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A finite-valued convex Hamiltonian equals its biconjugate

Statement

Let $n \geq 1$ and let $H : R^{n} \to R$ be convex (Convex and strictly convex functions on Euclidean convex sets), with Legendre transform $L$ as in The Legendre transform of a finite-valued convex Hamiltonian and the convention $p \cdot v - (+ \infty) : = - \infty$ . Then for every $p \in R^{n}$ $H (p) = \sup_{v \in R^{n}} (p \cdot v - L (v)),$ so that $H = L^{*} = H^{* *}$ is the biconjugate of $H$ . Equivalently, the Moreau envelopes $e_{λ} H (p) : = \inf_{q \in R^{n}} {H (q) + \frac{∣ p - q ∣^{2}}{2 λ}} (λ > 0)$ satisfy $e_{λ} H \leq L^{*} \leq H$ for every $λ > 0$ , and $e_{λ} H (p) \to H (p)$ as $λ ↓ 0$ for every $p$ . No superlinearity, differentiability, smoothness, coercivity or strict convexity of $H$ is assumed, and no choice principle is used; in particular the supporting-hyperplane route is not needed.

Facts & Assumptions

Given: An integer $n \geq 1$ , a convex function $H : R^{n} \to R$ , its Legendre transform $L (v) = \sup_{p} (p \cdot v - H (p))$ with the convention $p \cdot v - (+ \infty) : = - \infty$ , the biconjugate $L^{*} (p) = \sup_{v} (p \cdot v - L (v))$ , and the Moreau envelopes $e_{λ} H (p) = \inf_{q} {H (q) + ∣ p - q ∣^{2} / (2 λ)}$ for $λ > 0$ .

[F1]

$L (v) = \sup_{p \in R^{n}} (p \cdot v - H (p))$ is the least upper bound in $\overline{R}$ of the set ${p \cdot v - H (p) : p \in R^{n}}$ , and in the biconjugate the convention $p \cdot v - (+ \infty) : = - \infty$ is adopted (The Legendre transform of a finite-valued convex Hamiltonian).

[F2]

$H$ is convex: $H ((1 - t) x + t y) \leq (1 - t) H (x) + t H (y)$ for all $x, y \in R^{n}$ and $t \in [0, 1]$ (Convex and strictly convex functions on Euclidean convex sets).

[F3]

Every convex function on an open convex set is continuous on it; in particular $H$ is continuous on $R^{n}$ (A convex function on an open convex set is continuous).

[F4]

A nonempty subset of $R^{n}$ is compact if and only if it is closed and bounded, and every continuous real-valued function on a nonempty compact subset attains a maximum and a minimum there (For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent).

[F5]

Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , and on a nonempty subset of $R$ bounded in $R$ these agree with the real supremum and infimum (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ).

[F6]

If $S \subseteq R$ is nonempty and $ℓ = \inf S$ , then $ℓ \leq s$ for every $s \in S$ and $ℓ^{'} \leq ℓ$ for every lower bound $ℓ^{'}$ of $S$ (Greatest lower bound (infimum)).

Proof

technique · conjugate monotonicity plus an attained Moreau minimiser; the supporting inequality is derived from two-point convexity at the minimiser

1.1F1F5algebra

$L^{*} (p) \leq H (p)$ for every $p$ . Fix $p$ and $v$ . By [F1] and the least-upper-bound property of [F5], $p \cdot v - H (p) \leq L (v)$ , that is $p \cdot v - L (v) \leq H (p)$ ; if $L (v) = + \infty$ this reads $- \infty \leq H (p)$ by the convention of [F1], so the inequality holds in all cases. Hence $H (p)$ is an upper bound in $\overline{R}$ of ${p \cdot v - L (v) : v \in R^{n}}$ , and since $L^{*} (p)$ is the least upper bound of that set by [F1] and [F5], $L^{*} (p) \leq H (p)$ .

