Extension of convex functions from a hyperplane to a half-space

It is shown that a possibly infinite-valued proper lower semicontinuous convex function on ${\mathbb R}^n$ has an extension to a convex function on the half-space ${\mathbb R}^n\times[0,\infty)$ which is finite and smooth on the open half-space ${\mathbb R}^n\times(0,\infty)$. The result is applied to nonlinear elasticity, where it clarifies how the condition of polyconvexity of the free-energy density $\psi(Dy)$ is best expressed when $\psi(A)\to\infty$ as $\det A\to 0+$.


Introduction
The main purpose of this paper is to prove the following theorem, giving an extension of a possibly infinite-valued proper lower semicontinuous convex function on R n to a convex function on the half-space R n × [0, ∞) which is finite and smooth on the open half-space R n × (0, ∞).
Corollary 1 Let Φ : R n → (−∞, ∞] be a proper lower semicontinuous convex function.Then there exists a sequence Φ (j) of smooth convex functions on R n such that lim j→∞ Φ (j) (y) = Φ(y) for each y ∈ R n .
The theorem applies, for example, to the case when Φ is the indicator function i K of a nonempty closed convex subset K ⊂ R n , defined by With K = {0} a suitable smooth strictly convex extension is then given by φ(x, y) = θ(x, y) − x x+1 , where which follows as a special case of (4) (or (10)) below.(We note that with y momentum and x density the convexity of θ plays an important role in optimal transport, as noted in [5].) The theorem was motivated by the problem of proving the existence of energy minimizers in 3D nonlinear elasticity under the assumption of polyconvexity of the free-energy density.In [3] an apparently weaker version of the polyconvexity condition given in [2] was used.That this version is indeed weaker follows from Theorem 1, and this is explained in Section 3.

Proof of Theorem 1
We first show the existence of an extension φ satisfying (i), (ii), which in addition is (strictly) decreasing in x, giving two different proofs.The first proof is the more direct and provides a wide range of possible extensions, while the second uses infimal convolution and is convenient for proving the assertion in the theorem regarding strict convexity.
Proposition 1 Under the assumptions of Theorem 1 there exists a lower semicontinuous convex extension φ = φ(x, y) of Φ to [0, ∞) × R n that is finite for x > 0, decreasing in x, and such that lim x→0+ φ(x, y) = Φ(y) for each y ∈ R n .1st Proof.We first note that Φ is the supremum of a family of affine functions: for some nonempty set S ⊂ R n+1 .This is a standard result; see, for example, [8, Proposition 3.1], [12,Theorem 12.1].(In Remark 1 below we note that we can take the family of affine functions to consist of exact affine minorants, but this is not needed for the proof.) We claim that φ(x, y) := sup provides a suitable convex extension.Indeed by (2) φ(0, y) = Φ(y) for all y ∈ R n , and since it is the supremum of continuous affine functions φ is convex and lower semicontinuous.Given x > 0, y ∈ R n , by (3) there exists M (x, y) > 0 such that Hence for |α| + |b| > M (x, y) we have Therefore, since ψ ⩾ 0, φ(x, y) ⩽ max(1, |y|)M (x, y) < ∞, as required.φ(x, y) is nonincreasing in x, and can be made decreasing by adding −x to φ.
Let φ = Φ □ θ be the infimal convolution of Φ and θ with respect to y ∈ R n defined by The convexity of Φ and θ implies that the function is convex on R 2n+1 .Hence by [4,Prop. 8.26] φ(x, y) = inf y ′ h(x, y, y ′ ) is convex in (x, y).Since Φ is proper, there exists ȳ ∈ R n with Φ(ȳ) < ∞.Therefore for x > 0 we have that φ(x, y) ⩽ Φ(ȳ) Hence for so that φ(x, y) is finite, and thus by convexity continuous on (0, ∞) × R n .If φ were not lower semicontinuous there would exist a sequence (x j , y j ) → (0, y) and y ′ j with In particular the left-hand side of ( 14) is bounded, and so, using (12), y ′ j → y.Thus by the lower semicontinuity of Φ the left-hand side is greater than or equal to Φ(y), a contradiction.If x j → 0+ and y ∈ R n then by the lower semicontinuity Φ(y) ⩽ lim inf j→∞ φ(x j , y) ⩽ lim sup j→∞ φ(x j , y) ⩽ Φ(y), so that lim x→0+ φ(x, y) = Φ(y) as required.
Clearly φ(x, y) is nonincreasing in x.As defined it may not be decreasing (consider the case Φ ≡ 0), but φ(x, y) − x is decreasing in x and satisfies the other requirements.

