Linear Functional Equations and Convergence of Iterates

Axel Torshage

June 2012

Bachelor’sthesis 15 Credits Umeå University Supervisor - Yuriy Rogovchenko Abstract

The subject of this work is functional equations with direction towards linear functional equations. The …rst part describes function sets where iterates of the functions converge to a …xed point. In the second part the convergence property is used to provide solutions to linear functional equations by de…ning solutions as in…nite sums. Furthermore, this work contains some transforms to linear form, examples of functions that belong to di¤erent classes and corresponding linear functional equations. We use Mathematica to generate solutions and solve itera- tively equations. Contents

1 Introduction 3 1.1 Historical perspective ...... 3 1.2 Idea ...... 6 1.3 Notation...... 7 1.3.1 Common notation ...... 7 1.3.2 Kuczma’snotation ...... 7 1.3.3 New classes of functions ...... 8

2 Convergence of function iterates 9 2.1 Counterexample ...... 11 2.2 Standardsets ...... 11 2.3 Simple extension ...... 12 2.4 Extension of the function’sclass ...... 14

3 Linear Functional Equations 19 3.1 Homogenous linear functional equations ...... 19 3.1.1 Case (i) ...... 21 3.1.2 Case (ii) ...... 22 3.1.3 Case (iii) ...... 23 3.2 Criteria for Gn ...... 24 3.3 Particular solution ...... 26 3.3.1 Unique continuos solution ...... 26 3.3.2 Arbitrary constant solution ...... 30 3.3.3 Arbitrary functions solution ...... 32

4 Transformations 35 4.1 Transformation to simpler linear form ...... 35 4.2 Transformation of a general equation ...... 36

1 5 Examples 38 5.1 Convergence...... 38 0 5.1.1 A function from R [I] ...... 38 0 5.1.2 A function from S [I] ...... 39 0 5.1.3 A function from P [I] ...... 41 0 5.1.4 A function from T [I] ...... 42 5.2 Solutions to linear functional equations ...... 44 0 5.2.1 A function from R [I] ...... 44 0 5.2.2 A function from S [I] ...... 45 0 5.2.3 A function from P [I] ...... 46 0 5.2.4 A function from T [I] ...... 46 6 Conclusions 49

7 Appendix 51 1. Introduction

In simple words, a functional equation is much like a regular algebraic equation, though instead of unknown elements in some set we are interested in …nding a function satisfying our equation. A typical example of a functional equation is

' (x) + ' x2 = x; where one has to …nd the function ' de…ned in some given interval. This equation has a unique solution in [0; 1) and another one in (0; ) but no continuous solutions in [0; ) : In Section 3 we will see that this example1 is linear as well. 1 1.1. Historical perspective

Even though functional equations were known long time before 17th century, it took another two hundred years until mathematicians …rst tried to organize the notations and theory. Grégoire de Saint-Vincent’s (1584-1667) work is a good starting point in the subject. He had noticed that the area under hyperbolic 1 graphs, for example, g (x) = x ; can be described by a function with the properties ' (x)+' (y) = ' (xy). This is the property possessed by logarithmic functions which is easily shown by modern calculus, but in that time Grégoire used a geo- metric argument to obtain a functional equation, which would ultimately have a logarithmic solution. However, the subject of functional equations at that time was rather undevel- oped and we can date the real birth of the subject with the famous Augustin-Louis Cauchy (1789-1857). Despite being most famous for his work in mathematical analysis, Cauchy has provided the basis of functional equations as well. The Cauchy equation ' (x + y) = ' (x) + ' (y) is one of the most important functional equations and is the …rst property of a linear functional often used in analysis and linear algebra. Cauchy also gave name

3 Figure 1.1: Grégoire de Saint-Vincent (1584-1667) to the exponential counterpart ' (x + y) = ' (x) ' (y) and found the complete set of functions solving the d’Alambert equation

' (x + y) + ' (x y) = 2' (x) ' (y) :

We introduce one more mathematician’swork. Charles Babbage (1791-1871) is known as a pioneer within computer science. Babbage’s work is closer to the direction this paper has due to the fact that he studied various functional equations with only one variable instead of the two used in the examples above. One family of functional equations that Babbage introduced is of the type

G [x; ' (x) ;' (a1 (x)) ; ::::; ' (an (x))] = 0; where G; a1; ::::; an are known functions and ' is the unknown. This equation is of interest for us because by de…ning G in a certain way we can obtain a linear functional equation.

' (f (x)) g (x) ' (x) F (x) = 0: We conclude this part by giving some further contribution to Babbage. Before we Figure 1.2: Augustin-Louis Cauchy (1789-1857)

Figure 1.3: Charles Babbage (1791-1871) x

x 4 2 2 4

Figure 1.4: Solutions of the Babbage equation. focus the linear functional equation, we shed some light on the Babbage equation. Though not linear, the following functional equation has some iterative values that we may …nd interesting in this work:

' (' (x)) x = 0: (1.1) Equation (1.1) is known as Babbage equation and involves an iteration of the unknown function ': The set of solutions to this equation is C C x ' (x) = x; ' (x) = C x; ' (x) = ;' ; x 1 + Dx where C and D are arbitrary real constants. Here we plot four solutions to equation (1.1), corresponding to the values C = 7 and D = 4:

1.2. Idea

The subject of functional equations has many similarities with that of di¤erential equations. Just as in the case of di¤erential equations the appearance of the functional equation is crucial for the solution method. One di¤erence is that for real-valued functional equation one uses function iterates to …nd a solution. Convergence of these functions is an important property that allows to …nd a solution by the methods described in this paper. In this project, we discuss only linear functional equations. 1.3. Notation

We start by de…ning some notations that we will use in the sequel. Some of these notations might seem unrelated to the subject, but many of them describe quite important properties of functional equations. Much of the notation used in this paper coincides with the notation used in Marek Kuczma’smonograph [1], which is the main source of information in this work.

1.3.1. Common notation De…nition 1. We denote the most common interval as I; where I can be any real interval that …nite or in…nite. If nothing else is indicated, all functions are de…ned as f : I I; and x I respectively. When a function is said to converge, it is understood! that it converges2 in the interval I: De…nition 2. Cn [I] denotes the n times di¤erentiable functions in the interval I: In this work we will at most times only demand that the functions are continuous, meaning that f C0 [I]. 2 1.3.2. Kuczma’snotation De…nition 3. For a given f (x) and an interval I such that f : I I; we de…ne the n-th iterate as !

0 n+1 n f (x) = x; f (x) = f (f (x)) ; x E; n N 0 : 2 2 [ f g Not to be confused with exponentiation, which will be denoted (f (x))n and likewise sin2 (x) = sin (sin (x)) ; not (sin (x))2 : De…nition 4. A …xed point I corresponding to a certain function f is a point such that f () = : 2 There are certain sets that are of special interest, the intervals or sets where our functions are de…ned. We will also de…ne some subsets of the set of functions de…ned in I. De…nition 5. The set Sn [I] Cn [I] is the set of functions such that (f (x) x)( x) > 0; x = ; (1.2) 6 and (f (x) )( x) < 0; x = : (1.3) 6 n n De…nition 6. R [I] denotes the subset of functions in S [I] that are strictly increasing.

De…nition 7. [I] represents all functions ' : I ; where is a given set (in ! this work, = R).

1.3.3. New classes of functions De…nition 8. P n [I] is the set of functions P n [I] Cn [I] satisfying (1.2) and (f (x) + x 2)( x) < 0; x = : (1.4) 6 De…nition 9. T n [I] Cn [I] is the set of functions with the property (1.2) and f (x) ((k + 1) kx) < 0; x < ; 8 (1.5) kf (x) ((k + 1) x) > 0; x > ; 8 where k is any constant larger then 0. 2. Convergence of function iterates

Convergence to a …xed point is a valuable property of functions in this topic, many theorems will demand that some functions in the linear functional equation behave in a certain manner. This chapter will discuss for which sets of functions we can ensure this convergence property. First, we show that for the two set n n S [I] and R [I] de…ned in Section 1.3 have convergence to a …xed point. Then, n n we show that the same is true for the sets P [I] and T [I]. There exist larger sets containing converging functions, but these are not considered in this work. We will …rst prove an important lemma.

Lemma 10. Suppose that f : I I is a continuous function. Assume also that, ! for all > 0; there exists a positive < 1 such that f (x) x0 < x x0 for j j n j j x I [x0 ; x0 + ] ; where might depend on : Then limn f (x) = x0: 2 n !1 Proof. We prove lemma by contradiction, suppose we have an x I such that 2 i 1 f (x) I [x0 ; x0 + ] ; for all i 1: (2.1) 2 n Then the assumptions of the lemma ensure

n n 1 n 2 f (x) x0 < f (x) x0 < f (x) x0 j j n < < x x0 : (2.2) j j Then (2.2) yields

n n lim f (x) x0 < lim x x0 = 0: n n !1 j j !1 j j n Hence, limn f (x) [x0 ; x0 + ] ; but this contradicts (2.1). Therefore, for each x I; !1there must2 exist an N 1 such that 2 N 1 f (x) [x0 ; x0 + ] 2

9 If we can prove that

f :[x0 ; x0 + ] [x0 ; x0 + ] ; (2.3) ! then n 1 f (x) [x0 ; x0 + ] ; n N; 2 and letting 0; we have the convergence. Suppose the opposite to (2.3), f (x) ! 2 I [x0 ; x0 + ] for some x [x0 ; x0 + ] : If x = x0; by continuity of f and n 2 using the fact that I [x0 ; x0 + ] is an open set, there must exist a c > 0 de- n pending on ; such that for all x [x0 c; x0 + c], f (x) I [x0 ; x0 + ] and by 2 2 n the assumptions of the lemma, we can …nd an 2 > 0 such f (x) x0 < x x0 j j j j for some < 1:Thus, f (x) x0 < x x0 and since x [x0 ; x0 + ], we have j j j j 2 that f (x) [x0 ; x0 + ] which contradicts the assumption. Thus we should 2 have x = x0; but we have shown that f (x) I [x0 ; x0 + ] is not possible if 6 2 n x = x0: Thus, we have shown that (2.3) hold, hence we have proven convergence. 6

Lemma 11. Suppose that f is continuous in I and, for any > 0; there exists n an x0 I such that limn f (x) = x0; for all x [x0 ; x0 + ] I: Then, 2 !1 2 = x0 is a …xed point of f and is the only …xed point in [x0 ; x0 + ] : Proof. We use the fact that f is continuous to obtain

n n 1 x0 = lim f (x) = f lim f (x) = f (x0) ; n n !1 !1 thus, x0 is a …xed point . Suppose now we have a second …xed point 2 2 [x0 ; x0 + ] : Then, we have n 2 = f (2) = f (f (2)) = f (2) : Letting n , we have ! 1 n lim f (2) = 2 = x0; n !1 6 which contradicts the the assumptions of the proof. Thus, x0 is the only …xed point : Note that Lemma 10 and Lemma 11 together yields a special case of Banach …xed-point theorem. 2.1. Counterexample

One might get an idea that only (1.2) is enough to ensure convergence. Therefore, we show by a counterexample that there exists a function that satis…es (1.2) but f n (x) does not converge to : Consider the function

f (x) = 2x: For this function, we have the …xed point = 0 as the only …xed point. We will show now that this function satis…es (1.2):

(f (x) x)( x) = ( 3x)( x) = 3x2 > 0 x = 0: 8 6 On the other hand, f n(x) = ( 2)n(x) = as n ; for any x = 0: This shows that we cannot ensure convergence! 1 by 6 (1.2). However,! 1 adjusting6 the demands by adding either of (1.3), (1.4) or (1.5), we can establish convergence.

2.2. Standard sets

n n We start by examining the sets de…ned by (1.2) and (1.3), namely S [I] and R [I]. n n n Since R [I] is a subset of S [I] ; it is enough to prove convergence for S [I] to n n ensure the same behavior for R [I]. On the other hand, S [I] is a subset of even larger sets with the convergence property, but we will use more complicated methods in those proofs. The proof in this subsection is from [1, Theorem 0.4, p. 21].

n n Theorem 12. Suppose that f S [I] : Then, limn f (x) = : Furthermore, the sequence f n (x) is strictly monotonic2 for x = : !1 6 n Proof. For x = ; the claim is trivial because if we have limn f (x) = ; is a …xed point by Lemma 11, so we suppose that x < : The proof!1 for the case < x is similar. From (1.2) we have

(f (x) x)( x) > 0: Since x < , we conclude that

f (x) x > 0 x < f (x) : () By (1.3), we have (f (x) )( x) < 0: Hence, f (x) < 0 f (x) < : () Combining these results, we …nd that

x < f (x) < :

Since f (x) < ; we can repeat this argument for f (x) n times to …nd that

n n+1 f (x) < f (x) ; n N 0 : 2 [ f g n So the sequence f (x) i=0 is monotonic (strictly increasing) for x < : In the case where < x;f it isg strictly decreasing instead. To prove that this sequence converges to ; we simply need note

x < f (x) < 0: It is easy to see that f (x) < x : j j j j For any x0 < ; > 0 and any x in the interval [x0; ] I; we can …nd a positive constant < 1 such that

f (x) < x : j j j j

Since f is de…ned in [x0; ] I; by applying Lemma 10, we have the convergence.

