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COMPOSITION OF FUNCTIONS
KANDASAMY MUTHUVEL
(Received 26 March 1999 and in revised form 11 August 1999)
Abstract.We prove that iffandgare functions from the reals into the reals such that the composition ofgwithfis continuous andfis both Darboux and surjective, thengis continuous. We also prove that continuous and Darboux can be interchanged in the above statement.
Keywords and phrases. Darboux functions, continuous functions, composition of functions.
2000 Mathematics Subject Classification. Primary 26A15; Secondary 54C30.
1. Introduction. Throughout the paper,f andgare functions from the reals into the reals. The composition ofgwithf is denoted byg◦f. It is easy to find examples to show that ifg◦fis continuous, then neitherfnorgis continuous. The problem be- comes more interesting if some restrictions are put onf. For example, a well-known theorem states that ifg◦f is continuous, andf is both continuous and surjective, theng is continuous. However, it follows from our example that if “continuous” is replaced by “open” or “Darboux” in the above statement, then the new statements are no longer true. In this paper, we prove some interesting results concerning composi- tion of functions. For example, we generalize the above well-known result by showing that ifg◦fis continuous, andfis both Darboux and surjective, thengis continuous.
This also improves [3, Theorem 3] which states that “ifg◦f is continuous,f andg are Darboux, andfis surjective, thengis continuous.” We also prove that continuous and Darboux can be interchanged in the above statement of our result, i.e., ifg◦f is Darboux andfis both continuous and surjective, thengis Darboux.
Definition1.1. A functionf from the reals into the reals is called Darboux if it has the intermediate value property, i.e., ifaandbare real numbers witha < band f (a)≠f (b), then for any real numberybetweenf (a)andf (b)there exists a real numberxbetweenaandbsuch thatf (x)=y.
Note that every continuous function is Darboux, but not every Darboux function is continuous.
Theorem1.2. Letf andgbe functions from the reals into the reals and letf be surjective.
(i) Ifg◦fis continuous andf is Darboux, thengis continuous.
(ii) Ifg◦f is Darboux andf is continuous, thengis Darboux.
Proof of(i). Assume, to the contrary, that the functiongis discontinuous at a real numbera, from the right. Sincef is surjective, there are real numbersb andc such thatf (b)=aandf (c) > a. We consider the following two cases.
214 KANDASAMY MUTHUVEL
Case1(c < b). LetX= {x < b:f (x) > a}and letdbe the supremum ofX. Then c≤d≤b. Iff (d) > a, then, sincefis Darboux anda=f (b) < f (d), there exists a real numbertin the nonempty open interval(d,b)such thatf (t) > a, which contradicts thatdis the supremum ofX. Hence,f (d)≤aandd∉X. Consequently, there exists a sequence(dn)inXsuch thatdn< dfor allnand(dn)converges tod.(∗)Sincegis discontinuous atafrom the right, there exists a sequence(xn)converging toasuch thatxn> afor allnand no subsequence of(g(xn))converges tog(a). Note that(xn) converges toa, for alln, a < xn anda < f (dn). Hence, there exists a subsequence (yn)of(xn)such thata < yn< f (dn). Sincefis Darboux andf (d)≤a < yn< f (dn), we haveyn=f (tn)for sometnin the nonempty open interval(dn,d). Hence,(tn) converges tod. Sinceg◦f is continuous atd, the sequence((g◦f )(tn))=(g(yn)) converges to(g◦f )(d). Consequently, by(∗), g(f (d))≠g(a). Sincef (dn) > a >
f (d)for allnandfis Darboux, there exists a real numbersnin the nonempty open interval(dn,d)such thatf (sn)=a. Hence,(sn)converges tod. Again, sinceg◦f is continuous atdand(g◦f )(sn)=g(f (sn))=g(a), we haveg(a)=g(f (d)), which is impossible.
Case2(c > b). LetX= {x > b:f (x) > a}and letdbe the infimum ofX. As shown in Case 1, it is not hard to show that this case is impossible.
Thusgis continuous atafrom the right. Similarly,gis continuous atafrom the left. Thusgis continuous.
Proof of(ii). Letaandbbe real numbers anda < b. Supposeyis a real number betweeng(a)andg(b). Sincefis surjective,f−1(a)andf−1(b)are nonempty sets.
Letc∈f−1(a)andd∈f−1(b). LetHbe the intersection of the setf−1(a)and the closed interval havingcanddas end points, and letKbe the intersection of the set f−1(b)and the closed interval havingcanddas end points. Sincef is continuous, HandKare disjoint nonempty compact sets. It is easy to see that there existh∈H and k ∈K such that |h−k| = dist(H,K)=min{|s−t|: s∈ H and t∈K}. Since (g◦f )(h)=g(f (h))=g(a),(g◦f )(k)=g(b), yis betweeng(a)andg(b), andg◦f is Darboux, there exists a real numberr betweenhandksuch that(g◦f )(r )=y.
