.PRfiPERTiES OF ‘L p DERIVATIVES Thesis for the Degree of Ph. D” MIDI-WAN. STATE UNIVERSFTY MICHAEL JON EVANS 197.0 (HES'S 0-169 This is to certify that the thesis entitled Properties of Lp Derivatives presented by Michael Jon Evans has been accepted towards fulfillment of the requirements for PhoDo degree in Mathematics C Q amigo? Major professor Date Max 15' 1970 runs-numb?! LIBRARY Michigan Stem Univ rsicy ....,.. r w I" K" ‘ Zi- A" ABSTRACT PROPERTIES OF Lp DERIVATIVES BY Michael Jon Evans Suppose a real valued function f, defined on an interval I, possesses an ordinary derivative f'(x) at each point of I. Then f' need not be a continuous function. However, A. Denjoy, J.A. Clarkson, and Z. Zahorski have shown that f' possesses the following four properties, defined here for an arbitrary function g. l. A function g is said to be of Baire class one if g is the pointwise limit of a sequence of continuous functions. 2. A function 9, defined on an interval I, is said to have the Darboux property if, on every subinterval [a,b] of I, 9 takes on all intermediate values. 3. A function g is said to have the Denjoy property if for every open interval (a,b), g—1((a,b)) either is empty or has positive measure. Michael Jon Evans 4. A function g is said to have the Zahorski property if for every open interval (a,b),x in g-1((a,b)), and {In} a sequence of closed intervals converging to x with -1 lg ((a.b>) n Inl = o for every n, then III ' ____fl__ _ 11m d(x,I ) — 0' n"0° n where IEI denotes the Lebesgue measure of a set E, and d(x,In) = inffl x - yI : y E In}. H;W. Oliver showed more generally that if a function f has a kth Peano derivative fk(x) at each point of an interval, then fk has properties 1, 2, and 3 listed above. C.E. Weil showed, furthermore, that f has property 4. k In Chapter I of this paper we look at the h definition of a kt Lp derivative for a function as given by A.P. Calderon and A. Zygmund. This type of differentiation is a generalization of the Peano type. In Chapter II it is shown that if a continuous L function f possesses a kth Lp derivative fkp(x) Michael Jon Evans at each point of an interval, then f:p has the above listed properties 1, 2, 3, and 4. We also prove some other properties of f:p dealing with its relationship to approximate and ordinary derivatives. In Chapter III we show that if we assume only that f is measurable and possesses a kth Lp derivative at each point of an interval, L then fkp still has property 1. PROPERTIES OF Lp DERIVATIVES BY Michael Jon Evans A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1970 To Mother ii and Dad ACKNOWLEDGEMENTS I wish to take this opportunity to express my sincere appreciation to Professor Clifford E. Weil for suggesting the prOblems studied in this thesis and for his helpful guidance throughout its preparation. Thanks also go to Mary Starr for typing the original copy of this thesis. Finally, thanks to my wife, Barbara, for her patience and encouragement. iii Chapter I. Chapter II. Chapter III. BIBLIOGRAPHY TABLE OF CONTENTS Introduction and Definitions . th . . Properties of k Lp Derivatives of Continuous Functions. . . . . . th h A Property of k Lp and kt Approximate Peano Derivatives of Measurable Functions . . iv 11 68 80 Chapter I. Introduction and Definitions Suppose a real valued function f, defined on an interval I, possesses an ordinary derivative f'(x) at each point of I. Although f' need not be continu— ous under these circumstances, it does possess certain properties which make it closely related to continuous functions. Consider the following definitions of these properties. Definition 1.1. A function g is said to belong to Baire class one if g is the pointwise limit of a sequence of continuous functions. Definition 1.2. A function 9, defined on an interval I, is said to have the Darboux property if, on every subinterval [a,b] of I, g takes on all inter- mediate values, i.e. if g(a) # g(b), and if c lies between 9(a) and g(b), there is a number d, a < d < b, for which g(d) = c. Definition 1.3. A function g is said to have the Denjoy property if for every open interval (a,b), g-1((a,b)) either is empty or has positive measure. Definition 1.4. A sequence of closed intervals {In} is said to converge to a point x if x is not in the union of the In and if every neighborhood of x contains all but a finite number of the intervals I . n Definition 1.5. A function g is said to have the Zahorski prOperty if for every Open interval (a,b), x in g‘1((a,b)) and {In} a sequence of closed intervals converging to x with lg-1((a.b)) n Inl = o, for every n, then d(x,I ) = 0’ where [E] denotes the Lebesgue measure of a set E, and d(x,In) = inf[Ix-yl : y 6 In}. If f' is the derivative of f at each point of an interval I, it is well known that f' .is of Baire class one and that f' has the Darboux property. Although f' need not be continuous, Denjoy and Clarkson have shown that f' has the Denjoy property (see [6] and [3]). Zahorski refined this property in [19] and proved that f' possesses what we have named the Zahorski property. These four properties do not classify derivatives. Indeed, they are possessed by more general types of derivatives. Consider the following definition. Definition 1.6. A function f, defined on an interval I, is said to have a kth Peano derivative at x0, k = 1,2,..., if there exist numbers fl (x0) , f;2 (x0) , . . . , fk (x0) such that hk k f(x0+h) — f(xo) - hfl(XO)-...— ET fk(x0) = o(h) as h fl 0. The numbers fl(x0),...,fk(xo) can easily be shown to be unique, and if f has a kth Peano deriva- tive fk(x0) at x0, then it also has an nth Peano derivative fn(xo). n = 1,2,...,k—1. Notice that f1(xo) = f'(x0), the ordinary first derivative. If f has an ordinary kth derivative, f(k)(xo), at x0, then Taylor's theorem shows that fk(x0) exists and (k) equals f (x ). However fk(xo) may exist without 0 f(k)(xo) existing. Example 1.1 Let f(x) = x3 sin i for x # O, and f(0) = 0. Then fl(0) = f'(O) = O, and since 4s 3 h sin = o(hz), l h (2) (0) as h d O, we have f2(O) = 0. However, f does not exist. Suppose now that a function f, defined on an h interval I possesses a kt Peano derivative at each point of I. Denjoy [5] and Oliver [13] have shown that under these circumstances fk belongs to the first Baire class. Oliver also has shown that fk has the Darboux property and the Denjoy property. Weil [18] has given another proof that fk has the Denjoy property, as well as proving that f has the Zahorski property. k Calderon and Zygmund [2] have introduced a generalization of the kth Peano derivative as follows: Definition 1.7. A function f, defined on an interval I, is said to have a kth Lp derivative at x l g p < m, k = 1,2,..., if there exist numbers 0’ Lp Lp Lp fO (x0), fl (x0),...,fk (x0) such that 1 L L k L - l h _ p p _ __ p p p {h f0 |f(xo+t) f0 (xO)-tfl (x0) ...