Assume $u$ is harmonic in $U^{+}$, $u\equiv0$ on $\partial U^{+}\cap\mathbb{R}^{n}_{+}$, and $u\in\mathscr{C}^{2}(\bar{U}^{+})$, where $U$ is the open ball $B_{1}(0)$ of radius $1$ about the origin in $\mathbb{R}^{n}$, $U^{+}$ being the upper half-ball: $U^{+}:=U\cap\partial\mathbb{R}^{n}_{+}.$
(This is problem #2.5.-something in Evans PDE text).
We want to show (under the assumed regularity of $u$) that the odd extension of $u$ into $U^{-}$ provides us a with a harmonic function on all of $U$. That is, if $v=u$ in $U^{+}$, $v=0$ on $\partial U\cap\mathbb{R}^{n}_{\pm}$, and $v=-u(-x)$ in $U^{-}$, then $v$ is harmonic and $\mathscr{C}^{2}$ in all of $U$.
Okay, it is obvious $v$ is $\mathscr{C}^{2}$ in the separated sets $U^{+}$ and $U^{-}$. Since $u$ is $\mathscr{C}^{2}$ upto the boundary of $U^{+}$ (in particular upto $\bar{U}\cap\mathbb{R}^{n}_{+}$), then it is also clear that $v$ is $\mathscr{C}^{2}$ in all of $\bar{U}$. We also see that $v$ satisfies the mean-value-properties in $U^{+}$ and $U^{-}$, and also on $U\cap\mathbb{R}^{n}_{\pm}$ 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 $u$ are satisfied on $U^{+}$, $U^{-}$ and $\partial U\cap\mathbb{R}^{n}_{+}$, yes. But only when viewed individually. How do you use the fact that $v\in\mathscr{C}^{2}(\bar{U})$ to then show that the mean-value property is satisfied in all of $U$ (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 $x\in U^{+}$ with radius sufficiently large to intersect all three sets, but sufficiently small to remain in $U$).
I will reiterate this: every proof I have seen does not make explicit reference to the $\mathscr{C}^{2}$ regularity of $v$. If the mean-value property can be demonstrated without $\mathscr{C}^{2}$ regularity, then all one needs is $\mathscr{C}$ regularity (not even differentiability) of $v$ in order to conclude $v$ 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 $u$ being $\mathscr{C}^{2}$ in $\bar{U}^{+}$, and thus $v$ $\mathscr{C}^{2}$ in $\bar{U}$?
NOTE: In part (b) of this problem, Evans drops the hypothesis that $u$ is $\mathscr{C}^{2}$ upto the boundary, only that $u\in\mathscr{C}^{2}(U^{+})\cap\mathscr{C}(\bar{U})$. But the suggested proof is entirely different: apply the Poisson integral formula for harmonic functions on a disc. Indeed, one solves the problem
$$\left\{\begin{array}{rl}
\Delta w=0&\text{in}\;U\\
w=g&\text{on}\;\partial U,\end{array}\right.$$
where $g(x)=u(x)$ on the upper boundary and $g(x)=-u(-x)$ on the lower boundary. The solution is given by the Poisson integral formula, and computing $w(x^{+})$ where $x^{+}\in\mathbb{R}^{n}_{+}\cap U$, we find $w(x^{+})=0$. From uniqueness, we conclude that $w(x)=v$ as above (the odd extension of $u$), 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!
Best Answer
The following property (sometimes dubbed as local mean value property): $$\forall x\ \exists \delta_x\ \text{s.t.}\ \forall \delta<\delta_x,\ u(x)=\frac{1}{\lvert B(x, \delta)\rvert}\int_{B(x, \delta)} u(y)\, dS$$ is equivalent to harmonicity. The claim can be proved in terms of this property.
EDIT 2018.
We want to check that $v$ satisfies the local mean value property, $v$ being the odd extension to $U$ of the harmonic function $u$, initially defined on $U^+$ only. If $x\in U^+$, then there is a $\delta_x>0$ such that $B(x,\delta)\subset U^+$. Since $u$ is harmonic on this ball, there is no problem in checking the mean value property here. The same thing happens if $x\in U^-$.
Finally, if $x$ lies on the boundary $\partial U^+\cap\partial U^-$, then for all balls $B(x, \delta)$ we have that $$ \int_{B(x, \delta)}v(y)\, dS = \int_{B(x,\delta)\cap\mathbb R^n_+}u(y)\, dS +\int_{B(x,\delta)\cap \mathbb R^n_-}-u(-y)\, dS =0. $$ Since $u(x)=0$ by assumption, the mean value property is satisfied again.