- Suppose that a convex domain D in the plane with smooth boundary has the following property:
There is a point P in D such that for every arc J contained in the boundary of D, the harmonic measure of J
as seen from P, equals the angle which J subtends, as seen from P, divided by 2π.
Show that D
must be a circle and that the point P is the center.
- Show there is a convex domain D in the plane with smooth boundary
and DISTINCT points P, Q in D for which the following is true:
For every arc J contained in the boundary of D, the harmonic measure of J
as seen from P, equals the angle which J subtends as seen from Q divided by 2π.
The questions 3a) and 3b) are UNSOLVED:
- a) Describe all such domains.
b) Is there an analogous example of a convex domain in R3 ?
Problem 2 can also be generalized as follows:
- Characterize all
simply connected domains D having the property that there exist
finitely many points P1,...,Pn and Q in D such that for every arc
J on the boundary of D the angular measure of J as seen from Q
(divided by 2π) equals a linear combination (independent of J)
of the harmonic measures of J at the points P1,...,Pn
The characterization can be formulated for example in terms of
a conformal map (one-to-one analytic function) from the unit
disk onto D. (This problem is a kind of miniresearch which can be
further developed in different directions.)
Remark: If D is an open, bounded domain in the plane with boundary ∂D the harmonic measure of an arc J in ∂D at a point P of D
is defined as the value at P of the solution u(x,y) to Laplace's equation Δu = 0, with boundary condition u = 1 on J and u = 0 otherwise on ∂D.
Here is another, perhaps more elementary algebraic problem:
4. A Hadamard matrix is, by definition, a matrix whose entries are
from the set {-1,1} and whose rows are mutually orthogonal. Suppose
we have an n x n Hadamard matrix and the upper left
p x q section
has all entries +1. Prove that pq cannot exceed n.
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