Let $u$ be a non-negative subharmonic function on the unit ball in $\Bbb{R}^n$. Does it follow that there exists a radial limit (including limits of infinity or negative infinity) along almost every line through the origin.

  • $\begingroup$ Is this relevant? en.wikipedia.org/wiki/… $\endgroup$ – András Bátkai Sep 12 '13 at 19:58
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    $\begingroup$ Positive harmonic functions have radial limits a.e. Bounded subharmonic functions also have. So the question makes sense. $\endgroup$ – Alexandre Eremenko Sep 13 '13 at 4:34

The answer is no, even in dimension $2$. Consider a simply connected region $D$ in the unit disc which approached the boundary with infinite spiraling (that is such region where a continuous branch of the $\arg z$ exists and tends to $+\infty$ as $|z|\to 1$ in the region). Let $u=Re f$, where $f$ is a conformal map of $D$ onto the right half-plane, so that every boundary point of $D$ inside the unit disc goes to the imaginary axis. Alternatively, $u$ is the positive harmonic function in $D$ zero on the part of the boundary of $D$ which is inside of the unit disc.

Extend this $u$ to the rest of the disc by setting it to $0$. The resulting subharmonic function has no radial limit on any radius (upper limit is $+\infty$ and lower is $0$).

Similar examples exist in any dimension. Function $u$ exists in any dimension by Martin's construction, for example: $$u(z)=\lim_{|\zeta|\to 1} g(z,\zeta)/g(z_0,\zeta),$$ where $g$ is the Green function of $D$, and $|\zeta|\to 1$ in $D$.

Just take $D$ which intersects every radius nicely.

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