Suppose is the fundamental solution of laplace equation.

We know it has property

We can define the Newtonianial potential of

is well defined. We have the following propery of

**Thm:** Suppose is a bounded domain, . Assume is bounded and local integrable in , then is and

**Proof:** Fix , let and .

Since is uniformly bounded in , by the Lebesgue dominating theorem, is differentiable and

Considering , we will prove if

where .

For any , . So

Similarly for , we also have

Considering , we have

Combing the fact about , we get

So

Applying the fact that is differentiable and the fact about

Let , we get exists and

Next we shall prove is continuous for , then

By the dominating theorem

where . So is continuous everywhere.

**Remark:** This revised version also have errors. It is a shame. What do you mean is differentiable? Don’t you notice that the integral domain of also has relation with ?

To prove this theorem rigorously and neatly, we need the following lemma from calculus

**Lemma:** Suppose is a sequence of differntiable function on . If converges to another function uniformly( is finite somewhere) and converges uniformly to in then is differentiable and .

**Proof:** Let and , , .

Define

Let Since

which is in . By the Lebesgue Differentiable theorem, is differntiable and

So converges to uniformly on any compact subset of . And it is easy to prove converges to uniformly on any compact subset of . So by the lemma, we have is differentiable and has

## Comments

Why is it that we can only conclude uniform convergence on compact sets at the end, and not uniform convergence on all of R^n?

I think we do not define Nf on all R^n. we need to make some assumption of f in order to define the convolution.

Hmm..but doesn’t the statement of the theorem say that Nf is in C^1(R^n)? Also, in the third line of your proof, you define the domains of integration to be subsets of Omega, so I don’t think there’s a problem in defining the convolution for all of R^n.

btw is your source from Gilbarg and Trudinger? I ask because there’s a very similar derivation in that book and they also conclude uniform convergence in compact subsets of R^n. It’s a rather minor issue in the context of this proof imo and doesn’t change the result but still just curious…

Thank you for pointing that out. I was sloppy. Here the second proof is mostly copied from GT’s book.

Firstly, Nf is define for whole R^n, but f is defined in a bounded domain(my previous comment just means this).

Secondly, if x is strictly away from \Omega, then Nf is differential obviously. The convolution has no singularity near x.

Last, I think the convergence can be made uniformly on R^n. Take a larger domain than \Omega and D_{x}w_\epsilon will converge to DNf. Outside that domain, you have control obviously. I think that works, but I did not really write it down.