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Littlewood’s one circle problem. (English) Zbl 0804.31001

It is shown that Littlewood’s one circle problem has a negative answer, i.e., there exists a continuous bounded function \(f\) on the unit disk \(U\) such that \(f\) is not harmonic, but nevertheless for every \(x\in U\) the equality \[ f(x)= {\textstyle {1\over {2\pi}}} \int_ 0^{2\pi} f(x+ r(x) e^{it}) dt \] holds for some \(0< r(x)< 1-\| x\|\). – This is achieved by the construction of a subset \(B\) of the unit circle \(U\) (having very small area) and a random walk on \(U\) consisting of jumps from each \(x\in U\) to a concentric annulus \(A_ x\subset U\) having the following properties: (i) The inner and outer radius of \(A_ x\) is a continuous function of \(x\), (ii) starting at 0 the random walk stays in \(B\) and converges to \(\partial U\) with probability almost one, (iii) starting at some point \(x_ 0\in U\setminus B\) close to 0 it hits \(B\) only with small probability. Then the probability that the random walk hits \(B\) arbitrarily close to the boundary is the desired function \(f\). It cannot be harmonic since \(f(0)- f(x_ 0)\) is almost 1.

MSC:

31A05 Harmonic, subharmonic, superharmonic functions in two dimensions
60G50 Sums of independent random variables; random walks
60J65 Brownian motion
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