I've been taught in class that if $\phi(x,y)$ is a harmonic function and $f(z)$ is a holomorphic function viewed as a function $\mathbb{R}^2\rightarrow\mathbb{R}^2$ then $\phi \circ f$ is also a harmonic function.
My question is how is that possible considering the following example:
We can consider solutions on the upper half-plane with boundary conditions: $\phi(x,0)=0$.
We then take a holomorphic function mapping the the interior of the unit circle to the upper half-plane (say a möbius transformation). Since the unit circle is mapped to the boundary of the plane, $\phi\circ f$ is harmonic on the unit disk with bundary zero everywhere.
By the maximum modulus principle, $\phi\circ f=0$ everywhere in the unit disk. but $\phi(x,y)=y$ is a harmonic function with the required boundary conditions that doesn't satisfy this equality.
How to resolve this apparent discrepancy?
Note that boundary of upper half plane is not just $x$-axis, $\infty$ is also a boundary point.
Under the biholomorphism between unit disc and upper half plane, unit circle is mapped to boundary of upper half plane i.e. $x$-axis $\cup${$\infty$}. Hence boundary condition you have given is not on every point of boundary. If you give same condition at $\infty$ then by Maximum modulus principle for unbounded domain, $\phi =0$ everywhere.