The proof is complete if we show thata < f (r ) < b. To prove this, suppose, to the contrary, thatf (r )≤a. Becausef is Darboux and f (r )≤a < f (k), a=f (x) for somexbetweenr andk, orx=r. Consequently,x∈Hand|x−k|<|h−k|, which contradicts that|h−k| =dist(H,K). Hence, f (r ) > a. Similarly, f (r ) < b. Because a < f (r ) < bandg(f (r ))=y, gis Darboux.
Note that [1, Theorem 1] is the following corollary.
Corollory1.3. Ifg◦fis continuous andf is both continuous and surjective, then gis continuous.
[3, Theorem 3] states that ifg◦f is continuous,f is surjective, andf and gare Darboux, thengis continuous. Note that our theorem shows that “gis Darboux” is not needed in [3, Theorem 3].
Remark1.4. If “continuous” and “Darboux” are interchanged in statement (i) of Theorem 1.2, we get statement (ii) of Theorem 1.2. If “Darboux” is replaced by “open”
in statement (i) of Theorem 1.2, then the new statement “Ifg◦fis continuous, andf is both open and surjective, thengis continuous” is true. Proof of this proposition is straightforward. If “Darboux” is replaced by “open” in statement (ii) of Theorem 1.2, then the new statement “Ifg◦fis open, andfis both continuous and surjective, then gis open” is true. This is an exercise in [2]. It follows from the following examples that if “continuous” is replaced by “Darboux” or “open” in statements (i) and (ii) of Theorem 1.2, then the new statements are no longer true.
Example1.5. There are functionsfandgfrom the reals into the reals such thatf is surjective,fis both open and Darboux, andg◦fis both open and Darboux, butgis neither open nor Darboux. For, letf:R→Rbe a function that maps every nonempty open interval ontoR. Such a function can be easily constructed by transfinite induc- tion. Letg(1)=2, g(2)=1, andg(x)=xwhen 1≠x≠2.
It is interesting to compare the following example with (i) and (ii) of Theorem 1.2.
Example1.6. There are functionsf andgfrom the reals into the reals such that gis surjective,g◦fis continuous, andgis both Darboux and open, butfis neither Darboux (hencef is not continuous) nor open.
For, letg:R→Rbe a function that maps every nonempty open interval ontoR.
Then there exist two distinct real numbersaandb such thatg(a)=g(b)=0. Let f:R→Rbe a function with two-element range{a,b}.
Remark1.7. For an injective functiong, it is not hard to show that the statements gis continuous,gis open, andgis Darboux are equivalent. Using the fact thatg−1 is continuous and the composition of continuous functions is continuous, Darboux functions is Darboux, and open functions is open, we have the following.
Letfandgbe functions from the reals into the reals and letgbe injective.
(i) Ifg◦fis continuous andgis Darboux, thenf is continuous.
(ii) Ifg◦fis Darboux andgis Darboux, thenf is Darboux.
(iii) Ifg◦f is open andgis open, thenfis open.
Example1.8. There are functionsf andgfrom the reals into the reals such that g◦f is both open and continuous, andgis continuous, butf is neither Darboux nor open.
For, letf (x)=x+1 whenx≥ −1 andf (x)=xwhenx <−1. Letg(x)= −x2−x whenx <0 andg(x)=xwhenx≥0. Thenf is neither Darboux nor open, because
f (−2,0)
=(−2,−1)∪[0,1). (1.1)
However, iff and gare functions from the reals into the reals such thatg◦f and gare both open and continuous, thenf is open and continuous. This follows from Remark 1.7 and the fact that ifg is both open and continuous, theng is injective (otherwiseghas a local extreme value on some open interval and consequentlygis not open).
Acknowledgement. I would like to thank the referee for the useful comments on the original version of the paper.
216 KANDASAMY MUTHUVEL References
[1] K. Ciesielski and S. B. Nadler, Jr.,An absorption property for the composition of functions, Real Anal. Exchange18(1992/93), no. 2, 420–426. MR 94f:54027. Zbl 791.54020.
[2] J. Dugundji,Topology, Allyn and Bacon Inc., Boston, Mass., 1978. MR 57#17581.
[3] R. Kellum,Compositions of Darboux-like functions, Real Anal. Exchange23(1997/98), no. 1, 211–216. MR 99b:26003.
Kandasamy Muthuvel: Department of Mathematics, University of Wisconsin Oshkosh, Oshkosh, Wisconsin54901-8601, USA
E-mail address:[email protected]
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