1.2F2F3F4algebra

A linear-growth lower bound for $H$ . By [F3] the function $H$ is continuous, and the closed unit ball $\overline{B} (0, 1)$ is nonempty, closed and bounded; by [F4] it is compact and $H$ attains on it a minimum $m \in R$ and a maximum $M \in R$ . Put $D : = \max {∣ M - H (0) ∣, ∣ m - H (0) ∣}$ , so that $∣ H (u) - H (0) ∣ \leq D$ for every $∣ u ∣ \leq 1$ . For $∣ q ∣ \geq 1$ put $u : = q / ∣ q ∣$ , so that $u = (1 / ∣ q ∣) q + (1 - 1 / ∣ q ∣) \cdot 0$ is a convex combination of $q$ and $0$ ; convexity [F2] gives $H (u) \leq (1 / ∣ q ∣) H (q) + (1 - 1 / ∣ q ∣) H (0)$ , hence $H (q) \geq ∣ q ∣ (H (u) - H (0)) + H (0) \geq - ∣ q ∣ ∣ H (u) - H (0) ∣ - ∣ H (0) ∣ \geq - (D + ∣ H (0) ∣) ∣ q ∣$ . For $∣ q ∣ \leq 1$ we have $H (q) \geq m \geq - (D + ∣ H (0) ∣)$ . Thus with $C : = D + ∣ H (0) ∣ + 1$ we get $H (q) \geq - C (1 + ∣ q ∣)$ for every $q \in R^{n}$ .

1.3F2F3F6algebra

$e_{λ} H (p) \to H (p)$ as $λ ↓ 0$ , for fixed $p$ . The competitor $q = p$ gives $e_{λ} H (p) \leq H (p)$ for every $λ > 0$ . Fix $ε > 0$ . By continuity of $H$ at $p$ [F3] there is $δ > 0$ with $∣ H (q) - H (p) ∣ \leq ε$ whenever $∣ q - p ∣ \leq δ$ . With $r : = ∣ q - p ∣$ and $φ (q) : = H (q) + r^{2} / (2 λ)$ : if $r \leq δ$ then $φ (q) \geq H (p) - ε$ ; if $r \geq δ$ , put $u : = (q - p) / r$ , so that $p + δ u = (1 - δ / r) p + (δ / r) q$ and convexity [F2] gives $H (p + δ u) \leq (1 - δ / r) H (p) + (δ / r) H (q)$ , that is $H (q) \geq H (p) + (r / δ) (H (p + δ u) - H (p)) \geq H (p) - (ε / δ) r$ . Hence for $r \geq δ$ we have $φ (q) \geq H (p) - (ε / δ) r + r^{2} / (2 λ)$ , and the right-hand side is minimised over $r \geq δ$ at $r = δ$ whenever $λ \leq δ^{2} / ε$ , with value $H (p) - ε + δ^{2} / (2 λ) \geq H (p) - ε / 2$ . Therefore $H (p) - ε$ is a lower bound of the set ${φ (q) : q \in R^{n}}$ whenever $0 < λ \leq δ^{2} / ε$ , and since $e_{λ} H (p)$ is its greatest lower bound [F6] we get $e_{λ} H (p) \geq H (p) - ε$ for all such $λ$ . As $ε > 0$ was arbitrary, ${lim inf}_{λ ↓ 0} e_{λ} H (p) \geq H (p)$ , which with the upper bound gives $e_{λ} H (p) \to H (p)$ .

2.1step 1.2F3F4F6algebra

The Moreau infimum is attained. Fix $p$ and $λ > 0$ and put $φ (q) : = H (q) + ∣ p - q ∣^{2} / (2 λ)$ . By [F3] $φ$ is continuous on $R^{n}$ , and by the bound of step 1.2, $φ (q) \geq - C (1 + ∣ q ∣) + ∣ p - q ∣^{2} / (2 λ)$ , whose right-hand side tends to $+ \infty$ as $∣ q ∣ \to \infty$ because $∣ p - q ∣ \geq ∣ q ∣ - ∣ p ∣$ ; hence there is $R > ∣ p ∣$ with $φ (q) > φ (p)$ for every $∣ q ∣ \geq R$ . The set $S : = {q \in R^{n} : φ (q) \leq φ (p)}$ is then nonempty (it contains $p$ ), contained in the closed ball $\overline{B} (0, R)$ , and closed (it is the preimage under the continuous $φ$ of a closed interval); being closed and bounded it is compact by [F4], so $φ$ attains on $S$ a minimum at some $q_{*} \in S$ by [F4]. For $q \notin S$ we have $φ (q) > φ (p) \geq φ (q_{*})$ , so $q_{*}$ is a global minimiser of $φ$ on $R^{n}$ , and by the definition of the infimum [F6] $φ (q_{*}) = \inf_{q} φ (q) = e_{λ} H (p)$ .

