A harmonic function degenerate in one direction 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 $u^{-1}(0) \cap \lvert Du \rvert^{-1}(0)$?
 A: 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.
