A WHITNEY EXTENSION THEOREM FOR CONVEX FUNCTIONS …dazagrar/articulos/AMC11Extension... · A...

A WHITNEY EXTENSION THEOREM FOR CONVEX

FUNCTIONS OF CLASS C1,ω.

DANIEL AZAGRA AND CARLOS MUDARRA

Abstract. Let C be a compact subset of Rn (not necessarily convex),f : C → R be a function, and G : C → Rn be a continuous function, withmodulus of continuity ω. We provide necessary and sufficient conditionson f , G for the existence of a convex function F ∈ C1,ω(Rn) such thatF = f on C and ∇F = G on C. We also give a geometrical applicationconcerning a characterization of compact subsets K of Rn which can beinterpolated by boundaries of C1,1 convex bodies with prescribed outernormals on K.

1. Introduction and main results

Throughout this paper, by a modulus of continuity ω we understand aconcave, strictly increasing function ω : [0,∞) → [0,∞) such that ω(0+) =0. In particular, ω has an inverse ω−1 : [0, β)→ [0,∞) which is convex andstrictly increasing, where β > 0 may be finite or infinite. Furthermore, ω issubadditive, and satisfies ω(λt) ≤ λω(t) for λ ≥ 1, and ω(µt) ≥ µω(t) for0 ≤ µ ≤ 1. It is well-known that for every uniformly continuous functionf : X → Y between two metric spaces there exists a modulus of continuityω such that dY (f(x), f(z)) ≤ ω (dX(x, z)) for every x, z ∈ X.

Let C be a compact subset of Rn, f : C → R a function, and G : C → Rn amapping with modulus of continuity ω. Assume that the pair (f,G) satisfies

|f(x)− f(y)− 〈G(y), x− y〉| ≤M |x− y|ω (|x− y|) (W 1,ω)

for every x, y ∈ C (in the case that ω(t) = t we will also denote this conditionby (W 1,1)). Then a well-known version of the Whitney extension theoremfor the class C1,ω holds true (see [5, 9]), and we get the existence of a functionF ∈ C1,ω(Rn) such that F = f and ∇F = G on C.

It is natural to ask what further assumptions on f , G would be necessaryand sufficient to ensure that F can be taken to be convex. In a recent paper[1], we solved a similar problem for the class of C1 convex functions on Rn,

Date: July 14, 2015.2010 Mathematics Subject Classification. 54C20, 52A41, 26B05, 53A99, 53C45, 52A20,

58C25, 35J96.Key words and phrases. extension of functions, convex function, C1,1 function.C. Mudarra was supported by Programa Internacional de Doctorado de la Fundacion La

Caixa–Severo Ochoa. Both authors were partially supported by Grant MTM2012-34341.

1

2 DANIEL AZAGRA AND CARLOS MUDARRA

and also, under the additional assumption that C be convex, the correspond-ing problem for C∞ smooth convex functions. We refer to the introductionof [1] and the references therein for background. Here we will only mentionthat results of this nature have interesting applications in differential geom-etry, PDE theory (such as Monge-Ampere equations), nonlinear dynamics,and quantum computing, see [2, 3, 4, 12] and the references therein.

Let us introduce one condition which, we have found, together with(W 1,ω), is necessary and sufficient for a function f : C → R to have aconvex extension F of class C1,ω(Rn) such that ∇F = G on C. For amapping G : C ⊂ Rn → Rn we will denote

M(G,C) := supx,y∈C, x6=y

|G(x)−G(y)|ω(|x− y|)

.

We will say that f and G satisfy the property (CW 1,ω) on C if either G isconstant, or else there exists a constant η ∈ (0, 1/2] such that

f(x)− f(y)− 〈G(y), x− y〉 ≥ η|G(x)−G(y)|ω−1(

1

2M|G(x)−G(y)|

),

(CW 1,ω)for all x, y ∈ C, where

M = M(G,C).

Throughout this paper we will assume that M > 0, or equivalently that Gis nonconstant. We may of course do so, because if M = 0 then our problemhas a trivial solution (namely, the function x 7→ f(x0) + 〈G(x0), x − x0〉defines an affine extension of f to Rn, for any x0 ∈ C).

In the case that ω(t) = t, we will also denote this condition by (CW 1,1).That is, (f,G) satisfies (CW 1,1) if and only if there exists δ > 0 such that

f(x)− f(y)− 〈G(y), x− y〉 ≥ δ |G(x)−G(y)|2, (CW 1,1)

for all x, y ∈ C.Our main result is the following.

