Given $z \in \mathbb{R}^n$ and $x \in \mathbb{R}$, we are given the following map $f: \mathbb{R}^{n+1} \rightarrow \mathbb{R}$, and $\alpha_{i,j}(x) \in L^{\infty}(\mathbb{R})$ :

$$f(x,z) = \sum_{i,j=1}^{n} \alpha_{i,j}(x) z_i z_j$$

I would like to check the map $z \mapsto f(x,z)$ is convex (for almost every $x \in \mathbb{R})$ in the case the ellipticity condition is satisfied.


1 Answer 1


For $z, w \in \Bbb R^n$, $ 0 < \lambda < 1$, and $1 \le i, j \le n$ is:

$$ (\lambda z_i + (1-\lambda) w_i)(\lambda z_j + (1-\lambda) w_j) - \lambda z_i z_j - (1-\lambda) w_i w_j \\ = -\lambda (1-\lambda) (z_i z_j - z_i w_j - z_j w_i + w_i w_j) \\ = -\lambda (1-\lambda) (z_i - w_i)(z_j - w_j) \, . $$

It follows that for every $x$, satisfying the ellipticity condition,

$$ f(x, \lambda z + (1-\lambda)w) - \lambda f(x, z) - (1-\lambda) f(x, w) \\ = -\lambda (1-\lambda) \sum_{i,j=1}^{n} \alpha_{i,j}(x) (z_i - w_i)(z_j - w_j) \\ \le -\lambda (1-\lambda) \eta \Vert z-w \Vert^2 $$

and that is strictly negative if $z \ne w$, so that $z \mapsto f(x,z)$ is strictly convex.

  • $\begingroup$ Thanks for the time to read my question and answer it! I edited a minor typo, but all was comprehensible and neat! I could not get the factoring you made! Thanks a lot again! $\endgroup$ May 21, 2021 at 6:44

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