ON PREPONDERANT DIFFERENTIABILITY OF TYPICAL CONTINUOUS FUNCTIONS

Transcription

1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 14, Number 3, March 1996 ON PREPONDERANT DIFFERENTIABILITY OF TYPICAL CONTINUOUS FUNCTIONS L. ZAJÍČEK (Communicated by C. D. Sogge) Abstract. In the literature, several definitions of a preponderant derivative exist. An old result of Jarník implies that a typical continuous function on [0, 1] has a (strong) preponderant derivative at no point. We show that a typical continuous function on [0, 1] has an infinite (weak) preponderant derivative at each point from a c-dense subset of (0, 1). 1. Introduction Let C denote the set of continuous real functions defined on [0,1] furnished with the metric of uniform convergence. As usual, when we say a typical f C has a certain property, we mean that the set of f C with this property is residual in C. References concerning differentiability properties of typical continuous functions can be found in [] and [11]. The first results of this type ([1], [7]) imply that a typical f C has a derivative at no point. A result of Jarník [5] (cf. [11], Theorem J, (iii)) immediately implies that a typical f C has a preponderant derivative (finite or infinite) at no point of (0,1), if we use a strong definition of a preponderant derivative (cf. Definition 1 below) which was used e.g. by Bruckner [3]. The same result holds true if we consider a slightly weaker definition of a preponderant derivative used in [8]. The main result of the present article says that the situation changes if we use another weaker notion of a preponderant derivative (cf. Definition below), which is still stronger than the notion of a preponderant derivative used by Denjoy [4]. Namely we prove that a typical function f C has this weaker preponderant derivative f pr (x) = at each point x from a c-dense subset of (0,1). Note that the infiniteness of the preponderant derivative is essential, since a theorem of Jarník [5] (cf. [11], Theorem J, (ii)) immediately implies that a typical f C has at no point a finite preponderant derivative (any existing definition of a preponderant derivative can be used here). Further note that a slightly weaker version of the main result of the present article is stated without a proof in [11] (Theorem 4, (ii)). The main tools for the proof of our result is the Banach-Mazur game and an unusual monotonicity theorem of [6]. Received by the editors March 15, 1994 and, in revised form, August 3, Mathematics Subject Classification. Primary 6A4. Key words and phrases. Preponderant derivative, typical continuous function, Banach-Mazur game. Supported by Research Grants GAUK 363 and GAČR c 1996 American Mathematical Society

2 790 L. ZAJÍČEK. Preliminaries If a>b, then the symbol (a, b) will denote the open interval (b, a). The symbol µ will denote the Lebesgue measure on R. The open ball in a metric space P with the center c P and the radius r>0 is denoted by B(c, r). The definition of the strong preponderant derivative used in [3] and mentioned in the Introduction is the following. Definition 1. We say that A is a strong preponderant derivative of a function f at a point a R if there exists a measurable set E R such that lim h 0+ µ(e (a, a + h)) h > 1, lim µ(e (a h, a)) h 0+ h > 1 and f(x) f(a) lim = A. x a,x E x a We shall work with the following weaker definition. Definition. We say that a function f has at a point a the preponderant derivative f pr(a) =A if there exists a measurable set E R such that f(x) f(a) lim = A x a,x E x a and µ(e I) µ(i) > 1 for every interval I containing a. One of the main tools of the present article is the Banach-Mazur game. It is the following infinite game between two players. Let P be a metric space and let Q P be given. In the first step the first playerchoosesanopenball B(g 1,δ 1 ). In the second step the second player chooses an open ball B(f 1,ε 1 ) B(g 1,δ 1 ), in the third step the first player chooses an open ball B(g,δ ) B(f 1,ε 1 ), and so on. If B(f i,ε i ) Q, i=1 then the second player wins. In the opposite case the first player wins. We shall need the following theorem essentially due to Banach. Theorem BM. The second player has the winning strategy in the Banach-Mazur game if and only if Q is a residual subset of P. A proof of this theorem can be found in [10] in the case P =(0,1) ; the proof in the general situation (cf. [9]) is essentially the same. The second essential tool is the following basic lemma which is an easy conseuence of the main result of [6].

