Assume is harmonic in and , where is the open ball of radius about the origin in , being the upper half-ball:
(This is problem #2.5.-something in Evans PDE text).
We want to show (under the assumed regularity of ) that the odd extension of into provides us a with a harmonic function on all of . That is, if in , on , and in , then is harmonic and in all of .
Okay, it is obvious is in the separated sets and . Since is upto the boundary of (in particular upto ), then it is also clear that is in all of . We also see that satisfies the mean-value-properties in and , and also on because of the odd symmetry.
Here's my problem, and of all the proofs I have seen, this is overlooked. The mean-value properties of are satisfied on , and , yes. But only when viewed individually. How do you use the fact that to then show that the mean-value property is satisfied in all of (not just the three aforementioned sets when the spherical averages are restricted to the individuals sets). In other words, how do you justify the extending of a spherical average across the three sets (say at a point with radius sufficiently large to intersect all three sets, but sufficiently small to remain in ).
I will reiterate this: every proof I have seen does not make explicit reference to the regularity of . If the mean-value property can be demonstrated without regularity, then all one needs is regularity (not even differentiability) of in order to conclude is harmonic (it is easy to prove that a continuous function which satisfies the mean-value property at every point in an open set is harmonic there). But if this were the case, then why would Evans (and other texts where the problem is posed) be insistent on requiring being in , and thus in ?
NOTE: In part (b) of this problem, Evans drops the hypothesis that is upto the boundary, only that . But the suggested proof is entirely different: apply the Poisson integral formula for harmonic functions on a disc. Indeed, one solves the problem
where on the upper boundary and on the lower boundary. The solution is given by the Poisson integral formula, and computing where , we find . From uniqueness, we conclude that as above (the odd extension of ), and the theorem is proved.
Anyway, if anyone could help me fill in the details of the mean-value property argument in the first part, I would appreciate it!