ON A WHITNEY EXTENSION PROBLEM FOR CONVEX FUNCTIONS

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1 ON A WHITNEY EXTENSION PROBLEM FOR CONVEX FUNCTIONS DANIEL AZAGRA AND CARLOS MUDARRA Abstract. Let C be a compact convex subset of R n, f : C R be a convex function, and m {1, 2,..., }. Assume that, along with f, we are given a family of polynomials satisfying Whitney s extension condition for C m, and thus that there exists F C m (R n ) such that F = f on C. It is natural to ask for further (necessary and sufficient) conditions on this family of polynomials which ensure that F can be taken to be convex as well. We give satisfactory solutions to this problem in the cases m = 1 and m =. Moreover, in the case m = 1 we also solve the problem for not necessarily convex compacta C, and we give a geometrical application concerning a characterization of compact sets which can be interpolated by boundaries of C 1 convex bodies. 1. Introduction and main results Let C be a closed subset of R n, and m N. The famous Whitney Extension Theorem [25] provides a necessary and sufficient condition, for a function f : C R and a family of functions {f α } α m defined on C (what we might call the would-be derivatives of f) satisfying f = f 0 and (1.1) f α (x) = β m α f α+β (y) (x y) β + R α (x, y) β! for all x, y C and all multi-indices α such that α m, to admit a C m extension F to all of R n such that D α F = f α on C for all α m. The condition, which in this paper we will denote (W m ), is that R α (x, y) (1.2) lim x y 0 x y m α = 0, uniformly on compact subsets of C, for every α m. If, instead of the functions f α, for every y C we are given a polynomial P y : R n R with degree(p y ) m and P y (y) = f(y) (what we might call the Date: January 30, Mathematics Subject Classification. 54C20, 52A41, 26B05, 53A99, 53C45, 52A20, 58C25, 35J96. Key words and phrases. extension of functions, convex function, differentiable function. C. Mudarra was supported by Programa Internacional de Doctorado de la Fundación La Caixa Severo Ochoa. Both authors were partially supported by Grant MTM

2 2 DANIEL AZAGRA AND CARLOS MUDARRA would-be Taylor polynomial of f of order m at y), then Whitney s condition (W m ) can be reformulated by saying that (1.3) lim δ 0 + ρ m(k, δ) = 0 for each compact subset K of C, where we denote ρ m (K, δ) = sup{ Dj P y (z) D j P z (z) y z m j : j = 0,..., m, y, z K, 0 < y z δ}. If this condition is met, then Whitney s theorem provides us with a function F C m (R n ) such that D j F (y) = D j P y (y) for every j = 0,..., m and y C; see [8, Theorem , p. 225]. In other words, each P y is the Taylor polynomial of order m of F at y. In the case m =, Whitney s theorem states that if we are given a function f : C R, together with a family of functions {f α } α (N {0}) n satisfying (1.1) and (1.2) for each m N, then there exists F C (R n ) such that F = f 0 and D α F = f α on C for every α (the converse is obviously true as well). The polynomial version of this statement is as follows. Let us call {Py m } y C,m N {0} a compatible family of polynomials for C extension of a function f defined on C, where Py m is a polynomial of degree up to m such that Py m (y) = f(y), if for every k > j the polynomial Py j is the Taylor polynomial of order j at y of the polynomial Py k. In other words, {Py m } y C,m N {0} is compatible if for every k > j the polynomial Py j is obtained from Py k by dropping all of its homogeneous terms of order greater than j. With this terminology, Whitney s extension theorem for C tells us that whenever we are given a compatible family of polynomials {Py m } y C,m N {0} for f such that for each m N the subfamily {Py m } y C satisfies Whitney s condition (1.3), then there is a function F C (R n ) such that Py m is the Taylor polynomial of order m of F at y (which we will denote Jy m F ), for every y C and m N. Again, the converse is obviously true. In recent years there has been great interest in finding sharper forms of Whitney s extension theorem, in constructing continuous linear extension operators with nearly optimal norms, and in extending these results to other spaces of functions such as Sobolev spaces, see [3, 4, 9, 10, 11, 20, 17, 12, 21] and the references therein. Returning to Whitney s theorem, it is natural to wonder what further conditions (if any) on those families of polynomials would be necessary and sufficient to ensure that F can be taken to be convex whenever f is convex. Besides its basic character, one should expect that a solution to this problem would find interesting applications in problems of differential geometry (see Theorem 1.11 below for an obvious example) and of partial differential equations (such as the Monge-Ampère equations). Let us begin by making a couple of general observations concerning solvability of our extension problem. Firstly, if C is not assumed to be compact,

3 SMOOTH CONVEX EXTENSIONS OF CONVEX FUNCTIONS 3 it is known that our problem has a negative solution. Indeed, there exists an unbounded closed convex subset C of R 2 and a C convex function f : C R which has no continuous convex extension to all of R 2, see [22, Example 4]. A modification of this example, which we will present in Section 4 below, shows that the obstruction persists even if we require that f have a strictly positive Hessian on a neighbourhood of C. See also [5, 24], which show that there are infinite-dimensional Banach spaces X, closed subspaces E X and continuous convex functions f : E R which have no continuous convex extensions to X. Secondly, if we do not require that C be convex, then the problem gets geometrically complicated, for the following reason. There are several possible, nonequivalent, definitions of convex functions defined on non-convex domains (see [26] for a study of three of them) but, no matter how one defines convexity of such functions, the problem cannot be solved just by adding further analytical conditions on the would-be Taylor polynomials of f and disregarding the global geometry of the graph of f. To see why this is so, let us consider the following example: take any four numbers a, b, c, d R with a < b < 0 < c < d, and define C = {a, b, 0, c, d} and f(x) = x for x C. Since C is a five-point set it is clear that, no matter what polynomials of degree up to k 1 are chosen to be the differential data of f on C, the function f will satisfy Whitney s extension condition (W k ) for every k N. Hence there are many C 1 (even many C ) functions F with F = f on C. But none of these F can be convex on R, because, as is easily checked, any convex extension g of f to R must satisfy g(x) = x for every x [a, d], and therefore g cannot be differentiable at 0. This example also shows that, from a purely analytical point of view, our problem is different in nature from the extension problems dealt with in the mentioned papers [3, 4, 9, 10, 11, 20, 17, 12, 21], and is surely closer to the classical Whitney s extension theorem [25]. Fortunately, there is evidence that the geometrical obstructions shown by these examples no longer exist when C is assumed to be compact and convex. In particular, it is clear that if f is convex on a compact convex set C and is C 1 on a neighbourhood of C then m(f)(x) := max y C {f(y)+ f(y), x y } defines a Lipschitz, convex function on all of R n (and that, in the case when C has nonempty interior, m(f) happens to be the minimal convex extension of f to R n ). Therefore, at least in a first approach to the problem, it seems reasonable to assume that C is convex and compact, which we will do in what follows until further notice 1, and ask ourselves if our extension problem can always be solved in this relatively simple case. Extension problems related to the one we are dealing with have been considered by M. Ghomi [14] and by M. Yan [26]. A consequence of their results is that, under the assumptions that 1 In the special case m = 1, even for non-convex domains C, we have found global geometrical conditions which, along with (W 1 ), are necessary and sufficient for the existence of convex functions F C 1 (R n ) such that F = f on C; see Theorem 1.9 below.

