---
tag: TPC99RZ316
kind: definition
name: Rank-one convexity
msc: [49J45, 26B25]
defines: GK4P8G0PB2
uses: [07GR3TNW4H, 0ZC9YTXZRY]
sources:
  - url: https://en.wikipedia.org/wiki/Quasiconvexity_(calculus_of_variations)
---

## Definition

Let $m, n \ge 1$ and let $f \colon \R^{m \times n} \to \R$.
For $a \in \R^m$ and $b \in \R^n$, write $a \otimes b$ for the $m \times n$ matrix with entries $a_i b_\alpha$; it has [[rank|0ZC9YTXZRY]] at most one.
The function $f$ is *rank-one convex* if for every $\xi \in \R^{m \times n}$, $a \in \R^m$ and $b \in \R^n$ the function
$$t \mapsto f(\xi + t \, a \otimes b), \quad t \in \R,$$
is [[convex|07GR3TNW4H]].

## 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.