Sets where Lip f is infinite and lip f is finite
aa r X i v : . [ m a t h . C A ] J u l Sets where Lip f is infinite and lip f is finite. Bruce Hanson, Department of Mathematics,Statistics and Computer Science,St. Olaf College, Northfield, Minnesota 55057, USAemail: [email protected]
Abstract.
We characterize the sets E ⊂ R such that there exists a continuousfunction f : R → R with lip f finite everywhere and Lip f infinite precisely on E .
1. Introduction
Throughout this note we assume that f : R → R is continuous.Then the so-called “big Lip” and “little lip” functions are defined asfollows:Lip f ( x ) = lim sup r → + M f ( x, r ) and lip f ( x ) = lim inf r → + M f ( x, r ) , where M f ( x, r ) = sup { | f ( x ) − f ( y ) | r : | x − y | ≤ r } . We also define L ∞ f = { x ∈ R : Lip f ( x ) = ∞} and l ∞ f = { x ∈ R :lip f ( x ) = ∞} .Results about the big Lip function date back to the early days of the20th century. For example, the proof of the well-known Rademacher-Stepanov Theorem (see [ ? ]), which states that f is differentiable almosteverywhere on the complement of L ∞ f , is almost 100 years old. (See[ M ] for an elegant proof of this result.) On the other hand, the littlelip function is a more recent phenomenon. As far as I know, the firstreference to the little lip function occurs in a 1999 paper by Cheeger,[ CH ]. Another early reference occurs in ([ BC ]), where the authorsshow that the sets L ∞ f and l ∞ f can differ greatly. There they constructa continuous function f : R → R such that L ∞ f = R , but | l ∞ f | = 0 Mathematics Subject Classification.
Key words and phrases. lip, Lip.
[email protected] (here | S | denotes the Lebesgue measure of S ). In fact, in their examplelip f ( x ) = 0 on R \ E , where | E | = 0. In ([ H1 ]) the author of this noteshows that it is possible to make the exceptional set E have Hausdorffdimension 0.On the other hand, it is impossible to construct a function f : R → R such that L ∞ f = R and l ∞ f = ∅ . Balogh and Cs¨ornyei showed ([ BC ])that if l ∞ f = ∅ , then every non-degenerate interval I ⊂ R contains a setof positive measure on which f is differentiable. It follows that in thiscase | L ∞ f ∩ I | < | I | for each non-degenerate interval I . This motivatesthe following definition: Definition . A subset E of R is trim if | E ∩ ( a, b ) | < b − a forall open intervals ( a, b ) . It is straightforward to show that L ∞ f is a G δ set for any continuousfunction f . Thus, if l ∞ f = ∅ , then L ∞ f is a trim G δ set. In this note weshow that this gives a characterization of trim G δ sets: Theorem . Suppose that f : R → R is continuous and l ∞ f = ∅ .Then L ∞ f is a trim G δ set. Conversely, if E ⊂ R is a trim G δ set, thenthere exists a continuous function f : R → R such that L ∞ f = E and l ∞ f = ∅ . Moreover, f may be constructed so that lip f ( x ) = 0 for all x ∈ E , as well. Now suppose that f is a monotone function. Then f is differentiablealmost everywhere so | L ∞ f | = 0. In this case, we have the followingresult: Theorem . Suppose that f : R → R is continuous and mono-tonic. Then L ∞ f is a G δ set of measure zero. Conversely, if E ⊂ R isa G δ set with measure zero, then there exists a continuous, monotonicfunction f : R → R such that L ∞ f = E and l ∞ f = ∅ . Moreover, f maybe constructed so that lip f ( x ) = 0 for all x ∈ E . The layout of this paper is the following: In section 2 we presentour notation and state a few useful lemmas. Sections 3 and 4 containthe proofs of Theorems 1.2 and 1.3, respectively. In the final sectionwe offer a few open problems that are connected to Theorems 1.2 and1.3.Other interesting recent results concerning the big and little lipfunctions can be found in [
BHMV1, BHMV2, BHMV3, H2 ] ETS WHERE Lip f IS INFINITE AND lip f IS FINITE. 3
2. Definitions and Lemmas
Definition . Given a continuous function f : R → R , we defineLip + f ( x ) = lim sup r → + sup ≤ y − x ≤ r | f ( x ) − f ( y ) | r , and Lip − f ( x ) = lim sup r → + sup ≤ x − y ≤ r | f ( x ) − f ( y ) | r . Note that Lip f ( x ) = max { Lip + f ( x ) , Lip − f ( x ) } . We leave theproof of the following simple lemma up to the reader. Lemma . Suppose that g and h are Lipschitz on [ a, b ] and that g ( x ) ≤ f ( x ) ≤ h ( x ) for all x ∈ [ a, b ] and A = { x ∈ [ a, b ] : g ( x ) = h ( x ) } . Then (1)
Lip f ( x ) < ∞ for all x ∈ A ∩ ( a, b ) and Lip + f ( a ) < ∞ if a ∈ A and Lip − f ( b ) < ∞ if b ∈ A . Definition . Given a closed interval I = [ a, b ] and n ∈ N , wedefine (2) Φ n,I ( x )) = (cid:26) n min { x − a, b − x } if x ∈ [ a, b ]0 if x / ∈ [ a, b ] Definition . Let Z = { z j : j ∈ Z } ⊂ ( a, b ) . We say that Z is n -close on ( a, b ) if (3) a < z j < z j +1 < b for all j ∈ Z , (4) lim j →∞ z j = b and lim j →−∞ z j = a, and (5) z j +1 − z j < n min { z j − a, b − z j +1 } for all j ∈ Z . Definition . Given an open set U and a countable set Z ⊂ U ,we say that Z is n -close on U if Z ∩ ( a, b ) is n -close on ( a, b ) for eachcomponent ( a, b ) of U . Definition . Suppose that Z = { z j : j ∈ Z } is n -close on [ a, b ] .Then we say that f : [ a, b ] → R is zig-zag of order n with respect to Z on [ a, b ] if f ( a ) = f ( b ) and (6) f ( z j ) = (cid:26) f ( a ) if j is even f ( a ) + Φ n, [ a,b ] ( z j ) if j is odd BRUCE HANSON, DEPARTMENT OF MATHEMATICS, STATISTICS AND COMPUTER SCIENCE, ST. OLAF COLLEGE, NORTHFIELD, MINNESOTA 55057, USA EMAIL:
[email protected] and (7) f is linear on [ z j , z j +1 ] for all j ∈ Z . Definition . Suppose that f is linear on [ a, b ] and U is an openset such that | U ∩ [ a, b ] | < b − a and { a, b } ∩ U = ∅ . Then we define g = g ( f, [ a, b ] , U ) on [ a, b ] as follows: g ( x ) = f ( a ) + | [ a, b ] \ U ) ∩ [ a, x ] || [ a, b ] \ U | ( f ( b ) − f ( a )) . The proof of the following lemma is straightforward and left to thereader.
Lemma . Suppose that f is linear on [ a, b ] and U is an open setsuch that | U ∩ [ a, b ] | < b − a and { a, b } ∩ U = ∅ . Let g = g ( f, [ a, b ] , U ) .Then (8) g is Lipschitz on [ a, b ] , (9) g is constant on each interval ( r, s ) contained in [ a, b ] ∩ U, (10) g ( a ) = f ( a ) and g ( b ) = f ( b ) , and (11) || g − f || ∞ ≤ | f ( b ) − f ( a ) | . Definition . Given an open set U and a closed interval [ a, b ] ,we say that U is trim in [ a, b ] if | U ∩ [ a, b ] | < b − a and { a, b } ∩ U = ∅ .Furthermore, if F = { [ a n , b n ] : n ∈ N } is a collection of pairwise non-overlapping closed interevals, we say that U is trim on F if U is trimin [ a n , b n ] for all n ∈ N . Lemma . Let n ∈ N . Suppose that Z = { z i : i ∈ Z } is n -closeon [ a, b ] and F is zig-zag of order n with respect to Z on [ a, b ] . Then (12) | F ( z i ) − F ( z i +1 ) | ≥ n | z i − z i +1 | for all i ∈ N Furthermore, let F = { [ z i , z i +1 ] : i ∈ Z } and assume U = ⊔ i ∈ J ( a i , b i ) = ⊔ i ∈ J I i ⊂ ( a, b ) is trim on F and G : [ a, b ] → R satisfies (13) G ( a ) = G ( b ) = F ( a ) and (14) G | [ z i ,z i +1 ] = g ( F, [ z i , z i +1 ] , U ) for all i ∈ Z . ETS WHERE Lip f IS INFINITE AND lip f IS FINITE. 5
Define H ( x ) = G ( x ) + X i ∈ J Φ n, [ a i ,b i ] ( x ) for all x ∈ [ a, b ] . Then (15) H ( x ) ≤ G ( a ) + 2Φ n, [ a,b ] ( x ) for all x ∈ [ a, b ] . Proof.