4 4 DANIEL AZAGRA AND CARLOS MUDARRA m 2 and that f has a strictly positive Hessian on the boundary C, there always exists F C m (R n ) such that F is convex and F = f on C. See also [13, 15, 7] for related problems. Of course, strict positiveness of the Hessian is a very strong condition which is far from being necessary, and it would be desirable to get rid of this requirement altogether, if possible. However, some other assumptions must be made in its place, at least when m 3 as, already in one dimension, there are examples of C 3 convex functions g defined on compact intervals I which cannot be extended to C 3 (J) convex functions for any open interval J containing I. Such an example is g(x) = x 2 x 3 defined for x I := [0, 1 3 ]. Therefore, we should look for conditions on the derivatives of f on C that are necessary and sufficient to guarantee that f has a C m convex extension F to all of R n. Now, observe that any such function F will satisfy that D 2 F (x)(v) 2 0 for every x R n, v S n 1, and therefore, if m 2 is finite, the Taylor polynomial of the second derivative D 2 F = D 2 f at points y C will also satisfy 0 D 2 F (y + tw)(v 2 ) = D 2 f(y)(v) 2 + t D 3 f(y)(w, v 2 ) + + where tm 2 (m 2)! Dm f(y)(w m 2, v 2 ) + R m (t, y, v, w), R m (t, y, v, w) lim t 0 + t m 2 = 0 uniformly on y C, w, v S n 1. Then we will also have ) 1 lim inf (D 2 t 0 + t m 2 f(y)(v) tm 2 (m 2)! Dm f(y)(w m 2, v 2 ) 0 uniformly on y C, w, v S n 1. This of course means that for every ε > 0 there exists t ε > 0 such that D 2 f(y)(v) 2 + t D 3 f(y)(w, v 2 ) + + tm 2 (m 2)! Dm f(y)(w m 2, v 2 ) εt m 2 for all y C, v, w S n 1, 0 < t t ε. We will abbreviate this by saying that f satisfies condition (CW m ) on C. On the other hand, if C is a convex compact set with nonempty interior (what is usually called a convex body) and f : C R is a convex function which has a (not necessarily convex) C m extension to an open neighbourhood of C, then the same argument shows that f automatically satisfies (CW m ) on the interior of C. Conversely, if f satisfies (CW m ) on the interior of C then it immediately follows, using Taylor s theorem, that D 2 f(x) 0 for all x in the open convex set int(c), hence f is convex on int(c), and by continuity we infer that f is also convex on C. These observations show that if C is a convex compact subset of R n, m N, m 2, and f : C R is a function then:

5 SMOOTH CONVEX EXTENSIONS OF CONVEX FUNCTIONS 5 (1) A necessary condition for f to have a C m convex extension to all of R n is that f satisfies both (W m ) and (CW m ) on C. (2) In the event that one can show that this requirement is sufficient as well, then one also has that a necessary and sufficient condition for a convex function f : C R to have a C m convex extension to all of R n is that f satisfy (W m ) on C, and (CW m ) on the boundary C. In this paper we will show that, if f : C R is a convex function which satisfies conditions (W k ) on C, and (CW k ) on C, for every k N, k 2, then there indeed exists a convex extension F C (R n ) of f. In view of the preceding remarks, this result provides a satisfactory solution to our extension problem in the case m =. Before stating our result precisely, let us see how one can equivalently write the condition (CW m ) in terms of the would-be Taylor polynomials of f. Definition 1.1. Let m N, m 2. We will say that f, together with a family of polynomials {P m y } y C of degree up to m such that P m y (y) = f(y), satisfy the condition (CW m ) provided that for every ε > 0 there exists t ε > 0 such that tm 2 D 2 Py m (y)(v) 2 +td 3 Py m (y)(w, v 2 )+ + (m 2)! Dm Py m (y)(w m 2, v 2 ) εt m 2 for all y C, v, w S n 1, 0 < t t ε. Thanks to Whitney s extension theorem this definition of (CW m ) coincides with the one we gave above. Recalling that every k-homogeneous polynomial Q : R n R determines a unique k-linear form A on R n such that A(x,..., x) = Q(x) for every x R n by means of the polarization formula A(x 1, x 2,..., x n ) = 1 n 2 n ε 1 ε n Q ε j x j, n! ε j =±1 one can also write the condition (CW m ) in terms of the would-be partial derivatives f α of f appearing in Whitney s original statement. This leads to somewhat complicated formulas which we will not need to use, so we leave them to the reader s care, as we do with the proof of the following. Remark 1.2. Let f and {Py m+1 } y C satisfy (CW m+1 ). Then f and {Py m } y C satisfy (CW m ) too, where each Py m is obtained from Py m+1 by dropping its (m + 1)-homogeneous term. Our first main result is as follows. Theorem 1.3. Let C be a compact convex subset of R n. Let f : C R be a function, and let {Py m } y C,m N be a compatible family of polynomials for C extension of f. Then f has a convex, C extension F to all of R n, with Jy m F = Py m for every y C and m N, if and only if {Py m } y C,m N satisfies (W m ) and (CW m ) on C for every m N. j=1

6 6 DANIEL AZAGRA AND CARLOS MUDARRA Moreover, if C has nonempty interior and f : C R is convex, then f has a convex, C extension F to all of R n, with Jy m F = Py m for every y C and m N, if and only if {Py m } y C,m N satisfies (W m ) on C, and (CW m ) on C, for every m N. For m 2 finite and n 2, we do not know if condition (CW m ) is sufficient for a convex function f : C R to have a C m convex extension to R n. Nevertheless, for dimension n = 1, we do know that this is indeed the case, and moreover, since the boundary of C has only two points and there are only two directions in which to differentiate, the condition (CW m ) can be very much simplified. Proposition 1.4. Let I be a closed interval in R, and m N with m 2. Let f : I R be a convex function of class C m in the interior of I, and assume that f has one-sided derivatives of order up to m, denoted by f (k) (a + ) or f (k) (b ), at the extreme points of I. Then f has a convex extension of class C m (R) if and only if the first (if any) non-zero derivative which occurs in the finite sequence {f (2) (b ), f (3) (b ),..., f (m) (b )} is positive and of even order, and similarly for {f (2) (a + ), f (3) (a + ),..., f (m) (a + )}. The easy proof is left to the reader s care. In the special case when condition (CW k ) is satisfied with a strict inequality for some k, the problem also becomes much easier to solve, because in this situation f must be convex on a neighbourhood of C (see subsection 3.9 below). Let us also note that in this case f automatically satisfies (CW p ) for all the rest of p s. Proposition 1.5. Let m N { }, m 2. If f C m (R n ) satisfies (CW k ) with a strict inequality on C for some k 2 (meaning that there are some η > 0 and t 0 > 0 such that D 2 f(y)(v) 2 + t D 3 f(y)(w, v 2 ) + + tk 2 (k 2)! Dk f(y)(w k 2, v 2 ) ηt k 2 for all y C, v, w S n 1, 0 < t t 0 ), then f satisfies (CW p ) on C for every p {2,..., m}, if m is finite, and for every p N with p 2, if m =. We leave the easy verification to the reader s care. As a straightforward consequence of the final part of the proof of Theorem 1.3 we will obtain the following. Corollary 1.6. Let m N { }, m 2. Let C be a convex compact subset of R n, and let f : C R be a convex function having a (not necessarily convex) C m extension to an open neighbourhood of C. If for some η > 0, t 0 > 0, and k N with 2 k m one has D 2 f(y)(v) 2 + t D 3 f(y)(w, v 2 ) + + tk 2 (k 2)! Dk f(y)(w k 2, v 2 ) ηt k 2

