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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.

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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

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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 Carathéodory's 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.

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