Note that (12) follows directly from the definition of F and(5). Let x ∈ [ a, b ]. Since equality holds in (15) when x = a or x = b ,we may assume that x ∈ ( a, b ). Choose j ∈ Z such that x ∈ [ z j , z j +1 ].We only consider the case where j is even and z j − a ≤ b − z j +1 . (Theother cases are handled by a similar argument.) Note that the lastinequality implies that(16) Φ n, [ a,b ] ( z j ) ≤ Φ n, [ a,b ] ( y ) for all y ∈ [ z j , z j +1 ] . Since j is even, it follows from the definition of G that(17) G ( x ) ≤ F ( z j +1 ) = G ( a ) + Φ n, [ a,b ] ( z j +1 ) . Note also that X i ∈ J Φ n, [ a i ,b i ] ( x ) ≤ Φ n, [ z j ,z j +1 ] ( x ) , and so we get(18) H ( x ) ≤ G ( a ) + Φ n, [ a,b ] ( z j +1 ) + Φ n, [ z j ,z j +1 ] ( x ) . Now, using the fact that Z is n -close on [ a, b ], the definition ofΦ n, [ a,b ] and the assumption that z j − a ≤ b − z j +1 , we get(19) x − z j ≤ z j +1 − z j ≤ n ( z j − a ) = 12 n Φ n, [ a,b ] ( z j ) . From (19) it follows that(20) Φ n, [ a,b ] ( z j +1 ) ≤ Φ n, [ a,b ] ( z j ) + 12 n ( z j +1 − z j ) ≤ (1 + 14 n )Φ n, [ a,b ] ( z j )and then using (19) and (16), we have(21) Φ n, [ z j ,z j +1 ] ( x ) ≤ n ( x − z j ) ≤ n Φ n, [ a,b ] ( z j ) ≤ n Φ n, [ a,b ] ( x ) . Now (15) follows from (18), (20), and (21) and the proof of the lemmais complete. (cid:3)
BRUCE HANSON, DEPARTMENT OF MATHEMATICS, STATISTICS AND COMPUTER SCIENCE, ST. OLAF COLLEGE, NORTHFIELD, MINNESOTA 55057, USA EMAIL:
3. Proof of Theorem 1.2
Proof.
Let E be a trim G δ set. It suffices to show that for every a < b with a, b ∈ R \ E we can construct f : [ a, b ] → R with lip f = 0on E ∩ ( a, b ) and such that L ∞ f ∩ ( a, b ) = E ∩ ( a, b ), Lip + f ( a ) < ∞ and Lip − f ( b ) < ∞ . We assume without loss of generality that( a, b ) = (0 ,
1) and E ⊂ (0 , f : [0 , → R such that(22) L ∞ f = E (23) Lip + f (0) < ∞ and Lip − f (1) < ∞ and(24) lip f = 0 on E. Let F = (0 , \ E . Because E is trim, it follows that F is densein (0 ,
1) and that given any collection F of pairwise non-overlappingclosed intervals whose endpoints are in F , we can choose an open set U such that U is trim on F and E ⊂ U .We first make use of the trimness of E and the denseness of F to construct a sequence of open sets { U n } , countable sets { Z n } , andcollections of closed intervals {F n } which will be used to construct f . Let { V n } ∞ n =1 be a collection of open sets such that V n +1 ⊂ V n forall n ∈ N and E = ∩ ∞ n =1 V n and define U = (0 , Z = { z j } j ∈ Z such that Z ⊂ F and Z is 1-close on U and let F = { [ z j , z j +1 ] } j ∈ Z . Now choose U = ⊔ j ∈ J ( a ,j , b ,j ) so that U is trim on F and E ⊂ U ⊂ V . We proceed by choosing Z = ∪ j ∈ J Z ,j where Z ,j = { z ,j,i : i ∈ Z } is 2-close on ( a ,j , b ,j ) for each j ∈ J . We alsorequire that Z ⊂ F . Then we define F = { [ z ,j,i , z ,j,i +1 ] : j ∈ J , i ∈ Z } .Making use of the trimness of E and the denseness of F we proceedwith an inductive argument and choose a sequence of open sets { U n } ,countable sets { Z n } and collections of closed intervals {F n } satisfyingthe following:(25) E ⊂ U n ⊂ V n for n ∈ N . (26) U n = ⊔ j ∈ J n ( a n,j , b n,j ) for n ≥ n ≥ Z n = ∪ j ∈ J n Z n,j , where(27) Z n,j = { z n,j,i : i ∈ Z } is n -close on ( a n,j , b n,j ) for j ∈ J n ETS WHERE Lip f IS INFINITE AND lip f IS FINITE. 7 (28) ∪ ∞ n =1 Z n ⊂ F (29) F n = { [ z n,j,i , z n,j, ( i +1) ] : j ∈ J n , i ∈ Z } for n ≥ U n +1 is trim on F n for each n ∈ N . Note that for each n ∈ N we have(31) ∪ [ a,b ] ∈F n [ a, b ] = U n and(32) Z n ⊂ U n \ U n +1 . We now begin the construction of f . Define g = 0 on [0 ,
