We want to find harmonic functions $w$ in (say) $\mathbb{R}^2$ that are zero on $\{y=0\}$ with the linear growth bound \begin{equation} \sup_{\mathbb{B}_R} |w| \leq C(1+R) \end{equation} where $C>0$ independent of $R$, are the only solutions to such a problem $w(x,y)=ay$ for some $a \in \mathbb{R}^2$?
I cannot think of a method to even start this at the moment as the usual tools (like mean value property or Harnack) do not seem to give anything of use. I feel like it is possible an application of Harnacks inequality in a clever way might do it?
The Poisson kernel for balls with centre $0$ in $\mathbb{R}^n$ is given by
$$P(x,\zeta) = \frac{1}{\lVert\zeta\rVert\omega_{n-1}} \cdot \frac{\lVert\zeta\rVert^2 - \lVert x\rVert^2}{\lVert \zeta - x\rVert^n},$$
and for any function $u$ harmonic on the open ball with radius $R$ and continuous on the closed ball, in particular for entire harmonic functions, we have the integral representation
$$u(x) = \int_{\lVert\zeta\rVert = R} u(\zeta)P(x,\zeta)\,d\sigma(\zeta)$$
for $\lVert x\rVert < R$, where $\sigma$ is the surface measure on the boundary sphere.
All partial derivatives of $u$ can then be obtained by differentiating under the integral, i.e.
$$D^\alpha u(x) = \int_{\lVert\zeta\rVert = R} u(\zeta)D_x^\alpha P(x,\zeta)\,d\sigma(\zeta)$$
for every multi-index $\alpha \in \mathbb{N}^n$.
For any compact $K\subset \mathbb{R}^n$, the Poisson kernel is asymptotically equal to $\dfrac{1}{\omega_{n-1}\lVert\zeta\rVert^{n-1}}$ as $\lVert\zeta\rVert\to\infty$, uniformly on $K$. Differentiating with respect to $x$, one verifies that $D_x^\alpha$ is asymptotic to
$$C_{\alpha}\cdot \frac{1}{\lVert\zeta\rVert^{n-1+\lvert\alpha\rvert}}$$
uniformly on $K$. Thus if we have a growth restriction
$$\lvert u(x)\rvert \leqslant C\cdot \lVert x\rVert^k$$
for $\lVert x\rVert \geqslant R_0$, we find
$$\lvert D^\alpha u(x)\rvert \leqslant \int_{\lVert\zeta\rVert = R} C\lVert\zeta\rVert^k \frac{C_\alpha}{\lVert\zeta\rVert^{n-1+\lvert\alpha\rvert}} \,d\sigma(\zeta) \leqslant \tilde{C}\cdot R^{k-\lvert\alpha\rvert}$$
for $x\in K$ and all $R\geqslant R_0$. If $\lvert\alpha\rvert > k$, then letting $R\to\infty$ shows that $D^\alpha u \equiv 0$ on $K$. If $K$ has non-empty interior, the real analyticity of harmonic functions implies that $D^\alpha u \equiv 0$ on $\mathbb{R}^n$. Letting $\alpha$ run through all multi-indices of order $k+1$ then shows that $u$ is a polynomial of degree $\leqslant k$, since all partial derivatives of order $k+1$ vanish identically.
Here, we have a bound with $k = 1$, so $w$ must be a polynomial of degree $\leqslant 1$. The condition that $w(x,0) = 0$ forces the constant term and the $x$ term to vanish, so $w(x,y) = a\cdot y$ for some $a$.