Lipschitz differences and Lipschitz functions

by

M. Balcerzak, Z. Buczolich, and M. Laczkovich

Abstract. Let f be continuous, and assume that Δ_{h}f(x) = f(x + h) − f(x) is a Lipschitz function for every element h of a given set A. We investigate those sets A for which this condition implies that f is Lipschitz. We study this question for continuous functions with bounded differences defined either on the reals or on the circle group.

Introduction. Let f : R → R be continuous and such that the difference function Δ_{h}f(x) = f(x + h) − f(x) is bounded for every h ∈ R. In a recent paper [T] S. I. Trofimchuk proved that if Δ_{h}f is uniformly continuous for every h ∈ R then f is also uniformly continuous. In this note we prove that in this theorem uniform continuity can be replaced by the Lipschitz property. More exactly, we investigate the following question. Suppose that f is continuous and Δ_{h}f is Lipschitz for every h belonging to a given subset, A, of R. We show that this condition implies that f is Lipschitz if and only if A cannot be covered by a proper F_{σ} group of R. We also discuss the analogous problem for uniform Lipschitz functions and for functions defined on the circle group T = R/Z.

We shall use the following notation. We denote N = {1, 2, ...}. Let G be any of the groups R or T. If A, B ⊂ G then we denote A + B = {a + b : a ∈ A, b ∈ B}. The sets A − B and −A are defined similarly. If k ∈ N, the k-fold sum A + ... + A is denoted by kA. By closed (open) intervals in T we mean closed (open) connected sets. For every L > 0 we denote by Lip_{L} the set of functions f : G → R satisfying

|f(x) − f(y)| ≤ L|x − y|

for every x, y ∈ G. In case of G = T, by |x| we mean min{|x|, 1 − |x|}, when we identify T with [0, 1). We put Lip = ⋃_{L>0} Lip_{L}. For H ⊂ G, the closure and the Lebesgue outer measure of H are denoted by cl H and |H|.

The identity Δ_{h_{1}+h_{2}}f(x) = Δ_{h_{1}}f(x + h_{2}) − Δ_{h_{2}}f(x) gives

Lemma 0.1. Assume that L_{1}, L_{2} > 0, f : G → R, B_{1}, B_{2} ⊂ G. If Δ_{h}f ∈ Lip_{L_{i}} for every h ∈ B_{i} (i = 1, 2) then Δ_{h}f ∈ Lip_{L_{1}+L_{2}} for h ∈ B_{1} + B_{2}.

It is well-known that if F_{1}, F_{2} ⊂ G are closed sets of positive measure then the interior of F_{1} + F_{2} is non-empty. This easily implies

Lemma 0.2. For every A ⊂ G the following statements are equivalent:

(i) kA is nowhere dense for every k ∈ N; (ii) |cl(kA)| = 0 for every k ∈ N.

1. Functions defined on the circle group T

Theorem 1.1. Let L > 0 and let A be a subset of T such that A = −A. Then the following statements are equivalent.

(i) If f : T → R is continuous and Δ_{h}f ∈ Lip_{L} for each h ∈ A, then f is Lipschitz. (ii) There is an n ∈ N such that nA is dense in T.

Proof. Suppose (ii), and let f : T → R be a continuous function such that Δ_{h}f ∈ Lip_{L} for every h ∈ A. By Lemma 0.1, this implies that Δ_{h}f ∈ Lip_{nL} for a set of h’s everywhere dense in T. Since f is continuous, we have Δ_{h}f ∈ Lip_{nL} for every h ∈ T, that is

|f(x + h) − f(x) − f(y + h) + f(y)| ≤ nL|x − y|  (1)

for every x, y and h. Using ∫_{T}(f(x + h) − f(y + h))dh = 0 and (1) we obtain

|f(y) − f(x)| = |∫_{T}[f(x + h) − f(x) − f(y + h) + f(y)] dh| ≤ nL|x − y|,

and this proves the implication (ii)⟹(i). To prove the converse we need the following lemma.