3 ON PREPONDERANT DIFFERENTIABILITY 791 Lemma 1. Let ϕ be a continuous function on [a, b],let S be a countable set ϕ(b) ϕ(a) and let b a >K. Then there exists x (a, b) \ S for which (1) µ{z (x, y) : ϕ(z) ϕ(x) >K}> 1 (y x) for all y (x, b] and () µ{z (y, x) : ϕ(z) ϕ(x) >K}> 1 (x y) for all y [a, x). Proof. Choose ε > 0 such that ϕ(b) ϕ(a) > (K + ε)(b a) and consider the function f(x) :=(K + ε)x ϕ(x). Since f(b) < f(a) and the conditions (B), (C), (D) of Theorem 1 from [6] are clearly satisfied for α =1/, we obtain that (A) does not hold. Conseuently there exists x (a, b) \ S such that (3) µ{z (x, y) :(K+ε)z ϕ(z)>(k+ε)x ϕ(x)}< 1 (y x) for all y (x, b] and (4) µ{z (y, x) :(K+ε)z ϕ(z)>(k+ε)x ϕ(x)} 1 (x y) for all y [a, x). It is easy to see that (3) immediately implies (1). Since f is continuous, it is easy to prove that (4) implies (). We shall also need the following two easy lemmas. Lemma. Let f be a measurable real function, let x R and let a n x and b n x be strictly monotone seuences such that, for all natural numbers n, µ{z (y, x) : f(z) f(x) whenever y [a n,a n+1 ] or y [b n+1,b n ]. Then Proof. Fix a natural n. The function >n}> x y f pr(x) =. 1 h(y) = µ{z (y, x) :f(z) f(x) >n} x y is clearly continuous on [a n,a n+1 ] [b n+1,b n ].Conseuently there exists an ε n > 0 such that µ{z (y, x) : f(z) f(x) >n}> x y (ε n +1/) for each y [a n,a n+1 ] [b n+1,b n ]. Therefore we can find h n > 0 such that µ{z (y, x) \ ( h n,h n ): f(z) f(x) >n}>(1/) x y for each y [a n,a n+1 ] [b n+1,b n ]. Now we put f(z) f(x) E =(,a 1 ) (b 1, ) ({z : >n}\( h n,h n )). z x n=1 It is easy to prove that E has the properties from Definition (with a = x and A = ).

4 79 L. ZAJÍČEK Lemma 3. Let 0 p<a<b 1, K > 0 and f C. Suppose that, for all y [a, b], µ{z (p, y) : f(z) f(p) >K}> 1 z p (y p). Then there exists 0 <η<a p such that the conditions f f <η, p p <η and y [a, b] imply µ{z (p,y): f (z) f (p ) z p >K}> 1 (y p ). Proof. Suppose that no such η exists. Then there exist seuences fn C, p n [0, 1], y n [a, b] such that fn f C, p n p and in (5) µ((p n,y n) A n ) 1 (y n p n ), where A n = {z [0, 1] : f n(z) fn(p n) z p >K}. n We can and will suppose that y n y, wherey [a, b]. Putting we know that A = {z [0, 1] : f(z) f(p) z p >K}, (6) µ(a (p, y)) > 1 (y p). But for each fixed z (p, y) obviously and therefore f n(z) f n(p n) z p n f(z) f(p) z p (7) A (p, y) lim inf (A n (p, y)). n Using (7) and (5) we obtain µ(a (p, y)) lim inf n µ(a n (p, y)) which contradicts (6). lim sup n lim sup n (µ(a n (p n,y n )) + p n p + y n y ) 1 (y n p n)= 1 (y p),