1] and f to be zig-zag of order 1 with respect to Z on U = (0 , n ≥ g n so that(33) for all [ a, b ] ∈ F n − , g n | [ a,b ] = g ( f n − , [ a, b ] , U n )and(34) g n ( x ) = f n − ( x ) for all x ∈ [0 , \ U n − and we define f n so that for all j ∈ N (35) f n is zig-zag of order n with respect to Z n,j on [ a n,j , b n,j ]and(36) f n ( x ) = g n ( x ) for all x ∈ [0 , \ U n . Now define(37) h = 2Φ , [0 , and for each n ≥ h n = g n + X j ∈ J n n, [ a n,j ,b n,j ] . We note that from (33)–(36), the definition of Φ n and (11) inLemma 2.8 we have:(39) || f n − g n || ∞ ≤ n and(40) || g n − f n − || ∞ ≤ n − . . BRUCE HANSON, DEPARTMENT OF MATHEMATICS, STATISTICS AND COMPUTER SCIENCE, ST. OLAF COLLEGE, NORTHFIELD, MINNESOTA 55057, USA EMAIL:
Moreover, from (38) and the definitions of f n and g n it follows that(41) g n ≤ f n ≤ h n . It follows from Lemma 2.8 that for every n ≥ g n and h n are Lipschitz on [ a, b ] for every [ a, b ] ∈ F n − and from (33) and (34) we deduce that(43) g n = f n − on ([0 , \ U n − ) ∪ Z n − for all n ≥ . Using (43) and (36), we get(44) f n = f n − on ([0 , \ U n − ) ∪ Z n − for all n ≥ n ∈ N and for all k ≥ n f k = f n on Z n . It follows from (39) and (40) that the sequence { f n } converges uni-formly to a function f on [0 ,
1] and since each f n is clearly continuouson [0 , f is continuous on [0 ,
1] as well. We need toshow that L ∞ f = E , Lip + f (0) < ∞ , Lip − f (1) < ∞ and lip f = 0 on E . We first note that (44) implies that for every n ∈ N we have f = f n on [0 , \ U n . Next, using Lemma 2.10, it is straightforward to showthat h n +1 ≤ h n for n ∈ N and from the construction of the g n s it is nothard to see that g n ≤ g n +1 for all n ∈ N . Therefore, by (41), we have(46) g n ≤ f ≤ h n for all n ∈ N . Using (46) in the case n = 1, we have that 0 ≤ f ≤ , [0 , and itfollows easily that Lip + f (0) ≤ − f (1) ≤ n ≥ a, b ] ∈ F n − . Then from (42), (46) and Lemma2.2 we conclude that Lip f is finite on ( a, b ) \ U n and Lip + f ( a ) < ∞ and Lip − f ( b ) < ∞ . Using (31), we conclude that Lip f is finite on U n − \ U n and unfixing n we get that Lip f is finite on ∪ ∞ n =2 ( U n − \ U n ) =(0 , \ E .We next show that Lip f = ∞ on E . We begin by observing thatfrom (44) and (32) it follows that f = f n on Z n for all n ∈ N . Let x ∈ E . For each n ∈ N choose [ a n , b n ] ∈ F n such that x ∈ [ a n , b n ] andlet r n = b n − a n . Then using (12) from Lemma 2.10, we see that M f ( x, r n ) ≥ | f ( b n ) − f ( a n ) | r n = | f n ( b n ) − f n ( a n ) | r n ≥ n − . Since r n →
0, we conclude that Lip f ( x ) = ∞ , as desired.It remains to show that lip f ( x ) = 0 for all x ∈ E . Let x ∈ E ,so x ∈ U n for all n ∈ N . For each n ≥ j ( n ) so that x ∈ ( a n,j ( n ) , b n,j ( n ) ) ⊂ U n . For notational convenience we let a n = a n,j ( n ) ETS WHERE Lip f IS INFINITE AND lip f IS FINITE. 9 and b n = b n,j ( n ) so we have x ∈ ( a n , b n ) ⊂ U n for all n ≥
2. Fix n ≥ g n is constant on [ a n , b n ]. Thenusing (38) and (46), we deduce that(47) g n ( a n ) ≤ f ( y ) ≤ g n ( a n ) + 2Φ n, [ a n ,b n ] ( y ) for all y ∈ [ a n , b n ] . Letting s n = min { x − a n , b n − x } , we see that M f ( x, s n ) ≤ n − . Since s n →
0, it follows that lip f ( x ) = 0 and we are finished with the proofof Theorem 1.2. (cid:3)
4. Proof of Theorem 1.3
Proof.
