The Riemann mapping theorem states, that any simply connected domain $U \subset \mathbb C$ can be conformally mapped to the open unit disk $D$. I.e. there is a Diffeomorphism $\Psi: D \to U$ such that $\Psi^\* g\_{U}=e^{u}g\_{D}$ where $u\in C^\infty(D)$ and $g$ denotes the Euklidean metric on $U$ respectively $D$.


My question is now whether the "inverse" is true, i.e.: given a function $u\in C^\infty(D)$, is there a domaine $U$ and a diffeomorphism $\Psi: D \to U$ such that $\Psi^\* g\_{U}=e^{u}g\_{D}$ holds? 


If it doesn't hold in general: Under what conditions on $u$ does it hold? 



Inverse "Riemann mapping"
It does not hold in general, as Benoit Kloeckner explained. More precisely,
there are two obstacles, one
local and one global. The local obstacle is the Gaussian curvature, $-e^{-2u}\Delta u$.
It must be zero for the pull-back of the Euclidean metric (which means that $u$ must
be harmonic). But there is also a global
obstacle. For a given metric of zero curvature on the disc, there exists a conformal local 
homeomorphism to the plane such that your metric is the pull-back of the Euclidean metric.
However this local homeomorphism is not necessary a global homeomorphism. There are Euclidean surfaces conformally equivalent to the unit disc but not isometric
to any region in the plane. 