Theorem 1.1. Let ω be a modulus of continuity. Let C be a compact (notnecessarily convex) subset of Rn. Let f : C → R be an arbitrary function,and G : C → Rn be continuous, with modulus of continuity ω. Then f hasa convex, C1,ω extension F to all of Rn, with ∇F = G on C, if and only iff and G satisfy the conditions (W 1,ω) and (CW 1,ω) on C.

In particular, for the most important case that ω(t) = t, we have thefollowing.

Corollary 1.2. Let C be a compact (not necessarily convex) subset of Rn.Let f : C → R be an arbitrary function, and G : C → Rn be a Lipschitzfunction. Then f has a convex, C1,1 extension F to all of Rn, with ∇F = Gon C, if and only if f and G satisfy the conditions (W 1,1) and (CW 1,1) onC.

C1,1 CONVEX EXTENSIONS OF CONVEX FUNCTIONS 3

It is worth noting that our proofs provide good control of the modulus ofcontinuity of the gradients of the extensions F , in terms of that of G. In fact,assuming η = 1/2 in (CW 1,ω), which we can fairly do (see Proposition 2.2below), there exists a constant k(n) > 0, depending only on the dimensionn, such that

M(∇F,Rn) ≤ k(n)M(G,C).

Since convex functions on Rn are not bounded (unless they are constant),the most usual definitions of norms in the space C1,ω(Rn) are not adequateto estimate convex functions whose gradients are uniformly continuous withmodulus of uniform continuity ω. In this paper, for a convex function F :Rn → R we will denote

(1.1) ‖F‖1,ω = |F (0)|+ |∇F (0)|+ supx,y∈Rn, x 6=y

|∇F (x)−∇F (y)|ω(|x− y|)

.

With this notation, and assuming 0 ∈ C and η = 1/2 in (CW 1,ω), the proofof Theorem 1.1 shows that

‖F‖1,ω ≤ k(n) (|f(0)|+ |G(0)|+M(G,C)) .

Let us conclude this introduction with a geometrical application of Corol-lary 1.2 concerning a characterization of compact subsets K of Rn which canbe interpolated by boundaries of C1,1 convex bodies (with prescribed unitouter normals on K). Namely, if K is a compact subset of Rn and we aregiven a Lipschitz map N : K → Rn such that |N(y)| = 1 for every y ∈ K, itis natural to ask what conditions on K and N are necessary and sufficientfor K to be a subset of the boundary of a C1,1 convex body V such that0 ∈ int(V ) and N(y) is outwardly normal to ∂V at y for every y ∈ K. Asuitable set of conditions is:

(O) 〈N(y), y〉 > 0 for all y ∈ K;

(KW1,1) 〈N(y), y − x〉 ≥ η

M|N(y)−N(x)|2 for all x, y ∈ K;

(W1,1) 〈N(y), y − x〉 ≤M |x− y|2, for all x, y ∈ K,

for some M > 0, η ∈ (0, 12 ].Our main result in this direction is as follows.

Theorem 1.3. Let K be a compact subset of Rn, and let N : K → Rn be aLipschitz mapping such that |N(y)| = 1 for every y ∈ K. Then the followingstatements are equivalent:

(1) There exists a C1,1 convex body V with 0 ∈ int(V ) and such thatK ⊆ ∂V and N(y) is outwardly normal to ∂V at y for every y ∈ K.

(2) K and N satisfy conditions (O), (KW1,1) and (W1,1).

This result may be compared to [2], where M. Ghomi showed how toconstruct Cm smooth convex bodies V with prescribed strongly convex sub-manifolds and tangent planes. In [1, Theorem 1.11] we obtained a resultin this spirit which allowed us to deal with arbitrary compacta instead of


manifolds, and to drop the strong convexity assumption, in the special casem = 1. The above Theorem provides a similar result for the class of C1,1

bodies.Let us finally note that in the above results the assumption that C and K

be compact cannot be removed in general, see [8, Example 4], [10, Example3.2], and [1, Example 4.1].

2. Necessity

It is clear that condition (W 1,ω) is necessary in Theorem 1.1. Let us provethe necessity of condition (CW 1,ω). We will use the following.

Lemma 2.1. Let ω : [0,∞) → [0,∞) be a modulus of continuity, let a, b, ηbe real numbers with a > 0, η ∈ (0, 12 ], and define h : [0,∞)→ R by

h(s) = −as+ b+ ω(s)s.

Assume that b < ηaω−1(ηa). Then we have

h(ω−1 (ηa)

)< 0.

Proof. We can write

h(ω−1 (ηa)

)= −a(1− 2η)ω−1(ηa) + b− ηaω−1(ηa),

and the result follows at once. �

Proposition 2.2. Let f ∈ C1,ω(Rn) be convex. Then

f(x)−f(y)−〈∇f(y), x−y〉 ≥ 1

2|∇f(x)−∇f(y)|ω−1

(1

2M|∇f(x)−∇f(y)|

)for all x, y ∈ Rn, where

M = M(∇f,Rn) = supx,y∈Rn, x 6=y

|∇f(x)−∇f(y)|ω (|x− y|)

.

In particular, the pair (f,∇f) satisfies (CW 1,ω), with η = 1/2, on everycompact subset C ⊂ Rn.

Proof. Suppose that there exist different points x, y ∈ Rn such that

f(x)−f(y)−〈∇f(y), x−y〉 < 1

2|∇f(x)−∇f(y)|ω−1

(1

2M|∇f(x)−∇f(y)|

),

and we will get a contradiction.Case 1. Assume further that M = 1, f(y) = 0, and ∇f(y) = 0. Byconvexity this implies f(x) ≥ 0. Then we have

0 ≤ f(x) <1

2|∇f(x)|ω−1

(1

2|∇f(x)|

).

Call a = |∇f(x)| > 0, b = f(x), set

v = − 1

|∇f(x)|∇f(x),


and define

ϕ(t) = f(x+ tv)

for every t ∈ R. We have ϕ(0) = b, ϕ′(0) = −a, and ω is a modulus ofcontinuity of the derivative ϕ′. This implies that

|ϕ(t)− b+ at| ≤ tω(t)

for every t ∈ R+, hence also that

ϕ(t) ≤ h(t) for all t ∈ R+,

where h(t) = −at+ b+ tω(t). By assumption,

b <1

2aω−1

(1

2a

),

and then Lemma 2.1 implies that

f

(x+ ω−1

(1

2a

)v

)= ϕ

(ω−1

(1

2a

))≤ h

(ω−1

(1

2a

))< 0,

which is in contradiction with the assumptions that f is convex, f(y) = 0,and ∇f(y) = 0. This shows that

f(x) ≥ 1

2|∇f(x)|ω−1

(1

2|∇f(x)|

).

Case 2. Assume only that M = 1. Define

g(z) = f(z)− f(y)− 〈∇f(y), z − y〉for every z ∈ Rn. Then g(y) = 0 and ∇g(y) = 0. By Case 1, we get

g(x) ≥ 1

2|∇g(x)|ω−1

(1

2|∇g(x)|

),

and since ∇g(x) = ∇f(x)−∇f(y) the Proposition is thus proved in the casewhen M = 1.Case 3. In the general case, we may assume M > 0 (the result is trivialfor M = 0). Consider ψ = 1

M f , which satisfies the assumption of Case 2.Therefore

ψ(x)−ψ(y)−〈∇ψ(y), x−y〉 ≥ 1

2|∇ψ(x)−∇ψ(y)|ω−1

(1

2|∇ψ(x)−∇ψ(y)|

),

which is equivalent to the desired inequality. �

3. Sufficiency

Throughout this section C will denote a compact subset of Rn.

Lemma 3.1. Assume that f : C → R and G : C → Rn satisfy the conditions(W 1,ω) and (CW 1,ω) on C. We have the following inequalities:

(i) For all x, y ∈ C,0 ≤ f(x)− f(y)− 〈G(y), x− y〉 ≤ 〈G(y)−G(x), y − x〉.


(ii) If h ∈ Rn, x0, y ∈ C are such that

mC(f)(x0 + h) = f(y) + 〈G(y), x0 + h− y〉,then

f(x0)− f(y)− 〈G(y), x0 − y〉 ≤ 〈G(y)−G(x0), h〉.(iii) For h ∈ Rn, x0, y ∈ C such that

mC(f)(x0 + h) = f(y) + 〈G(y), x0 + h− y〉,we have that

|G(x0)−G(y)| ≤ 2M

ηω (|h|) ,

where we denote

(3.1) M = M(G,C) = supx,y∈C, x6=y

|G(x)−G(y)|ω (|x− y|)

.

Proof.(i) It is enough to use that f(y) ≥ f(x)+ 〈G(x), y−x〉 (which in turn followsfrom (CW 1,ω)).

(ii) By the definition of mC , we have that

f(x0) + 〈G(x0), h〉 ≤ f(y) + 〈G(y), x0 + h− y〉.Then, it is easy to deduce that

f(x0)− f(y)− 〈G(y), x0 − y〉 ≤ 〈G(y)−G(x0), h〉.(iii) By (ii) we have that

f(x0)− f(y)− 〈G(y), x0 − y〉 ≤ |G(y)−G(x0)||h|,and from condition (CW 1,ω) we immediately deduce that

|G(x0)−G(y)| ≤ 2M

ηω (|h|) .

�

In the sequel we assume that f : C → R and G : C → Rn satisfy theconditions (W 1,ω) and (CW 1,ω) on C. By the C1,ω version of Whitney’sExtension Theorem (see [5, 9]), we may further assume that f is extendedto a function of class C1,ω(Rn) with support contained in C + B(0, 1), andsuch that ∇f = G on C. Moreover, bearing in mind Lemma 3.1(i), it followsfrom the construction in [9, Chapter VI] that there is a constant c(n) > 0,depending only on the dimension n, such that

(3.2) M(∇f,Rn) = supx,y∈Rn, x 6=y

|∇f(x)−∇f(y)|ω (|x− y|)

≤ c(n)M,

where M is defined in (3.1). Since the case M = 0 is trivial, we also assumeM > 0. We will denote

mC(f)(x) = supy∈C{f(y) + 〈∇f(y), x− y〉}


for every x ∈ Rn. The function mC(f) : Rn → R is obviously convex andLipschitz. In the case when C has nonempty interior, mC(f) is the minimalconvex extension of f|C : C → R to all of Rn (however, if C has emptyinterior then there is no minimal convex extension operator). See [8, 1] formore information on mC(f).

Lemma 3.2. Under the above assumptions, we have:

|f(x)−mC(f)(x)| ≤(c(n) +

2

η

)Mω(d(x,C))d(x,C), for all x ∈ Rn.

Proof. Let x ∈ Rn, x0 ∈ C be such that |x−x0| = d(x,C), define h := x−x0(that is, x = x0 + h), and take a point y ∈ C such that mC(f)(x) =f(y) + 〈∇f(y), x− y〉. We have

|f(x)−mC(f)(x)| ≤|f(x)− f(x0)− 〈∇f(x0), x− x0〉|+ |mC(f)(x)− f(x0)− 〈∇f(x0), x− x0〉|.

Using inequality (3.2), the first term satisfies

|f(x)−f(x0)−〈∇f(x0), h〉| ≤M(∇f,Rn)ω(|h|)|h| ≤ c(n)Mω(d(x,C))d(x,C).

The second term can be estimated as follows:

0 ≤ mC(f)(x)− f(x0)− 〈∇f(x0), x− x0〉= f(y) + 〈∇f(y), x− y〉 − f(x0)− 〈∇f(x0), x− x0〉≤ f(y) + 〈∇f(y), x− y〉 − f(y)− 〈∇f(y), x0 − y〉 − 〈∇f(x0), x− x0〉= 〈∇f(y)−∇f(x0), x− x0〉 ≤ |∇f(y)−∇f(x0)||x− x0|

By Lemma 3.1 (iii) we have

|∇f(y)−∇f(x0)| ≤2M

ηω(|h|) =

2M

ηω(d(x,C)).

This implies that

0 ≤ mC(f)(x)− f(x0)− 〈∇f(x0), x− x0〉 ≤2M

ηω(d(x,C))d(x,C).

In particular, mC(f) is ω-differentiable on points x ∈ C, with ∇mC(f)(x) =∇f(x). 1 Therefore

|f(x)−mC(f)(x)| ≤(c(n) +

2

η

)Mω(d(x,C))d(x,C).

�

1Recall that a function ψ is said to be ω-differentiable on C provided there exists aconstant K > 0 such that |ψ(x)− ψ(y)− 〈∇ψ(y), x− y〉| ≤ K|x− y|ω (|x− y|) for everyy ∈ C and every x in a neighbourhood of C.


Our next step is to obtain a function ϕ of class C1,ω(Rn) which vanisheson C and is greater than or equal to the function |f −mC(f)| on Rn \ C.To this purpose we need to use a Whitney descomposition of an open setinto cubes and the corresponding partition of unity, see [9, Chapter VI] foran exposition of this technique. So let {Qk}k be a decomposition of Rn \ Cinto Whitney cubes, and for a fixed number 0 < ε0 < 1/4, consider thecorresponding cubes {Q∗k}k with the same center as Qk and dilated by thefactor 1+ε0. We next sum up some of the most important properties of thiscubes.

Proposition 3.3. The families {Qk}k and {Q∗k}k are sequences of compactcubes for which:

(i)⋃kQk = Rn \ C.

(ii) The interiors of Qk are mutually disjoint.(iii) diam(Qk) ≤ d(Qk, F ) ≤ 4 diam(Qk) for all k.(iv) If two cubes Ql and Qk touch each other, that is ∂Ql ∩ ∂Qk 6= ∅, thendiam(Ql) ≈ diam(Qk).

(v) Every point of Rn \ C is contained in an open neighbourhood which in-tersects at most N = (12)n cubes of the family {Q∗k}k.

(vi) If x ∈ Qk, then d(x,C) ≤ 5 diam(Qk).(vii) If x ∈ Q∗k, then 3

4 diam(Qk) ≤ d(x,C) ≤ (6 + ε0) diam(Qk) and inparticular Q∗k ⊂ Rn \ C.

(viii) If two cubes Q∗l and Q∗k are not disjoint, then diam(Ql) ≈ diam(Qk).

Here the notation Ak ≈ Bl means that there exist positive constants γ,Γ,depending only on the dimension n, such that γAk ≤ Bl ≤ ΓAk for all k, lsatisfying the properties specified in each case.

The following Proposition summarizes the basic properties of the Whitneypartition of unity associated to these cubes.

Proposition 3.4. There exists a sequence of functions {ϕk}k defined onRn \ C such that

(i) ϕk ∈ C∞(Rn \ C).(ii) 0 ≤ ϕk ≤ 1 on Rn \ C and supp(ϕk) ⊆ Q∗k.(iii)

∑k ϕk = 1 on Rn \ C.

(vi) For all multiindex α there exists a constant Aα > 0, depending onlyon α and on the dimension n, such that

|Dαϕk(x)| ≤ Aα diam(Qk)−|α|,

for all x ∈ Rn \ C and for all k.

The statements contained in the following Proposition must look fairlyobvious to those readers well acquainted with Whitney’s techniques, but wehave not been able to find an explicit reference, so we include a proof forthe general reader’s convenience.

Proposition 3.5. Consider the family of cubes {Qk}k asociated to Rn \ Cand its partition of unity {ϕk}k as in the preceding Propositions.


(i) Suppose that there is a sequence of nonnegative numbers {pk}k anda positive constant λ > 0 such that pk ≤ λ ω(diam(Qk)) diam(Qk),for every k. Then, the function defined by

ϕ(x) =

{ ∑k pk ϕk(x) if x ∈ Rn \ C,

0 if x ∈ C

is of class C1,ω(Rn) and there is a constant γ(n) > 0, depending onlyon the dimension n, such that

(3.3) M(∇ϕ,Rn) := supx,y∈Rn, x 6=y

|∇ϕ(x)−∇ϕ(y)|ω (|x− y|)

≤ γ(n)λ.

(ii) There exists a function ∆ ∈ C1,ω(Rn) such that ∆ = 0 on C and

ω(d(x,C))d(x,C)

6 + ε0≤ ∆(x) ≤ (6 + ε0)16

9ω(d(x,C))d(x,C), for all x ∈ Rn.

In addition, we have:

M(∇∆,Rn) = supx,y∈Rn, x 6=y

|∇∆(x)−∇∆(y)|ω (|x− y|)

≤ (6 + ε0)γ(n).

Proof. Let us start with the proof of (i). Since every point in Rn \C has anopen neighbourhood which intersects at most N = (12)n cubes, and all thefunctions ϕk are of class C∞, it is clear that ϕ ∈ C∞(Rn \C). Given a pointx ∈ Rn, by our assumptions on the sequence {pk}k, we easily have

ϕ(x) =∑Q∗

k3xpkϕk(x) ≤ λ

∑Q∗

k3xω(diam(Qk)) diam(Qk)ϕk(x).

Using Proposition 3.3 we have that diam(Qk) ≤ 43d(x,C) for those k such

that x ∈ Q∗k. Then, we can write

ϕ(x) ≤ λ∑Q∗

k3xω

(4

3d(x,C)

)4

3d(x,C)ϕk(x) ≤

λ

(4

3

)2

ω(d(x,C))d(x,C)∑Q∗

k3xϕk(x) =

(4

3

)2

λω(d(x,C))d(x,C).

In particular, due to the fact that ω(0+) = 0, the above estimation showsthat ϕ is differentiable on C with ∇ϕ = 0 on C. We next give an estimationfor ∇ϕ. Bearing in mind Proposition 3.4, we set

(3.4) A = A(n) := max{√nAα, |α| ≤ 2}.

Given a point x ∈ Rn, we have that

|∇ϕ(x)| ≤∑Q∗

k3xpk|∇ϕk(x)| ≤ A

∑Q∗

k3xpk diam(Qk)

−1 ≤ Aλ∑Q∗

k3xω(diam(Qk))

≤ A λ∑Q∗

k3xω

(4

3d(x,C)

)≤ 4AN

3λ ω(d(x,C)).


Summing up,

(3.5) |∇ϕ(x)| ≤ 4AN

3λ ω(d(x,C)) for every x ∈ Rn.

Note that the above shows inequality (3.3) when x ∈ Rn and y ∈ C. Hence,in the rest of the proof we only have consider the situation where x, y aresuch that x 6= y and x, y ∈ Rn \C. However, we will still have to separatelyconsider two cases. Let us denote L := [x, y], the straight line connecting xto y.

Case 1: d(L,C) ≥ |x − y|. Take a multiindex α with |α| = 1. Because inthis case the segment L is necessarily contained in Rn \ C, and thefunction ϕ is of class C2 on Rn \ C, we can write

(3.6) |Dαϕ(x)−Dαϕ(y)| ≤(

supz∈L|∇(Dαϕ)(z)|

)|x− y|.

Using Proposition 3.4 and the definition of A in (3.4), we maywrite

|∇(Dαϕ)(z)| ≤∑Q∗

k3zpk|∇(Dαϕk)(z)| ≤ A

∑Q∗

k3zpk diam(Qk)

−2 ≤ Aλ∑Q∗

k3z

ω(diam(Qk))

diam(Qk).

By Proposition 3.3, we have that (6 + ε0) diam(Qk) ≥ d(z, C) ≥d(L,C) ≥ |x − y| for those k with Q∗k 3 z and by the properties ofω, we have that

Aλ∑Q∗

k3z

ω(diam(Qk))

diam(Qk)≤ Aλ

∑Q∗

k3z

ω(|x−y|6+ε0

)|x−y|6+ε0

≤ (6 + ε0)ANλω(|x− y|)|x− y|

.

Therefore, by substituting in (3.6) we find that

|∇ϕ(x)−∇ϕ(y)| ≤ (6 + ε0)√nANλω(|x− y|).

Case 2: d(L,C) ≤ |x−y|. Take points x′ ∈ L and y′ ∈ C such that d(L,C) =|x′ − y′| ≤ |x− y|. We have that

|x− y′| ≤ |x− x′|+ |y′ − x′| ≤ |x− y|+ |x′ − y′| ≤ 2|x− y|,

and similarly we obtain |y− y′| ≤ 2|x− y|. Hence, if we use (3.5) weobtain

|∇ϕ(x)−∇ϕ(y)| ≤ |∇ϕ(x)|+ |∇ϕ(y)| ≤ 4AN

3λ (ω(d(x,C)) + ω(d(y, C)))

≤ 4AN

3λ(ω(|x− y′|) + ω(|y − y′|)

)≤ 8AN

3λω(2|x− y|) ≤ 16AN

3λω(|x− y|).

If we call

γ(n) = max

{4AN

3, (6 + ε0)

√nAN,

16AN

3

}= (6 + ε0)

√nAN,


we get (3.3).Now let us prove the part (ii). Define the sequence

pk = ω((6 + ε0) diam(Qk)) diam(Qk), for all k.

We have that pk ≤ (6 + ε0)ω(diam(Qk)) diam(Qk) and, by (i), the functiondefined by

∆(x) =


0 if x ∈ Cis of class C1,ω(Rn), vanishes on C, and satisfies M(∇∆,Rn) ≤ (6 + ε0)γ(n)(that is, the second part of (ii)). Given a point x ∈ Rn \ C, we know that(6 + ε0) diam(Qk) ≥ d(x,C) if x ∈ Q∗k. Then

∆(x) =∑Q∗

k3xpk ϕk(x) ≥ ω(d(x,C))d(x,C)

6 + ε0

∑Q∗

k3xϕk(x) =

ω(d(x,C))d(x,C)

6 + ε0.

On the other hand, from Proposition 3.3, we have that diam(Qk) ≤ 43d(x,C)

if x ∈ Q∗k. This yields

∆(x) =∑Q∗

k3xpk ϕk(x) ≤ (6 + ε0)ω

(4

3d(x,C)

)(4

3d(x,C)

)

≤ (6 + ε0)16

9ω(d(x,C))d(x,C),

hence the first inequality in (ii) is proved. �

Let us continue with our proof. Consider the numbers

pk := supx∈Q∗

k

{|f(x)−mC(f)(x)|}, for every k.

By Lemma 3.2, |f(x) − mC(f)(x)| ≤ M∗ω(d(x,C))d(x,C), where M∗ =(c(n) + 2η−1)M, and if k is such that x ∈ Q∗k, the last term is smaller thanor equal to

M∗ω((6+ε0) diam(Qk))((6+ε0) diam(Qk)) ≤ (6+ε0)2M∗ω(diam(Qk)) diam(Qk).

This shows that

pk ≤ (6 + ε0)2M∗ω(diam(Qk)) diam(Qk) for every k,

and, by Theorem 3.5, the function defined by

ϕ(x) =


0 if x ∈ C

is of class C1,ω(Rn) and

M(∇ϕ,Rn) ≤ γ(n)(6 + ε0)2M∗ = γ(n)(6 + ε0)

2(c(n) + 2η−1)M.