7 SMOOTH CONVEX EXTENSIONS OF CONVEX FUNCTIONS 7 for all y C, w, v S n 1, 0 < t t 0, then there exists a convex function F C m (R n ) such that F = f on C. The easiest instance of application of this Corollary is of course when f has a strictly positive Hessian on C, in which case we recover the aforementioned consequence of the results of M. Ghomi s [14] and M. Yan s [26]. Let us finally consider the case m = 1. Observe that Whitney s condition (W 1 ) is equivalent to the following one: given a function f : C R and a continuous mapping G : C R n, we have that f(z) f(y) G(y), z y lim = 0, uniformly on compact K C. z y 0 + z y In this case Whitney s theorem gives us a function F C 1 (R n ) such that F (y) = f(y) and F (y) = G(y) for every y C. The solution to our extension problem for the C 1 class depends on whether or not C has empty interior. When C is a convex body and f : C R is convex and satisfies Whitney s extension condition (W 1 ), it turns out that f always has a convex C 1 extension to all of R n, with no further assumptions on the would-be derivatives of f. Theorem 1.7. Let C be a compact convex subset of R n with non-empty interior. Let f : C R be a convex function, and G : C R n be a continuous mapping satisfying Whitney s extension condition (W 1 ) on C. Then there exists a convex function F C 1 (R n ) such that F (y) = f(y) and F (y) = G(y) for every y C. However, if int(c) is empty then, in order to obtain differentiable convex extensions of f to all of R n, (W 1 ) must be complemented by the following geometrical condition (CW 1 ) : f(x) f(y) = G(y), x y = G(x) = G(y), for all x, y C. Theorem 1.8. Let C be a compact convex subset of R n. Let f : C R be a convex function, and G : C R n be a continuous mapping. Then f has a convex, C 1 extension F to all of R n, with F = G on C, if and only if f and G satisfy (W 1 ) and (CW 1 ) on C. As a matter of fact, in the case m = 1 we will be able to solve our extension problem for non-convex domains as well, just by adding another geometrical condition. If C is a compact (not necessarily convex) subset of R n, f : C R is an arbitrary function, and G : C R n is a continuous function, we will say that f and G satisfiy condition (C) provided that (C) : f(x) f(y) G(y), x y for all x, y C. Theorem 1.9. Let C be a compact (not necessarily convex) subset of R n. Let f : C R be an arbitrary function, and G : C R n be a continuous mapping. Then f has a convex, C 1 extension F to all of R n, with F = G on C, if and only if f and G satisfy the conditions (C), (W 1 ), and (CW 1 ) on C.

8 8 DANIEL AZAGRA AND CARLOS MUDARRA In the particular case when C is finite, the above result provides necessary and sufficient conditions for interpolation of finite sets of data by C 1 convex functions. Notice that in this case condition (W 1 ) is automatically satisfied and plays no role. Corollary Let S be a finite subset of R n, and f : S R be a function. Then there exists a convex function F C 1 (R n ) with F = f on S if and only if there exists a mapping G : S R n such that f and G satisfy conditions (C) and (CW 1 ) on S. The above results can be used to establish their convex-body counterparts. Namely, if K is a compact subset of R n and we are given a continuous function N : K R n such that N(y) = 1 for every y K, it is natural to ask what conditions on K and N are necessary and sufficient for K to be a subset of the boundary of a C m smooth convex body V such that 0 int(v ) and N(y) is outwardly normal to V at y for every y K. We next solve this problem for m = 1. The pertinent conditions are, in this case: (O) N(y), y > 0 for all y K; (K) N(y), x y 0 for all x, y K; (KW 1 ) N(y), x y = 0 = N(x) = N(y) for all x, y K; (W 1 N(y), x y ) lim = 0, uniformly on x, y K. x y 0 + x y Our main result for convex bodies reads as follows. Theorem Let K be a compact subset of R n, and let N : K R n be a continuous mapping such that N(y) = 1 for every y K. Then the following statements are equivalent: (1) There exists a C 1 smooth convex body V with 0 int(v ) and such that K V and N(y) is outwardly normal to V at y for every y K. (2) K and N satisfy conditions (O), (K), (KW 1 ) and (W 1 ). We conclude with the counterpart of Corollary 1.10 for convex bodies. Corollary Let S be a finite subset of R n containing 0 in the interior of its convex hull. Then there exists a C 1 convex body V such that 0 int(v ) and S V if and only if there exists a mapping N : S R n such that S and N satisfy conditions (O), (K) and (KW 1 ). The rest of the paper is organized as follows. In Section 2 we will prove Theorems 1.7, 1.8, 1.9 and The proof of Theorem 1.3 and Corollary 1.6 will be provided in Section 3. Finally, in Section 4 we will make some remarks and present some examples related to the above results. 2. The C 1 case Theorem 1.7 is a consequence of Theorem 1.8 and of the following result.

9 SMOOTH CONVEX EXTENSIONS OF CONVEX FUNCTIONS 9 Lemma 2.1. Let f C 1 (R n ), C R n be a compact convex set with nonempty interior, x 0, y 0 C. Assume that f is convex on C and Then f(x 0 ) = f(y 0 ). f(x 0 ) f(y 0 ) = f(y 0 ), x 0 y 0. Proof. Case 1. Suppose first that f(x 0 ) = f(y 0 ) = 0. We may of course assume that x 0 y 0 as well. Then we also have f(y 0 ), x 0 y 0 = 0. If we consider the C 1 function ϕ(t) = f (y 0 + t(x 0 y 0 )), we have that ϕ is convex on the interval [0, 1] and ϕ (0) = 0, hence 0 = ϕ(0) = min t [0,1] ϕ(t), and because ϕ(0) = ϕ(1) and the set of minima of a convex function on a convex set is convex, we deduce that ϕ(t) = 0 for all t [0, 1]. This shows that f is constant on the segment [x 0, y 0 ] and in particular we have f(z), z 0 z 0 = 0 for all z, z 0, z 0 [x 0, y 0 ]. Now pick a point a 0 in the interior of C and a number r 0 > 0 so that B(a 0, r 0 ) int(c). Since C is a compact convex body, every ray emanating from a 0 intersects the boundary of C at exactly one point. This implies that (even though the segment [x 0, y 0 ] might entirely lie on the boundary C), for every a B(a 0, r 0 ), the interior of the triangle a with vertices x 0, a, y 0, relative to the affine plane spanned by these points, is contained in the interior of C; we will denote relint ( a ) int(c). Let p 0 be the unique point in [x 0, y 0 ] such that a 0 p 0 = d (a 0, [x 0, y 0 ]) (the distance to the segment [x 0, y 0 ]), set v 0 = a 0 p 0, and denote v a := a p 0 for each a B(a 0, r 0 ). Thus for every a B(a 0, r 0 ) we can write v a = u a + v 0, where u a := a a 0 B(0, r 0 ), and in particular we have {v a : a B(a 0, r 0 )} = B(v 0, r 0 ). Claim 2.2. For every z 0, z 0 in the relative interior of the segment [x 0, y 0 ], we have f(z 0 ) = f(z 0 ). Let us prove our claim. It is enough to show that f(z 0 ) f(z 0 ), v a = 0 for every a B(a 0, r 0 ) (because if a linear form vanishes on a set with nonempty interior, such as B(v 0, r 0 ), then it vanishes everywhere). So take a B(a 0, r 0 ). Since z 0 and z 0 are in the relative interior of the segment [x 0, y 0 ] and relint ( a ) int(c), we have that there exists t 0 > 0 such that z 0 + tv a, z 0 + tv a int(c) for every t (0, t 0 ]. If we had f(z 0 ) f(z 0), v a > 0 then, because f is convex on C and f(z 0 ) = f(z 0 ) = 0, f(z 0 ), z 0 z 0 = 0, we would get f(z 0 + tv a ) = f(z 0 + z 0 z 0 + tv a ) f(z 0), z 0 z 0 + tv a = f(z 0), tv a, hence f(z 0 + tv a ) lim f(z f(z 0 + tv a ) t 0 + t 0), v a > f(z 0 ), v a = lim, t 0 + t a contradiction. By interchanging the roles of z 0, z 0, we see that the inequality f(z 0 ) f(z 0), v a < 0 also leads to a contradiction. Therefore f(z 0 ) f(z 0), v a = 0 and the Claim is proved.

10 10 DANIEL AZAGRA AND CARLOS MUDARRA Now, by using the continuity of f, we easily conclude the proof of the Lemma in Case 1. Case 2. In the general situation, let us consider the function h defined by h(x) = f(x) f(y 0 ) f(y 0 ), x y 0, x R n. It is clear that h is convex on C, and h C 1 (R n ). We also have h(x) = f(x) f(y 0 ), and in particular h(y 0 ) = 0. Besides, using the assumption that f(x 0 ) f(y 0 ) = f(y 0 ), x 0 y 0, we have h(x 0 ) = 0 = h(y 0 ), and h(x 0 ) h(y 0 ) = h(y 0 ), x 0 y 0. Therefore we can apply Case 1 with h instead of f and we get that h(x 0 ) = h(y 0 ) = 0, which implies that f(x 0 ) = f(y 0 ). From the above Lemma it is clear that (CW 1 ) is a necessary condition for a convex function f : C R to have a convex, C 1 extension to all of R n, and also that if the function f satisfies (W 1 ) and int(c) then f automatically satisfies (CW 1 ) as well. It is also obvious that Theorem 1.8 is an immediate consequence of Theorem 1.9. Thus, in order to prove Theorems 1.7, 1.8, and 1.9 it will be sufficient to establish the if part of Theorem Proof of Theorem 1.9. Because f satisfies (W 1 ), according to Whitney s Extension Theorem, there exists f C 1 (R n ) such that, on C, we have f = f and f = G. Thus we may and do assume in what follows, for simplicity of notation, that f is of class C 1 (R n ), with f = G on C, and that f satisfies conditions (C) and (CW 1 ) on C. Let us consider the function m(f) : R n R defined by m(f)(x) = sup{f(y) + f(y), x y }. y C Since C is compact and the function y f(y)+ f(y), x y is continuous, it is obvious that m(f)(x) is well defined, and in fact the sup is attained, for every x R n. Furthermore, if we set K := max y C f(y) then each affine function x f(y) + f(y), x y is K-Lipschitz, and therefore m(f), being a sup of a family of convex and K-Lipschitz functions, is convex and K-Lipschitz on R n. Moreover, we have m(f) = f on C. Indeed, if x C then, because f satisfies (C) on C, we have f(x) f(y) + f(y), x y for every y C, hence m(f)(x) f(x). On the other hand, we also have f(x) m(f)(x) because of the definition of m(f)(x) and the fact that x C. (In the case when C is convex and has nonempty interior, it is easy to see that if h : R n R is convex and h = f on C, then m(f) h. Thus,

11 SMOOTH CONVEX EXTENSIONS OF CONVEX FUNCTIONS 11 in this case, m(f) is the minimal convex extension of f to all of R n, which accounts for our choice of notation. However, if C is convex but has empty interior then there is no minimal convex extension operator. We refer the interested reader to [22] for necessary and sufficient conditions for m(f) to be finite everywhere, in the situation when f : C R is convex but not necessarily everywhere differentiable.) If the function m(f) were differentiable on R n, there would be nothing else to say. Unfortunately, it is not difficult to construct examples showing that m(f) need not be differentiable outside C (even when C is convex and f satisfies (CW 1 ), see Example 4.3 in Section 4 below). Nevertheless, a crucial step in our proof is the following fact: m(f) is differentiable on C, provided that f satisfies conditions (C) and (CW 1 ) on C. Lemma 2.3. Let f C 1 (R n ), let C be a compact subset of R n (not necessarily convex), and assume that f satisfies (C) and (CW 1 ) on C. Then, for each x 0 C, the function m(f) is differentiable at x 0, with m(f)(x 0 ) = f(x 0 ). Proof. Notice that, by definition of m(f) we have, for every x R n, f(x 0 ), x x 0 + m(f)(x 0 ) = f(x 0 ), x x 0 + f(x 0 ) m(f)(x). Since m(f) is convex, this means that f(x 0 ) belongs to m(f)(x 0 ) (the subdifferential of m(f) at x 0 ). If m(f) were not differentiable at x 0 then there would exist a number ε > 0 and a sequence (h k ) converging to 0 in R n such that (2.1) m(f)(x 0 + h k ) m(f)(x 0 ) f(x 0 ), h k h k ε for every k N. Because the sup defining m(f)(x 0 + h k ) is attained, we obtain a sequence (y k ) C such that m(f)(x 0 + h k ) = f(y k ) + f(y k ), x 0 + h k y k, and by compactness of C we may assume, up to passing to a subsequence, that (y k ) converges to some point y 0 C. Because f = m(f) on C, and by continuity of f, f, and m(f) we then have f(x 0 ) = m(f)(x 0 ) = lim k m(f)(x 0 + h k ) = lim k (f(y k) + f(y k ), x 0 + h k y k ) = f(y 0 ) + f(y 0 ), x 0 y 0, that is, f(x 0 ) f(y 0 ) = f(y 0 ), x 0 y 0. Since x 0, y 0 C and f satisfies (CW 1 ), this implies that f(x 0 ) = f(y 0 ). And because m(f)(x 0 )

12 12 DANIEL AZAGRA AND CARLOS MUDARRA f(y k ) + f(y k ), x 0 y k by definition of m(f), we then have m(f)(x 0 + h k ) m(f)(x 0 ) f(x 0 ), h k h k f(y k ) + f(y k ), x 0 + h k y k f(y k ) f(y k ), x 0 y k f(x 0 ), h k h k f(y k ) f(x 0 ), h k f(y k ) f(x 0 ) = f(y k ) f(y 0 ), h k from which we deduce, using the continuity of f, that lim sup k in contradiction with (2.1). m(f)(x 0 + h k ) m(f)(x 0 ) f(x 0 ), h k h k 0, = Now we proceed with the rest of the proof of Theorem 1.9. Our strategy will be to use the differentiability of m(f) on C in order to construct a (not necessarily convex) differentiable function g such that g = f on C, g m(f) on R n, and lim x g(x) =. Then we will define F as the convex envelope of g, which will be of class C 1 (R n ) and will coincide with f on C. Let us define H(x) = f(x) m(f)(x) + 2d(x, C) 2, where d(, C) stands for the distance from x to C. Claim 2.4. H is differentiable on C, with H(x 0 ) = 0 for every x 0 C. Proof. The function d(, C) 2 is obviously differentiable, with a null gradient, at x 0, hence we only have to see that f m(f) is differentiable, with a null gradient, at x 0. Since m(f)(x 0 ) = f(x 0 ) by Lemma 2.3, the Claim boils down to the following easy exercise: if two functions h 1, h 2 are differentiable at x 0, with h 1 (x 0 ) = h 2 (x 0 ), then h 1 h 2 is differentiable, with a null gradient, at x 0. Now, according to Whitney s approximation theorem [25], we can find ϕ C (R n \ C) such that ϕ(x) H(x) d(x, C) 2 for every x R n \ C. Let us define ϕ : R n R by ϕ = ϕ on R n \ C and ϕ = 0 on C. Claim 2.5. The function ϕ is differentiable on R n. Proof. It is obvious that ϕ is differentiable on int(c) (R n \ C). We only have to check that ϕ is differentiable on C. If x 0 C we have ϕ(x) ϕ(x 0 ) x x 0 = ϕ(x) H(x) + d(x, C)2 x x 0 x x 0 0

13 SMOOTH CONVEX EXTENSIONS OF CONVEX FUNCTIONS 13 as x x 0 0, because both H and d(, C) 2 vanish at x 0 and are differentiable, with null gradients, at x 0. Therefore ϕ is differentiable at x 0, with ϕ(x 0 ) = 0. Next we define g := f + ϕ. The function g is differentiable on R n, and coincides with f on C. And, for x R n \ C, we have g(x) f(x) + H(x) d(x, C) 2 = f(x) + f(x) m(f)(x) + d(x, C) 2 m(f)(x) + d(x, C) 2. This shows that g m(f). On the other hand, we know that m(f) is Lipschitz on R n, and therefore m(f) may decay only linearly at infinity, while d(, C) 2 grows quadratically at infinity. Hence the above inequality also implies that lim x g(x) =. Now we will have to use a differentiability property of the convex envelope of a function ψ : R n R, defined by conv(ψ)(x) = sup{h(x) : h is convex, h ψ} (another expression for conv(ψ), which follows form Carathéodory s Theorem, is n+1 n+1 conv(ψ)(x) = inf{ λ j ψ(x j ) : λ j 0, λ j = 1, x = j=1 j=1 n+1 λ j x j }, see [19, Corollary ] for instance). The following result is a restatement of a particular case of the main theorem in [18]; see also [16]. Theorem 2.6 (Kirchheim-Kristensen). If ψ : R n R is differentiable and lim x ψ(x) =, then conv(ψ) C 1 (R n ). If we define F = conv(g) we thus get that F is convex on R n and F C 1 (R n ). It only remains to check that F = f on C. This is easy: as m(f) is convex on R n and m(f) g, we have that m(f) F on R n by definition of conv(g). On the other hand, since g = f on C we have, for every convex function h with h g, that h f on C, and therefore, for every y C, j=1 F (y) = sup{h(y) : h is convex, h g} f(y) = m(f)(y). This shows that F (y) = f(y) for every y C, and concludes the proof of Theorem Proof of Theorem (2) = (1): Set C = K {0}, choose a number α > 0 sufficiently close to 1 so that 0 < 1 α < min N(y), y y K (this is possible thanks to condition (O), continuity of N and compactness of K; notice in particular that 0 / K), and define f : C R and G : C R n by f(y) = { 1 if y K α if y = 0, and G(y) = { N(y) if y K 0 if y = 0.

14 14 DANIEL AZAGRA AND CARLOS MUDARRA By using conditions (K), (KW 1 ) and (W 1 ), it is straightforward to check that f and G satisfy conditions (C), (CW 1 ) and (W 1 ). Therefore, according to Theorem 1.9, there exists a convex function F C 1 (R n ) such that F = f and F = G on C. Moreover, from the proof of Theorem 1.9, it is clear that F can be taken so as to satisfy lim x F (x) =. If we define V = {x R n : F (x) 1} we then have that V is a compact convex body with 0 int(v ) (because F (0) = α < 1), and F (y) = N(y) is outwardly normal to {x R n : F (x) = 1} = V at each y K. Moreover, F (y) = f(y) = 1 for each y K, hence K V. (1) = (2): Let µ C be the Minkowski functional of C. By composing µ C with a C 1 convex function φ : R R such that φ(t) = t if and only if t 1/2, we obtain a function F (x) := φ(µ C (x)) which is of class C 1 and convex on R n and coincides with µ C on a neighborhood of C. By Theorem 1.9 we then have that F satisfies conditions (C), (CW 1 ) and (W 1 ) on C. Since µ C (y) is outwardly normal to C at every y C and F = µ C on C, we have F (y) = N(y) for every y C. Taking into account that K V, this is easily checked to imply that N and K satisfy conditions (K), (KW 1 ) and (W 1 ). On the other hand, because 0 int(c) and µ C (x) is outwardly normal to C for every x C, we have that µ C (x), x > 0 for every x C, and in particular condition (O) is satisfied as well. 3. The C case In this section we will prove Theorem 1.3. By using Whitney s extension theorem we may and do assume that f C (R n ), and f satisfies condition (CW m ) on C for every m N. We may also assume that f has a compact support contained in C + B(0, 2) Idea of the proof. Let us give a rough sketch of the proof so as to guide the reader through the inevitable technicalities. We warn the reader, however, that what we now say we are going to do is not exactly what we will actually do, but something slightly different. Our proof could be rewritten to match this sketch exactly, but at the cost of adding further technicalities, which we do not feel would be pertinent. This proof has two main parts. In the first part we will estimate the possible lack of convexity of f outside C: by using the conditions (CW m ), a Whitney partition of unity, and some ideas from the proof of the Whitney extension theorem in the C case, we will construct a function η C (R) such that η 0, η 1 (0) = (, 0], and min v =1 D 2 f(x)(v) 2 η (d(x, C)) for every x R n. In the second part of the proof we will compensate the lack of convexity of f outside C with the construction of a function ψ C (R n ) such that ψ 0, ψ 1 (0) = C, and min v =1 D 2 ψ(x)(v) 2 2η (d(x, C)). Then, by setting F := f + ψ we will conclude the proof of Theorem 1.3. There are many ways to construct such a function ψ. The essential point is to write C as an intersection of a family of half-spaces, and then to make a weighted sum, or an integral, of suitable convex functions composed with the

15 SMOOTH CONVEX EXTENSIONS OF CONVEX FUNCTIONS 15 linear forms that provide those half-spaces. If the sequence of linear forms is equi-distributed, in the weighted sum approach, or if one uses a measure equivalent to the standard measure on S n 1, in the integral approach, then the different functions ψ produced by these methods will have equivalent convexity properties. See [2] for an instance of the weighted sum approach, and [15, Proposition 2.1] for the integral approach. Of course our situation is more complicated than that of these references, as we need to find quantitative estimations of the convexity of ψ outside C which are good enough to outweigh our previous estimations of the lack of convexity of f outside C. It turns out that, in the present C case, this goal can be achieved with either method of construction of ψ. Here we will follow the integral approach of Ghomi s in [15, Proposition 2.1], as it will lead us to easier calculations First lower estimates for the Hessian of f: the numbers {r m } m. Given x R n \ C, v = 1, t := d(x, C), let y be the unique point of C with the property that d(x, C) = x y. Take w = x y/ x y. We have y = x + tw. By Taylor s Theorem, we can write D 2 f(x)(v) 2 =D 2 f(y)(v) 2 + t D 3 f(y)(w, v 2 ) + + tm 2 (m 2)! Dm f(y)(w m 2, v 2 ) + tm 2 [ D m f(y + sw)(w m 2, v 2 ) D m f(y)(w m 2, v 2 ) ], (m 2)! for some s [0, t]. Since f satisfies the condition (CW ) m, there exists a positive number r m, independent of y, v and w, for which D 2 f(y)(v) 2 + rd 3 f(y)(w, v 2 ) + + rm 2 (m 2)! inf Dm f(y)(w m 2, v 2 ) <r r m r m 2 Thus, for 0 < t r m, D 2 f(x)(v) 2 tm tm 2 [ D m f(y + sw)(w m 2, v 2 ) D m f(y)(w m 2, v 2 ) ]. (m 2)! On the other hand, if s [0, t], we can write D m f(y + sw)(w m 2, v 2 ) D m f(y)(w m 2, v 2 ) D m f(x + sw) D m f(y), where we denote A := sup ui S n 1 A(u 1,..., u m ), for every m-linear form A on R n. Moreover, the above expression is smaller than or equal to ε m (t) := sup D m f(z) D m f(z ). {z R n, z C, z z t} Since D m f is uniformly continuous, there is r m > 0 such that if 0 < r r m, then ε m (r) 1 2 (in fact we have lim r 0 + ε m(r) = 0). Therefore, if we suppose 0 < t min{r m, r m}, we obtain D 2 f(x)(v) 2 tm 2 2 We have thus proved the following. tm 2 (m 2)! ε m(t) t m 2.

16 16 DANIEL AZAGRA AND CARLOS MUDARRA Proposition 3.1. Given m N if f C m (R n ) and f satisfies (CW m ) then there is a number r m > 0 such that, whenever d(x, C) r m, we have D 2 f(x)(v) 2 d(x, C) m 2, for all v S n A Whitney partition of unity on (0, + ). For all k Z, we define the closed intervals [ 3 I k = [2 k, 2 k+1 ], Ik = 4 2k, 9 ] 8 2k+1. Obviously (0, + ) = k Z I k. We note that I k and Ik have the same midpoint and l(ik ) = 3 2 l(i k), where l(i k ) = 2 k denotes the length of I k. In other words, the interval Ik is I k expanded by the factor 3/2. Proposition 3.2. The intervals I k, Ik satisfy: 1. If t Ik, then 3 4 l(i k) t 9 4 l(i k). 2. If Ik and I j are not disjoint, then 1 2 l(i k) l(i j ) 2l(I k ). 3. Given any t > 0, there exists an open neighbourhood U t (0, + ) of t such that U t intersects at most 2 intervals of the collection {Ik } k Z. This is a special case of the decomposition of an open set in Whitney s cubes, see [23, Chapter VI] for instance. In the one dimensional case things are much simpler and, for instance, it is easy to see that one may replace the number N = 12 in [23, Proposition VI.1.2, p. 169] with the number 2. Anyhow, dealing with the number 12 instead of 2 would have no harmful effect in our proof. We now relabel the families {I k } k and {Ik } k, k Z, as sequences indexed by k N, so we will write {I k } k 1 and {Ik } k 1. For every k 1, we will denote by t k and l k the midpoint and the length of I k, respectively. Next we recall how to define a Whitney partition of unity subordinated to the intervals Ik. Let us take a bump function θ 0 C (R) with 0 θ 0 1, θ 0 1 on [ 1/2, 1/2]; and θ 0 0 on R \ ( 3 4, 3 4 ). For every k, we define the function θ k by ( ) t tk θ k (t) = θ 0 t R. l k It is clear that θ k C (R), that 0 θ k 1, θ k 1 on I k and that θ k 0 on int(ik )c. Now we consider the function Φ = k 1 θ k defined on (0, + ). Using Proposition 3.2, every point t > 0 has an open neighbourhood which is contained in (0, + ) and intersects at most two of the intervals {Ik } k. Since supp(θ k ) Ik, the sum defining Φ has only two terms and therefore Φ is of class C. For the same reason, Φ(t) = Ik t θ k(t) 2, for t > 0. On the

