Rank-one convexity

Remarks

Example: for $m = n = 2$, $f = \det$ is rank-one convex. By the matrix determinant lemma, $\det(\xi + t \, a \otimes b) = \det \xi + t \, b^T \operatorname{adj}(\xi) \, a$, which is affine in $t$. But $\det$ is not convex: along $\xi(t) = \operatorname{diag}(t, -t)$ it equals $-t^2$.

Non-example: for $m = n = 1$, $f(\xi) = -\xi^2$ is not rank-one convex, since $t \mapsto -t^2$ is not convex.

For $m = 1$ or $n = 1$ every matrix has rank at most one, so rank-one convexity is ordinary convexity. If $f$ is $C^2$, it is rank-one convex exactly when $\sum_{i,j,\alpha,\beta} \frac{\partial^2 f}{\partial \xi^i_\alpha \partial \xi^j_\beta}(\xi) \, a_i a_j b_\alpha b_\beta \ge 0$ for all $\xi$, $a$ and $b$, the Legendre–Hadamard condition.