Lipschitz sums of convex functions

Marianna Csörnyei^{∗} and Assaf Naor ^{†}

Abstract

In this paper we give a geometric characterization of the convex subsets of a Banach space with the property that for any two convex continuous functions on this set, if their sum is Lipschitz, then the functions must be Lipschitz. We apply this result to the theory of Δ-convex functions.

1 Introduction

This paper deals with the following problem: let K be a convex subset of a Banach space X. Let h_{1}, h_{2} : K → ℝ be convex continuous functions such that h_{1} + h_{2} is Lipschitz. Does this necessarily imply that h_{1} and h_{2} are themselves Lipschitz? More precisely, we are interested in the geometric properties of K which imply such a statement. Under some mild assumptions, we give here a necessary and sufficient condition for K to have such a property.

Throughout this paper, all Banach spaces are real. If X is a Banach space, x ∈ X and r > 0 we use the notation B(x, r) = {y ∈ X; ∥x − y∥ ≤ r} and S(x, r) = {y ∈ X; ∥x − y∥ = r}. We also denote B_{X} = B(0, 1) and S_{X} = S(0, 1). X^{∗} denotes the dual space of X. For any two sets A, B we denote by d(A, B) the distance between A and B.

The method of proof forces us to impose some assumptions on K. Apart from the natural assumptions 0 ∈ K and \overline{span}(K) = X, we essentially assume that either K − K or its complement is nowhere dense. This will be formulated more precisely later, but we prefer to begin by stating the theorem in a simpler, somewhat less general form. We do not presently know if the result is true for a general convex set which is not contained in any hyperplane.

∗Supported by the Hungarian National Foundation for Scientific Research, grant # F029768.

†Partially supported by the Clore Foundation and the EU grant HPMT-CT-2000-00037.

Theorem 1 Let X be a Banach space and suppose that K is a convex subset of X for which

• 0 ∈ K;

• \overline{span}(K) = X;

and either

• K is locally weakly compact

or

• K has nonempty interior.

Then the following are equivalent:

(i) If h_{1}, h_{2} : K → ℝ are continuous convex functions such that h_{1} + h_{2} is Lipschitz, then both h_{1} and h_{2} must be Lipschitz.

(ii) There exists a constant 0 < c ≤ 2 such that for every x ∈ K, 0 < r < diam K and y ∈ S_{X} there are u, v ∈ K ∩ B(x, r) such that u − v = cry.

(iii) There is a c > 0 such that for every x^{∗}, y^{∗} ∈ S_{X^{∗}} and for every x, y ∈ K with y^{∗}(x) < y^{∗}(y) there are u, v ∈ K such that y^{∗}(x) ≤ y^{∗}(u) ≤ y^{∗}(v) ≤ y^{∗}(y) and |x^{∗}(u) − x^{∗}(v)| > c(y^{∗}(y) − y^{∗}(x)).

Remark 2 We remark that the implication (ii) ⟹ (i) requires only the assumptions 0 ∈ K and \overline{span}(K) = X. Thus in our later applications, we will be able to state results for arbitrary convex subsets of Banach spaces satisfying these two assumptions.

If K is bounded and has nonempty interior, then (ii) must be satisfied. Indeed, by convexity, if B(x_{0}, r_{0}) ⊂ K for some x_{0} and r_{0}, then K ∩ B(x, r) contains a segment of length r_{0}r/diam K in each direction. It is also easy to see that every convex set K which contains an infinite open cone satisfies condition (ii).

If dim X < ∞, then every convex set K with \overline{span}(K) = X and 0 ∈ K has nonempty interior. It is also easy to see that if dim X < ∞, and if a convex set K ⊂ X has nonempty interior then conditions (i)–(iii) of Theorem 1 imply that K is either bounded or contains an infinite open cone:

Corollary 3 Conditions (i)–(iii) of Theorem 1 hold for any convex set K with nonempty interior, which is either bounded or contains an infinite open cone.

Suppose dim X < ∞ and K is a convex set for which 0 ∈ K and \overline{span}(K) = X. Then (i)–(iii) of Theorem 1 are satisfied by K if and only if either K is bounded or K contains an infinite cone C for which \overline{aff}(C) = X.

In contrast, in Remark 13 we give an example of an open unbounded convex set K in an infinite dimensional Banach space, which fulfills conditions (i)–(iii) but does not contain a half-line. We also give an example of an open convex set K which contains an infinite cone C such that \overline{span}(C) = X, but for which conditions (i)–(iii) fail.

For any convex set K, x ∈ K and r > 0 set

K_{x,r} = K ∩ B(x, r).

It is easy to see that if K has nonempty interior or if K_{x,r} is weakly compact for every x ∈ K, r > 0, then K_{x,r} − K_{x,r} has nonempty interior or is weakly compact, respectively. Every weakly compact set is either nowhere dense or has nonempty interior.

In fact, our proof uses only the following fact:

(∗) For every x ∈ K, r > 0 and y ∉ K_{x,r} − K_{x,r} there exists a z arbitrarily close to y and there exists z^{∗} ∈ S_{X^{∗}} which separates z and K_{x,r} − K_{x,r}.

By the Hahn-Banach Theorem, (∗) is satisfied if and only if K_{x,r} − K_{x,r} is either nowhere dense or has nonempty interior. We prove, in fact, the following strengthening of Theorem 1:

Theorem 4 Let X be a Banach space and suppose that K is a convex subset of X for which

• 0 ∈ K;

• \overline{span}(K) = X; and either

• K_{x,r} − K_{x,r} is nowhere dense for every x ∈ K and r > 0

or

• K_{x,r} − K_{x,r} has nonempty interior for every x ∈ K and r > 0.

Then the following are equivalent:

(i) If h_{1}, h_{2} : K → ℝ are continuous convex functions such that h_{1} + h_{2} is Lipschitz, then both h_{1} and h_{2} must be Lipschitz.

(ii) There exists a constant 0 < c ≤ 2 such that for every x ∈ K, r < diam K and y ∈ S_{X} there are u, v ∈ K ∩ B(x, r) such that u − v = cry.

(iii) There is a c > 0 such that for every x^{∗}, y^{∗} ∈ S_{X^{∗}} and for every x, y ∈ K with y^{∗}(x) < y^{∗}(y) there are u, v ∈ K such that y^{∗}(x) ≤ y^{∗}(u) ≤ y^{∗}(v) ≤ y^{∗}(y) and |x^{∗}(u) − x^{∗}(v)| > c(y^{∗}(y) − y^{∗}(x)).

A rather unexpected consequence of our result is the following fact:

Corollary 5 Suppose that K is an unbounded open set which satisfies the hypotheses of Theorem 4. If H ⊂ X is convex and K ⊂ H, then H satisfies (i).

Proof: By Remark 2, all we have to do is to check that condition (ii) for K implies condition (ii) for H. Fix some x ∈ H, r > 0, y ∈ S_{X} and z ∈ K.

If r ≥ d(x, z) then, using (ii) for K, there are

u, v ∈ B(z, r) ∩ K ⊂ B(x, 2r) ∩ H

with u − v = cry. Then of course u′ = (u + x)/2 and v′ = (v + x)/2 are in B(x, r) ∩ H and u′ − v′ = (c/2)ry.

On the other hand, if r < d(x, z) then, using (ii) for K, there are

u, v ∈ B(z, d(x, z)) ∩ K ⊂ B(x, 2d(x, z)) ∩ H

such that u − v = cd(x, z)y. Now, the points

u′ = x + \frac{r}{2d(x, z)}(u − x) and v′ = x + \frac{r}{2d(x, z)}(v − x)

are in B(x, r) ∩ H and

u′ − v′ = \frac{crd(x, z)}{2d(x, z)}y = (c/2)ry.

Therefore H satisfies (ii) for c/2.

Before passing to the proof of Theorem 4, we discuss an application of this result to the theory of Δ-convex mappings.

Let X and Y be Banach spaces, and K ⊂ X convex. A mapping F: K → Y is called Δ-convex if there is a continuous convex function f: K → ℝ such that for every y^{∗} ∈ S_{Y^{∗}}, y^{∗} ∘ F + f is continuous and convex. Such an f is called a control function for F.

Such functions are known to have many interesting properties. For instance, it is known that the Δ-convex mappings from ℝ^{n} to ℝ form an algebra, and that the composition of two Δ-convex mappings is itself Δ-convex. Δ-convex mappings have applications in optimization theory, approximation theory, and the theory of Gâteaux differentiability of convex functions. For more information we refer to the papers [VZ], [DVZ] and the references therein. The geometric importance of Δ-convex mappings may be appreciated from a result of M. Cepedello Boiso [C1, C2] (see also the book [BL]), which states that a Banach space X is superreflexive if and only if every realvalued Lipschitz function on X can be uniformly approximated by Δ-convex functions.

The properties of a control function f for a Δ-convex function F impose restrictions on the behavior of F. For instance, it is proved in [VZ] that F is differentiable (Fréchet or Gâteaux) at some point whenever f is. One can ask: is F Lipschitz whenever f is? Theorem 1 and a simple Banach-Steinhaus argument give the following improvement of Theorem 18 in [DVZ]:

Theorem 6 Let X and Y be Banach spaces and K ⊂ X a convex set satisfying (ii). If F: K → Y is Δ-convex with Lipschitz control function f, then F is Lipschitz.

Proof: For each y^{∗} ∈ S_{Y^{∗}}, the functions g_{1} = y^{∗}∘F +f and g_{2} = −y^{∗}∘F +f are continuous and convex on K, and g_{1} + g_{2} is Lipschitz. By Remark 2, K satisfies also (i), so that g_{1} and g_{2} are Lipschitz. Consequently y^{∗} ∘ F is Lipschitz for each y^{∗} ∈ Y^{∗}. The Banach-Steinhaus theorem now implies that F is Lipschitz.

The example in Remark 13 shows that this is a strict improvement of the result of [DVZ].

Acknowledgment We would like to express our sincere gratitude to Libor Veselý for suggesting the problem to us and pointing out that our results imply Theorem 6.

2 Proof of Theorem 4

Let K ⊂ X be a convex set which satisfies the assumptions of Theorem 4. We break the proof of Theorem 4 down into several lemmas. We prove the theorem by showing that (ii) ⟹ (i), (iii) ⟹ (ii) and ¬(iii) ⟹ ¬(i).

For every pair of vectors x, y ∈ X \ {0} define

α(0, x, y) = min (d (\frac{x}{∥x∥}, ℝy) , d (\frac{y}{∥y∥}, ℝx)) .

For any p ∈ X put α(p, x, y) = α(0, x − p, y − p).

Similarly, if x, y ∈ X and p ∈ X \ {x, y} then we define:

β(p, x, y) = \frac{∥x − y∥}{max(∥p − x∥, ∥p − y∥)}.

Obviously β(p, x, y) ≥ α(p, x, y).

We begin by showing that condition (ii) of Theorem 4 implies an apparently stronger geometric condition.

Lemma 7 Assume that condition (ii) of Theorem 4 holds for some constant 0 < c ≤ 2. Then for every x ∈ K, r < diam K and y ∈ S_{X} there are u, v ∈ K ∩ B(x, r) such that:

(1) u − v = \frac{c^{2}r}{8+c}y;

(2) d(x + ℝy, u + ℝy) ≥ \frac{cr}{10};

(3) β(x, u, v) ≥ α(x, u, v) ≥ \frac{c^{3}}{120}.

Proof: Define r′ = r/(1 + c/8). Note that there is some z ∈ K ∩ B(x, r′) such that d(z, x + ℝy) ≥ cr′/4. Indeed, if y^{∗} is a norm-one functional with y^{∗}(y) = 0 then for every w ∈ S_{X} with y^{∗}(w) > 1/2, by our assumption, there are a, b ∈ K ∩ B(x, r′) for which a − b = cr′w. Then

d(a, x + ℝy) + d(b, x + ℝy) ≥ |y^{∗}(a − x)| + |y^{∗}(b − x)| ≥ y^{∗}(cr′w) > cr′/2,

thus max{d(a, x + ℝy), d(b, x + ℝy)} ≥ cr′/4.

Fix any such z. Using the assumption again, there are

u, v ∈ K ∩ B(z, cr′/8) ⊂ K ∩ B(x, r)

for which

u − v = c(cr′/8)y = \frac{c^{2}r}{8 + c}y.

Now

d(x + ℝy, u + ℝy) ≥ \frac{cr′}{4} − \frac{cr′}{8} = \frac{cr}{8 + c} ≥ \frac{cr}{10}.

For any q ∈ x + ℝ(u − x) write q = λu + (1 − λ)x and define

q′ = x + \frac{v − q}{1 − λ} = u + \frac{v − u}{1 − λ}.

Clearly q′ is on the line u + ℝy, thus ∥q′ − x∥ ≥ cr/10. Therefore

∥v − q∥ = ∥q′ − x∥|1 − λ| = ∥q′ − x∥ · \frac{∥u − q∥}{∥u − x∥} ≥ \frac{cr}{10} · \frac{∥u − v∥ − ∥v − q∥}{∥u − x∥}.

Hence

∥v − q∥ ≥ \frac{\frac{cr}{10} · ∥u − v∥}{∥u − x∥ + \frac{cr}{10}} ≥ \frac{\frac{cr}{10} · \frac{c^{2}r}{10}}{r + \frac{cr}{10}} ≥ \frac{c^{3}r}{120}.

Finally,

\frac{∥v − q∥}{∥v − x∥} ≥ \frac{c^{3}}{120},

which by symmetry proves (3).

Lemma 8 Assume that f, g: K → ℝ are such that h_{1} = g + f and h_{2} = g − f are continuous convex functions, and g is Lipschitz with constant L. Then for every line ℓ which intersects K in at least two points and for every x, y, z, w ∈ ℓ ∩ K, x ≠ y, z ≠ w

\frac{|f(x) − f(y)|}{∥x − y∥} − \frac{|f(z) − f(w)|}{∥z − w∥} ≤ 4L.

Proof: Without loss of generality we may assume that ℓ = ℝ. Put u = min(x, y, z, w) and v = max(x, y, z, w). Then for any a, b ∈ [u, v], a ≠ b and i = 1, 2

D^{+}h_{i}(u) ≤ \frac{h_{i}(a) − h_{i}(b)}{a − b} ≤ D^{−}h_{i}(v),

where D^{+} and D^{−} denote the right and left derivatives, respectively. In particular, 0 ≤ D^{−}h_{i}(v) − D^{+}h_{i}(u), and so

0 ≤ (D^{−}h_{2}(v)−D^{+}h_{2}(u))+(D^{−}h_{1}(v)−D^{+}h_{1}(u)) = 2D^{−}g(v)−2D^{+}g(u) ≤ 4L.

It follows that either 0 ≤ D^{−}h_{1}(v) − D^{+}h_{1}(u) ≤ 2L, or 0 ≤ D^{−}h_{2}(v) − D^{+}h_{2}(u) ≤ 2L. If 0 ≤ D^{−}h_{1}(v) − D^{+}h_{1}(u) ≤ 2L, then

4L ≥  (D^{−}h_{1}(v) + L) − (D^{+}h_{1}(u) − L)
≥ ^{\frac{(h_{1}(x)−h_{1}(y))−(g(x)−g(y))}{x−y}} − ^{\frac{(h_{1}(z)−h_{1}(w))−(g(z)−g(w))}{z−w}}
=  \frac{f(x)−f(y)}{x−y} − \frac{f(z)−f(w)}{z−w}.

If 0 ≤ D^{−}h_{2}(v) − D^{+}h_{2}(u) ≤ 2L, then a similar calculation gives the same bound. Proof that (ii) ⟹ (i): Suppose that h_{1}, h_{2} are continuous convex functions on K, for which h_{1}+h_{2} is Lipschitz. We write h_{1} = g+f and h_{2} = g−f. Then g is Lipschitz with constant L, say. Our goal is to prove that f is also Lipschitz.

Fix two points p_{1}, p_{2} ∈ K and some ε > 0. Since f is continuous at p_{1} and p_{2}, there is a 0 < δ ≤ ∥p_{1} − p_{2}∥/4 such that |f(p_{1}) − f(q)| < ε for every q ∈ B(p_{1}, δ) and |f(p_{2}) − f(q)| < ε for every q ∈ B(p_{2}, δ). Let ℓ be a line which intersects K in at least two points. Take any p which is in the relative interior of the interval K ∩ ℓ. Clearly max{∥p − p_{1}∥, ∥p − p_{2}∥} ≥ 2δ, so that we may assume that ∥p − p_{1}∥ = d ≥ 2δ.

Applying Lemma 7 for x = p, r = d − δ and y a unit vector in the direction of ℓ, we find u, v ∈ K ∩B(p, d−δ) which satisfy conclusions (1), (2) and (3) of Lemma 7. For every a ∈ B(p, d − δ) ∩ K we have ∥p_{1} − a∥ ≥ δ. By convexity, this implies that a′ = p_{1} + \frac{δ}{∥a−p_{1}∥}(a − p_{1}) is in K. Lemma 8 applied to the line connecting a and p_{1} shows that for any two points b, c on this line

\frac{|f(b) − f(c)|}{∥b − c∥} ≤ \frac{|f(a′) − f(p_{1})|}{∥a′ − p_{1}∥} + 4L ≤ \frac{ε}{δ} + 4L

holds. By Lemma 7 the distance between p and u and the distance between p and v is at least c(d − δ)/10 ≥ cd/20. On the other hand, the distance between p_{1} and any of the points p, u, v is at most 2d. Hence

|f(u) − f(p)| ≤ |f(u) − f(p_{1})| + |f(p) − f(p_{1})|
≤ (\frac{ε}{δ} + 4L) · 4d
≤ (\frac{ε}{δ} + 4L) · \frac{80∥u − p∥}{c}.

That is,

\frac{|f(u) − f(p)|}{∥u − p∥} ≤ \frac{80}{c} (\frac{ε}{δ} + 4L) ,

and, similarly,

\frac{|f(v) − f(p)|}{∥v − p∥} ≤ \frac{80}{c} (\frac{ε}{δ} + 4L) .

In addition to the above, we know that α(p, u, v) ≥ c^{3}/120.

Now we use the continuity of f to find some p′ = p + λy ∈ K ∩ ℓ where λ > 0 is small enough to ensure that

max (\frac{|f(u) − f(p′)|}{∥u − p′∥}, \frac{|f(v) − f(p′)|}{∥v − p′∥}) ≤ \frac{100}{c} (\frac{ε}{δ} + 4L) .

Let s be the intersection point of the line segments between p and u, and p′ and v (note that p′ − p and u − v have the same direction, therefore this intersection point exists). By Lemma 8

\frac{|f(p) − f(s)|}{∥p − s∥} ≤ \frac{|f(p) − f(u)|}{∥p − u∥} + 4L ≤ \frac{100}{c} (\frac{ε}{δ} + 5L) ,

and similarly, with p′ replacing p and v replacing u. Hence if λ is small enough then

\frac{|f(p′) − f(s)|}{∥p′ − s∥} ≤ \frac{100}{c} (\frac{ε}{δ} + 6L) .

By choosing λ to be small enough, we can also ensure that α(s, u, v) = α(s, p, p′) ≥ c^{3}/200. Hence

|f(p) − f(p′)| ≤ |f(p) − f(s)| + |f(p′) − f(s)|
≤ 2 max (∥p − s∥, ∥p′ − s∥) · \frac{100}{c} · (\frac{ε}{δ} + 6L)
= \frac{∥p − p′∥}{β(s, p, p′)} · \frac{200}{c} · (\frac{ε}{δ} + 6L)
≤ \frac{∥p − p′∥}{α(s, p, p′)} · \frac{200}{c} · (\frac{ε}{δ} + 6L)
≤ \frac{40000∥p − p′∥}{c^{4}} (\frac{ε}{δ} + 6L) .

This proves that f is Lipschitz on each line ℓ with Lipschitz constant

\frac{40000}{c^{4}} (\frac{ε}{δ} + 6L) .

Proof that (iii) ⟹ (ii): It is convenient to distinguish three cases:

Case 1: K is bounded. Case 2: K is unbounded, and there is an x^{∗} ∈ S_{X^{∗}} for which

sup_{x∈K} |x^{∗}(x)| < ∞.

Case 3: For every x^{∗} ∈ S_{X^{∗}} we have

sup_{x∈K} |x^{∗}(x)| = ∞.

Proof for Case 1: In this case condition (ii) of Theorem 4 is equivalent to the fact that for some ε > 0, K contains a line segment of length ε in each direction. Indeed, (ii) clearly implies the existence of such line segments. On the other hand, if there are segments of length ε in each direction, then by convexity for every x ∈ K and r < diam K the ball B(x, r) contains a segment of length εr/diam K in every direction.

Fix any y^{∗} ∈ S_{X^{∗}} and x, y ∈ K such that y^{∗}(x) < y^{∗}(y). Assume that (ii) does not hold. Then there are unit vectors {y_{n}}_{n=1}^{∞} ⊂ S_{X} and a sequence of positive numbers {ε_{n}}_{n=1}^{∞} tending to zero such that K ∩ (K + ε_{n}y_{n}) = ∅ for every n. In other words, ε_{n}y_{n} ∉ K − K. Our assumption (∗) on K ensures that there is a z_{n} of norm arbitrarily close to ε_{n} which can be separated from K − K. Therefore there is an x_{n}^{∗} ∈ S_{X^{∗}} such that x_{n}^{∗}(u) − x_{n}^{∗}(v) ≤ 3ε_{n} for every u, v ∈ K. But condition (iii) implies the existence of u, v ∈ K for which

c(y^{∗}(y) − y^{∗}(x)) < x^{∗}(u) − x^{∗}(v) ≤ 3ε_{n},

which is a contradiction when n is large enough. Proof for Case 2: In this case it is easy to see that (iii) cannot hold. Indeed, since K is unbounded, we can find y^{∗} ∈ S_{X^{∗}} for which sup_{x∈K} |y^{∗}(x)| = ∞. Then for every c > 0 there are x, y ∈ K such that

c(y^{∗}(y) − y^{∗}(x)) > 2 sup_{x∈K} |x^{∗}(x)| ≥ sup_{u,v∈K} |x^{∗}(u) − x^{∗}(v)|.

Proof for Case 3: Assume first that (ii) fails for some constant 0 < c < 1/3. In other words, there is 0 < c < 1/3, x ∈ K, y ∈ S_{X} and r > 0 for which

cry ∉ K_{x,r} − K_{x,r}.

Without loss of generality, we may assume that x = 0. Hence, for every ε > 0 we can find z^{∗} ∈ S_{X^{∗}} such that z^{∗}(u) − z^{∗}(v) ≤ (c + ε)r for all u, v ∈ K ∩ B(0, r), that is, by choosing ε < 1/3 − c and replacing c by c + ε we have a z^{∗} ∈ S_{X^{∗}} such that z^{∗}(u) − z^{∗}(v) ≤ cr for all u, v ∈ K ∩ B(0, r). In particular, for every u ∈ K ∩ B(0, r) we have |z^{∗}(u)| ≤ cr. Define

A^{+} = {x ∈ X; z^{∗}(x) > cr and z^{∗}(x) > c∥x∥} ,
A^{−} = {x ∈ X; z^{∗}(x) < −cr and z^{∗}(x) < −c∥x∥} .

We claim that both A^{+} and A^{−} are disjoint from K. Indeed, if z^{∗}(x) > cr and z^{∗}(x) > c∥x∥ for some x ∈ K, then there is a λ > 1 such that z^{∗}(x) > λcr and z^{∗}(x) > λc∥x∥. Then λcrx/|z^{∗}(x)| ∈ K ∩ B(0, r), so |z^{∗}(λcrx/|z^{∗}(x)|)| ≤ cr, which is a contradiction. Similarly for A^{−}.

By the Hahn-Banach Theorem, since A^{+}, A^{−} are convex with nonempty interior, we can find x^{∗}, v^{∗} ∈ S_{X^{∗}} and α ≥ 0 ≥ β for which

inf_{u∈A^{+}} x^{∗}(u) ≥ α ≥ sup_{u∈K} x^{∗}(u), inf_{u∈K} v^{∗}(u) ≥ β ≥ sup_{u∈A^{−}} v^{∗}(u).

Since K is not contained in any hyperplane, we have x^{∗} ≠ v^{∗}. Let δ > 0, and take x, z ∈ B(0, 1 + δ) for which x^{∗}(x) = z^{∗}(z) = 1. Note that z^{∗}(u) ≤ c∥u∥ for any u ∈ ker(x^{∗}), since otherwise for λ large enough and u′ = λu ∈ ker(x^{∗}) we would have z^{∗}(u′) ∈ int(A^{+}), a contradiction. Since z = x^{∗}(z)x + (z − x^{∗}(z)x) and z − x^{∗}(z)x ∈ ker(x^{∗}), we have for δ sufficiently small,

1 = z^{∗}(z) = x^{∗}(z)z^{∗}(x) + z^{∗}(z − x^{∗}(z)x)
≤ x^{∗}(z)z^{∗}(x) + c∥z − x^{∗}(z)x∥
≤ x^{∗}(z)z^{∗}(x) + 3c.

It is also easy to see that λz ∈ A^{+} if λ is large enough and δ is small enough, therefore x^{∗}(z) ≥ 0. Then, since c < 1/3 and 1 ≤ x^{∗}(z)z^{∗}(x) + 3c, it is immediate to see that x^{∗}(z) > 0 and z^{∗}(x) > 0. Therefore

1 ≤ x^{∗}(z)z^{∗}(x) + 3c ≤ (1 + δ)z^{∗}(x) + 3c,

that is, z^{∗}(x) ≥ (1 − 3c)/(1 + δ) > 1 − 4c if δ is small enough. Hence, for any u ∈ S_{X}

c(2 + δ) ≥ c∥u − x^{∗}(u)x∥ ≥ |z^{∗}(u − x^{∗}(u)x)|
= |z^{∗}(u) − x^{∗}(u)z^{∗}(x)|
= |z^{∗}(u) − x^{∗}(u) + x^{∗}(u)(1 − z^{∗}(x))|
≥ |z^{∗}(u) − x^{∗}(u)| − 4c.

Thus ∥z^{∗} − x^{∗}∥ ≤ c(6 + δ) for every δ small enough, therefore ∥z^{∗} − x^{∗}∥ ≤ 6c. Similarly, we find that ∥z^{∗} − v^{∗}∥ ≤ 6c. Let

c_{0} = ∥x^{∗} − v^{∗}∥ ≤ 12c,

take an arbitrary point x_{1} ∈ K, and define

y^{∗} = \frac{x^{∗} − v^{∗}}{c_{0}} ∈ S_{X^{∗}}.

By assumption, no functional is bounded on K, and yet

y^{∗}(x) ≤ \frac{α − β}{c_{0}} ∀x ∈ K.

Therefore for every t < y^{∗}(x_{1}) we can find an x_{2} ∈ K for which y^{∗}(x_{2}) = t. Fix x_{2} with

y^{∗}(x_{2}) = 2y^{∗}(x_{1}) − \frac{α − β}{c_{0}} .

Since v^{∗}(x) ≥ β, we have x^{∗}(x) ≥ β +x^{∗}(x_{2})−v^{∗}(x_{2}) for any x ∈ K for which y^{∗}(x_{2}) ≤ y^{∗}(x). Thus, for any y_{1}, y_{2} ∈ K for which y^{∗}(x_{2}) ≤ y^{∗}(y_{i}) ≤ y^{∗}(x_{1}), α ≥ x^{∗}(y_{i}) ≥ β + x^{∗}(x_{2}) − v^{∗}(x_{2}) is satisfied. That is,

|x^{∗}(y_{1}) − x^{∗}(y_{2})| ≤ α − β − x^{∗}(x_{2}) + v^{∗}(x_{2})
= α − β − y^{∗}(x_{2})c_{0}
= (2y^{∗}(x_{1}) − y^{∗}(x_{2}))c_{0} − y^{∗}(x_{2})c_{0}
= 2c_{0}(y^{∗}(x_{1}) − y^{∗}(x_{2}))
≤ 24c(y^{∗}(x_{1}) − y^{∗}(x_{2})).

Summarizing, we have proved that if condition (iii) is satisfied with a constant c < 1/72, then condition (ii) must be satisfied with constant c/24. But if (iii) is satisfied for some c then it is also satisfied for every c′ < c, which completes our proof.

This finishes the proof of the implication (iii) ⟹ (ii).

For later purposes, we record here that from the proof of Case 2 it follows that if K is unbounded and (iii) fails for some constant c, then there are x^{∗}, y^{∗} ∈ S_{X^{∗}} and x, y ∈ K such that (iii) fails for c, x, y, x^{∗}, y^{∗} and y^{∗} is not bounded on K. By symmetry, we can assume that sup_{x∈K} y^{∗}(x) = +∞.

It remains to prove that if (iii) fails then there are continuous functions f, g: K → ℝ such that f is not Lipschitz, g is Lipschitz and h_{1} = g + f, h_{2} = g − f are convex.

We begin by defining some auxiliary functions:

Lemma 9 Fix 0 < ε < 1/2, β > 0 and L > 1. Let α = ε^{2}β and γ = e^{L}β. Then there are continuous functions f, g: ℝ^{2} → ℝ for which

(A) g + f, g − f are both convex on ([−α, α] × ℝ) ∪ (ℝ × (ℝ \ [0, 2γ]));

(B) |f_{y}| < ε, |g_{x}|, |g_{y}| < 40ε on ([−α, α] × ℝ) ∪ (ℝ × (ℝ \ [0, 2γ]));

(C) |f(x, y)| ≤ αL on [−α, α] × ℝ and f(x, y) = 0 on ℝ × (ℝ \ [β, 2γ − β]);

(D) f_{x}(x, γ) = L and |f_{x}(x, y)| ≤ L for every x, y ∈ ℝ.

(Here f_{x}, f_{y}, g_{x}, g_{y} denote the partial derivatives of f and g.)

Proof: Denote the sets {y ≤ β}, {β ≤ y ≤ γ}, {γ ≤ y ≤ 2γ − β} and {2γ − β ≤ y} by S_{1}, S_{2}, S_{3} and S_{4}, respectively. Let

f(x, y) = \begin{cases}0 & y ∈ S_{1} \\ x log(y/β) & y ∈ S_{2}, \end{cases}

and extend f to ℝ × S_{3} and ℝ × S_{4} by f(x, y) = f(x, 2γ − y). Clearly (C) and (D) hold. Let g = p + q + r, where

p(x, y) = \begin{cases}0 & y + |x| ≤ β \\ \frac{ε(y+|x|−β)^{2}}{α} & y ≤ β ≤ y + |x| \\ \frac{εx^{2}}{α} & y ∈ S_{2} \end{cases}

q(x, y) = \begin{cases}0 & y ∈ S_{1} \\ −\frac{α log(y/β)}{ε} & y ∈ S_{2} \end{cases}

r(x, y) = \begin{cases}0 & y ∈ S_{1} \\ 12ε(y − β) & y ∈ S_{2} \\ 12ε(2y − β − γ) & y ∈ S_{3} \\ 12ε(3y − 3γ) & y ∈ S_{4}, \end{cases}

and extend p, q to the whole plane by p(x, y) = p(x, 2γ − y), q(x, y) = q(x, 2γ − y). These are well-defined continuous functions.

It is easy to check that the restriction of g to ℝ × S_{1} is convex, and thus, as the restriction of f to ℝ×S_{1} is 0, it follows that g +f and g −f are convex on ℝ×S_{1}. It is also straightforward to verify that the restrictions of g+f and g−f to {−α ≤ x ≤ α, y ∈ S_{2}} are convex: Indeed, on this set g+f and g−f may be written as the sum of linear functions together with the functions h^{+} and h^{−}, respectively, where h^{±}(x, y) = ±x log y + εx^{2}/α − (α log y)/ε. On calculating, the second partial derivatives of h^{±} we find h_{xx}^{±} = 2ε/α > 0 and

h_{xx}^{±}h_{yy}^{±} − (h_{xy}^{±})^{2} = \frac{2\frac{ε}{α}(\frac{α}{ε} ∓ x) − 1}{y^{2}} = \frac{2 ∓ 2εx/α − 1}{y^{2}} ≥ \frac{1 − 2ε}{y^{2}} > 0,

that is, the functions h^{±} are convex on ℝ×S_{2}. This also shows that g+f and g − f are convex on the sets ℝ × S_{4} and {−α ≤ x ≤ α, y ∈ S_{3}}. Therefore, in order to show that g ± f are convex on ([−α, α] × ℝ) ∪ (ℝ × (ℝ \ [0, 2γ])), it only remains to check their convexity around the points of the segments {y = β}, {y = γ} and {y = 2γ − β}.

At the points (x, y) ∈ ([−α, α] × ℝ) ∪ (ℝ × (ℝ \ [0, 2γ])) we have

f_{x}(x, y) = \begin{cases}0 & y ∈ S_{1} \\ log(y/β) & y ∈ S_{2} \end{cases}

|f_{y}(x, y)| ≤ \frac{α}{β} < ε

|p_{x}| ≤ 2ε, |p_{y}| ≤ 2ε, q_{x} = 0, |q_{y}| ≤ \frac{α}{εβ} = ε, r_{x} = 0

r_{y} = \begin{cases}0 & y ∈ S_{1} \\ 12ε & y ∈ S_{2} \\ 24ε & y ∈ S_{3} \\ 36ε & y ∈ S_{4}. \end{cases}

Since |f_{y}|, |p_{y}|, |q_{y}| are at most 2ε, and r_{y}|_{S_{i+1}} − r_{y}|_{S_{i}} = 12ε, it follows that

(g_{y} ± f_{y})|_{S_{i+1}} − (g_{y} ± f_{y})|_{S_{i}} ≥ 0.

From this it easily follows that g ± f are also convex around the points of the required segments. This also shows that (B) is satisfied. Proof that (i) ⟹ (iii): Suppose that (iii) does not hold. For every n fix c_{n} > 0, x_{n}, y_{n} ∈ K and x_{n}^{∗}, y_{n}^{∗} ∈ S_{X^{∗}} such that c_{n} → 0, y_{n}^{∗}(x_{n}) < y_{n}^{∗}(y_{n}) and for every n and for every u, v ∈ K with y_{n}^{∗}(x_{n}) ≤ y_{n}^{∗}(u) ≤ y_{n}^{∗}(v) ≤ y_{n}^{∗}(y_{n}) we have

|x_{n}^{∗}(u) − x_{n}^{∗}(v)| ≤ c_{n}(y_{n}^{∗}(y_{n}) − y_{n}^{∗}(x_{n})).

For simplicity we use the notation

u_{n}(x) = x_{n}^{∗}(x) − x_{n}^{∗}(x_{n})
v_{n}(x) = y_{n}^{∗}(x) − y_{n}^{∗}(x_{n})
φ_{n}(x) = (u_{n}(x), v_{n}(x)) ∈ ℝ^{2}
α_{n} = c_{n}(y_{n}^{∗}(y_{n}) − y_{n}^{∗}(x_{n}))
η_{n} = \frac{y_{n}^{∗}(y_{n}) − y_{n}^{∗}(x_{n})}{2}.

Using this notation, we know that 0 < η_{n} and

u_{n}(x_{n}) = v_{n}(x_{n}) = 0, v_{n}(y_{n}) = 2η_{n},

α_{n} = 2c_{n}η_{n},

and for all x ∈ K,

0 ≤ v_{n}(x) ≤ 2η_{n} ⟹ −α_{n} ≤ u_{n}(x) ≤ α_{n}.  (⋆)

If K is bounded, then α_{n} = 2c_{n}η_{n} ≤ c_{n}diam K, and so α_{n} → 0. Since u_{n} and v_{n} are both affine, φ_{n}(K) must be convex, and as φ_{n}(x_{n}) = (0, 0) ∈ φ_{n}(K), it is clear that (⋆) remains true if we replace α_{n}, η_{n} by Nα_{n}, Nη_{n} for any N ≥ 1. If K is unbounded then, as we noticed after the proof that (iii) ⟹ (ii), for every N > 1 there is an x ∈ K for which v_{n}(x) = Nη_{n}. That is, replacing α_{n} and η_{n} by Nα_{n} and Nη_{n} if it is necessary, we obtain the following lemma:

Lemma 10 Suppose that condition (iii) of Theorem 4 fails. Then for every constant c > 0 there is a sequence of positive numbers c_{n} with c_{n} → 0 as n → ∞ and there are sequences of positive numbers α_{n}, η_{n}, such that:

(a) 0 ≤ v_{n}(x) ≤ 2η_{n} ⟹ −α_{n} ≤ u_{n}(x) ≤ α_{n} ∀x ∈ K;

(b) α_{n} = 2c_{n}η_{n};

(c) there is an x ∈ K for which v_{n}(x) = η_{n};

(d) At least one of the following is satisfied:

either (d1 ) α_{n} → 0
or (d2 ) cα_{n} − ∥x_{n}∥ → ∞.

In order to complete the proof that (i) ⟹ (iii), it is suffices to prove the following lemma:

Lemma 11 Assume that for every c > 0 there are sequences of positive numbers α_{n}, η_{n}, c_{n} with c_{n} → 0, for which conditions (a)-(d) of Lemma 10 are satisfied. Then for every 0 < ε < 1 and δ, L > 0 there are continuous functions F, G: K → ℝ such that:

• G + F and G − F are convex;

• Lip(G) ≤ ε;

• 2L ≥ Lip(F) ≥ L; and • either (d1) holds and sup |F | ≤ δ, or (d2) holds and F (x) = 0 for every

∥x∥ ≤ δ.

Indeed, by Lemma 11, for every sequence ε_{n} → 0, L_{n} → ∞ and δ_{n} > 0 we can find F_{n}, G_{n}: K → ℝ and points a_{n}, b_{n} ∈ K such that G_{n} ±F_{n} are convex, Lip(G_{n}) ≤ ε_{n}, Lip(F_{n}) ≤ 2L_{n}, |F_{n}(b_{n}) − F_{n}(a_{n})| > L_{n}∥b_{n} − a_{n}∥, and either (d1) holds and sup |F_{n}| ≤ δ_{n} or (d2) holds and F_{n}(x) = 0 for every ∥x∥ ≤ δ_{n}. We can also assume that for a fixed point x^{0} ∈ K, G_{n}(x^{0}) = 0 for every n. Then, by choosing ∑_{n} ε_{n} < ∞, g = ∑_{n} G_{n} is Lipschitz. By requiring ∑_{n} δ_{n} < ∞ in case (d1) and δ_{n} → ∞ in case (d2), we can also ensure that f = ∑_{n} F_{n} exists and is continuous. Since the sum of convex functions is convex, g + f and g − f are convex. If we also require that ∑_{i<n} 2L_{i} < L_{n}/3 for every n, and ∑_{i>n} 2δ_{i} < (L_{n}/3)∥b_{n} − a_{n}∥ for every n if (d1) holds, or , δ_{i} > max(∥a_{n}∥, ∥b_{n}∥) for every i > n if (d2) holds, then, in the first case,

|f(b_{n})−f(a_{n})| > L_{n}∥b_{n} −a_{n}∥−∑_{i<n} 2L_{i}∥b_{n} −a_{n}∥−2 ∑_{i>n} δ_{i} ≥ (L_{n}/3)∥b_{n} −a_{n}∥,

and, in the second case

|f(b_{n}) − f(a_{n})| = | ∑_{i=1}^{n} F_{i}(b_{n}) − F_{i}(a_{n})| > L_{n}∥b_{n} − a_{n}∥ − ∑_{i<n} 2L_{i}∥b_{n} − a_{n}∥
> (2L_{n}/3)∥b_{n} − a_{n}∥.

In either case, we deduce that f is not Lipschitz and (i) ⟹ (iii), as required.

Before passing to the proof of Lemma 11, we prove the following simple result. This is the only place in our proof at which the assumption 0 ∈ K is crucial, since the statement is clearly false for the set {1} × ℝ ⊂ ℝ^{2}.

Lemma 12 For every η, ϱ > 0, x ∈ K and w ∈ S_{X} there are y, z ∈ B(x, ϱ)∩ K such that

∥\frac{y − z}{∥y − z∥} − w∥ < η.

Proof: Since for every u, v in B(0, ϱ) ∩ K and x ∈ K the vectors

u′ = \frac{ϱ}{∥x∥ + ϱ}u + \frac{∥x∥}{∥x∥ + ϱ}x

and

v′ = \frac{ϱ}{∥x∥ + ϱ}v + \frac{∥x∥}{∥x∥ + ϱ}x

are in B(x, ϱ) ∩ K and (u′ − v′)/∥u′ − v′∥ = (u − v)/∥u − v∥, it is enough to prove the lemma for x = 0.

Since \overline{span}(K) = X, there are positive numbers a_{1}, ..., a_{n}, b_{1}, ..., b_{m} and there are u_{1}, ..., u_{n}, v_{1}, ..., v_{m} ∈ K such that

∥\frac{∑_{i=1}^{n} a_{i}u_{i} − ∑_{j=1}^{m} b_{j}v_{j}}{∥ ∑_{i=1}^{n} a_{i}u_{i} − ∑_{j=1}^{m} b_{j}v_{j}∥} − w∥ < η.

Then for M = max (∑_{i=1}^{n} a_{i}, ∑_{j=1}^{m} b_{j}), the vectors u = ∑_{i=1}^{n} a_{i}u_{i}/M and ∑_{j=1}^{m} b_{j}v_{j}/M are in K, and satisfy the claim of the lemma.

Proof of Lemma 11: Set

c = \frac{1}{(ε/80)^{2}} ,

and choose n to be so large that

c_{n} < \frac{(ε/80)^{2}}{2e^{L+2ε}} .

In addition, in case (d1), for n large enough we know that

α_{n} < \frac{δ}{L + 2ε},

and in case (d2), for n large enough,

\frac{α_{n}}{(ε/80)^{2}} − ∥x_{n}∥ > δ.

Fix an n for which these inequalities are satisfied, and set η = η_{n}, α = α_{n},

β = \frac{α_{n}}{(ε/80)^{2}} ,

γ = e^{L+2ε}β < η, and u(x) = u_{n}(x), v(x) = v_{n}(x), φ(x) = (u(x), v(x)). Let f and g be the functions given by Lemma 9 when the parameters are ε/80, L + 2ε, α, β and γ. Define

F (x) = f(φ(x)), G(x) = g(φ(x)).

Since γ < η, from (a) of Lemma 10 we can see

φ(K) ⊂ ([−α, α] × ℝ) ∪ (ℝ × (ℝ \ [0, 2γ])).

Thus G + F and G − F are continuous convex functions. We also know from (C) of Lemma 9 that if (d1) holds, then sup |F | ≤ α(L + 2ε) < δ, and if (d2) holds, then F (x) = 0 for every x with v(x) ≤ β. Indeed, if v(x) > β then, ∥x∥ ≥ β − ∥x_{n}∥ > δ, that is, F (x) = 0 for every x with ∥x∥ ≤ δ. We also see from (B), (D) of Lemma 9 that Lip(g) < ε and that Lip(F) ≤ Lip(f) ≤ \sqrt{L^{2} + ε^{2}} ≤ 2L.

By continuity and the fact that γ < η, by applying (c) of Lemma 10 we can find an x = x^{0} for which v(x^{0}) = γ. By Lemma 12, for every 0 < ϱ, η < 1 we can find y, z ∈ B(x^{0}, ϱ) ∩ K for which x_{n}^{∗}(y − z) > (1 − η)∥y − z∥. If ϱ is small enough then v(y) and v(z) are close to v(x^{0}) = γ. From (B) of Lemma 9 we get

|f(u(z), v(y)) − f(u(z), v(z))| < 40(ε/80)|v(y) − v(z)| < (ε/2)∥y − z∥,

and if ϱ is small enough then (D) of Lemma 9 gives

|f(u(y), v(y)) − f(u(z), v(y))| > ((L + 2ε) − ε)|u(y) − u(z)|
= (L + ε)|x_{n}^{∗}(y) − x_{n}^{∗}(z)|
≥ (L + ε)(1 − η)∥y − z∥.

Therefore

|F (y) − F (z)| = |f(u(y), v(y)) − f(u(z), v(z))| ≥
≥ |f(u(y), v(y)) − f(u(z), v(y))| − |f(u(z), v(y)) − f(u(z), v(z))| ≥
≥ ((L + ε)(1 − η) − ε/2)∥y − z∥.

For η small enough this gives Lip(F) > L, and the lemma is proved.

Remark 13 Let X = ℓ_{1}, and let e_{1}, e_{2}, … be the standard basis of ℓ_{1}. Let

C_{0} = \{∑_{j} λ_{j}e_{j} ∈ ℓ_{1} : 0 ≤ λ_{j} ≤ \frac{λ_{1}}{j} \}

C_{n} = {∑_{j} λ_{j}e_{j} ∈ ℓ_{1} : 0 ≤ λ_{j}, ∑_{j} λ_{j} ≤ 2λ_{n} ≤ 2n}

D = {∑_{n=1}^{∞} a_{n} : a_{n} ∈ C_{n}, ∑_{n=1}^{∞} a_{n} ∈ ℓ_{1}}

K_{0} = {x ∈ ℓ_{1} : d(x, C_{0}) < 1}
K_{1} = {x ∈ ℓ_{1} : d(x, D) < 1}.

