Mařík, Jan

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Institute of Mathematics AS CR, v. v. i.

Mařík, Jan: Scholarly worksJan MaříkDerivatives and closed setsActa Math. Hungar. 43 (1-2) (1984), 25–29Persistent URL: http://dml.cz/dmlcz/502137Terms of use:© The Hungarian Academy of Sciences, 1984Institute of Mathematics of the Czech Academy of Sciences provides access to digitized documentsstrictly for personal use. Each copy of any part of this document must contain these Terms of use.This document has been digitized, optimized for electronic delivery andstamped with digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://dml.czActa Math. Hung. 43(1--2) (1984), 25--29. DERIVATIVES AND CLOSED SETS J. MAI~J[K (East Lansing) In their article [1] G. Petruska and M. Laczkovich proved (among other things) that a function defined on a perfect set S and differentiable relative to S can be extended to a function differentiable on the whole real line R. This note contains an elementary proof  of a more general theorem where the set S is supposed only to be closed in R. NOTATION. The word function means a mapping to R = ( - * %  oo). Let a E S c R  and let F be a function. If  SA(a, b)~ 0 for each b>a, we define F~ + (a) = lira (F(x)-F(a)) / (x-a)  (xES, x \ a) provided that this limit exists. We define analogously the meaning of  F}-(a) and Us(a). (Note that Us(a) may exist even if F's+(a) is undefined.) The symbols F'+(a), F'-(a) and F'(a) will have the usual meaning (i.e. F'+(a)=F'R+(a) etc.). Points in R•  will be denoted by ( . ,  .). 1. Let a, bER, a<b and let J=[a, b]. Let ~o and ~k be functions continuous on S. Let q~ be convex, ~ concave, r on {a, b}. Set s=(q~(b)-cp(a))/(b-a). Let ~, fl, M, NER, q~'+(a)=<a<--~'+(a), t~'-(b)~=fl<-q~'-(b), M < m i n  (a, fl, s), max (a, t ,  s)<N. Then there is a function G continuously differentiable on J such that G'+(a)=a, G'-(b)=fl, M < G ' < N  on (a,b) and that, for each xE(a,b), G(x)=~p(a)+ s(x-a)  or q~(x)<a(x)<r PROOF. We may assume that ~p=~=0  on {a, b}. Then s=0 .  Let e=(a+b)]2, We construct a function H continuously differentiable on J such that H '+(a)=cq H = 0  on (c,b), M < H ' < N  on (a,b) and that, for each xE(a,b), H ( x ) = 0  or ~p(x)<H(x)<~,(x). If  a=0 ,  we choose H = 0  on aT. Now let, e.g., a>0 .  Choose an ~ E ( 0 , - M )  and set # (x)=r  We have ==</~(a)=#(a+). There is an axE(a, e) such that ~ increases on (a, az). There is an a=E(a, al)and a function p continuous and decreasing on [a, a2] such that a(a~-a)<n(al-a,), LI 2 p(a)=a ,  p < p  on (a, a2) and p(a2)=0. Since f p<=(a , -a )<~(a l -a , ) ,  there is a g ~1 Gj~ function q continuous on [a.,, as) such that 0<=q-<_e, f q--- f p and that q = 0  on a s  a x {az, al}. Set h=p on [a, a~), h = - q  on [a.~, ad, h = 0  on (al, b] and H ( x ) =  f h a for each xEJ. It is easy to see that - e < - H ' ( x ) < ~  and O~H(x)<~(x) for each xE(a, b). A c t a  Ma~hernar H u n g a t i c a  43, 1934 26 J. MARIK In an analogous way we construct a function K continuously differentiable on J such that K = 0  on (a, c), K'-(b)=/~, M < K ' < N  on (a,b) and that, for each xE(a, b), K(x)=0  or q~(x)<K(x)<~b(x). Now it suffices to take G=H+K. 2. Let a, b and J be as in 1. Let P be a function on J such that the derivatives ~=P'+(a) ,  [3=-P'-(b) exist. Set s=(P(b)-P(a))/(b-a).  Let M, NER, M <  -<min (c~,/3, s), max (c~, fl, s)<N. Then there is a .function G continuously dif- ferentiable on Y such that the graph of G is contained in the convex hull of  the graph of  P and that G'+(a)=~, G'-(b)=/~, G=P on {a, b} and M-<G'<N on (a, b). PROOF. Let 9 and T be functions continuous on a r such that r  on {a, b}, r is convex, T is concave, ~ '+(a)=  T ' - ( b ) =  - co, T '+(a)= ~ ' - ( b ) =  o~. Set Po=(PV~)AT.  