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Extended-real convexity and lower semicontinuity are characterized by the epigraph

Statement

Let $d \geq 1$ and let $f : R^{d} \to \overline{R}$ use the conventions of Proper, convex, and lower-semicontinuous extended-real functions on Euclidean space. Then $f$ is convex if and only if $epi f$ is a convex subset of $R^{d + 1}$ , and $f$ is lower semicontinuous if and only if $epi f$ is closed.

Facts & Assumptions

Given: $d \geq 1$ , the extended-real conventions and epigraph of the Statement, and the Euclidean topology.

[F1]

Proper, convex, and lower-semicontinuous extended-real functions on Euclidean space: convexity is the extended chord inequality; lower semicontinuity means that every weak sublevel set is closed.

[F2]

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: a set is open when every point has a positive-radius metric ball contained in the set; it is closed exactly when its complement is open. Here the metrics are Euclidean.

Proof

technique · direct

1.1givenF1algebra

Suppose $f$ is convex and take $(x, r), (y, s) \in epi f$ . For $t \in [0, 1]$ , the chord inequality gives $f ((1 - t) x + t y) \leq (1 - t) f (x) + t f (y) \leq (1 - t) r + t s$ , so $(1 - t) (x, r) + t (y, s) \in epi f$ . Thus the epigraph is convex.

1.2F1algebra

Conversely, if the epigraph is convex and $f (x), f (y)$ are finite, then $(x, f (x))$ and $(y, f (y))$ lie in it; their convex combination lies in it and gives the chord inequality. If either value is $+ \infty$ and its coefficient is positive, the right side is $+ \infty$ and the inequality is automatic; at $t = 0$ or $t = 1$ it is equality.

1.3F1F2construct

Suppose $epi f$ is closed and fix $a \in R$ . For $n = d, d + 1$ write $d_{n} (u, v) = (\sum_{j = 1}^{n} (u_{j} - v_{j})^{2})^{1 / 2}$ for the Euclidean metric. The slice map $i_{a} (x) = (x, a)$ satisfies $d_{d + 1} (i_{a} (x), i_{a} (y)) = d_{d} (x, y)$ . If $f (x) > a$ , then $(x, a) \notin epi f$ ; closedness gives $ε > 0$ such that the $d_{d + 1}$ -ball of radius $ε$ around $(x, a)$ misses $epi f$ . Whenever $d_{d} (x, y) < ε$ , the distance identity puts $(y, a)$ in that ball, so $f (y) > a$ . Thus the complement of ${y : f (y) \leq a}$ is open, and this weak sublevel set is closed.

2.1givenF1F2construct∎

Conversely suppose every weak sublevel set is closed and let $(x, r) \notin epi f$ , so $r < f (x)$ . Choose a finite real $a$ with $r < a < f (x)$ when $f (x)$ is finite, and choose $a = r + 1$ when $f (x) = + \infty$ . The set $N = {y : f (y) > a}$ is an open neighborhood of $x$ , as it is the complement of a closed weak sublevel set. Choose $δ > 0$ with $B_{d} (x, δ) \subseteq N$ and $η > 0$ with $r + η < a$ . If $d_{d + 1} ((x, r), (y, s)) < \min (δ, η)$ , then $y \in N$ and $s < r + η < a < f (y)$ , so $(y, s) \notin epi f$ . The epigraph's complement therefore contains a ball around each of its points and is open, so the epigraph is closed.

Extended-real convexity and lower semicontinuity are characterized by the epigraph

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources