ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Séria I: PRACE MATEMATYCZNE XXIII (1983)
M. Sin h a r o y (Howrah)
Remarks on Darboux and Mean Value properties of approximate derivatives
Abstract. This paper concerns real valued functions of a real variable. It is known that an approximately continuous function is of Baire type 1 and is Darboux ; also, a finite approximate derivative is of Baire type 1, Darboux and has the Mean Value property. In this paper, these properties are proven for functions satisfying similar but more general conditions, mostly for unilateral limits and derivatives of certain generalized kinds.
1. Introduction. In this paper, we are interested in real valued functions of a real variable. It has long been known that a finite bilateral approximate derivative satisfies Rolle’s theorem (and, hence, also Mean Value theorem) (Khintchine [5]), and that it is of Baire type 1 and is Darboux (that is, has the Intermediate Value property) (Khintchine [5], Tolstoff [7]). Goffman and Neugebauer [3] have given elegant proofs of these results, making use of the fact that the limit of a convergent interval function is of Baire type 1 (Gleyzal [2], section 3, p. 226). Also, Denjoy [1], the initiator of the notions of approximate continuity and derivative, himself proved that an approximately continuous function is of Baire type 1 and is Darboux. We ask if these properties are also true for unilateral approximate limits and derivatives. The purpose of this paper is to show that these properties are indeed possessed by functions satisfying similar but much more general conditions.
Recently, Sarkhel and De [6] have introduced the notions of proximal limit, continuity and derivative, which are substantially more general than those in the approximate sense, and they have found important applications of these notions in the theory of generalized integrals ; also, they have shown that a proximally continuous function is Darboux. The theorems proved" in this paper (Section 3) even apply to these concepts. Indeed, Theorem 1 implies that a finite proximal limit (see Section 2) is Darboux, Theorem 2 (resp. Theorem 3) implies that a finite unilateral proximal limit (resp. deriva
tive) is of Baire type 1, Theorem 4 implies that a finite unilateral proximal derivative of a proximally continuous function is Darboux under certain simple conditions (which are also necessary), and Theorem 5 implies that a finite unilateral proximal derivative of a proximally continuous function has the Mean Value property if it has the Darboux property.
2. Preliminaries. Throughout, we shall use the following notations: R
= the real line ; \E\ = the и-dimensional outer Lebesgue measure of a set E a Rn ; d~ (E , x) = the left upper metric density» (*), and etc., of a set E c= R at a point x; and I = (a , b) a fixed bounded open interval in R.
Besides the ordinary and approximate extreme limits, we shall deal with certain generalizations of them. It will be convenient to have a systematic scale of notations for all of them.
With every x e R , we first associate the following families of ‘ignorable’
sets :
S0(x — ) = {E cz R\ E n (y, x) — 0 for some y < x}, (x — ) = {E <= R\ d~ (E , x) = 0},
S3(x — ) = {E czR\ (E , x) = 0 and d~ (E , x) < 1}, S4(x — ) = { EczR\ d~ (E, x) < 1},
$2(x ~ ) = { E e S 3(x — )\ E u F e S 3(x —) for all F e S 3{x — )}.
The families S,(x + ) are defined analogously; also, we define S,(x) = S,(x —) n 5,(x + ), i = 0, 1,2 , 3, 4.
Now given a function/: / ->R, a point x e l and a family S of subsets of R associated with the point x, we define
5 -lim inf/ = sup {reR\ {yel\ f{y ) < r }e S ] and
S — lim sup / = inf {r e R\ {yel\ f (y) > r} e 5}.
Thus, the ordinary and approximate extreme unilateral limits o f/ at x are the above limits when S e {S 0(x — ), S0(x + )} and S6{S'1(x — ), S1(x + )}, respectively. The extreme unilateral proximal limits of / at x, as defined in [6], are precisely the above limits when S e {S2(x — ), S2 (* + )}, and the extreme unilateral proximal derivatives are the extreme unilateral proximal limits of the usual incrementary ratio. The extreme bilateral limits are defined by taking S — <Sf (x).
• For i = 0, 1,2 , 3, we have S/x — ) — lim in f/ ^ Si + 1 (x — ) — lim inf/
and St + i (x — ) — lim sup/ < 5 ,(x — ) — lim sup/ (because St{x — )<=Si + l (x — )), and 5, (x — ) — lim inf / ^ S/x —) —lim sup / (because £ e S f(x — ) implies that I\E ф St (x — )). However, we may have S4 (x — ) — lim sup / < S4 (x — ) —
— lim inf/. These, and similar observations for the right-hand extreme limits, will make clear the extents of the theorems we shall prove.
( l ) d (E , x) = lim sup (\E r\(x — h, x)\/h).
h-o +
In the proofs of our theorems, we shall make use of the result of 'Gleyzal and of the following lemmas, of which the first one is a variant form of Lemma 2.3 in [6]. Also, we shall use certain well-known results on density of sets (see, e.g., [4]).
Lemma 1. Let A cz / be such that (i) А eS4(x — ) for every x e B = I\A and (ii) B e S 4{x + ) for every x eA . Then inf A ^ sup В .
Proof. Let us put
A0 = {xeA j d + (B, x) = 0} and B0 = {xeB\ d~ (A, x) = 0}.
Since an arbitrary set has density 0 or 1 a.e. on R, (i) and (ii) imply that
|/1\A0| = \B\B0\ = 0. Hence, if x0eA 0 and y0e B 0, where x0 < y0, then there exisfi for every e > 0, points Xj e A 0 h ( x 0, y0), yx e B Qn (x 0, y0) such that (1) \An(t, y0)\ < e\(t, y0)\ for all te (x u y0),
(2) IB n ( x 0, 01 < e|(x0, 01 for all te (x 0, y j .
Now suppose, to get a contradiction, that inf A < sup B. Then, by (i) and (ii), there exist points a0e A 0 and b0e B 0 such that a0 < b0. There
fore, applying (1) and (2) alternately, we can select successively points a x, bl5 a2, b2, ... such that, for each positive integer n, ane A 0, bne B 0, an_!
< a„ < bn < bn -1 4
(3) IA n (t, b„_ 0t < (l/и) l(f, b„_ 0i for all t e(a„, bn_ 0 and
(4) \Bn(a„, t)\ <{l/n)\{a„, t)\ for all te {a h, bn).
Put c = lim a„. Then both (3) and (4) hold with t = c. Consequently d + (A, c)
= 0 = (В, c), contrary to (i) and (ii), and this proves the lemma.
Rem ark. Incidentally, Lemma 1 implies a strong connectivity property of intervals in R, namely that an interval cannot be expressed as the union of two disjoint non-empty sets, either of which has bilateral upper density less than 1 at each point of the other.
Lemma 2. Let f : I - * R satisfy, for all x e l , the conditions (i) {yel\ f{y )
< / (x) }eS4(x + ) and (ii) S4(x — ) — lim sup/ ^ f{x ). Then f is non-de
creasing.
Proof. Fix J = \_a0, b0] с: I, and put
A = {xeJ\ / (x )^ / (ao )j, B = J\A.
Then, by (ii), T e S 4(x — ) for every x e B , ' and by (i), £ e S 4(x + ) for every
x eA . Since a0eA , it follows by Lemma 1 that В = 0 . Hencef ( b 0) ^ / K )>
proving the lemma.
Lemma 3. Let f : I -> R satisfy, for all x e l , the conditions
(i) {yel\ f { y ) > /(x )} eS4(x —) and (ii) S4(x + ) — lim inf/ ^ /(x ). Then f is non-decreasing.
Proof. Fix J = [a0, b0] c= /, and apply Lemma 1 to A = {xeJ\
f(x ) >i f ( b 0)} and В = J\A. 1 3. Theorems.
Th e o r e m 1. Let f , g : I -> R satisfy, for all x e l , the condition S4 (x) —
— lim sup / ^ g(x) ^ S4(x) —lim inf /. Then g is Darboux.
