Let $u(x, y)$ be a harmonic function on the upper half-plane $\mathbb{R}\times \mathbb{R}^+$, that is, $$\partial_x^2 u(x, y) + \partial_y^2 u(x, y) = 0$$ for $x \in \mathbb{R}, y>0$. Assume $u(x, 0)$ is smooth for $x \in \mathbb{R}$. In addition, we assume that both $u(x, 0)$ and $\partial_yu(x, 0)$ vanish at $x = 0$ to infinite order, i.e., for every $k \in \mathbb{Z}^+$, $$\lim_{x \to 0}\frac{u(x, 0)}{x^k} = \lim_{x\to 0}\frac{\partial_yu(x,0)}{x^k} = 0.$$

More explanation: As pointed out by Alexandre Eremenko, here we assume that $u(x, y)$ is continuous up to the boundary $\mathbb{R}\times \{0\}$, and we treat $-\partial_yu(x, 0)$ as the outer normal derivative of $u$ at the boundary point $(x, 0)$.

Question: can we conclude that $u(x, y)$ vanish to infinite order at the origin with respect to interior points? In other words, does the following limit hold for every $k \in \mathbb{Z}^+$? $$\lim_{\substack{(x, y) \to (0, 0)\\ x\in \mathbb{R}, y>0}}\frac{u(x, y)}{(|x|+|y|)^k} = 0.$$ If not, is there a counter-example?

  • $\begingroup$ (I think you mean "the upper half-plane $\mathbb R\times \mathbb R_+$" ) $\endgroup$ Aug 5, 2021 at 12:17
  • $\begingroup$ The physical intuition already suggests it can't be true. Imagine a soap film bounded by circular wire of radius 1, with an obstacle in the middle, given by a unit wire straight segment, located at some distance from the disk on a plane parallel to it. We expect the shape of the film not to be smooth near the segment, but more similar to a Canadian tent. $\endgroup$ Aug 5, 2021 at 12:26
  • $\begingroup$ You should add some conditions, explaining what $u(x,0)$ and $u_y(x,0)$ exactly are. Your function is defined only in the upper half-plane. Is it assumed to be continuous in the closed half-plane? $u_y$ is continuous in the closed half-plane? What about $u_x$? $\endgroup$ Aug 5, 2021 at 13:47
  • $\begingroup$ I just misses to see the condition on $u_y$ .. $\endgroup$ Aug 5, 2021 at 15:21
  • $\begingroup$ @PietroMajer As a Canadian, what is a Canadian tent? $\endgroup$ Aug 5, 2021 at 20:40

2 Answers 2


There is a counterexample. Consider the harmonic function $$u(x,y) = Re\left(e^{-1/z^2}\right) = e^{-\frac{x^2-y^2}{r^4}}\cos\left(\frac{2xy}{r^4}\right),$$ where $r^2 = x^2 + y^2$. We have that $$u(x,\,0) = e^{-1/x^2}$$ vanishes to infinite order in $x$, and that $$u_y(x,\,0) \equiv 0$$ since $u$ is even in $y$. However, $$u(x,\,x) = \cos(2/x^2)$$ does not vanish to infinite order.

  • $\begingroup$ This $u$ is not even bounded near the origin: $u(0, y) = e^{1/y^2}$, so what would $u_y(0, 0) = 0$ mean? $\endgroup$ Aug 5, 2021 at 14:59
  • $\begingroup$ @MateuszKwaśnicki: That is a fair point. As Alexandre Eremenko points out in the comments, it could be helpful for the question-asker to clarify what is meant by $u,\,\nabla u$ on $\{y = 0\}$. $\endgroup$ Aug 5, 2021 at 15:50

Assuming $u$ is smooth enough, the answer seems to be affirmative.

The $k$-th term in the power series of $u$ near the origin must be a solid harmonic polynomial $P_k$ of degree $k$, satisfying two independent conditions: $\partial_x^k P_k = 0$ and $\partial_x^{k-1} \partial_y P_k = 0$. The space of harmonic polynomials of degree $k$ is two-dimensional, so this essentially tells us that $P_k = 0$. Consequently, the power series of $u$ near $(0, 0)$ is zero, and hence all partial derivatives of $u$ vanish at the origin.

Edit: Here are some additional details. Suppose that $u$ is the Poisson integral of the boundary data $f$ (so, for example, it suffices to know that $u$ is bounded, or non-negative — this is a rather mild condition). Suppose, furthermore, that $f$ is infinitely smooth in a neighbourhood of $0$. Then it is an easy exercise to see that $u$ is infinitely smooth in a neighbourhood of $(0,0)$ (intersected with $\mathbb R \times [0, \infty)$, of course).

In particular, we can develop $u$ into the power series at $(0, 0)$. Of course, this power series need not be convergent, it is just a convenient formal way to speak about the partial derivatives of $u$. The $k$-th term of this power series, call it $P_k$, is a homogeneous polynomial of degree $k$, and using smoothness of $u$ it is easy to check that $P_k$ is a harmonic polynomial.

The remaining part of the argument is given in the original answer.

  • $\begingroup$ If we want to expand $u$ as harmonic polynomials near the origin, do we need $u(x, y)$ to be harmonic on the whole plane $\mathbb{R}^2$? I am not sure whether this would work when the origin lies on the boundary. $\endgroup$
    – Jacob Lu
    Aug 6, 2021 at 4:23
  • $\begingroup$ I added some details. Of course the power series need not converge to $u$ (consider $u(x,y)=\Re(\exp(-1/\sqrt{y-ix}))$), but it allows one to easily handle the partial derivatives of $u$. $\endgroup$ Aug 6, 2021 at 7:15
  • $\begingroup$ I see. This makes sense. If $u$ is harmonic on the upper half-plane and smooth up to the boundary, and vanishes at $(0, 0)$ up to infinity order from inside, do you think now $u$ is identically zero? $\endgroup$
    – Jacob Lu
    Aug 13, 2021 at 4:44
  • $\begingroup$ Yes, I think this is what I tried to write. $\endgroup$ Aug 15, 2021 at 22:10

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