5 ON PREPONDERANT DIFFERENTIABILITY Main result Theorem. For a typical f C the sets {x : f pr(x) = } and {x : f pr(x) = } are c-dense in (0, 1). Proof. Since the mapping h h is a homeomorphism of C on C and h pr(x) = iff ( h) pr (x) =, it is sufficient to prove that {x : f pr (x) = } is c-dense for a typical f C. Conseuently it is sufficient to prove that for each fixed interval (α, β) with rational endpoints the set Q := {f C : card{x (α, β) :f pr (x) = } = c} is residual. By Theorem BM it is sufficient to find a winning strategy for the second player in the Banach-Mazur game for P = C and Q. To describe this strategy, put S n = {0, 1} n and S = {0, 1} N,wheren Nand N = {1,,...}. By our strategy the second player will construct in his nth move not only a ball B(f n,ε n ) C but also numbers ξ n > 0, n >0 and n points {x s n : s S n } such that the following conditions (8)-(13) hold (where we have put a s n = xs n 50ξ n, b s n = xs n +50ξ n): (8) {[a s n,bs n ]:s S n} are pairwise disjoint subintervals of (α, β); if we fix an arbitrary s =(s 1,..., s n ) S n and denote x k = x (s1,...,s k) k, a k = a (s1,...s k) k and b k = b (s1,...,s k) k for 1 k n,then (9) [a n,b n ] (x n 1 4ξ n 1,x n 1 +4ξ n 1 ) for n>1; (10) f n is linear on [x n 5ξ n,x n +5ξ n ], constant on the intervals [a n,x n 5ξ n ],[x n +5ξ n,b n ] and f n(b n ) f n (a n ) = n > 10(n +1); b n a n if n>1, then for each x [x n 4ξ n,x n +4ξ n ] and f B(f n,ε n ) we have (11) µ{z (y, x) : f(x) f(z) x z >n 1}> 1 (x y) for y [a n 1,a n ] and (1) µ{z (x, y) : f(z) f(x) >n 1}> 1 (y x) for y [b n,b n 1 ] ; (13) B(f n,ε n ) B(g n,δ n ) and ε n < 1 ξ n. If we prove that the second player can play according to this strategy in all moves, we shall be done. In fact, (9) and (13) imply that ξ n 0, ε n 0, n=1 B(f n,ε n ) consists of one function f and the set n=1 [a(s1,...,sn) n,b (s1,...,sn) n ] consists of one point x s (a, b) foreach s=(s 1,s,...) S. Lemma, (9), (11) and (1)

6 794 L. ZAJÍČEK easily imply that f pr (x s)=. Since (8) implies that x s x s for s s,we obtain f Q. Thus suppose that, for a fixed natural number m, the second player has played m 1 moves and the first player has played m moves such that the conditions (8)-(13) hold for each n<m. We know that B(g m,δ m ) B(f m 1,ε m 1 ) (if m>1) and we can clearly suppose without any loss of generality that g m is a piece-wise linear function. Our task is to construct f m,ε m,ξ m, m and {x s m : s S m } such that the conditions (8)-(13) hold for n = m. If m = 1, then the conditions (9), (11) and (1) are trivially satisfied and the construction of f 1,ε 1,ξ 1, 1 and x (0) 1,x(1) 1 such that the conditions (8), (10) and (13) are satisfied for n = 1 is uite easy. If m>1, we shall consider an arbitrary (s 1,..., s m 1 ) S m 1 and the function g m on the interval [a m 1,b m 1 ]:=[a (s1,...,sm 1) m 1,b (s1,...,sm 1) m 1 ]. Since (13) implies g m f m 1 <ε m 1 < 1 ξ m 1 < 1 (b m 1 a m 1 ), we have g m (b m 1 ) g m (a m 1 ) b m 1 a m 1 > f m 1(b m 1 ) f m 1 (a m 1 ) b m 1 a m 1 1= m 1 1. Since g m is piece-wise linear, Lemma 1 ( used for ϕ = g m, [a, b] =[a m 1,b m 1 ], K= m 1 1 ) implies that we can choose two points x m (s1,...,sm 1,0),x m (s1,...,sm 1,1) in (a m 1,b m 1 ) such that, denoting any of them by x m, the following assertions hold : (14) g m is linear on a neighbourhood of x m, (15) for each y (x m,b m 1 ] we have µ{z (x m,y): g m(z) g m (x m ) m > m 1 1}> 1 (y x m) and (16) for each y [a m 1,x m ) we have µ{z (y, x m ): g m(z) g m (x m ) > m 1 1}> 1 m (x m y). Thus we have constructed for each s S m apoint x s m (denoted above as x m ). In the following we shall suppose that s =(s 1,..., s m ) is fixed and we shall put for brevity as above a m 1 = a (s1,...,sm 1) m 1,b m 1 =b (s1,...,sm 1) m 1, x m 1 = x (s1,...,sm 1) m 1, x m = x s m.

7 ON PREPONDERANT DIFFERENTIABILITY 795 Now we shall observe that (17) x m (x m 1 4ξ m 1,x m 1 +4ξ m 1 ). In fact, suppose that (17) does not hold. We can suppose without any loss of generality that x m x m 1,thecase x m x m 1 being uite symmetrical. Then x m [x m 1 +4ξ m 1,x m 1 +5ξ m 1 ] or x m [x m 1 +5ξ m 1,b m 1 ]. In the first case (10) yields f m 1 (x m ) f m 1 (x m 1 +4ξ m 1 ) =f m 1 (x m 1 ) (f m 1(b m 1 ) f m 1 (a m 1 )) =f m 1 (x m 1 )+40 m 1 ξ m 1. For each z [x m 1 +5ξ m 1,b m 1 ] we have by (10) f m 1 (z) =f m 1 (b m 1 )=f m 1 (x m 1 )+50 m 1 ξ m 1 and z x m 0ξ m 1 ; conseuently f m 1 (z) f m 1 (x m ) z x m Since g m f m 1 < 1 ξ m 1 we have 10 m 1ξ m 1 0ξ m 1 = m 1. g m (z) g m (x m ) z x m m 1 + ξ m 1 < m 1 +1 < m 1 1 0ξ m 1 for each z [x m 1 +5ξ m 1,b m 1 ]. This clearly contradicts (15) for y = b m 1. If x m [x m 1 +5ξ m 1,b m 1 ], we can obviously find y [x m 1 +5ξ m 1,x m 1 +50ξ m 1 ] such that y x m =10ξ m 1. For each z ( xm+y,y) we have z x m 5ξ m 1. Thus, since f m 1 (z) = f m 1 (x m ) and g m f m 1 < 1 5 ξ m 1,wehave g m (z) g m (x m ) m ξ m 1 5ξ m 1 = 1 5 < m 1 1. This clearly contradicts (15) (if y>x m ) or (16) (if y<x m ). Further observe that (14), (15) and (17) yield the existence of v = v s such that (18) g m is linear on [x m v s,x m +v s ] (x m 1 4ξ m 1,x m 1 +4ξ m 1 ) with a slope = s > m 1 1 and the intervals {[x s m v s,x s m +v s ]:s S m } are pairwise disjoint.

8 796 L. ZAJÍČEK Using (15), we can apply Lemma 3 to f = g m,p=x m, a = x m +v, b = b m 1, K = m 1 1 and obtain η = η s > 0 such that : (19) whenever h g m <η and x x m <η, then µ{z (x, y) : h(z) h(x) > m 1 1}> 1 (y x) for all y [x m + v, b m 1 ]. Using (16) and the symmetrical version of Lemma 3, we obtain η = η s > 0 such that : (0) whenever h g m < η and x x m < η, then µ{z (y, x) : h(z) h(x) > m 1 1}> 1 (x y) for all y [a m 1,x m v]. Now choose m so big and ξ m > 0sosmall(notethat m and ξ m do not depend on s) thatforeach s S m (1) m > max( s, 10(m +1)), () 50ξ m < min( ηs η s, ), m m (3) 50ξ m < s v s m and (4) 50 m ξ m <δ m. On account of (1) we obtain 5ξ m < 50ξmm s and (3) implies that Im s := [x s m 50ξ m m s,x s m + 50ξ m m s ] [x s m v s,x s m +v s ]. Conseuently (18) gives that {I s m : s S m} are pairwise disjoint intervals. On each interval I s m we now define the function f m by the following conditions: f m (x s m )=g m(x s m ), f m is linear on [x s m 5ξ m,x s m +5ξ m ] with the slope 10 m, f m (x)=g m (x s m )+50 mξ m for x [x s m +5ξ m,x s m + 50ξ m m s ] and

