Let $\Omega = \{x+iy : 0 < x < 1 \text{ and } y \in \mathbb R\}$. Suppose $u: \bar\Omega \to \mathbb R$ is continuous, harmonic on $\Omega$ and satisfies $u(z) \leq A \cosh(k \text{ Im}(z))$ throughout $\Omega$ for some $A > 0$ and $k \in (0, \pi)$. Show that $u$ satisfies $u(x + iy) \leq (1-x)A_0 + xA_1$ where $A_t = \sup\{u(t+iy):y \in \mathbb R\}$.
Here is my work so far:
For simplicity, assume $A_0, A_1$ are nonnegative. I am given that I should use the fact that, for any $h \in (0,\pi)$, the function $b_h(x+iy) := \cos(h(x-\frac{1}{2})\cosh(hy)$ is a nonnegative and harmonic function on $\Omega$. Choose $h$ so that $k < h < \pi$. For any $L > 0$, let $K_L := \{z: 0 <x<1, |y| \leq L \}$. Let $\epsilon > 0$ be fixed. Then by our choice of $h$, there exists $L_1 >0$ such that for all $L > L_1$, $A\cosh(kL) < \frac{1}{2}\epsilon\cos(h/2)\cosh(hL)$. Let $S_\text{left}, S_\text{right}, S_\text{top}, S_\text{bottom}$ be the four sides of $K_L$. Then $u(x + iy) \leq (1-x)A_0 + xA_1$ on $S_\text{left}, S_\text{right}$. My instructor suggested that I should try to get the bound $u(x + iy) \leq (1-x)A_0 + xA_1 + \epsilon b_h$ for all $x+iy \in S_\text{top}, S_\text{bottom}$ and let $\epsilon \to 0$. We know that for $x \in (0,1)$, since $u(z) \leq A \cosh(k \text{ Im}(z)), u(x \pm iL) \leq A \cosh(kL) < \frac{\epsilon}{2}\cos(h/2)\cosh(hL) \leq \frac{\epsilon}{2}\cos(h(x-1/2))\cosh(hL) = \frac{\epsilon}{2}b_h(x+iL)$. By the assumption that $A_0, A_1$ are nonnegative, we thus have that $u(x+iL) \leq (1-x)A_0 + xA_1 + \epsilon b_h(x+iL)$. Thus, by the maximum principle for harmonic functions, we have that for any $x+iy \in K_L, u(x+iy) \leq (1-x)A_0 + xA_1 + \epsilon b_h(x+iL)$.
However, because $L$ depends on $\epsilon$, we cannot simply state "let $\epsilon \to 0$ and the result follows." Indeed, as $\epsilon \to 0, L \to \infty$. I am looking for ways to fix so that I could have the solution as my instructor suggested. How can I modify my proof?