Assume that E is G δ with | E | = 0. We assume with-out loss of generality that E ⊂ (0 , Definition . Suppose that U is open, a < b and n ∈ N . Thenwe say that U is n -small on ( a, b ) if E ⊂ U and (48) | U ∩ [ a + b − a j +1 , a + b − a j ] | = | U ∩ [ b − b − a j , b − b − a j +1 ] | = b − a n + j +1 ∀ j ∈ N . Furthermore, if V is an open set and U is n -small on each componentof V , then we say that U is n -small on V . Note that if U is n -small on ( a, b ), then | U ∩ ( a, b ) | = 2 − n ( b − a ).Moreover, because | E | = 0, given any open set V such that E ⊂ V and n ∈ N , we can find an open set U which is n -small on V . Definition . Suppose that the open set U is n -small on ( a, b ) .Then we define g = h ( n, U, ( a, b )) on [ a, b ] as follows: (49) g ( x ) = 2 n · | U ∩ [ a, x ] | . The proofs of the following two lemmas are straightforward and leftto the reader.
Lemma . Suppose that U = ∪ ∞ j =1 ( a j , b j ) is n -small on ( c, d ) , g = h ( n, U, ( c, d )) ,f ( x ) = g ( x ) for all x ∈ [ c, d ] \ U, and g ( a j ) ≤ f ( x ) ≤ g ( b j ) for all x ∈ ( a j , b j ) and for all j ∈ N . Then for all x ∈ [ c, d ] we have (50)max { , − n +1 ( x − c + d } ≤ f ( x ) ≤ min { − n +1 ( x − c ) , − n ( d − c ) } [email protected] and letting r x = min { x − c, d − x } , we have (51) M f ( x, r x ) ≤ − n +2 for all x ∈ ( c, d ) . Lemma . Suppose that the open set V is trim in [ a, b ] and h islinear on [ a, b ] with slope m > and n ∈ N . Set α = | [ a, b ] \ V | anddefine f = f ( h, [ a, b ] , V, n ) on [ a, b ] as follows: f ( x ) = h ( a ) + ( ( m − − n +1 )( b − a ) α + 12 n ) | [ a, x ] \ V | + 12 n | [ a, x ] ∩ V | . Then f is increasing on [ a, b ] and (52) f ( a ) = h ( a ) and f ( b ) = h ( b ) , (53) f is linear with slope − n on each component of ( a, b ) ∩ V, and (54) f is Lipschitz on [ a, b ] . Definition . Suppose that U and V are open sets and V is trimon each component of U . Then we say that V is trim on U . Let V = (0 ,
1) and define { U n } and { V n } inductively to satisfy thefollowing conditions:(55) U n is n -small on V n for all n ∈ N (56) V n +1 is trim on U n for all n ∈ N (57) V n +1 ⊂ U n ⊂ V n for all n ∈ N (58) ∩ ∞ n =1 V n = E. For each n ∈ N we define U n to the the collection of all componentsof U n and V n to be the collection of all components of V n .We now begin the construction of a function f satisfying the con-clusions of the theorem. We begin by defining sequences of functions { f n } and { g n } on [0 ,
1] and then defining f = lim n →∞ f n = lim n →∞ g n .The g n s will be constructed to ensure that Lip f = ∞ on E and the f n s will be constructed to ensure that lip f = 0 on E . First of all,define f ( x ) = x on [0 , f n s and g n s inductivelyto satisfy the following for all n ∈ N : ETS WHERE Lip f IS INFINITE AND lip f IS FINITE. 11 (59) for all I = ( a, b ) ∈ V n , g n | I = h ( n, U n , I ) + f n ( a )and(60) g n = f n on [0 , \ V n (61) for all J ∈ U n , f n +1 | J = f ( g n , J , V n +1 , n + 1)and(62) f n +1 = g n on [0 , \ U n . The following lemma follows easily from the definitions above. Weleave the proof to the reader.
