# Vanishing rate of a harmonic function near a boundary point

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?

• (I think you mean "the upper half-plane $\mathbb R\times \mathbb R_+$" ) Aug 5, 2021 at 12:17
• 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. Aug 5, 2021 at 12:26
• 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 $u_x$? Aug 5, 2021 at 13:47
• I just misses to see the condition on $u_y$ .. Aug 5, 2021 at 15:21
• @PietroMajer As a Canadian, what is a Canadian tent? Aug 5, 2021 at 20:40

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.

• 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? Aug 5, 2021 at 14:59
• @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 $\{y = 0\}$. 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.

• 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. Aug 6, 2021 at 4:23
• 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 $u$. Aug 6, 2021 at 7:15
• 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? Aug 13, 2021 at 4:44
• Yes, I think this is what I tried to write. Aug 15, 2021 at 22:10