# How do harmonic function approach boundaries?

Suppose that $D$ is a domain in $\mathbb{R}^n$ (that is, an open, bounded and connected subset), and that $u$ is an harmonic function on $D$. Let $x_0$ be a point at the boundary of $D$.

Question (vague version): Can we say something about the limit $$\lim_{x \to x_0} u(x)?$$

I guess that the short answer is no: that limit might exist as well as not exist, which happens, for example, if we plug a jumping initial datum into Poisson's integral formula on the ball. Clearly, what is giving us trouble is the possibility that $x$ approaches the boundary point $x_0$ in a tangential way. So here's a first refinement:

Question (refinement 1): Suppose that $D$ has a smooth boundary, so that it makes sense to speak of a normal vector field $\nu$ on $\partial D$. Is it true that the limit along normal lines, that is $$\lim_{\varepsilon \to 0} u(x_0 - \varepsilon \nu(x_0))$$ exists?

More generally:

Question (refinement 2): Can we develop a notion of nontangential limit of some kind so that $$\text{non-t.-}\lim_{x\to x_0}u(x)=u(x_0)?$$

Thank you.

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Do you want these limits to exist at every point or at almost every point? And do you want to assume any kind of regularity, boundedness, or something? –  Lukas Geyer Dec 1 '12 at 0:11
Refinement 1 is stringent: smooth boundary, limit at all points. But what I want to say with refinement 2 is that I'm interested also in results of a different kind, such as limits at almost all points of the boundary and the like. –  Giuseppe Negro Dec 1 '12 at 0:13

Let $D$ be the unit disk in the complex plane, and let $G = D \setminus S$, where $S$ is the spiral given in polar coordinates by $r = 1-e^{-\theta}$, where $0 \le \theta < \infty$. (This is a spiral contained in $D$ which accumulates on the whole boundary $\partial D$.) By the Riemann Mapping Theorem there exists a conformal map $f:D \to G$, and by some standard facts about boundary behavior of conformal maps there exists $z_0 \in \partial D$ such that the image of the radius $\{f(rz_0): 0\le r<1$} accumulates on the whole boundary $\partial D$. Now this implies that $u(z) = \operatorname{Re} f(z)$ is a bounded harmonic function in the unit disk $D$ for which the image of the radius $\{u(rz_0)\}$ accumulates on the whole interval $[-1,1]$, so that in particular the radial limit at $z_0$ does not exist.