17 SMOOTH CONVEX EXTENSIONS OF CONVEX FUNCTIONS 17 other hand, every t > 0 must be contained in some I k, where the function θ k takes the constant value 1, so we have 1 Φ 2. These properties allow us to define, on (0, ), the functions θk = θ k Φ. These are C functions satisfying k θ k = 1, 0 θ k 1, and supp(θ k ) I k. Less elementary, but crucial, are the following properties; see [25, 23] for a proof. Proposition 3.3. For every j N {0}, there are positive constants A j, A j, A j such that for all t > 0 and all k N, 1. θ (j) k (t) A jl j k. 2. If t Ik, then Φ(j) (t) A j l j k. 3. (θk )(j) (t) A j l j k The sequence {δ p } p and the function ε. Let us consider the numbers r m of Proposition 3.1. We can easily construct a sequence {δ p } p of positive numbers satisfying δ p min{r p+2, δ p < δ p 1 2, p 2. 1 (p + 2)! }, p 1 Of course the sequence {δ p } p is strictly decreasing to 0. Now, for every k we define a positive integer γ k as follows. In the case that l k δ 1, we set γ k = 1. In the opposite case, l k < δ 1, we take γ k as the unique positive integer for which Finally let us define: { ε(t) = δ γk +1 l k < δ γk. (t) if t > 0 0 if t 0 k 1 tγ kθ k It is obvious that ε 1 (0) = (, 0], that ε > 0 on (0, + ) and that ε C (R \ {0}). To prove that ε C (R), we have to work a bit more. Lemma 3.4. The function ε is of class C (R). Proof. It is sufficient to prove that for all j N, ε (j) (t) lim = 0. t 0 + t To check this, fix j N {0} and η > 0 and take η t j := min{ 2B j 4 j (j + 1)!, δ j+5}, where B j = max{a l : 0 l j}. Recall that the numbers A l are those given by Proposition 3.3. Let 0 < t t j. Because t δ j+5 < δ 1, we can find a unique positive integer q such that

18 18 DANIEL AZAGRA AND CARLOS MUDARRA δ q+1 t < δ q, and due to the fact that {δ p } p is strictly decreasing, we must have q j + 4. Now, if k is such that t Ik, Proposition 3.2 tells us that l k 4 3 t < 2t 2δ j+5 < δ 1, and using the definition of γ k, we have δ γk +1 l k 4 3 t < 2t < 2δ q < δ q 1. The above inequalities imply that γ k + 1 > q 1, that is γ k q 1. In particular γ k j + 3. On the other hand, using Proposition 3.2 again, we obtain: δ γk > l k 4t 9 t 4 δ q+1 > δ q+3, 4 so γ k q + 2. If we use Leibnitz s Rule, we obtain ε (j) (t) = j ( ) j d l l dt l (tγ k )(θk )(j l) (t), k 1 l=0 and since γ k j + 3 for those k such that t Ik, we can write j j ε (j) (t) t = I k t l=0 ( ) j γk! l (γ k l)! tγk l 1 (θk )(j l) (t) Ik t l=0 j!γ k!t γ k l 1 A j l ll j k. Now, by Proposition 3.2 we know that l k 4 9 t 1 4t. Moreover, because γ k q + 2, we have γ k! (q + 2)! and the last sum is smaller than or equal to j j! (q + 2)! t γk l 1 A t l j j l 4 l j. Ik t l=0 Writing t γk l 1 = t 2 t γk l 3 t δ q t γk l 3, this sum is smaller than or equal to j j! (q+2)! tδ q t γk l 3 A t l j j l 4 l j j 4 j j!b j (q + 2)!δ q t γ k j 3 t. Ik t Ik t l=0 Noting that t δ j+5 < 1 and γ k j + 3, we must have t γk j 3 1. The construction of the sequence {δ p } p implies that (q + 2)! δ q 1, and using that the sum Ik t has at most 2 terms, we obtain ε (j) (t) t l=0 4 j (j + 1)j! 2B j t 4 j (j + 1)! 2B j t j η. The function ε does not have to be increasing, but nonetheless satisfies the following important property.

19 SMOOTH CONVEX EXTENSIONS OF CONVEX FUNCTIONS 19 Lemma 3.5. If 0 < t δ 4 and q N are such that δ q+1 t < δ q and s [t/2, t], then ε(2s) t q+2. Proof. First of all, we note that δ q+1 t 2s 2t < 2δ q < δ q 1, and in particular q 3. Let us suppose that 2s Ik. Using Proposition 3.2, δ γk +1 l k 4 3 (2s) < 2(2s) < 2δ q 1 < δ q 2, that is γ k q 2. If we use Proposition 3.2 again, δ γk > l k 4(2s) 9 (2s) 4 δ q+1 4 > δ q+3, and then γ k q + 2. Finally, note that 2s 2t < δ q 1 < δ 1 < 1, and due to the fact that γ k q + 2 for those k such that 2s Ik, we have that (2s)q+2 (2s) γ k. Therefore we easily obtain the desired inequality: t q+2 (2s) q+2 = (2s) q+2 θk (2s) (2s) γk θk (2s) = ε(2s). Ik 2s Ik 2s 3.5. The function ϕ. Next, we will first adapt the function constructed by Ghomi in [15, Proposition 2.1] to suit our purposes, and then we will find quantitative estimates for its Hessian. We begin by defining { ε(2t) ε(t) = t n+3 if t > 0 0 if t 0. Since ε C (R), with ε (j) (0) = 0 for all j N {0}, it is easily seen that ε C (R) as well. Now, let us consider the function { t s g(t) = 0 0 ε(r)dr ds if t > 0 0 if t 0. It is clear that g C (R). In addition, g 1 (0) = (, 0] and g (t) = ε(t) > 0 for all t > 0. In particular, g is convex on R and positive, with a strictly positive second derivative, on (0, + ). We may assume that 0 C. Now, for every vector w S n 1, define h(w) = max z C z, w, the support function of C (for information about support functions of convex sets, see [19] for instance). We also define the function φ : S n 1 R n R (w, x) φ(w, x) = g( x, w h(w)). Note that when x C, we have x, w h(w) for every w S n 1. Therefore φ(w, ) is a function of class C (R n ) that vanishes on C for every w S n 1. It is also easy to check that the function φ(w, ), being a composition of a

20 20 DANIEL AZAGRA AND CARLOS MUDARRA convex function with a non-decreasing convex function, is convex as well. Finally, we define the function ϕ : R n R as follows: ϕ(x) = φ(w, x) dw S n 1 for every x R n. Again it is easy to check that ϕ 1 (0) = C and ϕ is convex. Because φ(w, ) is in C (R n ), φ is continuous and S n 1 is compact, it follows from standard results on differentiation under the integral sign that the function ϕ is of class C (R n ) as well. One can also check easily that D 2 ϕ(x)(v) 2 = g ( x, w h(w)) w, v 2 dw. S n Lower estimates for the Hessian of ϕ. For given x R n \ C and v S n 1 we will now find a region W = W (x, v) of S n 1 of sufficient volume (depending only, and conveniently, on d(x, C)) on which we have good lower estimates for g ( x, w h(w)) w, v 2. This will involve a careful selection of angles and directions. Fix a point x R n \ C, let x C be the metric projection of x onto the compact convex C, and set u x = 1 x x C (x x C), and α x = d(x, C) d(x, C) + diam(c). Lemma 3.6. With the above notation we have x, u x h(u x ) = d(x, C) and d(x, C) x, w h(w) 1 d(x, C) 2 for all w S n 1 such that ŵ u x [ α x ] 3, αx 2. Here ŵ u x denotes the length of the shortest geodesic (or angle) between w and u x in S n 1. Proof. The fact that x, u x h(u x ) = d(x, C) is a straightforward consequence of the definition of h and u x. For the second part, given w S n 1 with ŵ u x [ ] α x 3, αx 2, let us denote θ = ŵ ux. Since C is compact, we can find ξ C such that h(w) = ξ, w. Using that x, u x h(u x ) = x x C

