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A Hopf--Lax minimiser satisfies the characteristic Euler relation at differentiability points

Statement

Let H:Rn→R be convex and superlinear with Legendre transform L, let u0:Rn→R be bounded and continuous, and let t>0, x∈Rn. Let y be a minimiser of φ(y):=u0(y)+tL((x−y)/t), which exists by Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers, put v:=(x−y)/t, and set u:=Qtu0. Then: (1) if u0 is differentiable at y and L is differentiable at v, then Du0(y)=DL(v); (2) if L is differentiable at v and u is differentiable at x, then Du(x)=DL(v). No choice principle is used.

Facts & Assumptions

Given: A convex superlinear H with Legendre transform L, bounded continuous u0, t>0, x∈Rn, a minimiser y of φ(y)=u0(y)+tL((x−y)/t), v=(x−y)/t, u=Qtu0, and the Euclidean norm ∣⋅∣.

[F1]

Qtu0(x)=inf⁡y∈Rn{u0(y)+tL((x−y)/t)}, and the infimum is attained under the present hypotheses (The Hopf--Lax operator and the Hopf--Lax formula, Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers).

[F2]

f is differentiable at a with derivative Df(a) exactly when f(a+h)=f(a)+Df(a)h+r(h) with r(h)/∣h∣→0 as h→0; in that case every directional derivative exists and equals Df(a)h (The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder, Directional derivatives and partial derivatives of a map U⊆Rm→Rn).

Proof

technique · expand the minimality inequality to first order in the direction of the perturbation
1.1F1F2algebra

First-order condition at a minimiser. Fix h∈Rn and τ∈(0,1); minimality of y at the point x gives u0(y)+tL(v)≤u0(y+τh)+tL(v−τh/t), where u0(y)+tL(v)=Qtu0(x) by [F1]. Assume u0 is differentiable at y and L at v. Then u0(y+τh)=u0(y)+τ⟨Du0(y),h⟩+ru(τh) and L(v−τh/t)=L(v)−τ⟨DL(v),h/t⟩+rL(−τh/t) with ru(τh)/∣τh∣→0 and rL(−τh/t)/∣τh∣→0 as τ↓0 by [F2]. Substituting and cancelling u0(y)+tL(v) gives τ⟨Du0(y),h⟩+ru(τh)≥τ⟨DL(v),h⟩−trL(−τh/t); dividing by τ>0 and letting τ↓0 gives ⟨Du0(y),h⟩≥⟨DL(v),h⟩.

2.1step 1.1F2algebra∎

The two equalities. The inequality of step 1.1 holds for every h∈Rn; applying it to −h as well gives ⟨Du0(y)−DL(v),h⟩≥0 and ⟨Du0(y)−DL(v),−h⟩≥0, that is ∣⟨Du0(y)−DL(v),h⟩∣≤0 for all h. Taking h=Du0(y)−DL(v) gives ∣Du0(y)−DL(v)∣2≤0, hence Du0(y)=DL(v), which is (1). For (2), minimality of y at the point x+τh gives u(x+τh)=Qtu0(x+τh)≤u0(y)+tL(v+τh/t)=u(x)+t(L(v+τh/t)−L(v)), and if L is differentiable at v and u at x then [F2] gives u(x+τh)=u(x)+τ⟨Du(x),h⟩+r1(τh) and t(L(v+τh/t)−L(v))=τ⟨DL(v),h⟩+r2(τh) with ri(τh)/(τ∣h∣)→0. Dividing by τ and letting τ↓0 gives ⟨Du(x),h⟩≤⟨DL(v),h⟩ for every h; applying this to −h yields Du(x)=DL(v) by the same argument as above, which is (2).

Remarks

  • Differentiability is assumed only where used. The minimiser exists by the localisation lemma, and the first-order conditions are obtained by perturbing the minimiser in a direction and expanding: no global smoothness of u0, L or Qtu0 is asserted, and in the convex-quadratic case H(p)=∣p∣2/2 a minimiser need not be unique when u0 is merely continuous.
  • Direction of the two relations. Part (1) relates the datum to the Lagrangian at the minimiser, part (2) relates the value function to the Lagrangian at the same minimiser; together they identify the slope of the minimising chord with the conjugate momentum.

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