- k: fk (x0)| at} = o(hk). as h ~ 0. As with the kth Peano derivative, the numbers L L L P P P fO (X0), fl (x0)....,fk (x0) can be shown to b: unique, and if f has a kth Lp derivative fkp(xo) at x0, then it also has an nth Lp derivative L f P(x ) at x for n = l,2,...,k—l. In this last n O 0 definition it appears that the value of f at x is O irrelevant. However, if we know that f has a kth Lp derivative everywhere on an interval I, then we can show that f(x) = f:p(x) for almost every x in I by the following reasoning. Call x a Lebesgue point of f if lim h f(x+t)dt = f(x) h-o I O :J‘IH By Lebesgue's theorem we know that almost every point in I is a Lebesgue point of f (see [11] for a detailed discussion.) Using the fact that f has a kth Lp derivative at each point x in I, it is not difficult to show that 1 11 Lp lim - I f(x+t)dt = f (x), 11 O O h~O L for each x in I. Hence fop(x) = f(x) for each Lebesgue point x of f. Another useful fact that is easy to show is that if f has a kth Peano derivative at a point x, then f has a kth Lp derivative at x for any 1.3 p < w. L and fk(x) = fkp(x). Furthermore if 1 g_q < p < m, and f has a kth Lp derivative at x, then f has th L L a k Lq derivative at x, and fkp(x) = fkq(x). One of the first differences to be noticed between the Peano and Lp types of differentiation h is that while the existence of the kt Peano deriva- tive of a function f at each point of an interval implies that f is continuous on that interval, the same is not true for the kth Lp derivative as the following example shows. This example also shows that the kth Lp derivative is a true generalization of the kth Peano derivative, i.e. a kth Lp derivative is not necessarily a kth Peano derivative. Example 1.2. Here we construct a function on the interval I = [0,1] which has a first Lp deriva— tive at each point of I (indeed it will have an ordinary derivative on (O,l]), but which is discon- tinuous at O. For each positive integer n, let __l_._1__1_ - n — 2n, 2n + 8Pn]. On In we W111 choose f so l) f has an ordinary derivative at each point of In' 2) “—334 = f(—lr-1 + 745;) = o. 2 2 8 . __ _ ..j; ._L_ _ . . 3) f+(2n) — f_(2n + 8Pn) — 0, where f+(or f_) denotes the right (or left) derivative at X, 4) max f(x) = l. xEIn 00 For X in I — U In, we let f(x) = 0. Since f(0) = 0, n=1 and since every right neighborhood of 0 contains infinitely many of the intervalS‘ In’ f is not conti— nuous at O. f clearly has an ordinary derivative at each point in (O,l]. Next we will show that L L fop(O) = flp(O) = O, i.e. we will show that l . h. p \i5 _ 11m f0 |f(t)| at] — 0. Let 0 < h < l, and choose the positive integer N so th Sue D‘II—l 1 1 that O, the set F€ = {x : IF(X) - Al‘g S} has xO as a point of dispersion. Definition 1.10. A function f is said to have an approximate derivative f (x0) at x if 5p 0 f (X) - f (x0) = f' (x ). ap x - x ap O O lim X "X Suppose that f is a real valued function, defined on an interval I, possessing an approximate derivative at each point of I. Tolstoff [15] has shown that under these circumstances fép is of Baire class one. Khintchine [8] has shown that Rolle's theorem holds for approximate derivatives. It then readily follows that fép has the Darboux property. The proofs of Khintchine and Tolstoff are long and complicated. A shorter proof of the fact that fép is of Baire class one and has the Darboux property has been given by Goffman and Neugebauer [7]. Marcus [10] has shown that fép has the Denjoy property. Weil [18] gave another proof that fép has the Denjoy property and, further— more, showed that it has the Zahorski property. Chapter II. Properties of kth L Derivatives of Continuous Functions The following theorem and its proof are due to Neugebauer [12]. Theorem 2.1. Let f be a measurable function defined on an interval I. Then at almost all points x at which f has a first Lp derivative, LP fl (x) = fap(x). Proof. Let x be a point at which f has a L first Lp derivative, i.e. there exist numbers fop(x) L and f1p(x) such that 1 h L L l {E f0|f(x+t) - fop(x) — t flp(x)lpdt}p = o(h). Let S > O, and set L L E = {t > o: lf(x+t) - fop(x) - t flp(x)l _>_ at} and L L E6 = {t < o: |f(x+t) - f0p(x) - t flp(x)|.2 eltl}. We need to show that ES and Fe have 0 as a point of dispersion. Let h > O and set Eh = E6 0 [0,h]. Then we have 1 L L _ {fl-1 jg If(x+t) - fop(x) — t flp(x) |Pdt}P ll 12 l L L l _>_ { HI lf(x+t) - fop(x) - t flp(x)|pdt}P Eh 1 2 { 1111‘]. eptp dt}p En 1 | _ _>.{11-1 fth eptp dt}P 1 _ '._EE ..LEQL}I5 I — i p+l 11 Eh ’ and this is o(h), i.e. l P - E {e1— LEEP Hi" h-o P Hence 1im IEhl _ O hfiO 11 — ' i.e. E has 0 as a point of dispersion. Similarly F has 0 as a point of dispersion. So L f(x+t) - fop(x) Lp 11map t = fl (X)- tdO If x also belongs to the Lebesgue set of f, then L L P = P = . - fO (x) f(x), and hence fl (x) fap(x). Since almost every x at which f has a first Lp derivative belongs to the Lebesgue set of f, the theorem is proved. 13 Remark. Suppose that in the statement of the above theorem measurability is replaced by continuity. Then since every point of continuity of a function is a Lebesgue point of the function, we have f(x) = f:p(x) for each point where f has a first Lp derivative. L Hence at each such point flp(x) = fép(x). Corollary 2.1. Let f be a measurable function, defined on an interval I, possessing a first L derivative at each point of I. Then ftp 1) belongs to Baire class one, 2) has the Darboux property, 3) has the Denjoy property, and 4) has the Zahorski property. Proof. In the above theorem we saw that L f(x+t) — fop(x) ap t t~O L 1im flp(x) L at each point x in I. Since f(y) fop(y) for almost every y in I, we have LP LP fO (x+t) - fO (x) L . = p llmap t fl (X) I t‘O L flp(x) A H) O II'I 'U V m - 'U 33 II 14 for each x in I. As pointed out in Chapter I, an approximate derivative satisfies 1), 2), 3) and 4). So the proof is complete. If we now consider higher order Lp derivatives, we obtain the following result. Theorem 2.2. Let f be a measurable function possessing a kth Lp derivative at each point of an L interval I. Then fkp(x) is the approximate derivative L of kal(x)L at almost every point x in I, i.e. P(x) =:§_’(f 1a)p (x) a.e. in I. Proof. Without loss of generality we may assume that III < m. Let s > 0 be given. By the corollary to theorem 9 in [2] there exist a closed set E C I with II - E] < e and a decomposition f = g + h, where 1) g E Ck(I), i.e. g has k continuous derivatives on I, 2) h(x) = O for all x E E, and 1 {éfglh(x+t)lpdt}p =o(hk) as h-+O for each x in E. 15 NOW let X be a point of density of E. If X + u E E, then by 3) Lp Lp hk_1(x+u) = hk_l(x) = O. So LP LP _ (x+u) — h (x) 1im hk 1 k—l = O, u~O u x+u€E LP i.e. (bk-1)ap (X) = 0. Hence LP LP LP fk (X) = 9k (x) + h.k (x) = g(k)(x) + O = (9‘k'l’)gp + (hifl);p L = \fk81)ap(x)' L P _ P . So fk (X) — (fk-1)ap(x) for almost every x E E, and since this is true for each 3 > O, we have LP LP fk (X) = (fk—l>ap (X) for almost every X E I. The conclusion of this theorem cannot be strengthened to an "everywhere type" result, even if we assume that f is continuous, as the following example shows. l6 3 . 