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A flat exposed face with many Fenchel minimizers

Example

Let $f$ be the indicator of $[- 1, 1]$ . The zero tilt $f (y) - 0 \cdot y$ has every point of the interval as a minimizer, and $f^{*} (0) = 0$ is attained at every such point. Equivalently, $0 \in \partial f (x)$ for every $x \in [- 1, 1]$ . Thus the supporting face at slope zero is the whole interval, and coercive Fenchel attainment need not be unique.

Facts & Assumptions

Given: The indicator $f (x) = 0$ for $x \in [- 1, 1]$ and $f (x) = + \infty$ otherwise.

[F1]

Proper, convex, and lower-semicontinuous extended-real functions on Euclidean space: the indicator of a nonempty closed convex set is proper, closed, and convex in the given conventions.

[F2]

Fenchel conjugate, biconjugate, and subdifferential: $f^{*} (p)$ is the supremum of $p y - f (y)$ over $y \in dom f$ , and $p \in \partial f (x)$ means $f (y) \geq f (x) + p (y - x)$ for all $y$ .

[F3]

Extended-real convexity and lower semicontinuity are characterized by the epigraph: a proper indicator with a closed convex epigraph is closed and convex in the local extended-real conventions.

[L1]

Fenchel inequality and its equality condition: equality in Fenchel's inequality at a finite point is equivalent to the subgradient condition and attainment of the conjugate supremum.

[L2]

A coercive Fenchel tilt attains its minimum and conjugate supremum: if the closed proper convex tilt is coercive, it has a finite global minimum and the conjugate supremum is attained, without a uniqueness conclusion.

Verification

technique · direct calculation

1.1F1F2F3given

The epigraph is $[- 1, 1] \times [0, \infty)$ , a nonempty closed convex rectangle, and the domain $[- 1, 1]$ is nonempty; hence the indicator is proper, closed, and convex by [F1, F3]. At slope $p = 0$ , every $y \in [- 1, 1]$ contributes $0$ to the conjugate supremum, and no other $y$ is in the effective domain, so $f^{*} (0) = 0$ and every point of the interval attains this supremum.

2.1F2L1step 1.1algebra

Fix any $x \in [- 1, 1]$ . For every $y \in R$ , $f (y) \geq 0 = f (x) + 0 (y - x)$ : equality holds on the interval and the left side is $+ \infty$ outside it. Hence $0 \in \partial f (x)$ ; [L1] also gives $f^{*} (0) = 0 = 0 \cdot x - f (x)$ at each such $x$ .

3.1L2step 1.1algebra∎

The zero tilt is $g = f$ . It is coercive as an extended-real function: for any $M \in R$ , choosing $R = 2$ ensures $∣ x ∣ \geq R$ implies $g (x) = + \infty > M$ . Its minimum is $0$ , attained at every $x \in [- 1, 1]$ , so this example satisfies [L2] while exhibiting nonuniqueness and a whole flat supporting face.

Depends on

Dependency tree · 2 levels

Sources

R. T. Rockafellar, Convex Analysis, §23