Possible curvatures of the topological sphere

Question

Consider the family $\mathbb{S}$ of compact oriented surfaces homeomorphic to the 2-sphere $\mathcal{S} = S^2$. Consider arbitrary continuous mappings $k: \mathcal{S} \rightarrow \mathbb{R}$ which obey the condition $\int_\mathcal{S} k = 4\pi$. The latter looks like the Gauss-Bonnet condition for 2-spheres, and that's why I want to call those mappings “curvature-like”.

For which curvature-like mappings $k$ does exist a surface $S \in > \mathbb{S}$ which $k$ is the Gaussian curvature of?

In other words: For which curvature-like mappings $k$ does exist a surface $S \in \mathbb{S}$ with a homeomorphism $s: S \rightarrow \mathcal{S} $ such that $k \circ s$ equals the Gaussian curvature $\kappa: S \rightarrow \mathbb{R}$?

Background (and for comparison's sake): When one considers (plane) Jordan curves - which are homeomorphic to the 1-sphere $S^1$ - and continuous “curvature-like” mappings $k: S^1 \rightarrow \mathbb{R}$ which obey the condition $\int_{S^1} k = 2\pi$ – in accordance to Hopf's Umlaufsatz – then the additional condition for a curvature-like mapping to be a “real” curvature seems to be given by the four-vertex theorem.

Answer 1

Kazdan, Jerry L. and Warner, F. W., Existence and conformal deformation of metrics with prescribed Gaussian and scalar curvatures, Ann. of Math. (2), 101, 1975, pp. 317--331 prove that a $C^{\infty}$ function $K$ on the 2-sphere is the Gauss curvature of a $C^{\infty}$ Riemannian metric if and only if $K$ is positive at at least one point of the sphere. In the same paper, they prove that if $M$ is a compact manifold of dimension 3 or more, and $K$ is a $C^{\infty}$ function on $M$, and $K$ is somewhere negative then $K$ is the scalar curvature of a $C^{\infty}$ Riemannian metric, while is $K$ is nowhere negative, then $K$ is the scalar curvature of a $C^{\infty}$ Riemannian metric if and only if $M$ admits a Riemannian metric of constant positive scalar curvature.