3.1step 2.1F1F2algebra

Supporting inequality and finiteness of $L (v_{*})$ at a minimiser. Keep $p, λ, q_{*}$ and $φ$ from step 2.1, put $w : = p - q_{*}$ , $v_{*} : = w / λ$ , and fix $q \in R^{n}$ with $h : = q - q_{*}$ . For $0 < r \leq 1$ the point $q_{*} + r h$ belongs to $R^{n}$ and minimality of $q_{*}$ gives $φ (q_{*}) \leq φ (q_{*} + r h)$ ; convexity [F2] gives $H (q_{*} + r h) \leq (1 - r) H (q_{*}) + r H (q)$ . Subtracting and using $∣ w - r h ∣^{2} = ∣ w ∣^{2} - 2 r w \cdot h + r^{2} ∣ h ∣^{2}$ , these two inequalities yield $r (H (q) - H (q_{*})) \geq r w \cdot h / λ - r^{2} ∣ h ∣^{2} / (2 λ)$ ; dividing by $r > 0$ and letting $r ↓ 0$ gives $H (q) \geq H (q_{*}) + v_{*} \cdot h$ . Hence $q \cdot v_{*} - H (q) \leq q_{*} \cdot v_{*} - H (q_{*})$ for every $q$ , with equality at $q = q_{*}$ , so the least upper bound $L (v_{*})$ of [F1] equals $q_{*} \cdot v_{*} - H (q_{*}) \in R$ .

4.1step 2.1step 3.1F1F5algebra

$L^{*} (p) \geq e_{λ} H (p)$ . By [F1] and [F5], $L^{*} (p) \geq p \cdot v_{*} - L (v_{*})$ for the vector $v_{*}$ of step 3.1 (the value $L (v_{*})$ is real, so $p \cdot v_{*} - L (v_{*})$ is an ordinary real number and no convention is needed). Substituting $L (v_{*}) = q_{*} \cdot v_{*} - H (q_{*})$ from step 3.1 and $v_{*} = (p - q_{*}) / λ$ , we get $L^{*} (p) \geq (p - q_{*}) \cdot v_{*} + H (q_{*}) = ∣ p - q_{*} ∣^{2} / λ + H (q_{*}) \geq ∣ p - q_{*} ∣^{2} / (2 λ) + H (q_{*}) = e_{λ} H (p)$ , where the last equality is step 2.1.

5.1step 1.1step 1.3step 4.1∎

Conclusion. Steps 1.1 and 4.1 give $e_{λ} H (p) \leq L^{*} (p) \leq H (p)$ for every $λ > 0$ and every $p$ , and step 1.3 gives $e_{λ} H (p) \to H (p)$ as $λ ↓ 0$ ; hence $L^{*} (p) = \sup_{v} (p \cdot v - L (v)) = H (p)$ for every $p$ , that is $H = L^{*}$ .

Remarks

What replaces the subgradient theorem. The only existence input is the attained minimiser of the strictly convex perturbation $q \mapsto H (q) + ∣ p - q ∣^{2} / (2 λ)$ , obtained from continuity, a linear lower bound and compactness. The supporting inequality $H (q) \geq H (q_{*}) + v_{*} \cdot (q - q_{*})$ is then a two-point convexity computation at that minimiser, so no supporting-hyperplane theorem and no choice principle is consumed.
Sharpness of hypotheses. Neither superlinearity nor coercivity of $H$ is used: the quadratic penalty provides the coercivity, and the continuity of $H$ is a consequence of convexity and finite-valuedness by [F3].

Depends on

Used by

The Hopf--Lax formula solves the Hamilton--Jacobi Cauchy problem Theorem

Dependency tree · two levels

40 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Hung Vinh Tran, Hamilton--Jacobi Equations: Theory and Applications, 2020 preliminary author manuscript of AMS Graduate Studies in Mathematics 213 (complete text) (standard reference, not scraped)
Christian Clason, Nonsmooth Analysis and Optimization, lecture notes winter 2021/22, February 18, 2022 (standard reference, not scraped)