Lemma . For every n ∈ N we have the following: (63) f n and g n are Lipschitz on [0 , , (64) f n and g n are increasing on [0 , , (65) for all J = ( c, d ) ∈ V n , f n ( c ) ≤ g n ( x ) ≤ f n ( d ) ∀ x ∈ J , (66) for all I = ( a, b ) ∈ U n g n ( a ) ≤ f n +1 ( x ) ≤ g n ( b ) ∀ x ∈ I. It follows from (57) and (59) - (62) that for all k ≥ n ≥ f n ( c ) ≤ f k ( x ) ≤ f n ( d ) = f n ( c ) + d − c n ∀ x ∈ J = ( c, d ) ∈ V n and(68) f k ( x ) = f n ( x ) ∀ x ∈ [0 , \ V n . Therefore, we may conclude that(69) || f k − f j || ∞ ≤ max J ∈V n | J | n ∀ j, k ≥ n. Since max J ∈V n | J | n → n → ∞ , it follows that { f n } converges uni-formly to a function f on [0 , { g n } converges uniformly to f as well. Since each g n iscontinuous on [0 , f is also continuous on [0 , n ∈ N and for all k ≥ n , we have g n ( a ) ≤ g k ( x ) ≤ g n ( b ) for all x ∈ ( a, b ) ∈ U n , and therefore for n ∈ N we have(70) g n ( a ) ≤ f ( x ) ≤ g n ( b ) for all x ∈ ( a, b ) ∈ U n . [email protected]
Now note that since f = lim f n , (68) implies that(71) f ( x ) = f n ( x ) for all x ∈ [0 , \ V n . Furthermore, using (57), (62) and (71), we get(72) f ( x ) = g n ( x ) for all x ∈ [0 , \ U n . Let n ∈ N and J = ( c, d ) ∈ V n . Then U n is n -small on J , g n | J = h ( n, U n , J ) + f n ( c ) = h ( n, U n , J ) + f ( c ) , and f ( x ) = g n ( x ) for all x ∈ J \ U n by (72). Using (70), we can applyLemma 4.3 to conclude that (50) holds for all x ∈ J .From (50), (71) and (63) we deduce that Lip f is bounded on[0 , \ V n and therefore by (58) we get that Lip f < ∞ on [0 , \ E. It remains to show that Lip f = ∞ on E and lip f = 0 on E . Let x ∈ E .We first show that Lip f ( x ) = ∞ . For each n ∈ N choose ( a n , b n ) ∈U n such that x ∈ ( a n , b n ). From (59) and (49) we see that g n ( b n ) − g n ( a n ) = 2 n ( b n − a n ). However, by (72) we have that f ( a n ) = g n ( a n )and f ( b n ) = g n ( b n ) and therefore M f ( x, r n ) ≥ n − where r n is eitherequal to x − a n or b n − x . Letting n → ∞ , we conclude that Lip f ( x ) = ∞ . Finally, we show that lip f ( x ) = 0. For each n ∈ N choose ( c n , d n ) ∈V n such that x ∈ ( c n , d n ) and let r n = min { x − c n , d n − x } . Then from(51) we conclude that M f ( x, r n ) ≤ − n +2 . Letting n → ∞ we deducethat lip f ( x ) = 0 and we are done with the proof of Theorem 1.3. (cid:3)
5. Some Related Problems
Let l f = { x ∈ R : lip f ( x ) = 0 } . There are a number of openproblems concerning the relationship between the sets L ∞ f , l ∞ f and l f .For example, it would be interesting to characterize the sets E ⊂ R that satisfy any of the following conditions.(1) E ⊂ L ∞ f ∩ l f .(2) E ⊂ L ∞ f ∩ l f and l ∞ f = ∅ .(3) E = L ∞ f ∩ l f .(4) E = L ∞ f ∩ l f and l ∞ f = ∅ .(5) L ∞ f = R and l f = E .(6) L ∞ f = l f = E . ETS WHERE Lip f IS INFINITE AND lip f IS FINITE. 13
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