Lemma 1.2. Assume that A ⊂ T and |cl(kA)| = 0 for any k ∈ N. Then there is a closed set H ⊂ T such that |H| > 0 and H + cl(kA) is nowhere dense for any k ∈ N.

Proof. Denote the rationals in T by Q. Clearly, |Q − cl(kA)| = 0 for any k ∈ N. Let B = ⋃_{k∈N}(Q − cl(kA)), then |B| = 0. Choose a closed set H ⊂ T \ B such that |H| > 0. Then H + cl(kA) is closed. Suppose that x ∈ (H + cl(kA)) ∩ Q. Then there exist h ∈ H and y ∈ cl(kA) with h + y = x ∈ Q, that is, h = x − y ∈ Q − cl(kA) ⊂ B, contradicting h ∈ H ⊂ T \ B. This implies that Q ∩ (H + cl(kA)) = ∅, and hence the closed set H + cl(kA) is nowhere dense.

Now we turn to the proof of the implication (i)⟹(ii). We may assume that L = 1, and 0 ∈ A. Suppose that (ii) is not true; this easily implies that kA is nowhere dense for every k ∈ N. We shall construct a continuous non-Lipschitz function f : T → R such that Δ_{h}f ∈ Lip_{1} for each h ∈ A. We shall define f as ∫_{0}^{x} g(t) dt, where g : T → R is summable and ∫_{0}^{1} g(t) dt = 0. (In this proof we identify T with [0, 1).) Then f will be continuous on T and will satisfy f(0) = lim_{x→1−} f(x) = f(1).

By Lemma 1.2 we can choose a closed set H ⊂ T such that |H| > 0 and H + cl(kA) is nowhere dense for any k ∈ N. Put H_{−1} = ∅, H_{0} = H and H_{j} = H +cl(jA) = H_{j−1} +cl(A) for j = 1, 2, …. From 0 ∈ A it follows that H_{j−1} ⊂ H_{j}. Put H_{∞} = ⋃_{j∈N} H_{j}, then lim_{j→∞} |H_{∞} \ H_{j−1}| = 0 by |T| = 1. Let j_{0} = 1. If j_{k−1} is defined for a k ∈ N, choose j_{k} such that j_{k} > j_{k−1} and

|H_{∞} \ H_{j_{k}−1}| < 1/(k2^{k}).  (2)

For j_{k−1} < j ≤ j_{k} we put c_{j} = k. Thus by induction we have defined j_{k} for all k, and c_{j} for all j. We put g_{1}(x) = c_{j} if x ∈ H_{j} \ H_{j−1} (j ∈ N), and g_{1}(x) = 0 for x ∈ T \ H_{∞}. From (2) it follows that c = ∫_{T} g_{1} < ∞. Let g(x) = g_{1}(x) − c for x ∈ T, then ∫_{T} g = 0.

Let x ∈ H_{∞} and h ∈ A. Then y = x + h ∈ H_{∞}, and thus x ∈ H_{j_{x}} \ H_{j_{x}−1} and y ∈ H_{j_{y}} \ H_{j_{y}−1} with suitable j_{x} and j_{y}. If j_{x} ≤ j_{y}, then y = x + h ∈ H_{j_{x}} + cl(A) = H_{j_{x}+1}, and hence j_{y} = j_{x} or j_{y} = j_{x} + 1. Thus, in this case, |g(y) − g(x)| = 0 or |g(y) − g(x)| = |(c_{j_{x}+1} − c) − (c_{j_{x}} − c)| ≤ 1. If, on the other hand, j_{x} > j_{y} then, using A = −A, x = y − h, and interchanging the roles of x and y, we reach the same conclusion. Therefore, |g(x + h) − g(x)| ≤ 1 holds for any x ∈ H_{∞} and h ∈ A. If x ∈ T \ H_{∞} and h ∈ A then x + h ∈ T \ H_{∞}. Indeed, from x + h ∈ H_{∞} it follows that x + h ∈ H_{j} for some j ≥ 0, and then A = −A implies x = (x + h) − h ∈ H_{j} + A ⊂ H_{∞}, contradicting x ∈ T \ H_{∞}. Therefore |g(x + h) − g(x)| = c − c = 0 holds for any x ∈ T \ H_{∞} and h ∈ A. Thus |g(x + h) − g(x)| ≤ 1 for x ∈ T and h ∈ A. Let f(x) = ∫_{0}^{x} g(t) dt for x ∈ T. To show that Δ_{h}f ∈ Lip_{1} for h ∈ A, let x, d ∈ T be given. We have

|Δ_{h}f(x + d) − Δ_{h}f(x)| = |Δ_{d}f(x + h) − Δ_{d}f(x)| = |∫_{x}^{x+d}(g(t + h) − g(t)) dt| ≤ |d|.

That is, Δ_{h}f ∈ Lip_{1}. Observe that we may replace A by A ∪ {1/n : n ∈ N} ∪ {−(1/n) : n ∈ N}. Then A_{∞} = ⋃_{k∈N} cl(kA) is dense in T. Thus, for any subinterval J of T, we have 0 < |H_{∞} ∩ J| = |(H + A_{∞}) ∩ J|. Since the H_{j}’s are nowhere dense, there are infinitely many j’s for which |H_{j} \ H_{j−1}| > 0. Hence, putting S_{K} = {x ∈ T : |g(x)| > K} (K > 0), we have |S_{K}| > 0 for all K > 0. Since f′ = (∫_{0}^{x} g(t) dt)′ = g(x) almost everywhere on T, it follows that, for any K, the inequality |f′(x)| > K holds for almost every x ∈ S_{K}. Thus f cannot be Lipschitz and hence (i) does not hold. This completes the proof of Theorem 1.1.

Remark 1.3. Since Δ_{−h}f ∈ Lip_{L} follows from Δ_{h}f ∈ Lip_{L}, the assumption A = −A is natural. We show that this assumption cannot be deleted from the implication (i)⟹(ii) of Theorem 1.1.

Indeed, by a result of Haight [H], there exists an F_{σ} subset B of the positive real line such that B −B = R but kB has zero Lebesgue measure for any positive integer k. Choose closed compact sets F_{n} of measure zero such that B = ⋃_{n=1}^{∞} F_{n} and F_{1} ⊂ F_{2} ⊂ …. Since B − B = R, it follows that F_{n} − F_{n} contains an interval for a suitable n ∈ N. Taking this F_{n} “mod 1” we obtain a nowhere dense compact set A such that kA is nowhere dense for every k ∈ N and A − A contains an interval. It is easy to see, following the proof of Theorem 1.1, that if Δ_{h}f ∈ Lip_{L} for every h ∈ A then f is Lipschitz.

Next we turn to the non-uniform case, i.e. to the case when the difference functions are Lipschitz but not necessarily with the same constant.

Theorem 1.4. For every A ⊂ T the following statements are equivalent.

(i) If f : T → R is continuous and Δ_{h}f ∈ Lip for every h ∈ A then f is Lipschitz. (ii) There is no proper F_{σ} subgroup of T containing A.

Proof. (ii)⟹(i): Suppose (ii), and let f : T → R be a continuous function such that Δ_{h}f ∈ Lip for every h ∈ A. Put G = {h ∈ T : Δ_{h}f ∈ Lip} and G_{n} = {h ∈ T : Δ_{h}f ∈ Lip_{n}} for n ∈ N. Then G = ⋃_{n∈N} G_{n}. Since f is continuous, it is easy to verify that the sets G_{n} are closed and G is an F_{σ} set. The identities Δ_{−h_{1}}f(x) = f(x − h_{1}) − f(x) = −Δ_{h_{1}}f(x − h_{1}) and Δ_{h_{1}+h_{2}}f(x) = Δ_{h_{2}}f(x + h_{1}) − Δ_{h_{1}}f(x) show that G is a group. Since A ⊂ G, (ii) implies that G = T. Therefore, by the Baire category theorem, there exists n ∈ N such that G_{n} contains a subinterval of T. Then kG_{n} = T for some k ∈ N. By Theorem 1.1, this implies that f is Lipschitz.

(i)⟹(ii): Suppose that there exists an F_{σ} group C such that A ⊂ C ⊂ T and C ≠ T. Then we can choose nowhere dense closed sets C_{n} such that C = ⋃_{n∈N} C_{n}. Since C = −C, we may assume C_{n} = −C_{n}. Denoting D/n = {x ∈ T : |x| < 1/n, n · x ∈ D} for every D ⊂ T, we define B = {0} ∪ ⋃_{n∈N} C_{n}/n. Then B = −B and B is a nowhere dense closed set. Thus, for each k ∈ N, the set kB is a closed subset of ⋃_{n∈N} C/n, as kC = C. Since C is of first category, so is ⋃_{n∈N} C/n. Therefore kB is a closed set of first category and thus it is nowhere dense. By Theorem 1.1, there exists a non-Lipschitz and continuous function f for which Δ_{h}f ∈ Lip_{1} if h ∈ B. It is clear that the group generated by B contains all C_{n}’s and hence all of C. Thus Δ_{h}f ∈ Lip for h ∈ C; that is, (i) does not hold.

2. Functions defined on the real line

Theorem 2.1. Let L > 0 and let A be a bounded subset of R. Then the following statements are equivalent.

(i) If f : R → R is continuous, Δ_{h}f is bounded for every h ∈ R and Δ_{h}f ∈ Lip_{L} for each h ∈ A then f is Lipschitz.

(ii) There is an n ∈ N such that nA is dense in a nondegenerate interval.

Proof. (ii)⟹(i): Suppose (ii) and let f : R → R be continuous such that Δ_{h}f is bounded for every h ∈ R and Δ_{h}f ∈ Lip_{L} for each h ∈ A. Then there exist M > 0 and δ_{0} > 0 such that |Δ_{h}f(x)| ≤ M holds for every |h| ≤ δ_{0} and x ∈ R. Indeed, since f is continuous, the sets H_{K} = {h : |Δ_{h}f| ≤ K} are closed. By the Baire category theorem one of these sets contains an interval. Then, using H_{K} −H_{K} ⊂ H_{2K}, one can easily deduce the existence of a δ_{0} > 0 such that (−δ_{0}, δ_{0}) ⊂ H_{2K}. Thus we may put M = 2K.

Since Δ_{0}f(x) = 0 and Δ_{−h}f(x + h) = −Δ_{h}f(x), we may assume that 0 ∈ A and −A = A. For ℓ ∈ N let S_{ℓ} = {h : Δ_{h}f ∈ Lip_{ℓ}}. Since f is continuous, the sets S_{ℓ} are closed. By Lemma 0.1 we have kS_{ℓ} ⊂ S_{kℓ}. From A ⊂ S_{L} it follows that k_{0}A ⊂ k_{0}S_{L} ⊂ S_{k_{0}L}. Since S_{k_{0}L} is closed, cl(k_{0}A) ⊂ S_{k_{0}L} and hence S_{k_{0}L} contains a non-degenerate interval.

Put L_{1} = 2k_{0}L. We can choose a 0 < δ < min{M/L_{1}, δ_{0}} such that (−δ, δ) ⊂ S_{k_{0}L} − S_{k_{0}L} ⊂ S_{2k_{0}L}. Our aim is to show that f is Lipschitz. If this is not true then there are x, y ∈ R, x < y, such that

|f(y) − f(x)| > (3M/δ)|y − x|.

For a fixed n ∈ N consider the points x_{k} = x + k(y − x)/n (k = 1, …, n). Plainly, we can find k ∈ {1, ..., n} such that

|f(x_{k}) − f(x_{k−1})| > (3M/δ)|x_{k} − x_{k−1}|.

Thus there exist α ∈ (0, δ/4) and t ∈ R such that |Δ_{α}f(t)| > (3Mα)/δ. Since Δ_{h}f ∈ Lip_{L_{1}} for h ∈ (−δ, δ), we have

|Δ_{α}f(t + h) − Δ_{α}f(t)| = |Δ_{h}f(t + α) − Δ_{h}f(t)| ≤ L_{1}α < Mα/δ.

Therefore |Δ_{α}f(t + h)| > 2Mα/δ holds for each h ∈ (−δ, δ), and thus the sign of Δ_{α}f(t) and Δ_{α}f(t + h) is the same for any h ∈ (−δ, δ). Choose an integer m such that δ/(2α) < m < δ/α (it exists since α < δ/4). Putting h_{k} = kα for k = 0, 1, ..., m − 1, we get |Δ_{α}f(t + h_{k})| > 2Mα/δ. Since all differences are of the same sign, we have

2mMα/δ < ∑_{i=0}^{m−1} |Δ_{α}f(t + h_{i})| = |Δ_{mα}f(t)|.

Thus, for m > δ/(2α) we obtain M < 2mMα/δ < |Δ_{mα}f(t)|; a contradiction, since, by the choice of δ and by the assumption of the theorem, 0 < mα < δ implies |Δ_{mα}f(t)| ≤ M.

(i)⟹(ii): Suppose that (ii) is not true. Let ν denote the canonical homomorphism which maps R onto R/Z = T. Since A is bounded and kA is nowhere dense in R, it is easy to see that kB is nowhere dense in T, where B = ν(A) ⊂ T. Applying Theorem 1.1, we find a continuous non-Lipschitz function g : T → R such that Δ_{h}g ∈ Lip_{L} for each h ∈ B. Extending this function g from T onto R periodically, that is, taking f = g ∘ ν, we obtain a periodic continuous non-Lipschitz function f : R → R such that Δ_{h}f ∈ Lip_{L} for each h ∈ A ⊂ ν^{−1}(B). Since f is obviously bounded, Δ_{h}f is also bounded for each h ∈ R.

Remarks 2.2. 1. The condition on the boundedness of the differences Δ_{h}f cannot be deleted. Indeed, for f(x) = x^{2}, Δ_{h}f ∈ Lip_{2} for every h ∈ [0, 1], but f is not Lipschitz.

2. The boundedness of A was not used in (i)⟹(ii). On the other hand, we do not know whether or not (ii)⟹(i) is true for every A ⊂ R.

In the non-uniform case we obtain

Theorem 2.3. For every A ⊂ R, the following two statements are equivalent.

(i) If f : R → R is continuous, Δ_{h}f is bounded for every h ∈ R and Δ_{h}f ∈ Lip for each h ∈ A then f is Lipschitz.

(ii) There is no proper F_{σ} subgroup of R containing A.

Proof. (ii)⟹(i): Suppose (ii) and let f : R → R be a continuous function such that Δ_{h}f is bounded for every h ∈ R and Δ_{h}f ∈ Lip for each h ∈ A. Let G = {h ∈ R : Δ_{h}f ∈ Lip} and G_{n} = {h ∈ R : Δ_{h}f ∈ Lip_{n}} for n ∈ N. Then G = ⋃_{n∈N} G_{n} and G is an F_{σ} group containing A. Hence G = R. By the Baire category theorem, G_{n} is dense in an interval for some n ∈ N. Next it suffices to apply Theorem 2.1.

(i)⟹(ii): Assume that there exists an F_{σ} group C ≠ R containing A. Since C must be of the first category, we can choose closed nowhere dense sets C_{n} ⊂ [−n, n] such that C_{n} = −C_{n}, and C = ⋃_{n∈N} C_{n}. Denoting D/n = {x/n : x ∈ D} for every D ⊂ R, we define B = {0} ∪ ⋃_{n∈N}(C_{n}/n^{2}). Then B = −B is bounded, closed and nowhere dense. The rest of the proof is similar to the (i)⟹(ii) part of the proof of Theorem 1.3.

Acknowledgement. The first author thanks Z. Buczolich and M. Laczkovich for inviting him to the Eötvös University in December ’93 when the work on this paper started.

References

[H] J. A. Haight, Difference covers which have small k-sums for any k, Mathematika 20 (1973), 109-118.

[T] S.I. Trofimchuk, Unbounded functions with bounded and uniformly continuous differences on R (in Russian), Dopov./Dokl. Akad. Nauk. Ukraini 5(1993), 24-25.

babula.tex 1/26/1995 revised 21:13:23 2/13/1995