9 ON PREPONDERANT DIFFERENTIABILITY 797 f m (x) =g m (x s m ) 50 mξ m for x [x s m 50ξ m m s,x s m 5ξ m]. For x [0, 1] which belongs to no interval Im s, we put f m(x) =g m (x). Using (18) we easily obtain that f m C. On account of (18) and the definition of f m we easily see that f m g m 50 m ξ m. Conseuently (4) implies f m B(g m,δ m ). Thus we can choose an ε m > 0 such that (5) B(f m,ε m ) B(g m,δ m ), ε m < ξ m and ε m < min( ηs, ηs ) for each s S m. Now we have to show that the conditions (8) - (13) hold for n = m. By (1) we have [a s m,bs m ] Is m, which implies (8). In the following we shall suppose that an arbitrary s S m is fixed and we shall omit the index s. The condition (9) easily follows from (18), since [a m,b m ] I m [x m v, x m + v]. Since [a m,b m ] I m, the condition (10) follows immediately from the definition of f m. The condition (13) is contained in (5). Thus it is sufficient to prove (11) and (1). We shall prove (1) only, the proof of (11) being symmetrical. To prove (1), consider arbitrary x [x m 4ξ m,x m +4ξ m ], f B(f m,ε m ) and y [b m,b m 1 ]. We shall distinguish two possibilities : In the first case () gives y [x m + v, b m 1 ] and y [b m,x m +v). f m g m 50 m ξ m < η. and conse- Conseuently (5) implies f g m <η.by () also x x m <η uently (19) and (10) yield (6) µ{z (x, y) : f(z) f(x) >10m 1} > 1 (y x). In the second case we shall estimate f(z) f(x) separately for the cases (7) z [x m +5ξ m,x m + 50ξ m m ] and (8) z (x m + 50ξ m m,x m +v).

10 798 L. ZAJÍČEK In the case (7) we have f m (x) f m (x m +4ξ m )=f m (x m )+40 m ξ m, f m (z)=f m (x m )+50 m ξ m and conseuently f m (z) f m (x) > 10 m ξ m. Since we obtain f(z) f(x) z x z x< 50ξ m m > 10 mξ m ξ m 50ξ m m > 9 m +4ξ m 54 m In the case (8) observe that we have proved +4ξ m and f m f < 1 ξ m, = 6 > 1 6 ( m 1 1) > 10m 1 6 f m (x m + 50ξmm ) f m (x) > (x m + 50ξmm ) x 6, f m (z) f m (x m + 50ξmm ) = and conseuently z (x m + 50ξmm ) f m(z) f m(x) > 6.Since f m f < 1 ξ m and z x>45ξ m,wehave f(z) f(x) Conseuently we have > 6 1>10m >m. (9) µ{z (x, y) : f(z) f(x) >m} µ[x m +5ξ m,y]> 1 (y x). From (9) and (6) we obtain (1) which completes the proof. >m. References 1. S. Banach, Über die Differenzierbarkeit stetiger Funktionen, Studia Math. 3 (1931), A. M. Bruckner, The Differentiability properties of typical functions in C[a, b], Amer. Math. Monthly 80 (1973), MR 47: , Some observations about Denjoy s preponderant derivative, Bull. Math. Soc. Sci. Math. R. S. Roumanie1(69) (1977), MR 57:99 4. A. Denjoy, Mémoire sur la totalisation des nombres dérivées nonsommables, Ann. École Norm Sup. 33, 34 (1916, 1917), 17-36, V. Jarník, Sur la dérivabilité des fonctions continues, Publications de la Fac. des Sc. de L Univ. Charles 19 (1934), 9 pp. 6. J. Malý, D. Preiss, and L. Zajíček, An unusual monotonicity theorem with applications, Proc. Amer. Math. Soc. 10 (1988), MR 89e: S. Mazurkiewicz, Sur les fonctions non dérivables, Studia Math. 3 (1931), J. L. Leonard, Some conditions implying the monotonicity of a real function, Rev. Roumaine Math. Pures Appl. 17 (197), MR 46: J. Oxtoby, The Banach-Mazur game and Banach category theorem, in: Contributions to the Theory of Games III, Ann. of Math. Stud. 39 (1957), MR 0:64 10., Measure and category, Springer Verlag, New York, MR 81j: L. Zajíček, The differentiability structure of typical functions in C[0, 1], Real Analysis Exchange 13 ( ), 119, , 93. Department of Mathematical Analysis, Charles University, Sokolovská 83, Praha 8, Czech Republic address: Zajicek@karlin.mff.cuni.cz