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?
 A: 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.
A: 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.