We claim that ϕ ≥ |f −mC(f)|. Indeed, if x ∈ C, there is nothing to say. Ifx ∈ Rn \ C, by the definition of the pk’s we easily have

ϕ(x) =∑Q∗

k3xpk ϕk(x) ≥

∑Q∗

k3x|f(x)−mC(f)(x)|ϕk(x) = |f(x)−mC(f)(x)|.


Now, making use of Proposition 3.5(ii), we obtain a function ∆ ∈ C1,ω(Rn)such that ∆ vanishes on C,

∆(x) ≥ ω(d(x,C))d(x,C)

6 + ε0for all x ∈ Rn,

and M(∇∆,Rn) ≤ (6+ε0)γ(n). Let us consider the function g = f+ϕ+M∆.Our function g is clearly of class C1,ω(Rn) and satisfies g = f , ∇g = ∇f onC. Using that ϕ ≥ |f −mC(f)|, we easily obtain that

g(x) ≥ f(x) + |mC(f)(x)− f(x)|+M∆(x) ≥ mC(f)(x),

that is, g ≥ mC(f) on Rn. Since f has compact support and ϕ ≥ 0, we alsohave that

g(x) ≥ −‖f‖∞ +M∆(x) ≥ −‖f‖∞ +Mω(d(x,C))d(x,C)

6 + ε0,

which tends to +∞ as |x| goes to +∞. Hence lim|x|→+∞ g(x) = +∞. Wecan also estimate M(∇g,Rn) as follows,

M(∇g,Rn) ≤M(∇f,Rn) +M(∇ϕ,Rn) +M ·M(∇∆,Rn)

≤ c(n)M + γ(n)(6 + ε0)2(c(n) + 2η−1)M + (6 + ε0)γ(n)M

= µ(n, η)M,

where

(3.7) µ(n, η) := c(n) + γ(n)(6 + ε0)2(c(n) + 2η−1) + (6 + ε0)γ(n).

Finally, let us define F := conv(g), the convex envelope of the function g.Namely,

conv(g)(x) = sup{h(x) : h is convex , h ≤ g}, x ∈ Rn.In order to finish our proof we need to use the following result of Kirchheimand Kristensen on differentiability of convex envelopes (see [6] for the proof).

Theorem 3.6 (Kirchheim-Kristensen). If ψ : Rn → R is an ω-differentiablefunction on Rn such that lim|x|→+∞ ψ(x) = +∞, then conv(ψ) is a convex

function on class C1,ω(Rn), and

M(∇conv(ψ),Rn) ≤ 4(n+ 1)M(∇ψ,Rn).

As a matter of fact, Theorem 3.6 is a restatement of a particular case ofKirchheim and Kristensen’s result in [6], and the estimation on the modulusof continuity of the gradient of the convex envelope does not appear in theiroriginal statement, but it does follow, with some extra work, from theirproof.

According to Theorem 3.6, our function F is a convex function of classC1,ω(Rn). It only remains to see that F = f on C and ∇F = ∇f onC. Using the fact that g ≥ mC(f) and mC(f) is convex, we must haveF ≥ mC(f). On the other hand, because g = f on C, for every x ∈ C wehave F (x) ≤ g(x) = f(x) = mC(f)(x). Hence F = mC(f) = f on C. Inorder to show that ∇F (y) = ∇f(y) for every y ∈ C, we use the following


well-known criterion for differentiability of convex functions, whose proof isstraightforward and can be left to the interested reader.

Lemma 3.7. If φ is convex, ψ is differentiable at y, φ ≤ ψ, and φ(y) = ψ(y),then φ is differentiable at y, with ∇φ(y) = ∇ψ(y).

Since we know that mC(f) ≤ F , mC(f)(y) = f(y) = F (y) for all y ∈ C,and F ∈ C1(Rn), it follows from this criterion that

∇f(y) = ∇mC(f)(y) = ∇F (y) for all y ∈ C.As for the control of the modulus of continuity of the extension, since

M(∇g,Rn) ≤ µ(n, η)M (see the inequalities before (3.7)), Theorem 3.6 givesus

M(∇F,Rn) ≤ 4(n+ 1)µ(n, η)M.

Assuming η = 1/2, let us denote k(n) = 4(n+ 1)µ(n, 1/2). We have provedthat

M(∇F,Rn) ≤ k(n)M(G,C),

where k(n) > 0 only depends on the dimension n. In terms of the normdefined in (1.1), assuming 0 ∈ C and η = 1/2, we have that F (0) = f(0)and ∇F (0) = G(0), and therefore

‖F‖C1,ω = |F (0)|+|∇F (0)|+M(∇F,Rn) ≤ k(n) (|f(0)|+ |∇f(0)|+M(G,C)) .