1 Example 2.1. Let f(x) =‘{x 51“ g' 0 < X 3,1. 0 , X = O. L We will show that f2(x) (and hence f2p(x)) exists for each X, 0 g_x g_1, but f2(O) #’(fl)ép(0). we have 1 2 . 1 f'(x) = { —X cos §_+ 3x Sln X' 0 < X g_1, O , x = O . Also f"(X) exists routinely for O < X's 1, and since f(h) = h3 sin % = o(h2), we have f2(O) = 0. Now if (f1)ép(0) = 0, then for each s > O, the set E6 = [h : If'(h)l > 611} has 0 as a point of dispersion. We will show that this is not the case. Let F€ = [h : loos %l > e} for e > O. In particular consider F l :5 l l {h . lcos El >-—;} ll IIC8 (1 l) l nw+g ' nw-I . n 4 4 For a fixed positive integer N, l7 °° 1 |F_1_n[O'NTr-T_r__]|=IngNl 2 4 4 fl _ °° 1 \ — ngg l ( nv + I ' nw - I) I _ 4 4 =Z( 1 ) n=N 2n W - 8 1 °° 1 >—Z( 4 n=N nzw — E) 4 l °° 1 1 = 4 n§;l( nw + E ’ nw — E>l ‘ 2 2 _ l l _ 4 l [0' NW — E J I 2 80 F1 does not have 0 as a point of dispersion. fl Let c > O, and consider ’ 1 El =[h: lf'(h)| >[——-e)h} _-€ fl fl 1 . l l = {h : Icos — + 3h Sln —| >-—- - e} h h x/i For a sufficiently large N, l8 -€n[0, WJDFJ-OLO, #:1' E_1_ fl 2 fl So El does not have 0 as a point of dispersion. ——-- 6 fl Consequently, f2(O) # (fl ap(O). Although a kth Lp derivative of a continuous function need not be an approximate derivative, we will show that it has the four properties of derivatives mentioned in Chapter I. We will need a few preliminary results. In [13] Oliver introduced certain types of h mean value prOperties for kt Peano derivatives. We will find it beneficial to exploit these prOperties, defined here in terms of kth Lp derivatives. Definition 2.1. If f has a kth Lp derivative at each point of an interval I, we say f has properties M?’ j = O,1,...,k-l, if for each X and x+h in I, there is an x' between X and x+h such that: LP LP 0 + - c —000- f1 (X h) f1 (X) . ESL (k-j): k-j-l Lp f (x) L 1"]. = fkp(xl) . h (k-i-l l 19 The special case of property M? when the left side of this inequality is equal to O, we refer to as property R? . The proof of the following lemma is due to Oliver [13]. Lemma 2.1. Let an interval I and a positive integer k be fixed. 1) Let j be an integer from O to k-l. If every function which possesses a kth L derivative at each point of I has property R?’ then every such function also has property M: 2) If a function f has property Ri—l and properties M? , for each j = O,l,...,k—2, then f has property R? for each j = O,l,...,k—l. Proof. Statement 1) is proved in the usual manner by adding an appropriate polynomial to f. Suppose y and y+h belong to I. Set Lp LP hk-j—l p ) 15' (Y+h)'f' (Y)"°-‘ “YT“:— f _ (y g(x) = f(x) — 1 2| k_j (k 3 1). k 1 L..._ (k-j): x- )k 20 Then Lp LP LP hk—j-l Lp gj (y+h) - gj (Y) " hgj+l (Y)-ooo- (k-j‘l): gk-l(y) = OI k i.e. 9 satisfies the hypotheses of Rj ; and k—j-l L - h f p (y) kaifiLl} krl L L fip(y+h)—fjp(y)—... bk"j (k-j)3 Applying the conclusion of R? to g, i.e., replacing L X by X' and gkp(x') by O, the conclusion of M? follows for f. To prove statement 2), let j be an integer, O g_j g_k-2, and suppose x and x+h are two points in I with Lp Lp hk-j—l Lp O = fi_(x+h) - fj (X)-...- (k-j—l)! fk_1(x) 2‘22... (k-j)3 So __ __ Lp Lp hk—j-2 Lp . + - . —...- . L O: 1 f1(x h) f] (x) lk-J-Z): fk_2(x) _ f p (x) ._;R__ k-l (k'j) I Since f has property ME—l, there is an X' between X and x+h such that 21 Lp Lp hk-j-2 Lp . + — . —...— ———w————- fJ (x h) f1 (x). (k-J-2)1 fk-Z = pr (X') hij—l k-l ' (k-j-l)! Consequently,- LP P P LP 0 = fk-1(X ) ' fk—1(X) = fk—1(X ) ' fk-I‘X) . (X'—x) h (X' — x) h ° k-j k-j Since f has property Ri-l’ there exists an x" between x' and x such that fk(x") = 0. Hence f has property R: . Lemma 2.2. Suppose f has a kth Lp derivative at a point X. Then L . k-l f.p(x) . L 1im 5%— J‘g f(x+t) — Z ——?-—,-—-— t1 dt = f P(x) . 1. k th h 1=O Proof. Let c > 0 be given. Since f has a kth Lp derivative at X, there exists a 6 > 0 such that for o < lhl < 6 1 k p . — k 1. h f. (X) 1 I) p elhl {[3] f()|f(x+t) - 12g 1i: t dtl } < (k+1): Let h be such that O < Ihl < 6. Then L ' k-l f.p(x) - L HEEL)... I2 f(x+t) _ 23 1., t1 dt - hkfkp(x)l i=0 1' Lp ]< fi (x) = l-U-<%]L2—:- f2 f(x+t) - Z T- ti dtl i: 22 L k f.p(x) . k+l): h 1 1 311—,— lf(X+t) - Z ——.. t dt h f0 i=0 1. l L k f.p(x) . l . l h 1 1 p p kl. f t — , t dt s<+>{m.rol O for each x in I (or fkp(x) 2 O for each x in I), L 4) f is upper semicontinuous on I. P k—l Then kal is increasing on I (or nondecreasing on I). Proof. As a first case suppose that f:p(x) > O for all X in I. Let a and b belong to I with a < b. We want to show that f:p(a) < f:€l(b). Since fifl is upper semicontinuous on [a,b], it :ttains a maximum on [a,b]. Let X 6 [a,b). Since fkp(x) > O, and since by lemma 2.2 L . k—l f.p(x) . L lim 13%;— jg f(x+t) - Z —i.—,——— t1 dt = ka(x) > o h—«o h 1: 1' there exists an O < h' < b—x such that L k-l fip(x) i f(x+h') - Z) '—§T——* (h') > 0. i=0 ' So L k-2 f.p(x) [f(x+h') - E —1——— i=0 gmk 1 (k-1)£ uni] i! Lp - fk—l(X) > 0. Since f has pr0perty Mg—l, there is an X' such that X < X' < x+h' < b, and L k-2 f.p(x) i [f(x+h') —Z‘. ———-—-—1i. (h') J L i=0 . P I = f (X ). h')k l k—l (k—l)! Hence LP LP fk-1(X ) - fk-1(X) > OI i.e Lp Lp fk_l(x ) > fk_l(X). L So fkgl must attain its maximum at b, which implies Lp Lp Lp that fk_l(a) < fk_l(b), i.e. fk—l ls 1ncreas1ng on I. 24 L If fkp(X)‘2 O on I, then for any 3 > 0, set ( ) - f( ) + k h )Lp — pr k' 0 96 X — X e X . T en (ge k (x) — k (X) + .e > for all x 6 I. Hence if a and b belong to I and a < b, then by the above argument, L L (g€)kEl(a) < (g€)kfl(b). L L P u_ p . fk_1(a) + k. as < fk_l(b) + k. be, or L L kal (a) < kal (b) + k: (b‘a) e: and since 6 is arbitrary, LP LP fk_1(a) g fk_l(b). So fkgl is nondecreasing on I. Definition 2.2. As in [9] we define differences Dk(x,h;f) for a function f as follows: D1(X,h;f) = f(x+h) - f(x), and for k = 2,3,..., k-l Dk(X,h;f) = Dk_l(x,2h;f) — 2 Dk_l(x,h;f). When no confusion seems likely, we will write Dk(x,h) for Dk(x,h;f). The following lemma appears in [9] and is not difficult to prove by induction on k. 25 Lemma 2.4. For each k = 1,2,..., there exist numbers ak j’ j = O,l,...,k such that for any f, k . . _ 3—1 1) Dk(x,h,f) — ak'0f(x) +jEé ak,j f(x+2 h), k 2) Z) a . = O, i=0 1"] k i(j-l) 3) Z} 2 a . = O for i = 1,2,...,k-1, j=1 1"] k i(j—l) 4) Z) 2 akj>0 for i=k,k+l,-°°. J=l k . We set iL-= Z) ZkU-‘Dak .. k j=1 '3 Lemma 2.5. Suppose f is continuous on an interval I and has a kth Lp derivative at a point X in I. Then lim -————-——— D X ,t dt = f (X . h~0 hk+1 (3 k 0 k 0 Proof. As a first case suppose f(xo) = O. L Since f is continuous fop(x0) = 0 also. Let 6 > 0 be given. Then there exists a 5 > 0 such that if 0 < lhl < 6. then 26 L . k f.p(x ) 1 h — {WT fo|fj=lak.j XO k' k Xo .Xk 27 k . = (k+1):tk{ l% (E; .Z)[ak j f(x0+23‘1t) j: 1 LP k f. (x ) ._ . '3 i=1 ' LP k k f. (x ) k L 1 0 3—1 1]_ t p + jEQIak,3 IE: it (2 t) kzx fk (XOIdtII k . l h J g (k+1).xk{j§=31 h ‘I‘OIak, f(x0+2 t) LP k f. (x ) '3 1=l ' LP f. (x ) k . l h 1 O 3-1 1 + _I—I- ., Z a (2 t) k L _ t p k k k hI < (k+1)11 2: 8L i LP k-l f. (x) k . l h 1 O 1 j 1 1 + h (0 _ i: t :43 ak’j(2 )Idt 1—1 3—1 LP f (X‘) k 1 11 k 0 k 3-1 k + W I . t Z a (2 ) k L . t ‘ I k: fk (XOIIdtI 28 L k k f p(X ) h] 1 h k 0 k = N(X), fn_l(X)‘2 fn(X). 30 Lemma 2.6. Suppose f is continuous and possesses a kth Lp derivative everywhere on an interval I, with f:p(x) > O at each point in I. Then there exists a quasi-nonincreasing sequence {fn} of continuous functions such that L - _ P 11m fn(x) — fk_l(x) for each X E I. n-wo Proof. For each positive integer n define functions fn and gn byl nk n _ n 2 fn(x) — k. 2 xk—l IO Dk_l(x,t)dt, _l_ _ . n(k+1) 2n gn(x) _ (k+l). 2 (k f0 Dk(x,t)dt. Clearly fn and gn are continuous functions of X, and by lemma 2.5 LP lim fn(x) = fk_l(x) for each x in I, n—om and L 1im gn(X) = fkp(x) for each X in I. n-ooo Let an X0 in I be fixed. There exists a positive integer N(xo) such that for n > N,gn(xo) > 0. Now let n > N. 31 l -1 _ I (n_l)k 2n _l(x0) — fn(xo) — k.2 Xk—l IO Dk_l(x0,t)dt l , nk 2n - k.2 *k—l f0 Dk_l(xo,t)dt J; _ , nk—k+l 2n _ k.2 Ik_l f0 Dk_l(xo,2t)dt EL nk 2n — k.2 *k-l f0 Dk_l(xo,t)dt l n = k:2nk—k+l)\k_l[ $02 2t) Dk—1(XO' k—l - 2 Dk_l(XO,t)dt] J; I n _ k—l . n(k+l) 2 _ EE:E:T?;:I;;— [(k+l).2 Ak f0 Dk(XO,t)dt] k = —— g my 2 (k+l)xk n >0. So fn_l(xo) > fn(x0), and the proof is complete. 32 The following lemma and its proof are due to Saks [14]. Lemma 2.7. If [fn} is a quasi—nonincreasing sequence of continuous functions with fn(x) converging to f(x) for each x in an interval I, then for each closed set P, there exists an interval (a,b) such that (a,b) n P # O, and f, restricted to P, is upper semicontinuous on (a,b) n P. Proof. For each positive integer m, let Pm = [x 6 P: fn(x) g fn_l(x) for all n > m}. Since a) [fn] is quasi—nonincreasing, P = mpl Pm’ and since each fn is continuous, each Pm is closed. According to the Baire category theorem, there is an interval (a,b) and an integer mO such that ¢ # P n (a,b) C Pmo. On Pmo {fn : n > mo} 18 a nonincreasing sequence of continuous functions con— verging to f. So f is upper semicontinuous on Pm O and hence on (a,b) n P. The next lemma and proof are due to Oliver [13]. 33 Lemma 2.8. Let f be defined on an interval [a,b]. Let P be a closed subset of [a,b] and let f, restricted to P, be upper semicontinuous on P. For each component interval (c,d) of the complement of P, let f be nondecreasing and upper semicontinuous on the closure [c.d]. Then f is upper semicontinuous on [a,b]. Proof. If x does not belong to P, then f is upper semicontinuous at x by hypothesis. If x belongs to P and is isolated on the right (or left) from P, then X is the left (or right) endpoint of a component interval of the complement of P, and so is upper semicontinuous on the right (or left). If x belongs to P and is an accumulation point from the right of P, choose 6 > 0 so that f(y) < g(x) + e, for all y in P such that X g y g x + 6. Since X is an accumulation point from the right of P, 5 may be chosen so that x + 6 belongs to P. Let X' be any point such that X < x' g x + 5. If x' is in P, f(x') < f(x) + e by the choice of 6. If X' is in the complement of P, x' belongs to a component interval, say (c',d'), of the complement of P. Since x + 6 and d' both belong to P, d' g x + 6, so 34 that f(d') < f(x) + 8. But f is nondecreasing on [c',d'], so that g(x') g g(d') < g(x) + 6. So f is upper semicontinuous on the right at each accumulation point from the right of P. Similarly f is upper semicontinuous on the left at each accumulation point from the left of P. Hence f is upper semicontinuous on [a,b]. The next lemma has been proved for the kth Peano derivative by Oliver [13]. The proof given here for the kth Lp derivative is based on his proof. Like Oliver, we shall use without specific reference in this and subsequent proofs several well known results dealing with functions that are of Baire class one and/ or have the Darboux property. ‘We list these results here without proof (see [13] for a more complete list of such properties): If f is monotone on (a,b), and has the Darboux prOperty on [a,b], then f is monotone on [a,b]. If f is monotone and has the Darboux pro- perty on an interval, then it is continuous on the interval. 35 If f belongs to Baire class one on an interval, then the points of continuity of f are everywhere dense in the interval. If f belongs to Baire class one, then f has the Darboux property if, and only if, for each number a, the sets Ea = [X : f(x) 3 a}, and Ea = [X : f(x) 2 a} have closed connected components. Lemma 2.9. Let f be continuous on an interval I and have the following properties: 1) f has property Mg—l, h 2) f has a kt Lp derivative everywhere L 3) fkp(x) > O for each x in I L (or fkp(x) 2 O for each x in I) L 4) kal has the Darboux property. L Then fkgl is increasing (or nondecreasing) and continuous on I. L Proof. Suppose first that fkp(x) > 0 every— L where. Let G = [x 6 I: f p is upper semicontinuous k—l and increasing in some neighborhood of X]. G is clearly an open set. Let P = I - G. By lemmas 2.6 and 2.7 there exists an interval (a,b) such that 36 L (a,b) n P # O, and fkgl is upper semicontinuous on L (a,b) n P. If (c,d) c (a,b) —— P, then fkf’l is upper semicontinuous and increasing on (c,d). Since L it possesses the Darboux property, kal is actually increasing on [c,d]. L Now if we apply lemma 2.8, we see that fkfl is upper semicontinuous on (a,b). Therefore, by lemma 2.3 fkfl is increasing on (a,b), and so (a,b) C G. Hence P = ¢, and ka1 must be increasing and upper semicontinuous on I. Furthermore, since L P fk-l continuous on I. L has the Darboux property, fkgl is actually L The case where fkp(x) 2 O on I can now be obtained from the above case as in lemma 2.3. The following theorem was proved for the kth Peano derivative by Oliver [13]. Theorem 2.4. Let f be continuous and possess a kth Lp derivative at each point of an interval I. Then L l) fkp has the Darboux property 2) f has properties Mé' j = O,l,...,k-l. 37 Proof. We prove this theorem by induction on k. For k = l we have from theorem 2.1 together with its following remark and corollary that L L f p(x) = fép(x) for each X in I, and that flp has the Darboux property. Goffman and Neugebauer [7] have shown that if f has an approximate derivative everywhere on an interval, then f has prOperty Mg. So the theorem is proved for the case k = 1. Assume that the theorem is true for k-l. We want to show it is true for k. Since by theorem 2.3 L L fkp is of Baire class one, in order to show that fkp has the Darboux prOperty we need only Show that the L connected components of {x : fkp(x) 2 a} and L [X : fkp(x) g a} are closed for every number a. By considering the function g(x) = f(X) — f? xk, we may reduce this to showing that the connected components L L of {X : fkp(x) 2 O} and IX : fkp(x)_g O} are closed. L So suppose fkp(x) 2 O for all X in the interior L of an interval J. We must show that fkp(x) 2 O on .. L the closure J of J. By lemma 2.9, if fkp(x)_2 O L P k-l the interior of J. Since by the inductive hypothesis L L fkgl has the Darboux prOperty, fkgl is increasing on on the interior of J, then f is increasing on 38 J. Let c be the left endpoint of 3, and suppose L that fkp(c) < 0. Then by lemma 2.2 there is an h > 0 such that c + h belongs to J and L k-l fip(c) i f(c+h) - Z ———.-T——h < O. . 1. i=0 L k—2 fip(c) 1 f(c+h) - Z T—h L i=0 ' _ p So hk-l fk_1(c) < O. (k-l)! By the inductive hypothesis f has property Mg‘l, and so there exists an X' with c < x' < c + h, such that L k-2 fip(c) i f(c+h) - Z)-—rT——— h L i=0 1. p ' k—l = fk-l (X )o L. (k—l): Hence LP LP fk_l(x ) - fk_l(c) < o. L but this contradicts the fact that fkgl is increasing L on J. So we must have kalIC)-2 O. 39 If d is the right endpoint of J, we can similarly show that f:p(d) 2 0. So {X : f:p(x) 2 O} has closed connected components. Similarly we can show that the connected components of [X : f:p(X) 3.0) are closed, and hence pr has the k Darboux property. By lemma 2.1 we know that in order to show that f has properties M?’ j = O,l,...,k—l, it suffices to show that f has prOperty Rt-l’ So let x and x+h be such that LP LP fk_1(x+h) — fk_1(x) = O. L 0 Let J = [x,x+h]. If fkp is identically zero on J, L the interior of J, we are done. fkp cannot be 0 positive everywhere in J because that would imply L that fkgl is increasing on 3 as seen in lemma 2.9, L and since fkgl has the Darboux property, this would LP LP LP imply that fk_l(x+h) > fk_l(x). Similarly fk L . . 0 cannot be negative everywhere in J. So fkp must be 0 either identically zero on J, or assume both positive and negative values. But the latter situation implies L 0 that there is a point x' in J' where fkp(x') = O L since fkp has the Darboux property. 40 Lemma 2.10. Let f be measurable and have L a kth Lp derivative at O with fip(0) = O, L i = O,l,...,k-1, fkp(0) = A. Then lim -£LEL =-i¥ ap h k. h~0 Proof. we will show here that lu—J H a DJ “U H) ”55 II II» The proof that I—J P a DJ .U I‘h W2: II TI» is analogous. Let c > 0 be given, and let k E6 = It > 0: [f(t) — A'ETI > e tk}. We must show that E€ has 0 as a point of dispersion from the right. For any positive h let Eh = E6 0 [O.h]. Then we have r I1 - 54-9 }E’ l _ .E_ P }I> IEIoIfIt) AMI dt ZIhIEhIf(t) AMI dt 1 — f 6p tkp dt}p Eh h/ r—*- U'H 41 { 6p . IEhI }p kp+l h ' and as h d 0+, this must be o(hk), i.e. l e . IEhI k+£—D _ l/p 1”“ (T ‘0 (kp+1) h~0+ Hence lim 0, and so 13.1- .- h-0+ 1im Elhl =-i¥ . ap hk k. h-0+ Besides the use that is made of the above lemma in this chapter, it has some other interesting consequences, which will be explored in Chapter III. Lemma 2.11. Let 0 be a point of density of a set E. Then there are sequences {an} C E and [bn} c E such that 1) lim a = lim b = 0, n n ndm new 2) an+1 \ an and bn+l > bn for each n, 42 b and 3) lim —§ii = lim “+1 11"00 n n"’00 I1 Proof. Here we construct the sequence {an}. The sequence‘ {hm} can be constructed in a similar fashion. First we define a sequence Iém}. Choose 61 > 0, so that if 0 < h g 61' then IEO.h In EL > g h 3 In general, hav1ng chosen 6m. choose 6m+l so that m . O < 5m+l < m+2 6m, and so that 1f 0 < h 3-5m+1’ then h m+2 Now pick a E [ 2 6 6 I O E Such an a l 3 1’ 14 ° 1 exists since 2 [[0,61] 0 El > g 51 2 . 3 ' . Pick a2 6 [ 4 a1, 4 a1] 0 E. Such an a eXISts 2 since 3 ] < 2 > Z _ Z IIO'4 a1 n El > 4 a1 3 — 4 a1 . . 2 3 Keep this process up, ch0031ng an+1 E 4 an’Z an n E, until we come to the smallest integer kl such that akl g 62. Notice that 43 (MN \/ film 62 (by the choice of kl) > 63 (by the choice of 6 ). 3 . 3 4 Pick akl+l E [ 5 akl. § akl] O E. Such a number exists since 4 1 4 3 3 I[O.- a 0 El > < - a > — = - a 5 k1 5 k1 4 5 k1 The process is as follows: Having chosen a choose a E I 3 a g a ] O E n’ n+1 5 n’5 n ’ and keep this up until arriving at the smallest integer k such that a Again we easily see 2 k2 g_53. that ak2 > 64. In general, having found 6 < a g 6 m+2 k m+l by m the above process we choose m+2 m+3 ak +1 6 I m+4 ak ' m+4 ak I n E' m m m which we can do since F m+3 m+3 m+2 _ m+2 IIO ’ m+4 akm I O El> >< m+4) akm < m+3) — m+4 akm 44 ' m+2 m+3 Keep ch0051ng an+1 E [ m+4 an, m+4 an] 0 E until we come to the smallest integer km+l such that ak g_5m+2, and so on. m+l In this way we arrive at a sequence {an}. We clearly have for each n, and an+1 < an lim an = 0. Now fix an n, and choose the smallest n—ooo of the integers in the sequence {km};=l such that an g_ak . If ak g an+l’ then m m+l a m+2 < n+1 m+3 m+4 —- a —- m+4 ’ n and on the other hand, if akm+l > an+l' then m+3 _ §m+l)+2 < an+1 < {m+1)+3 _ m+4 m+5 _ (m+l)+4 a ‘—- (m+l)+4 — m+5 Now as n ~ w, m ~ m, and as m ~ m LIE—é”, and fig; ~ l. Hence 1im —§il- = l. 1'1'":> n 45 Lemma 2.12. Let f be measurable and have a L kth Lp derivative at O with fip(0) = O, L i = O,l,...,k-l, fkp(0) = A. If f is monotone on a right neighborhood of O and monotone on a left neighborhood of 0, then A = fk(0), the kth Peano derivative of f at 0. Proof. From lemma 2.10 we have lim_ ELEL =-fi¥ , h~0 h ° hEE where E is a set of density 1 at 0. As a first case suppose that f is monotone nondecreasing on a right neighborhood of 0. Let [an} C E 'be a sequence as described in lemma 2.ll. Let h be in the given right neighborhood of 0. Choose n so that a n+1 g h i an. Then f(an+1) f(h) f(an) ()k S hk S( )k’ an an+1 so < an+1)k . f(an-I-l) < fgh) < f(an) < an >k a k —- k ‘— k n (an+l) h (an) n+1 46 + . Now as h 4 O , n fl m and the two outSide members of the above inequality tend to 1‘ - 1 --%¥ . TE:— Hence JJJI £3%;-= 1% . +h 1" th All other cases are treated analogously, and we conclude that lim.£ihl - _ .3: h—o hk 1" I fk(0) = A. We are now in a position to prove a theorem . th . . th . which relates k Lp derivatives to k ordinary derivatives. This result will then be used to establish the Denjoy property and the Zahorski property for kth Lp derivatives of continuous functions. The theorem is patterned after a result by Oliver [13] in which he shows that if a kth Peano derivative fk is bounded either above or below on an interval, then f (k) fk(x) = (x), the ordinary kth derivative of f at x, for each x in the interval. 47 Theorem 2.5. Let f be continuous and possess a kth Lp derivative at each point of an interval I. L If fkp is bounded either above or below, then L fkp(x) = f(k)(x), the ordinary kth derivative of f at x, for each x in I. Proof. Assume that fkp is bounded below on L I; say fkp(x) > M for all x in I. (The proof for L the case where fkp is bounded above is analogous). Let xO be any point in I. we will show that the kth Peano derivative of f at xO exists and that L p _ . . . fk (x0) — fk(xo). Hav1ng done this for all X0 in I, we will use Oliver's result mentioned above to conclude ) exists for each x in I and that f(k)(x O O (k) L P _ fk (X0) — f (x0) . For a fixed X0 in I, let L 2 g(x) = f(x+x0) - (f(xo) + xflp(xo) + gT-f2(xo)+... k—l L k x f p (x ) + ET M) . + (k—l)! k-l Then L L 9(0) = glp(0) =...= gk§1(0) = o, 48 and if we let J represent the interval obtained by translating I in such a way that x0 is moved to the origin, then L L 9kp(X) = fkp(X) - M > O for all x E J. L L In particular set A = fkp(xo) — M = gkp(0). L p _ . In order to show that fk (x0) — fk(x0), 1t L suffices to show that gkp(0) = gk(0), i.e. it suffices to show that lim 113-)— = “A.“ . h~o h k' L Since gkp is positive on J, lemma 2.9 L together with theorem 2.4 imply that gkgl is non— decreasing on J. Suppose J = [a,b], and let L O _ . p yk_l — sup {y E J. 9k-1(Y) g 0}. o _ o 1 _ 0 Let Jk__l — [a,yk_l], Jk_l — [yk_l.b]. L p 1 Now gk_l(x) 2.0 on Jk-l’ and so by the same L argument used directly above, gk€2 is nondecreasing on J1 and similarl Lp is nonincreasin on J0 k-l' Y gk—2 g k—l' So let 49 L l yk_2 = sup [y 6 Jk-l = gk§2.g 0}. and O _ O Jk-Z "' [al Yk_2 J t l _ O O Jk-Z ‘ l:Yk-z' Yk-lJ 2 _ O l Jk_2 _ [Yk-l' Yk_2] 3 _ l Jk—Z [Yk—z' b] Then /' d . 0 non ecrea51ng on Jk—2' . . l nonincrea81ng on Jk-2 gk_3 IS < . 2 nondecreaSing on Jk—2 nonincreasin on J3 K g k—2 Continuing this process for a total of k—l times, we arrive at Zk-l intervals Ji, k—l k-l . 2 ‘1 i i = O,l,...,2 — 1, With J = L) J1. On each i=0 of these intervals g is monotone and continuous. The point 0 can belong to at most two of these intervals (with the possible exception of degenerate intervals). Hence g is monotone on a right neighborhood of O and monotone on a left neighborhood Of 0- 30 by 50 lemma 2.12 1im 9121. = i? th h ° which is what we wanted to show. In [18] Weil proved that a function g of Baire class one has the Denjoy property on an interval I if, for every subinterval J of I on which g is bounded either above or below, g restricted to J has the Denjoy property. Using this result along with the fact that an ordinary kth derivative has the Denjoy property, we have the following corollary to the last theorem. Corollary 2.5. If f is a continuous function having a kth Lp derivative at each point of an interval I, then fkp has the Denjoy prOperty on I. Theorem 2.6. Let f be continuous and possess a kth Lp derivative at each point of an interval I. L Then fkp has the Zahorski prOperty. Proof. Following along the lines of the proof in [18] we first consider a special case where Lp Lp Lp f0 (0) = fl (0) =...= fk (0) = o. 51 and assume that {In = [an, bnj} is a sequence of closed intervals, with positive endpoints, converging to O in such a way that, for each n, x in In implies fkp(x) 2_c, where c is a fixed positive number. Let n be a positive integer. Then the same inductive proof as that given in [18] will show that, for each positive integer j with 1 g_j g_k, there is a partition . . = b Jlm(]) n a = t. < ... < t of In such that m(j) ;_23 and, for each i = 1,...,m(j), one of the following holds for every t. .]: x in [tj,i-l' 3’1 L L ' . . p _ P ii — 3 L P and fk-j(tj,i-1)'2 O - LP LP c j 2(3): fk‘j (X) - fk-j (tj,i) g —(T) (tj,i-X) I L P and fk-j(tj,i) g_o L L ' ' . P _ P _ J; _ 3 L P and fk-j(tj,i-l)-i O 52 L L . 4(j) : fk§j(x) - fk€j(tj’i)‘2 C§%)(tj’i—X)J. _2 o. L P and fk-j(tj,i) In particular if we examine the situation when L j = k and recall that f0p = f, then we can say that, for each n, there is a partition, an = t0 < ... < tm = bn of In such that m.g 2k, and for each i = 1,...,m, one of the following holds for every x in [ti_1. ti]: (1) f(x) 2 (f?) (x-tH)k (2) f(x) 3 455—) (ti-wk c k (3) f(x) 3-‘(ET)(X'ti—1) (4) f(x) _>_ (if?) (ti-x)k. In case (1) or (3) holds on [ti_l,ti] we have t. a t k a 1 p p _g_ ’ i P P {Iti_llf(x)l dx} '2 (k1) ifti—1(X—ti_l) dx} kp+l = ( C ) (t1 t1 1) p ; 53 and in case (2) or (4) holds on [ti—l'ti] l l t. — t. _ {1; Iflpdx}P .>. a) {Li O (or [(x : f'(x) exists and lies in (a,b)}l > O). In his proof of these results Weil used only those properties of fk (or fép) which we have also shown L P k . proof we state the following theorem. to be possessed by f So without repeating Weil's Theorem 2.7. If f is a continuous function possessing a kth Lp derivative at each point of an interval I, and if, for some interval (a,b), L (fkp)-l((a,b)) is not empty, then (k) ]{x : f (x) exists and lies in (a,b)}l > O. 56 The following theorem is already known to hold for kth Peano derivatives and approximate derivatives (see [5] or [13], and [7] or [16]). Theorem 2.8. Let f be continuous and possess a kth Lp derivative at each point of an interval I. Then there is an everywhere dense, open set G C I at each point of which the ordinary kth derivative exists. Proof. Let U = [x e I: for every right neighborhood of x or for every left neighborhood of x, f:p is unbounded both above and below]. Let G be the complement of U. Then G = {x E I: there is a right neighborhood of x and a left neighborhood of L x in each of which f p is bounded above or below}. k (k) G is open in I, and by theorem 2.5 f (x) exists at each point x 6 G. Since f:p is in Biire class one, the set of points of continuity of fkp is everywhere dense. But each point in U is a point of discontinuity of fip; and so U can contain no interval. Hence G is everywhere dense. Next we show that in a sense this result is the best possible. 57 Theorem 2.9. If an interval I, an everywhere dense, open set G C I, and an integer k.2 l are given, there is a continuous function f which has a kth Lp derivative at every point of I and for which the set F of points where the ordinary kth derivative exists contains G, and furthermore, if S is an open set with G C S C F, then G = S. Proof. For k.2 2 Oliver constructs a function f which has a kth Peano derivative at each point of I, and hence a kth Lp derivative at each point of I, and which satisfies the theorem. Here we will treat the remaining case, namely k = 1. Denote by P the closed, nowhere dense set I — G. Let {(an,bn)}n=l denote the sequence of intervals contiguous to P. Fix a particular integer bn-an 3(bn-an) n. Let c = a + , and d = a +-——-———~—. n n 4 n n 4 Choose an integer mn‘Z 2 so that .______ _ 3P (2.9.1) m p < (cn an) , and choose an integer jn so that n n n n 2 2 58 and 1 1 b-——.--——'——-.—->d. n J m P] n 2n 2nn We shall first define f on (an,cn) as follows: For every integer jig jn, let I . = n,j [ l l l J a + —- a a + _—-' + . , n 23 n 23 2mnp] . On each In,j we define a function g so that l) g has an ordinary derivative at each point of I ., n,] l __ ___ l _ 2) g(an + j) g(an + 3 m p3) o, 2 2 n 2 l l 3) g'(a +-—T) - g'(a +-—- + .) - O, + n 23 n 23 zmnpj 4) max g(x) = 1. x61 . n,j For x E (an,cn) - U I., we set g(x) = 0. Then j=jn for any x E (an,cn), we set f(x) = (x—an)g(x). We set f(x) = O for x E [cn,dn]. Next we will similarly define f on (dn,bn). For every integer j.2 jn. 59 l l 1‘ let J 0 = [b - —T - O ’ b _ _1‘] o Hp] n 23 Zmnpj n 2] we define a function g so that 1) g has an ordinary derivative at each point of J ., n,j l _ ;L._.__£._ _ 2) g(bn j) — g(bn J m p3) — O, 2 2 n 2 I __:_L_. _ ' _i.._—_._—_l — 3) 9 (bn j) — 9+(bn j mp3) — o 2 2 n 2 4) max g(x) = 1. XEJ . 11,] For x E (dn,bn) — U Jj’ we set g(x) = 0. Then j=j n for x E (dn,bn), we set f(x) = (bn-X)g(x). Now f(x) has been defined for each x E (an,bn). We do this for each n and set f(x) = O for x 6 P to arrive at a function f defined on all of I. We first show that f is continuous on I. f is clearly continuous on each interval (an,bn). Now suppose an is a left endpoint of an interval (an,bn) contiguous to P. If h > O and sufficiently small, then [f(an+h) - f(an)l = f(an+h), 60 I ., then and if a + h 6 (a ,c ) - n n n n,j II C8 3 3n f(a +h) = 0; whereas if a + h E I ., for some n n n j j‘g jn' then f(an+h) = h§;(an+h) g_h, by condition 4) on 9. So f is continuous from the right at an. and similarly f is continuous from the left at each bn' Next suppose that x is an accumulation point of P from the right, and let h > 0. If x + h E P, [f(x+h) — f(x)| = |o - 0| = o; and if x + h E G, then x + h e (an,bn) for some n, and so x + h = an + h' for some 0 < h' < (bn—an); SO [f(x+h) — f(x)l = f(x+h) = f(an+h') g_h' < (bn-an). As h ~ 0+, (bn-an) ~ 0 since x is an accumulation point of P. So f is continuous from the right at x. Similarly f is continuous from the left at each point x which is an accumulation point of P from the left. Hence f is continuous on I. By the way f was defined, it clearly has an ordinary first derivative at each point in G. However, if S is another Open set such that G g S, then S must contain at least one of the endpoints an or bn 61 for some n. We now show that f is not differentiable in the ordinary sense at any endpoint an or bn. Fix an an. Choose a sequence {bk} such that hk d O as I . for each k. k q m, and an + hk E (an,cn) — n,j j ll C8 3n Then f(an+hk) - f(an) lim = lim '51 = O. k~m hk kem hk Now according to condition 4) which we placed on the function g defined on In j' we may find a number hj > 0 such that g(an+hj) = 1. Doing this for each j.2 jn’ we have a sequence {hj] with hj ~ 0 as f(a +h.) - f(a ) f(a + h.) 1im n %» n = lim n j“00 j jtm j = lim :')-'co j = lim 1 jtm So f does not have an ordinary derivative at an. Similary f does not have an ordinary derivative at any bn’ 62 In order to complete the proof, we must now show that f has a first Lp derivative at each point in I. As noted above, f has an ordinary derivative at each point x E G, and so it has a first Lp derivative at each point x E G. The following calculation shows that at each left end— point an of an interval contiguous to P, the right- L hand first Lp derivative flp(an) = 0. Let a n be such an endpoint, and let 0 < h <-—;— +'——J;T— . 3n mnPJn 2 2 . 1 1 Choose an integer J so that -— < h < Then 2J — 2J-l 1 (2.9.2) %{% jg [f(an+t) [pdt}p l 2J{2J fon- 1 it p p (f(an+t)) at} I/\ |/\ M q PM M q M L—) H ruf- 0.: (1" W WIH |/\ A) FL $ P. 5: L .:: WIH - (2 )P 2" mnp an 2 —1 2 22p+1 1' 2J _<_( 2p )p 73 (since In 22) 2 -1 2 < K11 So we have 1 {%fglf(an+t)lpdt}p = o(h), as h ~ 0+. Similarly it can be shown that at each right endpoint bn of an interval contiguous to P, L the left—hand first Lp derivative flp(bn) = 0. Next suppose that x E P is an accumulation point of P from the right. Let h > O, and let 34h) = [n : x < an, and bn < x + h}. There are two possible situations to consider here, namely x + h E P and x + h E G. If x + h E P, then { 1 1 (2.9.3) 1&1; jg|f(x+t)|pdt}P f:+h(f(t))pdt}p E 5H4 DflH SflH b . 1 Z f“(f(t))pdt}p n€.&(h) an P-Afi fine I/\ I/\ S 3‘! )—I F—A“ Una 311—! r—"-\ Una D‘H—J r—*-\ :I'IN ’J‘l )—- /-*-\ Dflw :3'I H A D‘IN :S‘H-I l-‘Lfi D‘IN 21/p 64 . l nEJM) j=jnUImj(f(mpdt+ fJn’j(f(t))pdt]}I—D o. l 3.291.) (.11): oo 1 .1. £201) 321 2,1,11ij 1 l .2...) ‘17:} l nEMh) (cn—an)3p}§ (by (2.9.1)) 1 . h3p}p h2 - l/p Zl/ph. On the other hand, if x + h E G, then there is an integer N(h) = N such that x + h E (aN,bN). So 65 1 1 f3lf -—l:-— + -——]-'—-.