Proof. Suppose, to get a contradiction, that g is not Darboux. Then there exist an interval J = [a0, b0] c= J and a number r lying between g(a0) and g(b0) such that гфд(Т). Now, let
A1 = {xeJ| g(x) < r} and A2 — {x e J\ g(x) > r}, E x= {xeJ\ f (x) ^ r} and E 2 = J \ E i .
Then, by hypothesis, £ ,e S 4(x), that is, d ± (Ei, x ) < 1 for every x e A t (i
= 1, 2). Since almost all points of a set are points of density of the set, it follows in particular that | £,• пД-| = 0. Hence, since Ex\j E 2 = J = At и A2, it follows again that ^ -б 54(х) for every x e A h where {/,;'} = {1, 2}.
Therefore, either g(a0) < r < g(b0), or g(b0) < r < g(a0), in both cases we have a contradiction to Lemma 1, which proves the theorem.
Th e o r e m 2. Let/, g : I -* R satisfy, for all x e / , the condition, S3(x + ) —
— lim inf/ = S3(x + ) — lim sup/ = g(x). Then g is o f Baire type1 relative to /.
Proof. We first show that / is measurable. Fix r eR , and put A= {xe/| /(x ) ^ r},
Bn = {xel\ f ( x ) > r-yl/n] (n = 1 ,2 ,...).
If c e l is any point of density of B„, then S3(c + ) —lim sup/ ^ r+ l/n , and hence, S3(c + ) —lim inf/ ^ r+l/n, so that Л е 5 3(с + ), whence d+(A, c) — 0.
Consequently A has density 0 a.e. on B„ ; and, hence, + has density 0 a.e. on
00 \
(J B„= I\A. Therefore, the set A is measurable, proving that / is n= 1
measurable.
We will now construct an interval function F converging to g on /, which by [2] proves the theorem. For any interval J = ( a 1, b l) with (alt 2bx—a x~\ a I, let F ( J ) denote the supremum of the numbers к for which there exists a Ô > 0 such that :
(5) |{xe(al5 f)| /( x ) ^ k}\ > 0\(alt r)|
for all te (2 bx—a x, b), and, for some txe(2bx—a x, b), (6) tx)\ f( x ) ^ k}\ > ( l - ^ K f l j , f j .
Now fix c e I and e > 0, and put
E x = (xe(c, b)I f ( x ) ^ g (c) e}, E2 = {xe{c, b)I f{ x ) < 0(c) + e}.
Let A{ = (c, b)\Ef, i = 1, 2. Since by hypothesis 53(c + ) —lim inf/ > 0(c)-e and S3(c + ) — lim sup/ < g(c) + s, so T ,e S 3(c + ), that is, d + (Ah c) < 1 and d+ (Ai, c) — 0. Consequently, since £, — (c , b)\Ai, we have d+ c) > 0 and d + (£,, c) = 1. So, there is re(0, such that, for all te(c, b),
(7) |£; n(c, 01 > 2r|(c, 01, i = 1, 2,N and there are points cl5 c2e(c, b), c2 < cl5 such that (8) . \ E i n ( c , C i ) \ > ( l - r 2)\(c,Ci)\, i = l , 2 .
Let h = r2(c2 — c), and let J = ( a 1, b 1) denote any subinterval of
(c — h , c + h ) r > I, J b c. Then 2bj — ax ^ 2(c + h ) —(c — j^i) = c + 3r2(c2 — c ) , whence 2b t —a t < c2 < cx < b since 0 < r < Now we have, for i = 1,2, (9) I£,• n (a ls c,-)| ^ n(c, cf)| > ( l - r 2)(c, - c ) by (8)
= (1 - r) {(Ci - a j + r (Ci - c ) - ( c - aO}
> (l-O K a !, с,)| V c - a x < h < t(с, c).
