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?

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