It is easy to see that K_{0} and K_{1} are open, unbounded convex sets, with \overline{span}(C_{0}) = X, C_{0} ⊂ K_{0} and \overline{span}(K_{1}) = X. However, K_{1} does not contain any half-line, and conditions (i)–(iii) of Theorem 4 fail for K_{0}, but hold for K_{1}.

Proof: It is easy to check that for every u = ∑ λ_{j}e_{j} ∈ B(0, n + 1) ∩ C_{0} we have 0 ≤ λ_{n} ≤ 1, and then B(0, n + 1) ∩ K_{0} does not contain any u, v with u − v = 4e_{n}, which contradicts (ii) for K_{0}.

For every x = d + z ∈ K_{1}, d ∈ D, ∥z∥ < 1, r > 0, n > max(∥d∥, r) and y = ∑_{j} y_{j}e_{j} ∈ S_{X} we have

2ne_{4n} + d + n ∑_{{j: y_{j}≥0}} y_{j}e_{j} ∈ C_{4n}

and

2ne_{4n} + d − n ∑_{{j: y_{j}<0}} y_{j}e_{j} ∈ C_{4n},

therefore

u = 2ne_{4n} + x + n ∑_{{j: y_{j}≥0}} y_{j}e_{j} ∈ K_{1} ∩ B(x, 3n)

and

v = 2ne_{4n} + x − n ∑_{{j: y_{j}<0}} y_{j}e_{j} ∈ K_{1} ∩ B(x, 3n),

and u − v = ny. Then u′ = x + (r/n)(u − x) and v′ = x + (r/n)(v − x) are both in K_{1} ∩ B(x, 3r) and u′ − v′ = ry. That is, (ii) is satisfied for K_{1} with c = 1/3.

It remains only to check that K_{1} does not contain any half-line. Assume instead that K_{1} contains a half-line of direction y = ∑ y_{n}e_{n} ∈ S_{X}. It is clear that y_{n} ≥ 0 for every n. Choose an N ≥ 4 for which

∑_{n=1}^{N−1} y_{n} > 2/3 and ∑_{n=N}^{∞} y_{n} < 1/3.

For every x ∈ K_{1} and λ ≥ 0, x + λy ∈ K_{1}. Indeed, fix some v ∈ K_{1} such that v + [0, ∞)y ⊂ K_{1}. Since K_{1} is open, there is an ε > 0 such that x + ε(x − v) ∈ K_{1}. By convexity:

x + λy = \frac{1}{1 + ε}(x + ε(x − v)) + \frac{ε}{1 + ε} (v + \frac{1 + ε}{ε} · λy) ∈ K_{1}.

In particular, 4N^{2}y ∈ K_{1}, and so 4N^{2} ∑_{n} y_{n}e_{n} = ∑_{n} a_{n} + b, where a_{n} = ∑_{j} λ_{jn}e_{j} ∈ C_{n} for every n and b = ∑_{j} b_{j}e_{j} ∈ B(0, 1). We also have 0 ≤ λ_{jn} for every j and n, and ∑_{j,n} λ_{jn} < ∞. Since C_{n} ⊂ B(0, 2n),

∑_{n=1}^{N−1} ∑_{j=1}^{N−1} λ_{jn} ≤ ∑_{n=1}^{N−1} 2n < N^{2}

follows. Hence

4N^{2} (∑_{j=1}^{N−1} y_{j} − ∑_{j=N}^{∞} y_{j}) = ∑_{n=1}^{∞} (∑_{j=1}^{N−1} λ_{jn} − ∑_{j=N}^{∞} λ_{jn}) + ∑_{j=1}^{N−1} b_{j} − ∑_{j=N}^{∞} b_{j}

≤ N^{2} − ∑_{n=N}^{∞} (∑_{j=N}^{∞} λ_{jn} − ∑_{j=1}^{N−1} λ_{jn}) + ∑_{j=1}^{N−1} b_{j} − ∑_{j=N}^{∞} b_{j},

where for every n ≥ N we have

∑_{j=N}^{∞} λ_{jn} − ∑_{j=1}^{N−1} λ_{jn} = 2 ∑_{j=N}^{∞} λ_{jn} − ∑_{j=1}^{∞} λ_{jn} ≥ 2 ∑_{j=N}^{∞} λ_{jn} − 2λ_{nn} ≥ 0

and

∑_{j=1}^{N−1} b_{j} − ∑_{j=N}^{∞} b_{j} ≤ 1.

Therefore

4N^{2} (∑_{j=1}^{N−1} y_{j} − ∑_{j=N}^{∞} y_{j}) ≤ N^{2} + 1.

This is a contradiction, since ∑_{j=1}^{N−1} y_{j} − ∑_{j=N}^{∞} y_{j} > 1/3.

Remark 14 Note that it does not make any difference if, instead of (i), we assume that for some (or every) k ≥ 2, and for every convex functions h_{1}, h_{2}, …, h_{k} the functions must be Lipschitz whenever their sum is Lipschitz.

Acknowledgments This work has been carried out while the second named author was visiting the mathematics department of the University College London. Both authors would like to express their sincere gratitude to Toby C. O’Neil for carefully reading the manuscript and making many invaluable corrections and suggestions.

References

[BL] Y. Benyamini and J. Lindenstrauss, Geometric Nonlinear Functional Analysis, vol. 1, Colloquium Publications (American Mathematical Society), Providence Rhode Island, 2000.

[C1] M. Cepedello Boiso, Approximation of Lipschitz functions by Δ-convex functions in Banach spaces, Israel J. Math. 106 (1998), 265-284.

[C2] M. Cepedello Boiso, On regularization in superreflexive Banach spaces by infimal convolution formulas, Studia Math. 129 (1998), 265-284.

[DVZ] J. Duda, L. Veselý and L. Zajíček, On D.C. Functions and Mappings, preprint 2001.

[VZ] L. Veselý and L. Zajíček, Delta-convex mappings between Banach spaces and applications, Dissertationes Math. (Rozprawy Mat.) 289 (1989).

Mathematics Subject Classification: 52A41, 26B35.

Keywords: convex functions, Lipschitz functions, Δ-convex mappings.

M. Csörnyei, Department of Mathematics, University College London, Gower Street, London, WC1E 6BT , United Kingdom. mari@math.ucl.ac.uk A. Naor, Department of Mathematics, Hebrew University, Givaat-Ram, Jerusalem, Israel. naor@math.huji.ac.il