Do discontinuous harmonic functions exist?

A function, $u$, on $\mathbb R^n$ is normally said to be harmonic if $\Delta u=0$, where $\Delta$ is the Laplacian operator $\Delta=\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}$. So obviously, according to this definition, $u$ must be twice differentiable and therefore continuous. However, if $u$ is at least continuous, being harmonic is equivalent to the condition that the average value of $u$ in any ball centered at a point $p$ is actually equal to $u(p)$: $$u(p)=\frac{1}{Vol(B_r(p))}\int_{B_r(p)}u\, dV.$$ My question is whether there are any integrable functions $u$ for which the above equation holds for all $r$ and $p$ but for which $u$ is not continuous.

• See also en.wikipedia.org/wiki/…. – Hans Lundmark Sep 1 '11 at 7:03
• I don't know much about this, but I would have said that a much stronger result holds. I would have said that a harmonic hyperfunction is automatically analytic... (I'm considering the usual laplacian on $\mathbb R^n$.) – Pierre-Yves Gaillard Sep 1 '11 at 9:02
• @Pierre: $\Phi_r$ in my answer is real analytic on the interior of $B_r(0)$. Therefore, with only a bit more work, my answer shows that $u$ is real-analytic. – robjohn Sep 1 '11 at 12:03
• Dear @robjohn: Thanks! I agree. I hadn't read the question carefully enough. The question doesn't ask explicitly if $u$ is harmonic, and, logically, the answers answer the question as asked. But it seems to me the question "is $u$ harmonic?" is very natural. Wikipedia says: "all locally integrable functions satisfying the (volume) mean-value property are infinitely differentiable and harmonic functions as well". A proof is a sketched. – Pierre-Yves Gaillard Sep 1 '11 at 13:27

On the interior of its domain, a harmonic function is infinitely differentiable (even with just the mean value property). However, on the boundary of its domain, it can limit to a discontinuous function.

Mean Value Property: $u(x)=\frac{1}{\Omega_nr^n}\int_{B_r(x)}u(v)\;\mathrm{d}v$ for any $r>0$ so that $B_r(x)\subset\mathrm{Dom}(u)$, where $\Omega_n$ is the volume of the unit sphere in $\mathbb{R}^n$.

Define $\displaystyle\phi_r(s)=\left\{\begin{array}{cl}e^{s^2/(s^2-r^2)}&\text{for }0\le s<r\\0&\text{for }s\ge r\end{array}\right.\hspace{.25in}$ then $\Phi_r(x)=\phi_r(|x|)$ is in $C_c^\infty(\mathbb{R}^n)$

We can average the Mean Value Property over a range of sphere radii: \begin{align} u(x)&=\frac{1}{\int_0^r\Omega_ns^n\phi_r^\prime(s)\mathrm{d}s}\int_0^r\int_{B_s(x)}u(y)\;\mathrm{d}y\;\phi_r^\prime(s)\;\mathrm{d}s\\ &=\frac{1}{\int_0^r\phi_r(s)n\Omega_ns^{n-1}\mathrm{d}s} \int_0^r\phi_r(s)\int_{S_s(x)}u(y)\;\mathrm{d}\sigma(y)\;\mathrm{d}s \\ &=\frac{1}{\int_{\mathbb{R}^n}\Phi_r(y)\mathrm{d}y} \int_{\mathbb{R}^n}\Phi_r(y-x)u(x)\;\mathrm{d}x\\ &=\frac{1}{\int_{\mathbb{R}^n}\Phi_r(y)\mathrm{d}y}\Phi_r*u(x) \end{align} Since $\Phi_r$ is in $C^\infty$, $u$ must be also.

Thus, the Mean Value Property implies $C^\infty$ on the interior of the domain.

• I should have seen this. I forgot that the convolution of an integrable function with a $C^\infty$ function is also $C^\infty$. – Cheerful Parsnip Sep 1 '11 at 11:10
• @Rohan: it is indeed integration by parts. Before $$v=\int_{B_s(x)}u(y)\;\mathrm{d}y$$ $$dw=\phi_r^\prime(s)\;\mathrm{d}s$$ and after $$w=\phi_r(s)$$ $$dv=\int_{S_s(x)}u(y)\;\mathrm{d}\sigma(y)\;\mathrm{d}s$$ The minus sign is cancelled by the integration by parts in the denominator out front. – robjohn Nov 22 '11 at 1:54

For fixed $r$, the measure of the symmetric difference of $B_r(x)$ and $B_r(y)$ goes to $0$ as $y$ goes to $x$. Therefore, by (local) absolute continuity of the integral (of a locally integrable function), $\int_{B_r(x)}udV-\int_{B_r(y)}udV$ goes to $0$ as $y$ goes to $x$. This implies that $u$ is continuous.

I would say "it depends". I was rather surprised by the following example, shown to me by a colleauge the other day. Dropping the assumptions that $u$ is locally integrable, strange things can happen!

Let $f(z) = \exp(-1/z^4)$, and let $u(x,y) = \Re f(x+iy)$. Then $u$ is clearly harmonic outside the origin. Extend $u$ to whole plane, by putting $u(0,0) = 0$. Then (outside the origin) $$u(x,0) = \exp(-1/x^4) \quad\text{and}\quad u(0,y) = \exp(-1/y^4).$$ It is not hard to see that both partial derivatives of $u$ exist at the origin, and in fact a little more work shows that the same holds for the second order partial derivatives. (However $u$ is of course not continuous at the origin, and neither is any of its derivatives.) All the derivatives above are in fact equal to $0$! In other words, the function $u$ satisfies "Laplace's equation" $$u''_{xx} + u''_{yy} = 0$$ everywhere on $\mathbb{R}^2$. A moment of contemplation also shows that $u$ fails to be integrable on any neighborhood of the origin, and hence does not define a distribution.

• For more about the particular case of $\mathbb{R}^2$ which has intimate connections with complex analysis, one should see this article in the AMM. – Willie Wong Jan 31 '12 at 12:44