Submanifolds lying on the boundary of a convex domain Let $M$ be a submanifold of $\mathbb R^n$. Call $M$ locally convex if locally $M$ is contained in the boundary of a convex domain of $\mathbb R^n$.
Is there any known condition that is equivalent to local convexity?
Some special cases are easy to treat. If the second fundamental form with respect to some normal direction is strictly positive definite, then $M$ is locally convex. Vice versa, if the second fundamental forms with respect to all normal directions are indefinite, $M$ is not locally convex.
Also the case of hypersurfaces is solved: $M$ is locally convex if and only if the second fundamental form, with respect to a smooth nowhere vanishing normal vector field, is positive semidefinite at every point.
 A: First another example (elaborating Petrunin's comment) of a curve in $\mathbb R^3$ that is not locally convex: consider $x \mapsto (x, x^3, x^5)$.  From the first derivative, any support plane at the origin needs to contain the $x$-axis.  From the third derivative, if there's a supporot plane it needs to
be the $xy$-plane. However, from the fifth derivative, the curve crosses this plane. Therefore, it is not locally
convex.  On the other hand, the curves $(x, x^3, x^4)$ and $(x, x^9, x^{132})$ are locally convex at the origin.
This example shows what you need to do: express the submanifold as the graph of a function
from the tangent bundle to the normal bundle.  Look at successive $k$-jets of the function.
Typically you can determine local convexity from the 2-jet, that is, the second fundamental
form.   If not, you can look at the higher and higher jets, that is, polynomial approximations
of greater and greater degree. If there is a linear functional on the normal bundle
that when composed with this polynomial is strictly positive in a neighborhood of the origin, then it's locally convex at that point.  If the submanifold is only $C^\infty$ with
infinite order contact to some subspace of the tangent space, you're out of luck with this
approach---even if you dutifully spend an infinite amount of time computing polynomial approximations,
you will never resolve it.  These polynomials, of course, can be expressed in terms of
the 2nd fundamental form and its covariant derivatives.
Depending on how the submanifold is defined near $x$, the question of local convexity
might be an algorithmically unsolvable question (depending on assumptions about the
form of the input data).
A: Well, you almost answer your question:
$M$ is locally convex if and only if at every point there is a local normal vector field, with respect to which the second fundamental form is positive semidefinite.
If there is such local vector field  $v_p$ for $p\in M$ then you can construct a convex body $x\in B$
$$\langle x, v_p\rangle\le \langle p, v_p\rangle.$$ 
For a convex hypersurface (not nesessary smooth) one can choose a smooth local vector field of supporting hyperplanes on $M$.
Then choose $v_p$ to be its normal... 
