In a paper I am studying, at a certain point the authors introduce a function $u\in C^1(B_1,\mathbb{R})$ which is harmonic in a dense open subset $U$ of $B_1$. From this, they seem to conclude that $u$ must be harmonic in all $B_1$. Is this fact true? Any hint on how to prove it? It should be elementary but I keep getting stuck. Thank you.
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2 Answers
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I don't think the claim is true.
Let $\mathfrak{C}$ be the ternary Cantor set in $[0,1]$; note that it is closed and nowhere dense, so its complement is open and dense. Let $C:[0,1]\to[0,1]$ be the Cantor function. It is continuous, and is differentiable on $(0,1)\setminus \mathfrak{C}$, where its derivative is $0$.
As $C$ is continuous, it is Riemann integrable, and the fundamental theorem of calculus applies: in particular, there exists some function $f:[0,1]\to\mathbb{R}$ such that $f\in C^1$ and $f' = C$.
This function $f$ is a $C^1$ function on $(0,1)$, which has a continuous (in fact vanishing) second derivative on an open, dense subset. Yet $f$ is not harmonic.
For higher dimensions, just let $g:B_1\to\mathbb{R}$ be given by $g(x_1, \ldots, x_n) = f(x_1)$.
There are some theorems concerning removable singularities for harmonic functions, but usually these require some amount of "isolation" of the singular points. For example:
If you know that $u\in C^1(B_1, \mathbb{R})$ and $u$ is harmonic on $B_1 \setminus \{x_1 = 0\}$, then $u$ must be harmonic on the whole ball by unique continuation.(The previous argument is only true if you make certain existence assumptions. So let's cross that out for now.) (The result however is true [using a different argument]; see Giorgio's comment below.)- If you know that $u$ is harmonic on $B_1\setminus \{0\}$ and that $u$ does not blow-up too fast at $0$ (compared to the Newton potential), then $u$ has a harmonic extension to the whole of $B_1$. This is a standard removable continuity result.
Maybe in the paper you are studying there are additional properties of $u$ that is used beyond what you stated in the question?
answered Dec 11, 2023 at 10:07
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4$\begingroup$ However seems to be true. We write Green's identitty for $u\Delta \phi-\phi \Delta u$ in the upper and lower ball (first with $|x_1| \geq \epsilon$ and then let $\epsilon \to 0$) and add. The boundary terms cancel and we get that $\int_B u \Delta \phi=0$, so that $u$ is harmonic in the distributional sence and apply Weyl's lemma. $\endgroup$ Commented Dec 11, 2023 at 13:57
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2$\begingroup$ @AlexandreEremenko: $u(x,y)=x$ is a non-constant harmonic function that depends on one variable only. (This is also essentially the function Willie constructs above.) $\endgroup$ Commented Dec 12, 2023 at 15:08
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3$\begingroup$ @AlexandreEremenko: As I understand it, Willie's function is $u(x,y)=m_jx+c_j$ on $a_j<x<b_j$, with the parameters adjusted such that $u\in C^1(\mathbb R^2)$ and $A=\bigcup (a_j,b_j)\times \mathbb R$ is dense and open. This function is harmonic on $A$ (but of course not on $\mathbb R^2$). $\endgroup$ Commented Dec 12, 2023 at 19:42
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2$\begingroup$ @AlexandreEremenko: I completely fail to understand your objection. I constructed a function that is harmonic on an open dense set (on each connected component of the open dense set it is in fact linear, and hence is harmonic). This function does not have all second partials outside of this dense set (it does not have $\partial^2_{x^1 x^1}$). $\endgroup$ Commented Dec 13, 2023 at 3:12
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3$\begingroup$ My answer is based on the following simple observations: given a function $f:\mathbb{R}\to\mathbb{R}$ and a function $g:\mathbb{R}^2\to\mathbb{R}$ with $g(x,y) = f(x)$. (a) If $f$ is harmonic on an open set $U$, then $g$ is harmonic on $U\times \mathbb{R}$. (b) If $U$ is open and dense in $\mathbb{R}$, then $U\times \mathbb{R}$ is open and dense in $\mathbb{R}^2$. Which of the two steps do you object to? $\endgroup$ Commented Dec 13, 2023 at 3:16
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Here is another example.
When $n=2$, let $E$ be a closed nowhere dense set of positive Lebesgue measure. Then it is easy to see that the function $$u(z)=\int_E\log|z-\zeta|d\zeta$$ is $C^1$. (Here $z$ and $\zeta$ are complex numbers.) Indeed, its convolution with an infinitely smooth kernel $k_t\to\delta$ is infinitely smooth, and as $t\to 0$, the convergence of the first derivatives is uniform.
When $n>2$, make the function independent of all but two variables.
answered Dec 12, 2023 at 14:49