Let $M$ be a Riemannian manifold of dimension $n$. Let $N\subset M$ be a subset with smooth boundary $\Sigma=\partial N$. If one assume the second fundamental form $II$ with respect to inner normal direction of $\Sigma$ is nonnegative. (here we use the convention that the second fundamental form of round sphere in $\mathbb R^n$ is positive respective to inner normal.) It is easy to show that for any point $x\in \Sigma$ there exists a neighborhood $U$ around $x$ in $M$ such that $U\cap N$ is convex in the sense that $p, q\in U$ then any short geodesic connecting $p$ and $q$ remains in $U\cap N$.


Is it true that for any two point $p, q\in N$ any short geodesic(realize the distance) connecting them remains in $N$? 



Positivity of second fundamental form implies global convexity?
From the comment of Willie Wong , the answer is NO.


Yet simpler example is $\mathbb{S}^1\times\mathbb{R}$, where you have locally convex discs which are not globally convex.


If you want "yes" as an answer, you have to assume bit more. For example simply connectedness plus sectional curvature $\le 0$ (It was proved by S. Alexander). 


An other related statement: if an immersed hypersurface is locally convex plus curvature is nonnegative then it bounds an immersed locally convex ball. (this was conjectured by S. Alexander and proved by Gromov independently). 