⊓ ⊔
Corollary 2 Assume in addition to the hypotheses of Proposition 1 that Φ is strictly convex on dom Φ.Then φ can be chosen so that in addition it is strictly convex on (0, ∞) × R n .
Proof We use the construction in the second proof of Proposition 1. Fix x > 0. Given y ∈ R n , by (12) the minimum of h(y, z) := Φ(z) + |y−z| 2 x for z ∈ R n is attained by some z = y ′ ∈ dom Φ, and the strict convexity of Φ on dom Φ implies that y ′ is unique.Given distinct y, ȳ ∈ R n let the corresponding unique minimizers be y ′ , ȳ′ respectively.For λ ∈ (0, 1) the strict convexity of h on Hence φ(x, y) is strictly convex in y.
To complete the proof we use the following lemma.
Proof g is convex.If g were not strictly convex then there would exist distinct pairs (x 1 , y 1 ), (x 2 , y 2 ) and λ ∈ (0, 1) with It follows from ( 17) and the convexity of ψ that ψ(λx , and since ψ is strictly convex we must have x 1 = x 2 .But then (17) contradicts the strict convexity of f (x, y) in y.
⊓ ⊔ To complete the proof of Theorem 1 we mollify φ as constructed in Proposition 1 with an x-dependent mollifier.Let ρ dx dy = 1, and define for (x, y) The integral is well defined since φ is convex on [0, ∞) × R n and thus bounded below by a linear function, and the convexity of φ also implies that φ is convex.Since φ is lower semicontinuous, by Fatou's Lemma (valid because φ is bounded below by a linear function) φ is lower semicontinuous.Furthermore φ(0, y) = Φ(y). ( Making the change of variables u from which it follows that φ is smooth for x > 0.
Finally, if Φ is strictly convex on dom Φ then by Corollary 2 we can suppose that φ is strictly convex on (0, ∞)×R n , so that φ is strictly convex on (0, ∞)× R n by (18).
Remark 2 It does not seem obvious how to construct a smooth extension φ(x, y) that is decreasing in x.This does not immediately follow from the fact that φ(x, y) is decreasing in x because the mollification (18) averages φ over a range of values of y ′ that grows with x.
Remark 3 If Φ is not strictly convex on dom Φ then the function φ cannot in general be chosen to be strictly convex on R n × (0, ∞).Indeed if Φ = 0 then φ can only depend on x.To see this let x > 0, y, y ′ ∈ R n and for ε > 0 note that where z := ε −1 (xy ′ − (x − ε)y), so that by convexity Letting ε → 0 we obtain φ(x, y ′ ) ⩽ φ(x, y).Interchanging y, y ′ we deduce that φ(x, y) = φ(x, y ′ ) as required.

Remark 4
An interesting open problem is to determine the pairs Φ 0 and Φ 1 of proper lower semicontinuous convex functions on R n which are such that there is a convex function φ that is finite on (0, 1) × R n and interpolates between Φ 0 and Φ 1 in the sense that φ(0, y) = Φ 0 (y), φ(1, y) = Φ 1 (y) for all y ∈ R n and lim The set of such pairs (Φ 0 , Φ 1 ) is clearly convex.In the case Φ 0 = 0, Remark 3 shows that the only possibility is that Φ 1 is constant, while in the case Setting C = {0, 1} × R n the problem is seen to be related to that of extending a convex function on C ⊂ R s to a convex function on its convex hull co(C).This is studied for C compact in [6] and for general C in [13] (but without any assertion of continuity of the extension as C is approached as in (29)).When C is compact and convex the question of extending a smooth convex function on C to a smooth convex function on R s is discussed in [1].
We give two examples of explicit constructions of convex extensions, using the two methods in the different proofs of Proposition 1.In neither example do we need to mollify φ since it is already smooth.
Example 1 Let Φ = i {0} be the indicator function of 0 as described in the introduction.Then dom ∂Φ = {0} and ∂Φ(0) = R n , so that (4) with S given by {(0, b) which is smooth for x > 0 if p is an even integer.In fact it is not hard to check that a more general convex extension which is smooth for x > 0 is given by where η : R n → R is convex, smooth, and such that lim Example 2 Let n = 1 and Then an elementary calculation shows that φ(x, y) := (Φ □ θ) (x, y)

Polyconvexity conditions
In this section we give an application of Theorem 1 to 3D nonlinear elasticity.Denote by M 3×3 the space of real 3 × 3 matrices.Consider an elastic body occupying a bounded open set Ω ⊂ R 3 in a reference configuration.The total free energy at a constant temperature corresponding to a deformation y : Ω → R 3 is given by where the free-energy density ψ : To help prevent interpenetration of matter it is usually assumed that which implies that if I(y) < ∞ then det Dy(x) > 0 for a.e.x ∈ Ω.
In order to prove existence of an absolute minimizer of I it is necessary to suppose, among other hypotheses, that ψ satisfies a suitable convexity condition.The convexity condition assumed in [2] (see [7] for a clear and more recent exposition) is that ψ is polyconvex, that is there is a convex function g : where cofA denotes the matrix of cofactors of A. Given δ > 0, define E δ = {(A, cofA, δ) : det A = δ}.Since, as is proved in [2,Theorem 4.3], the convex hull of follows from (36) that g(A, H, δ) < ∞ for all A, H ∈ M 3×3 and δ > 0.
Later, in [3] (see also [7]) it was observed that existence could be proved if one only assumes (35) and that (36) holds for a convex g : M 3×3 × M 3×3 × (0, ∞) → R.But it is not immediately obvious that this really is a weaker hypothesis.Applying Theorem 1 we see that it is.

⊓ ⊔
From the point of view of mechanics, Theorem 2 is unsatisfactory because the ψ constructed does not satisfy the physically necessary frame-indifference condition which is not used for the proofs of existence in [2,3].In addition one would like an example which is also isotropic, so that ψ(AQ) = ψ(A) for all Q ∈ SO(3), A ∈ M 3×3 + .