21 SMOOTH CONVEX EXTENSIONS OF CONVEX FUNCTIONS 21 and w u x θ, we have x, w h(w) = x, w u x + x x C + h(u x ) h(w) x, w u x + x x C + ξ, u x w = x ξ, w u x + x x C (diam(c) + x x C ) θ + x x C (diam(c) + x x C ) α x 2 + x x C 1 2 x x C. The other inequality, d(x, C) x, w h(w), is straightforward. Next we find the region W we need. Lemma 3.7. Given any v S n 1 with u x, v 0, there exists a vector w 0 = w 0 (x, v) S n 1 such that if we define W = {w S n 1 : ŵ w 0 [0, α x 12 ]}, then: 1. for every w W, we have u x w [ ] α x 3, αx 2 ; 2. for every w W we have w, v sin( αx 3 ); 3. there exists a constant V (n) > 0, independent of x and v, such that vol S n 1(W ) V (n)αx n 1. Proof. We prove 1 and 2 at the same time by studying two cases separately: Case (i): u x v. Take w 0 in the unit circle of the plane generated by the vectors u x, and v so that ŵ 0 u x = 5αx 12 and so that the arc in that circle joining u x with w 0 has the same orientation as the arc joining u x with v. Set W = {w S n 1 : ŵ w 0 [0, αx 12 ]} and let w W. First, recalling that the angles shorter than π give the usual distance between points of S n 1, we may use the triangle inequality to estimate and û x w û x w 0 + ŵ 0 w 5α x 12 + α x 12 = α x 2 û x w û x w 0 ŵ 0 w 5α x 12 α x 12 = α x 3, that is û x w [ ] α x 3, αx 2. It only remains to see that w, v sin(αx /3) for all w W. First, we easily check that v w 0 π 2 5αx 12. For an arbitrary w W, we have v w v w 0 + ŵ 0 w π 2 5α x 12 + α x 12 = π 2 α x 3. Therefore v, w = cos( v w) cos( π 2 αx αx 3 ) = sin 3. Case (ii): u x = v. Take w 0 in the sphere S n 1 such that ŵ 0 u x = 5αx 12. If we define W = {w S n 1 : ŵ w 0 [0, αx 12 ]}, following the same estimations as

22 22 DANIEL AZAGRA AND CARLOS MUDARRA in Case (i) we obtain û x w [ αx 3, αx 2 ] for every w W. And we easily have w, v = w, u x sin αx 3. We now prove 3. Since the standard measure on S n 1 is invariant under isometries we may assume that W = {w S n 1 : ŵ e 1 [0, αx 12 ]}, where e 1 = (1, 0,..., 0). The set W is an hyperspherical cap, and its volume is given by vol S n 1(W ) = vol(s n 2 ) αx/12 0 sin n 2 (β)dβ, where vol(s n 2 ) = 1 in the special case n = 2. But for angles β such that 0 β αx 12 π 3, it is clear that sin β 1 2β, and therefore where vol S n 1(W ) vol(s n 2 ) V (n) = αx/12 0 ( 1 2 β)n 2 dβ = V (n)α n 1 x, vol(s n 2 ) 12(n 1)(24) n 2 for n 2, and this completes the proof of the Lemma Convexity of f + ψ on a neighbourhood of C. Now, using the constant V (n) obtained in Lemma 3.7, define C(n) = V (n) 36(1 + diam(c)) n+1. Proposition 3.8. With the notation of subsection 3.4, consider the function H = f + 2 C(n) ϕ defined on Rn, and take r = δ 4. Then, for every x R n \ C such that t := d(x, C) r, and for every v S n 1, we have D 2 H(x)(v) 2 t q, where q N is the unique positive integer such that δ q+1 t < δ q. Proof. Fix x, t, v, q as in the statement. Since D 2 H(x)(v) 2 = D 2 H(x)( v) 2, we may suppose that v, u x 0, where u x = x x C / x x C and x C is the metric projection of x onto C. Take the angle α x and the set W = W (x, v) as in Lemmas 3.6 and 3.7 respectively. By the construction of ϕ, we have (3.1) D 2 ϕ(x)(v) 2 = ε( x, w h(w)) w, v 2 dw S n 1 ε( x, w h(w)) w, v 2 dw > 0, and for w W, Lemma 3.7 gives us that ŵ u x [ α x Lemma 3.6 says that, in this case, W t 2 x, w h(w) t δ 4. 3, αx 2 ] ; on the other hand

23 SMOOTH CONVEX EXTENSIONS OF CONVEX FUNCTIONS 23 Using Lemma 3.5 we obtain ε( x, w h(w)) = ε(2( x, w h(w))) ( x, w h(w)) n+3 t q+2 tq+2 ( x, w h(w)) n+3 t n+3 = tq t n+1. On the other hand, due to Lemma 3.7 the product v, w is greater or equal than sin( αx 3 ) for all w W. Combining the two preceding inequalities, we have tq D 2 ϕ(x)(v) 2 t n+1 sin2 ( α x 3 ) vol Sn 1(W ). Due to Part 3 of Lemma 3.7, the last term is greater or equal than t q t n+1 sin2 ( α x )V (n)αn 1 x. 3 Since αx < π αx 3, we have that sin( 3 ) 1 α x 2 3, so we obtain Moreover, we have D 2 ϕ(x)(v) 2 α x = tq αx 2 t n+1 36 V (n)αn 1 x = tq αx n+1 t n+1 36 V (n). t t + diam(c) t 1 + diam(c), because t r = δ 4 < 1. Gathering these inequalities, we get D 2 ϕ(x)(v) 2 tq t n+1 t n+1 36(1 + diam(c)) n+1 V (n) = C(n)tq. Finally, due to the construction of the sequence {δ p } p, we have d(x, C) = t < δ q r q+2, hence Proposition 3.1 ensures that D 2 f(x)(v) 2 t q. Therefore D 2 H(x)(v) 2 = D 2 f(x)(v) C(n) D2 ϕ(x)(v) 2 t q + 2t q = t q. Since ϕ = 0 on C, we have proved that H is a function of class C that coincides with f on C and such that H has a strictly positive Hessian on the set {x R n : 0 < d(x, C) r} Conclusion of the proof: convexity of f + ψ on R n. To complete the proof of Theorem 1.3 we only have to change the funcion H slightly. Proposition 3.9. There exists a constant a > 0 such that the function F := f + aϕ is C (R n ), coincides with f on C, is convex on R n, and has a strictly positive Hessian on R n \ C.