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If we let

$$\Omega_1=\left\{z \in \mathbb{C} : 0<\operatorname{Im}(z)< \pi\}, \quad \Omega_2=\{z \in \mathbb{C}: 0<\operatorname{Im}(z) \right\},$$

can we establish a one-to-one correspondence between $H(\Omega_1)$ and $H(\Omega_2)$ where $H(\Omega)$ represents the set of real-valued harmonic functions on $\Omega \subset \mathbb{C}$?

Thoughts:

Can we appeal to simple conformal mapping and say that, if we can establish a conformal map between $\Omega_1$ and $\Omega_2$ then the correspondence exists as the harmonic nature of the functions is preserved under such a map i.e. the composition of a holomorphic and harmonic map is harmonic? If so, what could this conformal map be?

If we apply $z \mapsto e^z$ initially, this transforms the 'strip'to a 'wedge', but how do we advance from here?

If such a method is incorrect, how else can we demonstrate the existence of the correspondence?

Any help would be greatly appreciated. Best, MM.

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If $f(z) = e^z$, then $f(\Omega_1) = \{e^{x+iy} \mid 0<y<\pi\} = \{z\in \mathbb C \mid |z|>0, \; 0< arg(z)<\pi \} = \Omega_2$. That's already the conformal map you are looking for. –  Sam Jan 14 '12 at 2:02
    
@Sam: Thanks! Great help. –  Mathmo Jan 14 '12 at 7:31
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up vote 0 down vote accepted

The Riemann Mapping theorem says there is a differentiable map between these two sets that is compositionally invertible. This just might do the job.

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