Obviously ~=P~§ fi=P~-(b). Let C and Co be the convex hulls of the graphs of P and P0 respectively. It is easy to see that CocC. Let ~o be the greatest convex function on J such that q~<=P0 and let ~k be the smallest concave function on J such that P0_-<~. Let C1 be the set of all points (x ,y)  such that xE(a,b) and that y=P(a)+s(x -a )  or ~o(x)<y<~O(x). Then C~CCo. Now we apply 1~ 3. Let S be a nonempty set dosed in R. Let A, BERU { -  ~, ~}. Let P be a function on R such that A<P' (x )<B for each xES and that a < (P(y)- -P(x)) / (y 'x)  < B, wl~enever x, yES, x ~  y. Then there is a function G differentiable on R such that G : P ,  G'=P'  on S and A < G ' < B  on R. PROOF. We may suppose that inf S = - % sup S . . . .  Let (a, b) be a compo- nent o f  R \ S  and let ~,/3, s be as in 2. There are M, NER such that A < M <  <rain (c~,/~, s), max (~,/3, s)<N<B. Construct a function G according to 2. In this way we define G on R \ S ;  further we set G=P on S. It is easy to see that G has the required properties. 4. Let xo, yo, sER. For each 7E(0, ~) define (1) W r = {(x, y)ERXR;  l y -yo-s (X-Xo) l  < 7(x-x0)}. Let eE(0, ~ ) a n d l e t  (xl, y,), (bl c)ErV,, 3Xi<=ab-xo . Then (2b-x~, 2c-yi)EWs,.  PROOF. We may suppose that x0=y0=0.  Then 6x~<=8b and hence 12C-y~- - S(2b - Xl)l <=2 lc -  sb [ + l y~ -  sxll <~(2b + x~) <=e( I Ob - 5x~) = 5~(2b - x~). REMARr:. The geometric meaning of W~ is obvious. To see the geometric meaning of assertion 4 the reader should realize that 3x~<=4b-xo means the same 4 as x~-xo<= ~-(b-xo)  and t h a t  (b, c) is the center of the segment with end points (x~,yl) and (2b-x1 ,  2c-yl) .  5. Let xo,Yo,sER. For each ~'E(0, oo) define Wr by (1). Let ~E(0, ~o)and let: ~x~, yt}, (b, c), <x~, y~)EW~, xx<b<x~, xER, 3lx-bl<-b-x~.  Let q=(Y~--Yx)/ /(xz--Xa ). Then (x, c+q(x-b))EWz, .  A c t a  Mathemat~ca  H u n g a r i c a  4~, 1984 DERIVATIVES AND CLOSED SETS 27 PROOF. We may suppose that x0 =y0=0. Set y=c+q(x-b) ,  Z= [x-b[(xl+x2)/ /(xs-xl). As 3[x-b[<min (xs-xl ,  b), we have  3Z<min (x~+x2, b(x~+x2)/ /(xs-xl)). If x2<-2b, then x~+x2<3b; if xs>2b, then < (b b)  < 3. Thus in either case Z < b .  Obviously [q-s] = ]Y2-SX2 - (yl - sxl)l/(x2- x~) <= e(xl + x2)/(x2 , xl); therefore ]y-xsl=[c-sb+(x-b)(q-s)] <= eb+~Z<2~b. Since x=b-(b-x)>2b/3 ,  we have [y--sxl<3ex. 6. Let S be a set closed in R. Let F be a function on S such that F~(X) is finite for each accumulation point x of S. Then there is a function H on ~R differentiable at each point of S such that H = F on S. PROOF. We may suppose that i n f S = - r  sup S=o~. Set A + = {xCS; S A ( x , y ) #  ~ for each y > x } ,  A - = { x E S ;  S O ( y , x ) #  0 for each y < x } ,  I + = A - ~ A  +, I - = A + ~ A  -, I = S ~ ( A + U A - ) .  Define a function f on S as follows: If b 6 A + U A - ( = S ~ I ) ,  set f(b)=F~(b). If bCI, find xl, x~ES such that S(~(xl,x2)={b} and set f (b) = (F(x2)- F(x~))/(x2-- x~). For each bCS define a set M b as follows: If bEA+f-IA -, let Mb={b}. If b~i+UI -, choose a db>0 such that either SN(b,b+3do)=fD or SO A ( b -  3db, b) = O and set M b = {x; 2b-xCS(~[b-d  b, b+db] }. If bEI, choose a db>0 such that SN(b-3do, b+3db)={b } and set Mb---- =[b-db, b+db]. Let M = U M  b (bCS). Obviously b~M b for each  bCS and M j ~ M b = ~ ,  whenever a, b~S, a#b. If (a, b) is a component of R ~ S ,  then M,(~(a, b)= 0 for each cCS~,{a, b}. Thus (a, b)~M=(a, b)~(M~UMb) which is Open. There- fore R ~ M = ( R ~ S ) ~ M  is open, M is closed. There is a unique function G on M with the following properties: G=F on S ; i f  x~Mb, b~I+UI -, then G(x)=2F(b)-F(2b-x); if x~M~, b~/, then G(x)= F(b) + (x-b)f(b). Let Xo~S. We shall prove that (2) (Xo) = f (xo). The case xoeA + is left to the reader. Now let x0~A + and let e~(0, oo). Set s=f(x0) (=F~(xo)). For each 7~(0, oo) define W~ by (1). There is a z>x  o such that (x, F(x))~W, for each x~S~(xo, z). There are z~, z2~S such that Xo<Z~<Z 3 and that O<z~-Xo<~(z~-xo) (so that Xo<Z~<Z~). Let x~M~(xo, Z~). If x~S, A c t a  Mathema~ica  Hunf farica 43, 198d 28 J. MARIK then, obviously, (x,G(x))EW~. Thus, let x r  and let (a,b) be the component of R ~ S  containing x. We have xo<a<x<b~_z 1. There are the following four possibilities: 1. xEMb, bE1-. Set xx=2b-x.  