— . Then 2 2 -1-{-1- (x+h(f(t))pdt}% <_l_[l_ aN+h'(f(t))pdt‘%’ h h aN —h' ..h' faN ] l j —l a + N rill—{El— ff 2 (f(t))pdt N l a +h' — 4.517(1) 1 (f(t))Pdt}P aN+ jN'l 2 l j —l l , a + N - efilri—é: faN 2 (f(t))pdt + o}P N 3 K-—;— (as in calculation (2.9.2)) 3N 2 _<_Kh' gKh. 67 Now if we let h~o+, with x+h es, then l %I% Ig|f(x+t)|p}pdtg[21/p +max(21/p,K)] h -O So the right—hand first Lp derivative of f at x is 0. Similarly it can be shown that if x is an accumulation point of P from the left, then the left-hand first Lp derivative of f at x is O. L Combining these results we have that flp(x) exists everywhere on I, and the theorem is proved. Chapter III. A Property of kth L and kth Approximate Peano Derivatives of Measurable Functions In this chapter we consider some further consequences of lemma 2.10. If we combine the notions of a kth Peano derivative and an approximate deriva— tive we have the following concept. Definition 3.1. We will say that a function f has a kth approximate Peano derivative at a point x k = 1,2,..., if there exist numbers f:p(xo). O, aP ap f2 (x0),...,fk (x0) such that k - .1. _ _ap __h_ap} 11map hk {f(xo+h) f(xo) hfl (x0) ... k! fk (x0) hfio Theorem 3.1. Let f be a measurable function possessing a kth Lp derivative at a point x0. Then L fkp(x0) is a kth approximate Peano derivative at (x ); specifically, O L ap f:p(xo) = (fop>k (X0)' 68 69 Proof. Set Lp Lp Lp Xk Lp , g(x) = fO (x—xo) - (f0 (x0) + xfl (x0) +...+ k? fk (xo)) . In order to prove the theorem, it suffices to show that L ap gkp(0) = gk (0)- But since L L g(o) = 919(0) =...= gkp<0) = o, it will suffice to show that lim HiEihl. = 0, ap h th and this follows immediately from lemma 2.10. We want to show that the kth approximate Peano derivative of a measurable function is of Baire class one if it exists on an interval, thus generalizing theorem 2.3. We need a couple of preliminary results. Definition 3.2. As in [9] we define differences Ak(x,h;f) for a function f, k = 1,2,... by k . Ak(x,h;f) = Z (—1)k‘3(]j?)f(x+jh — .1. kh). j=o 2 70 The following lemma is not difficult to prove using induction on k (see [1] or [9]). Lemma 3.1. Let 1 be any real number. Then Lemma 3.2. If f has a kth approximate Peano derivative at a point x, then for any fixed number 1 there is a set F().X) of density 1 at 0 such that A (X+1h,h:f) 1im k k = f:p(x). h~O h h€F(l,x) Proof. Let E be a set of density 1 at 0 such that aP k fi (X) i k f(x+h) - Z Th =O(h) i=0 as h e O, h E E. Let 1 be given, and define F(),x) = {h : )h + jh - ékh 6 E for each j = O,l,...,k}. Then F(),x) is of density one at O. Let c > 0 be given. There exists a 6 > 0 such that if h E E and |h| < 5, then 71 span 1 . [f(x+h) — Z) ——;T——'h | < |h i=0 ° k2 and furthermore if I is an interval containing 0 and III < 6. then Consequently we can find a O < 5'.g 6, such that if h e F(1,x) and [h] < 5', then 1 k f?p(x) 1 . . . l . l 1 (3.2.1) |f(x+)h+jh— —kh) — Z)-—-eT——(1+3 — —k) h | 2 ._ 1. 2 1—0 6 . l k k < k-l [6+3 ' fikl 'hl k2 for each j = O,l,...,k, and if I is an interval containing 0 with III < 6', then 1F11.x) n Il > 1 _ TIT 6* If we now consider the right hand side of (3.2.1L'we have raw 7: e . l k k e h _ |1+3 - -k| hl .: _ kzk 1 2 I k2k elhlk g; k n < k (n)|1l 2 n=0 72 elhlkmk if l1|>1 elhlk. if mgl. Let us first suppose that [1| > 1 and that h e F(),x) with lhl < 6'. Then h k k . - . l - f:p(x) = (it Z)(-l)k j(1?)f(x+)(h+jh — ikh) h j=O 3 ap — fk (x)| k . — . 1 g_ 4%- Z)(—1)k j(1?)[f(x+>(h+jh — -kh) . j 2 h j=O ap k f. (x) . 1 l 1 1 l: k . k £39m) . . - . 1 + _Jk Z (-l)k 3(3) Z} —L-1".——'().+j — 5k)lhl h j=o 3 1—0 ' .. ap fk (X) k k 3, Z} (.) e Illk j=o 3 ap k f. (x) k _. + '4? 1., hl Z)(—1)k j(Hj 1k)l . 1. . 2 h 1: j: _ aP fk (x) 73 g_2k|1|ke + If:p(x) — f:p(x)| (by lemma 3.1) = Zklxlke. In considering the case where '1] g_1 we similarly Obtain A (x+1h.h;f) k k k — fipm) < 2e, h if h e F(1.x) and |h| < 5'. In either case we have Ak (X+)\hl h; f) 1im k = f:p(x) h~O h hEF().X) Theorem 3.2. If f is measurable and has a kth approximate Peano derivative at each point of an interval I, then fip belongs to Baire class one. Proof. For each positive integer n and each integer p set and 74 I = [_ _31_ __1__] n 211+]. 211+]. For each point 1% E I, define 2 . Ak(X,h;f) n,p' hk > a} | > g(xn pl} 1 > 5 |1n1} . For each fixed n extend fn linearly to arrive at a continuous function fn on I. Let XO 6 I. We want to show that fn(xo) e f:p(xo). From lemma 3.2 we know that there is a set F(O,xo) of density 1 at 0 such that lim Ak(xo'h’f) = fap(X ) hfio hk k 0 hEF(O,XO) Set G = [%kh : h E F(O,XO)]. G clearly is of density 1 at 0. Let e > 0 be given. We shall find it convenient later to suppose that e is so small that 3 1 — (k+1)e > 4 and QL l - 262k — e > g . 75 Then there exists a 6 > 0 such that if we let k fip k E6 = {h: [f(x+h) - Z) hik|< llhl}, i=0 _ 1 A;k(x0,h f) ap k as _ { 5kh. |———;k—-———— fk (xo)l < 2 e}, l . l . ‘ and Fe_ I §J 1 - e, and ——§TfT—— > 1 — e for any interval I containing 0 with II) < 6. Now choose a positive integer N so large that -;E < 2. Let n > N, and find the unique 2 integer p so that 11 Bil 2n 1 - e l . k If we set B = fl Bj, then 77 1 . IB n 1 2n+1 ’ 2n+11‘ 2n—l > 1 - (k+1)e. then Furthermore, if 111 G B O [- 2n3+l . 2n+l]' . 1 x0 + 111 6 In,p’ and 111 + 311 — alch E E6 for each j = O,l,...,k. So performing calculations as in lemma 3.2 we have the following: 1) If lxl : 1. A (X + 1h, h; f) k 0 _ ap k I hk fk (x0)‘ < 2 e 2) If [1| > 1, A (x + 1h, h; f) k 0 _ ap k k I hk fk (x0)| < 2 (1| = (3k)k e1/2 If we set. C = (3k)k, then regardless of the absolute value of X we have A (x + 1h. h; f) k 0 -f:p(xo)l 1 n_];+l)€ . 2 So - A (X.h7f) 1 |{ §1 i and 1 — 2e2k — e > g . So we have the following: 79 A (x,h;f) l .I ._k______ |{2kh€In.l[x€I .l .619 n,p hk fk (x0)| 1/2 1 } 1 2 lIn’pl |>2 IInl- This then implies that f:p(x0) — cal/2 < fn< :%> < f:p(xo) + cal/2. In a similar manner we can find an N' such that for n > N' and p such that JL < x g_Eil 2n 0 2n we have f:p(xo) - c 81/2 << fn < B:%-) < f:p(x0) + c el/Z. We then let N0 = max(N,N') and have that for n> NO, lfn(Xo) - f:p(xo)| < eel/2. a 3P . Hence fn(xo) fk (x0), and the theorem is proved. Corollary 3.1. If f is measurable and has a kth Lp derivative at each point of an interval I, L then fkp belongs to Baire class one. B IBLIOGRAPHY BIBLIOGRAPHY J.M. Ash, "Generalizations of the Riemann derivative", Trans. Amer. Math. Soc., 126 (1967), pp. 181—199. A.P. Calderon and A. Zygmund, "Local properties of solutions of elliptic partial differential equations", Studia Math., 20 (1961), pp. 171—225. J.A. Clarkson, "A property of derivatives", Bull. Amer. Math. Soc., 53 (1947), pp. 124—125. A. Denjoy, [Memoire sur la totalisation des nombres derivés nonsommables", Ann. Sci. Ecole Norm., 33 (1916), pp. 127—222. , "Sur l'integration des/coefficients differentiels d'ordre superieur", Fund. Math., 25 (1935), pp. 320—328. , "Sur une propriété des fonctions derivées", L'Enseignement Mathematigue, 18 (1916), C. Goffman and C.J. Neugebauer, "On approximate derivatives”, Proc. Amer. Math. Soc., 11 (1960), pp. 962—966. A. 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