Also, for all r E(2bl —a l , b), we have 2|(c, t)\ — \(ax,t)\ = t + a l — 2 c > 2 b l —
— 2c > 0 ; and, hence, (7) implies that
(10) \Ei n ( a 1, 01 > r\(ax, 01 for / = 1, 2.
Considering (9) and (10) for E x, we get F (J) ^ g(c) — s. Further, we assert that F (J) ^ g(c) + e. For, if not, then (5) and (6) must hold for some к > g(c) + s and for some S > 0. Then, recalling that/is measurable, taking t
= c2 in (5) and using (9) for E 2, we easily get S < r; whereas (6) implies, by (10) for\E2, that Ô > r, leading to a contradiction. Thus \F(J) — g(c)\ ^ e.
Hence, for every c e l , F (J) -* g(c) as |J|->0, J s c .
Theorem 3. Let f , g: I - + R satisfy, for all x e l , the condition S3(x + ) ~ - lim inf f x = g (x) = S3 (x + ) - lim sup f x, where f x (y) s (/ (y) - f (x))/(y - x) for ye(x, b). Then g is o f Baire type 1 relative to I.
Proof. We first observe that/ is measurable. This can be proved as in the first part of the proof of Theorem 2.
We will construct an interval function F converging to g on I, which by [2] proves the theorem. For any interval J = ( a t, b x) with (ax, 2bx —a xJi c /, let F(J) denote the supremum of the numbers к for which there exists a 3 > 0 such that
(x, y ) e l x x l 2/ ( * ) - / ()0
> к
(*) x - У >3\Ix x I 2\
for every ordered pair of intervals I x — {a1, tx), I 2 = (ui, t2), where t2 < tx and tx, t2e(2bl —a l , b), and, for at least one pair of such intervals, the left- hand side of ( * ) exceeds (1—<5)|I x x/21.
Now fix c e I and в > 0, put
17 f. , / (X) - / ( C) ^ M 1 1 E x = <*e(c, b)I — — - — ^ g{c)-?e> , V f / l\I f ( X) ~ f (C) ^ Ч , 1 1 E2 = <xe{c, b)I — - — < g(c) + jB>.
Then by hypothesis At = (c, b)\E, e S 3(c + ), that is, d +(А{, с) < l and {Аь c) = 0; and, hence, d + (£;, c) > 0 and d + (E{, c) = 1, i = 1, 2.
Therefore, there is re(0, 4) such that, for all re(c, b),
(11) \Ei n(c,t)\>4r\(c,t)\, / = 1,2,
and there are
(12)
(13) (14) (15) (16) (17)
points Cj, c2, c3, c4 and c5 such that:
c < c5 < c4 < c3 < c2 < c x < b,
\Ex n (c 2, c x)I > (1 — r4)|(c, Cj)I, c3 <(2c + c2)/3,
IE 2 n (c 4, c3)\ > ( l - r 4)|(c, c3)|, c5 <(2c + c4)/3,
1^1 n(c, c5)I > ( l - r 4)|(c, c5)|.
Let h — r4(c5 — c), and let J —(al , b l ) denote any subinterval of (c — h , c + h ) n l , j 3 c . Then 2bx — a x^ 2(c + h) — (c — h) = c + 3/i < c5, since 0 < r < Also, for i = l , 3 , 5, we have
(18) C i - a x ^ C i-c + h = с , - с + г4(с5- е ) < (1+ r4)(cf-c ) . Now, if x e E x n ( c 2, c x) and y e £ 2 n ( c 4, c3), then
(19) № - f ( y ) = ( f ( x ) - f ( c ) ) - ( f ( y ) - f { c ) )
> ( д ( с ) - $ Е ) ( х - с ) - ( д ( с ) + $ е ) ( у - с )
> ( x - y ) ( g { c ) - e ) , since, by (14),
x + y — 2c
2 { x - y ) I + --- < 1 + C3-C
< 1.
x — у C 2 - C 3
Again, if x e £ 2 n (c4, c3) and y e E l n (c, cs), then (20) f ( x ) - f { y ) = ( f { x ) - f { c ) ) - ( f { y ) - f { c ) )
< (в (c) + ie) (x - c) - (g (c) - \ e) (y - c)
<(x~y)(g(c) + e), since, by (16),
x + y — 2c i - У ^ , i c s - c _ --- - - 1 H---< 2~\---< 1.
