Suppose $D\subset \mathbb C$ is a smoothly bounded domain and it contains the origin. Let $ds$ denote the arc length measure on $\partial D.$ I am interested in the following two inverse problems (perhaps related to certain moment problem):

A) Suppose for some polynomial $p$ there exists $n_0\in\mathbb N$ so that the following identity holds $$\int_{\partial D}\frac{p(z)}{z^n}ds_z=0$$ for all positive integer $n\geq n_0.$ Does it imply that $\partial D$ must be a disk?

A') With the same assumption as in (A) we have $$\int_{\partial D}\frac{p(z)}{(z-t)^n}ds_z=0 \,\,\,\,\text{for all}\,\,\, t\in\ D$$
holding for all $n\geq n_0$. Does this force $\partial D$ to be a disk?

In here $ds$ denotes the arc length measure.

  • $\begingroup$ As it stands the answer to (A) is clearly no, simply take $p=0$, and $D$ can be anything. $\endgroup$
    – Fan Zheng
    Mar 21 '15 at 0:53
  • $\begingroup$ Lets rule that out. $p=0$ is not really interesting. $\endgroup$
    – BigM
    Mar 21 '15 at 3:54

(A') is equivalent to (A).
Note that (for any given $p$ and $n$) $$F_n(t) = \int_{\partial D} \dfrac{p(z)}{(z-t)^n} ds_z$$ is an analytic function of $t \in D$, and its Maclaurin series is $$ F_n(t) = \sum_{k=0}^\infty {{k+n-1}\choose k} F_{n+k}(0) t^k $$ Thus if $F_n(0) = 0$ for all $n \ge n_0$, we also have $F_n(t) = 0$ for all $n \ge n_0$ and all $t \in D$.

EDIT: It's convenient to invert the domain: let $C = \{0\} \cup \{1/z: z \notin \overline{D}\}$, another smoothly bounded domain. There is a positive measure $d\mu$ on $\partial C$ (not, in general, arc length, but absolutely continuous with respect to it) corresponding to $ds$ on $\partial D$, and $\int_{\partial C} p(1/w) w^n\; d\mu(w) = 0$ for $n \ge n_0$. Now harmonic measure $d\omega_q(w)$ on $\partial C$ for the pole $q \in C$ would have this property for all $n > \deg(p)$ if either $q = 0$ or $p(1/q) = 0$. Is every positive measure on $\partial C$, absolutely continuous wrt arc length, that annihilates $p(1/w) w^n$ for sufficiently large $n$ a linear combination of these? Is it possible to find $D$ (other than a disk) such that $d\mu$ is a linear combination of harmonic measures?


[EDIT]: as pointed out in the comments below, I misunderstood the question (missing the "arc length measure" bit) so that the answer is silly.

Maybe I'm misinterpreting the question, please let me know if this is the case; but I think that the answer is "No". Indeed, by usual analysis in one complex variable, for every $C^1$ curve $\gamma$ that parametrizes your boundary $\partial D$ and every holomorphic function $f$ on $C^*$ you have $$ \int_\gamma f = 2\pi i \text{Res}(f,0). $$ If you take $f(z) = \frac{p(z)}{z^n}$ where $p(z) = \sum a_j z^j$ is your polynomial, then $\text{Res}(f,0) = a_{n-1}$ regardless of the form of your boundary. So I would say that for every boundary and every $p(z)$ you always get 0 provided $n_0 \geq \text{deg}(p)+2$.

The answer for the second one is the same, simply by a change of variables.

(and since you are only interested in very regular functions, namely meromorphic functions, if I remember correctly the $C^1$ condition on the boundary may be dropped, too)

  • 2
    $\begingroup$ In the question $ds$ is the arc length measure. $\endgroup$
    – BigM
    Mar 23 '15 at 13:41
  • $\begingroup$ I'm sorry, I didn't note that in the first version of the question, thanks for the added note. Would you mind pointing to a definition of the arc length measure? Because the only reference I found by googling is page 199 in "The Theory of Subnormal Operators" by John B. Conway (on google books) where it says that the arc-length measure on a (geometric) curve is defined precisely by taking any function $u$ and any parametrization $\gamma$ and imposing $\int_{\partial D}u d\mu = \int_0^1u(\gamma(y))|\gamma'(t)|dt$. This would give the usual notion, thus 0 as in my answer... $\endgroup$ Mar 23 '15 at 14:56
  • 2
    $\begingroup$ @MarcoSpinaci: This definition of arc length measure is correct. It differs from $\int f\, dz = \int f\gamma'\, dt$ (no absolute value here). For the circle, the difference (essentially) disappears. In a sense, this is what the question is about. $\endgroup$ Mar 23 '15 at 17:26
  • $\begingroup$ Ops, sorry, I was indeed careless in reading, answering and commenting, now I see why the answer could (should?) be yes. I'll stop adding nonsense and leave that to people that know what they're talking about! $\endgroup$ Mar 23 '15 at 18:56

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