A question of uniqueness Let $u$ an harmonique function on $\Omega=(a,b)\times (0,+\infty)$ and boundary conditions :
$\displaystyle  u(a,y)=u(b,y)=0,\quad\forall y\geq 0$
$\displaystyle u(x,0)=0,\,\lim_{y\to +\infty} u(x,y)=0 \quad \forall x\in (a,b)$
Can we conclude that  $\quad u=0$ on $\Omega$ ?
My adempt
Let $$\Omega_{R}=(a,b)\times (0,R),\forall R>0$$By IBP, i show that
$$\int_{\Omega_R}u\Delta u=\int_a^{b}u(x,R)\frac{du}{dy}(x,R)dx-\int_{\Omega_R}|\nabla u|^2 $$ Thus
$$\forall R>0,\quad \int_{\Omega_R}|\nabla u|^2=\int_a^bu(x,R)\frac{du}{dy}(x,R)dx$$
I need help to cointinuous ( For example to show  $\int_{\Omega}|\nabla u|^2=0$)
edit Continuing the initial reasoning, with $a=0$  and $b=\pi$ as suggested by A Ermenko
$\Big(\int_{\Omega_R}|\nabla u|^2\Big)^2=\Big(\int_0^{\pi}u(x,R)\frac{du}{dy}(x,R)dx\Big)^2\\
\leq \int_0^{\pi}u(x,R)^2dx\int_0^{\pi}\Big(\frac{du}{dy}(x,R)\Big)^2dx ,\mbox{by Cauchy–Schwarz inequality
}\\\\$
$\leq  \int_0^{\pi}u(x,R)^2dx \int_0^{\pi}|\nabla u|^2(x,R)dx \\\\$
$= \int_0^{\pi}u(x,R)^2dx\int_{]0,\pi[\times\{R\}}|\nabla u|^2 \\\\$
$\leq  \int_0^{\pi}u(x,R)^2dx \int_{\Omega_R}|\nabla u|^2,\mbox{because} ]0,\pi[\times\{R\}\subset\Omega_R$
Then $$\int_{\Omega_R}|\nabla u|^2\leq \int_0^{\pi}u(x,R)^2.$$ I can only conclude if $\displaystyle ||u(.,R)||_{L^2]0,\pi[}\to^{R\to\infty} 0$
 A: If you want the positive answer you should state your last condition more carefully.
For example, add that $u$ is bounded, or that $u(x+iy)$ tends to $0$ as $y\to\infty$
UNIFORMLY with respect to $x$.
As you presently stated, the answer is negative. I sketch the construction of a counterexample.

*

*There exists a non-zero entire function, real on the real line and such that $f(re^{i\theta})\to 0$ as $r\to+\infty$ for every $\theta$. (See, for example, my answer to this question, which explains how to construct $f$.)


*$v(z)=\Im f(z)$ is a non-zero harmonic function in the upper half-plane, equal to $0$ on
the real line and $u(re^{i\theta})\to 0$ as $r\to+\infty$ for every $\theta\in(0,\pi)$.


*Take without loss of generality $a=0,\; b=\pi$, then $e^{-i(z-\pi)}$
maps your strip $0<z<\pi$ into the upper half-plane, with a removed half-disk.
So the function $w(z)=v(e^{-i(z-\pi)})$ is harmonic, zero on infinite sides
of your half-strip and satisfies the property at $\infty$:
$w(x+iy)\to 0$ as $y\to+\infty$ for every $x\in(0,\pi)$. Notice that this function is unbounded. To satisfy the last requirement, that $u(x)=0$ for $0<x<\pi$,
set $u(z)=w(z)-w_1(z)$, where $w_1(z)$ is the solution of Dirichlet problem matching
the boundary values of $w$ on the finite part of the boundary and bounded in your strip.
Remark. Since every simply connected domain other than the plane is conformally equivalent to the unit disk, your question is equivalent to the following. Suppose that $u$ is harmonic in the unit disk and $\lim_{z\to\zeta} u(z)=0$ for all $\zeta\in\{ \zeta:|\zeta|=1\}\backslash\{1\}$, and moreover $u(z)\to 0$ along any non-tangential segment ending at $1$, that is $u(1-re^{i\theta})\to 0,\; r\to 0$ for every $\theta\in(-\pi/2,\pi/2)$. Does it follow that $u=0$? The answer is NO.