2 (x -y ) x - y c4- c 5
Consider first the intervals J x = ( a l , c x), J 2 = {ax, c3) and J 3 = {ax, (19), (13), (15) and (18), we have
(21) , (x, y ) e J, r J x J 2\ ---> g ( c ) - er , f ( x ) - f ( y ) ^ 4 x — y
^ \EX n (c2, Ci)HJS2 n (c4 1 - r 4
> 1+ r4
> (1 - r 3)\Jt x j 2\ (V 0 <
Similarly, by (20), (15), (17) and (18), we have
(22) , (x, y ) e J 2 x J 3| ---< g(c)+£. 7 7 1 f ( x ) ~ f { y ) ^ , ,
x - y > ( l - r 3)|J2 x J 3|.
Next, consider any two intervals I x = (ax, tx) and /2 = {ax, t2]
2b x — a x < t2 < tx < b. Put
dx = c + 3r(t2 — c) and d2 = c + r(t2 — c).
Then c < d2 < dx < t2. Also, by (11) we have, for / = 1,2, (23) \Ехп((1х, f j ^ \E{ n (c, ti)\-\(c, dx)I
> 4r|(c, Г^-ЗгКс, Г2)| > r|(c, tx)\, (24) . IEt n (c, d2)I > 4r |(c, d2)| = 4r2 |(c, r2)|,
and, since a x < c < b x < 2b x —a x < th
(25) t i~ a x = (ff- c ) + ( c - a 1) < (ti- c ) + (2bl - a i ) - b l < 2 (r ,-c ).
If xe'Ex n ( d x, tx) and y e E 2 n(c, d2), then we have
Сз)- ВУ
» ^з)1
Г < i).
I, where
9 — Prace Matematyczne 23.2
(26) / ( * ) - / (У) = ( f ( x ) - f ( c ) ) - ( f { y ) - f { c ) )
since
^ (g (с) - \ e)(x - c) - (g(c) + { e)(y - c)
> (x-y)(si(c)-fi),
•v + У — 2c l y - c d2- c
--- = 2 H--- - < 1 + - ,---Г
2(x —у) 2 x —у ^ dy — d2 Similarly, if x gE2 n(dy, ty) and y e E j n(c, d2), then (27) f ( x ) - f ( y ) < ( x - y ) ( g ( c ) + e).
Now, from (26), (23), (24) and (25) it follows that (x, y) G I у x/2
ÿ |£, n (d u f,)| -|£, n(c, d2)I > r3|/, x/,|;
and from (27), (23), (24) and (25) it follows that
> IE 2 n {dy ,ty)I • IEy n (c, rf2)| > r3 (/j X / 2j.
Considering (21) and (28), we get F(J) ^ g(c) —e. Also, considering (22) and (29), and arguing as in the last part of the proof of Theorem 2, we get F(J) ^ g(c) + v. Thus \F(J) — g(c)\ < e. Hence, for every c e l , F {J) ->g{c) as
|J| -> 0, J 3 c . t
Theorem 4. Let f , g : I —> R satisfy, for ail x e l , the conditions:
(i) (x — ) — lim inf/ ^ / (x) ^ S4 (x — ) — lim sup/,
(ii) S3(x + ) —lim inf/x = S3(x + ) —lim sup/x = g(x), where f x(y)
= { f ( y)~ f ix ))/(y —x) f or ye(x, b). Then the following conditions on / are equivalent :
(I) S0(x —) —lim inf D_/ < g(x) ^ S0(x —) —lim sup D~f.
(II) 50(x — ) — lim inf D +f ^ g(x) ^ S0(x —) —lim sup D+f.