Then x~ES, O<xt-b<=db~_(b-a)/3< 4 <(b-xo)/3, therefore 3xx<4b-xo, and xl<zl+(zl-xo)/3=xo+-~(zl-Xo)<Z2. Set c=F(b), yx=F(xx). We have (b,c), (xl, yt)EW~, x = 2 b - x l ,  G(x)=2c-yx so that, by 4, (x, G(x))E lrv6,. 2. xEMb, bEI. There is an xaESf')(b, ~) such that SO(b, xa)= O. Obviously xa~_z~. Then G(x)=F(b)+(x-b)f(b), O<b-x<=db<-(b-a)/3 so that by 5 with x l = a ,  q=f(b) etc. we have (x, G(x))EWa~. 3. xEM,, aEl +. Proceeding as in 1 we get (x, G(x))EWs~. 4. xEM,, aEI. Proceeding as in 2 we get (x, G(x))EWa~. This proves (2). Similarly, it can be shown that G'M~(Xo)=f(xo) for each x0E S. Now it suffices to choose for H the function that equals G on M and is linear on the closure of each component of R ~ M .  7. Let T be a closed set in R, V = R ~ T ,  Q c V  and let Q be isolatedin V. Let g be a function on Q. Then there is a function K differentiable on R such that KIn0  on TUQ, K'=O on T and K'=g on Q. PROOF. Let ~o be a function differentiable on R such that ~o=0 on {0}U kJ R ~ (  1, 1)), r ]~pt<l on R There is a co on ( -- , -- . function continuous R such that co=o) =0  on T and that ~o>0 on V. There are positive numbers 8~(qEQ) such that the intervals Jq=[q-eq, q+eq] are pairwise disjoint and that Jq=V for each q. Now let ~/q=min {og(x);xEJq}, Cq=max(1/~,lg(q)l/nq) and, for each xER, let K(x)  = ,~ ~ - cp(cq(x--q)). qEQ Cq Obviously lKl~_o~ on R. It is easy to see that K satisfies our requirements. R~MARK. The following assertion is a generalization of Theorem 5.5.3 in [1]. 8. Let S be a nonempty set closed in R. Let F and f be functions on S such that F's(x)=f(x ) for each accumulation point x of S. Let A, B E R U { - ~ ,  co}. Suppose that A<f (x )<B for each xES and that A<(F(y) -F(x) ) / (y -x)<B,  whenever x, yES and x ~ y. Then there is a function G differentiable on R such that G=F, G ' = f  on S and A<G'<B on R. PROOF. Let T be the set of all accumulation points of S. Let H be as in 6. By 7 there is a function K differentiable on R such that K = 0  on S, K ' = 0  on T and that K ' = f - H "  on S',,.T. Set P=H+K. Obviously P=F and P ' = f  on S. Now we apply 3. REMARK. It has been mentioned in [1] that there is a perfect set S and a func- tion F on S such that IF~(x)l-<_l for each xES and that G' is unbounded for each function G differentiable on R such that G=F on S. The following example shows a little more. Ac~a Mathemc~tlca Hunffar~ea 43, 1984 DERIVATIVES AND CLOSED SETS 29 Let l=xo>xl>... ,x,-*O, yn~-'~- gn- l - -X2n(gn- l - -Xn)  ( n =  l,  2, . . .).  It is easy to see that x.<y~<X._l. Set S =  [x.,y.] U {0}. Define a function F on S setting F ( 0 ) = 0  and F(x)=x~ for each xE[x.,y.]. Then S is perfect and F} = 0 on S. Now let G be a function diffcrentiable on R such that G = F on S, Then G(x._O-G(y.)=Xl_l-X~>2x.(x._l-XJ=2(x._l-y.) /x .  so that (G(x.-1)- - G ( y . ) ) / ( x ~ - l - y . ) ~  (n-*~). We see that G' is unbounded on (0, 1). Thus, we have constructed a perfect set S and a function on S twice (actually. infinitely many times) differcntiable relative to S that cannot be extended to a func- tion twice differentiable on R. Reference [1] G. Petruska and M. Laczkovich, Baire 1 functions, approximately continuous functions and derivatives, Acta Math. Acad. Sci. Hungar., 25 (1974), 189---212. (Received March 17. 1982) DEPARTMENT OF MATHEMATICS MICHIGAN STATE UNIVERSITY EAST LANSING, MICHIGAN 48824 USA Acta Mathe~atlca Hungar~ca 43, 1984 

Derivatives and closed sets

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Derivatives and closed sets

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Institute of Mathematics AS CR, v. v. i.