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Symmetric vectorization transports closed convex domains and subgradients

Statement

Let $C \subseteq S^{d}$ be a closed convex set of real symmetric matrices in the Frobenius metric, let $F : C \to R$ be convex, let $svec : S^{d} \to R^{m}$ be the symmetric vectorization isometry, and define $C^{'} = svec (C)$ and $f : C^{'} \to R$ by $f (z) = F ({svec}^{- 1} (z))$ . Then $C^{'}$ is closed and convex and $f$ is convex. If $G$ is a Frobenius subgradient of $F$ at $M \in C$ , meaning $F (N) \geq F (M) + ⟨ G, N - M ⟩_{F}$ for every $N \in C$ , then $svec (G)$ is a Euclidean subgradient of $f$ at $svec (M)$ .

Statement references

Symmetric vectorization preserves inner products and distances

Subgradient inequality on a convex domain

Euclidean inner product in coordinates

Convex functions

Convex sets

Closed sets and open complements in a metric topology

Metric balls

Bubeck, §3.1, printed pp. 262–266

Bertsekas, MIT 6.253 Lecture 15, printed pp. 200–205

Facts & Assumptions

Given: A closed convex set $C \subseteq S^{d}$ , a convex function $F : C \to R$ , a point $M \in C$ , and a Frobenius subgradient $G$ at $M$ .

[F1]

The map $svec$ is a linear bijection and preserves inner products and distances (Symmetric matrices are Euclidean under the Frobenius norm).

[F2]

A function is convex when it satisfies the convex-combination inequality, and a set is convex when it contains every convex combination of its points (Convex and strictly convex functions on Euclidean convex sets, A convex subset of $R^{m}$ contains every line segment between two of its points).

[F3]

A set is closed when its complement is open, and an open set contains a positive-radius ball around each of its points (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space).

[F4]

A subgradient supplies a global supporting affine lower bound on the convex function (Subgradients and the subdifferential of a convex function).

Proof

1.1givenF1F2construct

If $z_{0}, z_{1} \in C^{'}$ , choose $M_{0}, M_{1} \in C$ with $z_{i} = svec (M_{i})$ . For $t \in [0, 1]$ , linearity gives $t z_{0} + (1 - t) z_{1} = svec (t M_{0} + (1 - t) M_{1})$ , and the matrix on the right is in $C$ by [F2]. Thus $C^{'}$ is convex.

1.2givenF1F3

Let $z \notin C^{'}$ and set $M = {svec}^{- 1} (z) \notin C$ . Since $C$ is closed, [F3] gives $r > 0$ such that $B_{F} (M, r) \subseteq S^{d} ∖ C$ , where $B_{F} (M, r) = {A \in S^{d} : ∥ A - M ∥_{F} < r}$ . For any $w \in B_{2} (z, r)$ , isometry [F1] gives ${svec}^{- 1} (w) \in B_{F} (M, r)$ , hence $w \notin C^{'}$ . Therefore $B_{2} (z, r) \subseteq R^{m} ∖ C^{'}$ , so the complement of $C^{'}$ is open and $C^{'}$ is closed.

1.3givenF1F2algebra

For $z_{0}, z_{1} \in C^{'}$ with preimages $M_{0}, M_{1}$ , linearity yields ${svec}^{- 1} (t z_{0} + (1 - t) z_{1}) = t M_{0} + (1 - t) M_{1}$ . Applying convexity of $F$ from [given] proves $f (t z_{0} + (1 - t) z_{1}) \leq t f (z_{0}) + (1 - t) f (z_{1})$ , so $f$ is convex.

2.1givenF1F4∎

Let $z = svec (M)$ and take any $w \in C^{'}$ , with preimage $N \in C$ . The assumed subgradient inequality gives $f (w) = F (N) \geq F (M) + ⟨ G, N - M ⟩_{F}$ . By inner-product preservation [F1], this equals $f (z) + ⟨ svec (G), w - z ⟩_{2}$ ; hence $svec (G)$ is a Euclidean subgradient by [F4].

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Sources

Sébastien Bubeck, Convex Optimization: Algorithms and Complexity, Foundations and Trends in Machine Learning 8(3–4) (2015) (standard reference, not scraped)
Dimitri P. Bertsekas, Convex Analysis and Optimization, MIT 6.253 Complete Lecture Notes (standard reference, not scraped)