2.3. Simple extension

n We will show now the convergence for functions in the set P [I], but …rst we want n n to show that P [I] is a larger set than S [I] ; to be sure that (1.4) is a looser condition than (1.3).

Proposition 13. Sn [I] P n [I] : Proof. Since both sets have the property (1.2), it is enough to prove that (1.3)= (1.4) and thus, Sn [I] has the property (1.4). By the assumption, ) (f (x) )( x) < 0; x = : 6 We simply subtract ( x)2 on the left hand side of the inequality: (f (x) )( x) ( x)2 < 0: Therefore, (f (x) + x 2)( x) < 0; and we are done.

Theorem 14. Assume that f satis…es (1.2) and (1.5) and I = [a; 2 a] ; where n n a < might be in…nite. Then, f P [I] and limn f (x) = ; for all x: 2 !1 Proof. To prove the theorem, we must show that f : I I and thus showing n ! that f P [I] ; otherwise we cannot use these calculations. By (1.2) and (1.4), we have2 for x < ; x < f (x) < 2 x: Since x I; x > a and we have 2 a < f (x) < 2 a = f (x) I; ) 2 n n thus f P [I] :For x = ; the claim is trivial because if we have limn f (x) = 2 n !1 ; is a …xed point by Lemma 11. To show that limn f (x) = ; we therefore suppose that x < (for < x the proof is similar). Condition!1 (1.2) yields

(f (x) + x 2)( x) < 0; and thus, f (x) > x: (2.4) One the other hand, (1.4) provides the inequality

f (x) + x 2 < 0 f (x) < x: (2.5) () Combining the inequalities (2.4) and (2.5), we obtain

x < f (x) < x: (2.6) Suppose that f (x) 0; then, by (2.6), x < f (x) 0; and thus, x < f (x) : j j j j Suppose now that f (x) > 0; then 0 < f (x) < x and f (x) < x = x : j j j j j j Therefore, in both cases, f (x) < x ; and, for any interval [x0; 2 x0] I; we have that for each j > 0 can j …ndj a positivej < 1 such that f (x) < x ; for all x I [ ; + ] : j j j j 2 n By Lemma 10, we proved the convergence. n The fact that (1.2), (1.4) and I = [a; 2 a] implies that f P [I] is also shown in the corollary to Lemma 16. 2

2.4. Extension of the function’sclass

To enlarge the set for which we can establish convergence, we note that f (x) < x for x = is not the only condition that provides functions thatj converge j j j 6 in I: It is, for example, enough to show that f n (x) < x for some n N and all x = . This construction, however, isj rather abstractj j and toj check whether2 a function6 possesses this property is not something we will discuss in this paper. We will, however, discuss a special case that is more general and where it is rather easy to verify if a function satis…es the assumptions and thus converges, by us- n ing (1.5). We will, therefore, construct the set of functions T [I] and show that functions in this set converge. Proposition 15. P n [I] T n [I] : n Proof. The construction of (1.5) says that f T [I] provided that there exists a k > 0 such that condition (1.5) is satis…ed.2 Suppose f is a function verifying (1.5) for k = 1: Then, (1.5) states f (x) (2 x) < 0 x < ; and f (x) (2 x) > 0 x > : Multiplying both sides of the inequalities by ( x) ; we obtain (f (x) + x 2)( x) < 0 x < ; 8 (f (x) + x 2)( x) < 0 x > ; 8 n which is exactly (1.4), just divided in two cases. Thus, the functions f P [I] belong also to T n [I] and P n [I] T n [I] : 2 n Lemma 16. Assume that f satis…es (1.2) and (1.5). Then f T [I] if the interval I is has the form I = [a; (k + 1) ka] ; for all a < ; where2 k is the same as in the condition (1.5).

