sin(z)/z to a horizontal half-strip surrounding the positive real axis, and whose boundary does not pass through any zeros. Precompose with a Riemann map taking the disk to this strip to get the desired map. (This is similar to J.J.'s example as above, but I've divided by z to ensure a continuous extension.)

Edit. It is worth noting that by the F. And M. Riesz theorem, the set of zeros on the boundary has zero one-dimensional Lebesgue measure.

Lasse Rempe
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  • Does the theorem of F. and M.Riesz imply that we cannot have a non constant holomorphic function $f$ in the unit ball, continuous up to the boundary of the ball such that $f(e^{i \theta}) = 0$ for $\theta \in [0,\alpha]$? Or the theorem applies only in your example? Where can I find the proof? – Overflowian Jun 09 '17 at 08:21
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    The theorem states that a univalent function on the unit disc has non-tangential limits almost everywhere, and cannot have the same non-tangential limit at a positive set of angles. You can find the proof in many books; one possible way to prove it is via a simple extremal length argument. – Lasse Rempe Jun 15 '17 at 08:49
  • Do you think that your first example can be modified to create a map $\phi:\mathbb{D}\to U$ which is $C^\infty$ on $\overline{\mathbb{D}}$? –  May 17 '20 at 10:01
  • @Caffeine not in a straightforward manner - you cannot get a bi-holomorphic function with the property that you ask for and having infinitely many zeros on the boundary (since more than two smooth curves cannot touch without crossing). I think one can make an example with countably many zeros using the exponential map. I am sure one with uncountably many zeros should also exist, but one would need to think a bit more carefully. – Lasse Rempe May 21 '20 at 16:42
  • @LasseRempe-Gillen I am not sure I follow on the crossing part: what about $(x,x^2); (x,\frac{x^2}{2}); (x,-x^2)$ at $(0,0)$? –  May 21 '20 at 17:52
  • @Caffeine You are right, I misstated what I meant. :) I mean, if the boundary of a simply-connected domain is smoothly parameterised, then each point is accessible from at most two sides (otherwise there will be an access of opening angle less than 180 degrees, where the curve is definitely not smooth). Hope that makes sense. – Lasse Rempe May 22 '20 at 09:03
  • @LasseRempe-Gillen I think that the curve can be smooth if in the corner every derivative is $\underline{0}$, e.g. $(x\exp(-1/x^2),|x|\exp(-1/x^2))$ is a smooth parametrization of $(t,|t|)$. Unless by smooth you mean a curve with non-vanishing derivative. –  May 22 '20 at 10:08