Define the Newtonial potential in a bounded domain by

If is bounded and integrable on , then . And if is bounded and locally holder continuous in then and . But when is only continuous, is not necessarily secondly differentiable. Here is a counterexample.

Let be a homogeneous harmonic polynomial of degree 2. Suppose for some multi-index , for instance . Choose a cut-off function with when . Denote , and let as and divergent. Define

Prove is a continuous function but has no solution near the origin.

If , then . If , then there exists only one such that . Since is a harmonic polynomial and has compact support,

So . As , , so , which means is continuous at 0.

Let , then it is easy to prove is well defined and for and the unique such that ,

(1) means but , otherwise which does not exist.

Suppose there exists is a solution at , (2) means that on . Since is bounded, by the removable singularity theorem, is a harmonic function in , thus an analytic function. However this means is a function on . Contradiction.

Gilbarg Trudinger’s book. Chapter 4. Exercise 4.9

is not bounded in general, because can not be proved bounded directly. But we have . In fact suppose , some , then

Because ,

So as . Then one can use the Removable singularity theorem(Bochner), or the conclusion of 3.7 in GT’s book.

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## Comments

I think the equation 1 might be off. With w as you defined, we have that

$w = t_0^{-2}c_0P(t_0x) + \cdots + t_{k_0}c_{k_0}P(t_{k_0}x)$. When you take derivatives, the $P$ term shouldn’t go away.

No. It should go away. Cuz p is haromic polynomial of degree 2. Take P=x_1x_2 and \alpha=(1,1) for example.

You are right. I have changed the expression (1). Thank you very much.