Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
2  
It is sufficient to check the mean value property locally. That is, the following property is equivalent to harmonicity: $$\forall x,\ \exists \delta_x\ \text{s.t}\ \forall \delta<\delta_x, u(x)=\frac{1}{\lvert B(\delta)\rvert}\int_{B(x, \delta)} u\, dS$$ –  Giuseppe Negro Feb 13 '13 at 23:14
    
Hmm...in that case, why the regularity assumptions (in either parts)? All that's needed is continuity... –  Taylor Feb 14 '13 at 2:36
1  
I now see why satisfying the mean-value property locally is sufficient, for in the proof one smoothes out $u$ using the standard radial mollifier $\eta_{\delta}$, and then proceeds to show that $u$ is actually equal to $u_{\delta}$ in some subset of $U_{\delta}\subset U$ which converges to $U$ as $\delta\to0$. The proof requires continuity of $u$ because of the use of the co-area formula (specifically, polar coordinates); however, the point where the mean-value property of $u$ is used, it is only needed locally in a ball of radius $\delta$! Interesting. –  Taylor Feb 14 '13 at 2:55
    
Of course, once one has $u=u_{\delta}$, one also has $u\in\mathscr{C}^{\infty}(U_{\delta})$, and can easily prove by contradiction that $\Delta u\equiv0$ in $U_{\delta}$. If you post your comment as an answer, I can accept it. I suppose the answer to my question is just that the authors assume too much regularity...or just said $u$ is $\mathscr{C}^{2}(\bar{U})$ in order to immediately have $\Delta u$ (and $\Delta v$) both exist, so the proof by contradiction can proceed without a smoothing argument as above. –  Taylor Feb 14 '13 at 2:59
1  
I agree with all of your comments. IMHO the author just didn't want to mess with symbols like $C^2(U)\cap C^0(\bar{U})$. You do need continuity up to the boundary, though. As Lukas Geyer points out here, harmonic functions may approach boundaries in a weird way. –  Giuseppe Negro Feb 14 '13 at 16:13
add comment

3 Answers

up vote 1 down vote accepted

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.

share|improve this answer
add comment

I am now wondering about this question too. I just want to give some of my ideas. I don't think the local mean value property is sufficient for the harmonicity. This is because in the proof of "MVP implies harmonicity", we prove the regularity of $u$ by proving that $u(x) = u_\delta(x)$ for some $\delta$. This is true. BUT you may notice that if only local MVP is satisfied, we may not choose a uniform $\delta$ for every $x\in \Omega_\delta \subset \Omega$. The identity $u(x) = u_\delta$ is valid pointwisely with different $\delta$ for different $x$. Therefore we can not use $u_\delta(x)$ is smooth to say that $u$ is smooth.

share|improve this answer
    
Nobody said it was trivial to prove! But it is true that the local mean value property is equivalent to harmonicity. Right now I cannot remember the details, but I can point to a source which I find very good: axler.net/HFT.html –  Giuseppe Negro Mar 18 '13 at 16:50
1  
@Giuseppe Negro You are quite right. Thank you for the good reference. I took it for granted that we can just modify the proof of "MVP implies harmonicity" to show "local MVP implies harmonicity". –  Unknown Mar 19 '13 at 3:15
2  
For anyone who is also confused with this problem, see Theorem 1.24 in axler's book HFT. –  Unknown Mar 19 '13 at 3:16
    
I understand your point, and thanks for posting your answer, especially since it led to the citation of a free online text; I had no idea Sheldon Axler had other text books besides his linear algebra one, much less on harmonic functions! With respect to this problem, I think anyone that has read through the responses/question will easily be able to finish the proof with a continuity argument. In either case, for purposes of this problem, it is solved by part (b) using the Poisson kernel anyhow. –  Taylor Mar 21 '13 at 7:38
add comment

Me neither. Today my teacher of PDE gives me this question. At least, by the normal method presented in the book I own, local mean value property fails to justify the harmoniticity of the function.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.