This is a vague question because I'm not sure what I want to ask.
An ellipsoid has positive curvature everywhere, and bounds a convex subset of $\mathbb R^3$. What I want to say now is "It seems as though a surface with negative curvature can't be convex" but this isn't right, because the ellipsoid itself is not convex, and I could only get at what I was thinking of by observing that it bounds a subset of $\mathbb R^3$.
But this isn't quite what I want, because the ellipsoid is also the boundary of a non-convex subset of $\mathbb R^3$, namely the complement of the convex region. Well, the convex region is bounded, okay, so I can pick out the convex region I want that way, and say that if a surface is the boundary of a bounded subset $S$ of $\mathbb R^3$, and has positive curvature everywhere, then $S$ is convex. But this is not as general as what I want, because a paraboloid also has positive curvature everywhere, and it is the boundary of a convex subset of $\mathbb R^3$, but the claim of the previous sentence doesn't apply to it.
Maybe what I want is something like: let $P$ be a point on a surface, and let $C$ be a small closed curve on the surface that bounds a region $R$ of the surface that contains $P$. Then the subset of $\mathbb R^3$ bounded by $R$ and by (something?) is convex if and only if the Gaussian curvature at $P$ is positive. Or perhaps the condition is that the curvature at every point of $R$ must be non-negative. Or something.
It seems like there are some theorems here, but I don't know quite where to look for them. Are there theorems relating Gaussian curvature of a surface to the convexity of the region of which it is the boundary?
Is it at least true that the boundary surface of a convex subset of $\mathbb R^3$ has non-negative curvature everywhere that its curvature is well-defined? What I am looking for is something like the converse of that.