(Ill) g is Darhoux. i
Proof. Re (I)o(II). Indeed we have, for all x e l , (30) S0(x — ) — lim sup D~f — S0{x — ) — lim sup D+f
Re (30). Fix c e l . If к > S0(c — ) — lim sup D+f, then there is h > 0 such that D+f < к on (c — h, c), which together with (i) implies by Lemma 2 that kx — f( x ) is non-decreasing on (c — h, c), whence D~f ^ к on (c — h, c). Hence and
( 3 D S0(x — ) — lim inf D_ f = S0(x —) —lim inf D+f.
S0(c — ) — lim sup D~f ^ So(c — ) — lim sup D+f. The necessary reverse in
equality follows analogously, by Lemma 3, since the condition S3(x + ) —
— lim inf/д. = g(x) > — x implies that S4(x + ) —lim inf / ^ /(x).
In (30), replacing / by —/ we get (31).
Re (II) =s>(III). Suppose that a0, b0 e l , a0 < b0, and g(a0) Ф g(b0). Fix к between g(a0) and g(b0) exclusive, and put
G, = |xe/| g(x) > k throughout a neighbourhood of xj, G2 = { x e l I g(x) < к throughout a neighbourhood of x}.
We first observe that, if (alt bj) is any open interval contained in G,, where [al5 fe,] c /, then, by (i), (ii) and Lemma 2, f ( x ) - k x is non-decreasing on [flj, b j]; consequently g {ax) ^ к and, by (II), g(bx) ^ k. Similarly, for any open interval (a2, b^ <= G2, where [a2, b2] c ^ we have #(a2) ^ and, by (II), g(b2) < fc.
Now, the sets Gt and G2 are obvjously open and disjoint, and hence, by the preceding observations, (a0, b0)\(Gl u G 2) # 0 - So, for some closed inter
val J cz (a0, b0)> the set P = J\{GX u G2) is non-empty and closed. If P has an isolated point, c0 say, then c0 is a common end-point of two components of Gj uG2; consequently, by the definitions of the sets Gt and G2, and by the preceding observations, we must have g(c0) = k. If, on the other hand, P is perfect, then, since, by Theorem 3, g is of Baire type 1, g | P must be continuous at some point x0e P , giving g(x0) = k. Hence (II) implies (III).
Re (III)=>(II). If S0(c —) — lim sup D+f <g(c) for some c e l , then there exist numbers к and h (> 0) such that g ^ D +f < k < g ( c ) on (c — h, c), showing that g is not Darboux. Similarly, if S0(c —) —lim inf D+f > g(c) for some c e l , then g cannot be Darboux. Hence (III) implies (II), completing the proof.
Note. Referring to the proof of (30), we observe that, by (i), kx — f (x) is actually non-decreasing on (c — h, c], so that D ~ f{c )^ k , whence D~f(c) ^ S0(c —) — lim sup D+f Similarly, S0(c —) — lim inf D + f < D_/(c).
Therefore, a simpler sufficient condition for the function g to be Darboux is:
D _ f( x ) ^ g { x ) ^ D ~ f{ x ) for all x e l .
This is, however, not a necessary condition. For, consider the functions f g on ( — 1, 1) defined by
/(x) = x 2 sin (1/x) and g(x) = f'{x) if x < 0,
f( x ) = x and g{x) = \ if x ^ 0.
The reader may wish to verify that /, g satisfy hypotheses (i) and (ii) of Theorem 4, g is Darboux, and yet D~f (0) < 0(0).
Theorem 5. Let /, g be as in Theorem 4, and let g be Darboux. Then g
has the Mean Value property: for every subinterval [u, и] c /, there is at least one point ce(u, v) such that ( f (v)—f (u))/(v — u) = g(c).
Proof. Similar to part (2) of the proof of Theorem 2 in [3].
Acknow ledgem ent. The author is grateful to Dr. D. N. Sarkhel for his kind help and suggestions in the preparation of this paper.
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BELUR MATH. HOWRAH WEST BENGAL, INDIA