4. Proof of Theorem 1.2.

(2) =⇒ (1): Set C = K ∪ {0}. We may assume η > 0 small enough, andchoose α > 0 sufficiently close to 1, so that

0 < 1− α+η

M< min

y∈K〈N(y), y〉

(this is possible thanks to condition (O), Lipschitzness ofN and compactnessof K; notice in particular that 0 /∈ K). Now define f : C → R and G : C →Rn by

f(y) =

{1 if y ∈ Kα if y = 0,

and G(y) =

{N(y) if y ∈ K

0 if y = 0.

By using conditions (KW1,1) and (W1,1), it is straightforward to check thatf and G satisfy conditions (CW 1,1) and (W 1,1). Therefore, according toTheorem 1.2, there exists a convex function F ∈ C1,1(Rn) such that F = fand ∇F = G on C. Moreover, the proof of Theorem 1.2 indicates that Fcan be taken so as to satisfy lim|x|→∞ F (x) =∞. If we define V = {x ∈ Rn :

F (x) ≤ 1} we then have that V is a compact convex body of class C1,1 with0 ∈ int(V ) (because F (0) = α < 1), and ∇F (y) = N(y) is outwardly normalto {x ∈ Rn : F (x) = 1} = ∂V at each y ∈ K. Moreover, F (y) = f(y) = 1for each y ∈ K, hence K ⊆ ∂V .

(1) =⇒ (2): Let µV be the Minkowski functional of V . By composing µVwith a C∞ convex function φ : R → R such that φ(t) = |t| if and only if


|t| ≥ 1/2, we obtain a function F (x) := φ(µV (x)) which is of class C1,1 andconvex on Rn and coincides with µV on a neighborhood of ∂V . By Theorem1.2 we then have that F satisfies conditions (CW 1,1) and (W 1,1) on ∂V .On the other hand ∇µV (y) is outwardly normal to ∂V at every y ∈ ∂Vand ∇F = ∇µV on ∂V , hence we have ∇F (y) = N(y) for every y ∈ ∂V .These facts, together with the assumption K ⊆ ∂V , are easily checked toimply that N and K satisfy conditions (KW1,1) and (W1,1). Finally, because0 ∈ int(V ) and ∇µV (x) is outwardly normal to ∂V for every x ∈ ∂V , wehave that 〈∇µV (x), x〉 > 0 for every x ∈ ∂V , and in particular condition(O) is satisfied as well. �

References

[1] D. Azagra and C. Mudarra, On a Whitney extension problem for convex function,prepint, 2015, arXiv:1501.05226v4 [math.CA]

[2] M. Ghomi, Strictly convex submanifolds and hypersurfaces of positive curvature. J.Differential Geom. 57 (2001), 239–271.

[3] M. Ghomi, The problem of optimal smoothing for convex functions. Proc. Amer. Math.Soc. 130 (2002) no. 8, 2255–2259.

[4] M. Ghomi, Optimal smoothing for convex polytopes. Bull. London Math. Soc. 36 (2004),483–492.

[5] G. Glaeser, Etudes de quelques algebres tayloriennes, J. d’Analyse 6 (1958), 1-124.[6] B. Kirchheim, J. Kristensen, Differentiability of convex envelopes. C. R. Acad. Sci.Paris Ser. I Math. 333 (2001), no. 8, 725–728.

[7] T. Rockafellar, Convex Analysis. Princeton Univ. Press, Princeton, NJ, 1970.[8] K. Schulz, B. Schwartz, Finite extensions of convex functions. Math. Operationsforsch.Statist. Ser. Optim. 10 (1979), no. 4, 501–509.

[9] E. Stein, Singular integrals and differentiability properties of functions. Princeton, Uni-versity Press, 1970.

[10] L. Vesely, L. Zajıcek, On extensions of d.c. functions and convex functions. J. ConvexAnal. 17 (2010), no. 2, 427–440.

[11] H. Whitney, Analytic extensions of differentiable functions defined in closed sets,Trans. Amer. Math. Soc. 36 (1934), 63–89.

[12] M. Yan, Extension of Convex Function. J. Convex Anal. 21 (2014) no. 4, 965–987.

ICMAT (CSIC-UAM-UC3-UCM), Departamento de Analisis Matematico, Fac-ultad Ciencias Matematicas, Universidad Complutense, 28040, Madrid, Spain

E-mail address: [email protected]

ICMAT (CSIC-UAM-UC3-UCM), Calle Nicolas Cabrera 13-15. 28049 MadridSPAIN

E-mail address: [email protected]

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