Question. Let $u: B^3 \to \mathbf{R}$ be a harmonic function with $u(0) = 0$, $Du(0) = 0$, where its homogeneous harmonic blow-up is a polynomial $p = p(x,y)$ in two variables, so independent of $z$; in other words $p$ is a non-zero homogeneous harmonic polynomial so that \begin{equation} u(x,y,z) = p(x,y) + o( \lvert (x,y,z) \rvert^m), \end{equation} where $2 \leq m = \operatorname{deg} p$. Must $u$ be translation-invariant with respect to $z$? Can the origin be isolated in the singular set $u^{-1}(0) \cap \lvert Du \rvert^{-1}(0)$?

  • $\begingroup$ Could you define the homogenuous harmonic blow-up? $\endgroup$ 2 days ago
  • $\begingroup$ The answer to the first question is negative: a counterexample is $u(x,y,x)=P_m(x,y)+Q_n(x,y,z),$ where $P_m,P_n$ are homogeneous harmonic polynomials of degrees $n>m$, and $P_n$ depends on all three variables. $\endgroup$ 2 days ago
  • $\begingroup$ @AlexandreEremenko Of course, how embarrassing that I missed that - thanks for pointing this out! $\endgroup$
    – Leo Moos
    2 days ago
  • 1
    $\begingroup$ @AlexandreEremenko By the way, if you wanted to post your comments as an answer, I'd be happy to accept it! $\endgroup$
    – Leo Moos
  • 1
    $\begingroup$ @Leo Moos: My second comment was not correct, and I deleted it. Anyway, you solved the problem yourself, probably inspired by the first (trivial) comment. $\endgroup$ yesterday

1 Answer 1


The questions have been answered in the comments, I am just recording them here: Alexandre Eremenko pointed out that no, the function $u$ need not be translation-invariant, because the dependencies on $z$ could be 'hidden' inside a polynomial of higher degree, say \begin{equation} u = p(x,y) + q(x,y,z), \end{equation} with $\operatorname{deg} q > \operatorname{deg} p$.

This also gives a hint for the second question: the answer is yes, there exist examples of such $u$ that only have isolated singularity at the origin. The example given below $u$ is basically of the form above—with $q$ picked so as to have an isolated singularity at the origin—, except for the fact that one multiplies $q$ by a small constant to avoid introducing new singular points.

Specifically, pick a constant $\delta \in (0,1/3)$ and define \begin{equation} u(x,y,z) = x^2 - y^2 + \delta(2x^3 - 3xy^2 - 3xz^2). \end{equation} Then \begin{equation} Du(x,y,z) = (2x + \delta( 6x^2 - 3y^2 - 3z^2),-2y - 6\delta xy,-6\delta xz). \end{equation} At a critical point $(x,y,z)$:

  • from $D_y u = 0$ one finds that $y(1 + 3\delta x) = 0$. As $\delta < 3$, the second factor never vanishes if $\lvert x \rvert < 1$, so $y = 0$;
  • from $D_z u = 0$ one finds that either $x = 0$ or $z = 0$.
  • from $D_x u = 0$, if $x = 0$ then immediately $z = 0$. If instead $z = 0$ then $0 = D_x u = 2x + 6\delta x^2 = 2x(1 + 3 \delta x)$. Again, our choice of a sufficiently small $\delta$ means that $1 + 3 \delta x > 0$ on $B^3$, so $x = 0$.

Therefore $Du(x,y,z) = 0$ is equivalent to $(x,y,z) = 0$. Obviously $u(0,0,0) = 0$, so this is indeed the unique singular point.


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