n Proof. To prove f T [I] we only have to show f : I I since the other conditions we veri…ed2 due to assumption. Condition (1.5) gives! us that for x < f (x) ((k + 1) kx) < 0 and, since a x; f (x) < ((k + 1) kx) ((k + 1) ka) : (2.7) Condition (1.2) provides the lower bound as a x < f (x) : Thus, f (x) I for x < : For x > (1.5) yields 2 kf (x) ((k + 1) x) > 0: Since x (k + 1) ka; kf (x) > (k + 1) x (k + 1) ((k + 1) ka) ; (2.8) and we have the lower bound f (x) a: Assumption (1.2) provides the upper bound as f (x) < x (k + 1) ka; so we have established that f is bounded in I for x = ; and thus, f (x) I: Note that the fact that Lemma 16 tells nothing6 about the case when2 x = n does not matter because if we can show convergence of T [I] ; we know that is a …xed point by Lemma 10. Corollary 17. f P n [I] is de…ned in I if I = [a; 2 a] ; for all a < : 2 n f g n n Proof. From Proposition 15, we know that P [I] is the set of functions in T [I] that satis…es (1.5) with k = 1: By Lemma 16, f is de…ned in I = [a; 2 a] : n Theorem 18. Suppose that f T [I] and I = [a; (a + 1) kx] ; for some 2 n a < that might be in…nite. Then, limn f (x) = ; for all x: !1 Proof. From Lemma 16, we know that f is de…ned in I: We start by supposing that x < : Condition (1.5) yields

f (x) ((k + 1) kx) < 0: By rearranging terms, we have

f (x) < k kx = k ( x) : It follows from (1.2) that, for x < ; we have that x < f (x) ; so we can estimate f (x) below by x < f (x) < k ( x) : (2.9) We will …nd now the corresponding bounds for < x: Applying (1.5), we have

(kf (x) ((k + 1) x)) ( x) < 0: Hence, kf (x) ((k + 1) x) > 0; or, x < f (x) : k Condition (1.2) yields that we have that f (x) < x; so we can estimate f (x) above by x < f (x) < x : (2.10) k This concludes the proof of the …rst part of the theorem. We will show now that the boundaries get more tight for each iteration. We start by showing that in (2.9), (2.10) the bounds di¤er in sign. This is obvious since in (2.9) x < 0 and, for k > 0, k ( x) > 0: The same is true for (2.10) 1 since k > 0: The method we use to prove the convergence is to show that these boundaries shrink for each iteration, simply meaning that the lower estimate will increase and the upper will decrease. Suppose that x < ; then we have the inequalities (2.9). Since the boundaries di¤er in sign we have two di¤erent cases, x < f (x) < 0 or 0 < f (x) < k ( x) : There are actually three cases, there is also the case when f (x) = 0; but in that case we are sure that we have convergence due to Lemma 11. Suppose that x < f (x) < 0. Then, f (x) < ; so we can apply another iteration and have by (2.9)

f (x) < f (f (x)) < k ( f (x)) : Now, recalling that x < f (x) and k > 0; we get

x < f (x) < f (f (x)) < k ( f (x)) < k ( x) : Suppose now that 0 < f (x) < k ( x) ; then f (x) > ; (2.10) yields

f (x) < f (f (x)) < f (x) : k Applying (2.9) and using the fact that

1 f (x) < k ( x) ( f (x)) k > x ; () we have f (x) x < < f (f (x)) < f (x) < k ( x) : (2.11) k Thus, the boundaries shrink with each iteration when x < : It remains to show now the same for x < : In that case, we have the inequalities (2.10). Once again, we can use the fact that the limits di¤er in sign to be sure that either 1 0 < f (x) < x or ( x) k < f (x) < 0: For f (x) = 0; the convergence is trivial due to Lemma 11. We prove now that the boundaries shrink in both cases. Suppose that 0 < f (x) < x ; then < f (x) ; and we can use (2.10), f (x) f (x) > f (f (x)) > : k Since f (x) < x ; we can add the inequalities, f (x) x x > f (x) > f (f (x)) > > : k k

1 Therefore, new boundaries are stricter. Suppose now that ( x) k < f (x) < 0; then f (x) < ; and we can use (2.9):

f (x) < f (f (x)) < k ( f (x)) : 1 But then, using ( x) k < f (x) and x < k (f (x) ) ; or x > k ( f (x)) ; we obtain, x < f (x) < f (f (x)) < k ( f (x)) < x : (2.12) k Consequently, in both cases, the limits shrink under iteration. The reasoning in the two following cases are the same as that in Lemma 10 and are therefore left out of this text. If we use (2.11) inductively for any x [x0; ] I; (2.11) yields that for each > 0 there exists a positive < 1 such that2

n ( x) < f n (x) < nk (x ) : (2.13) Letting n ; note that (2.13) states that f n (x) converges to 0 and thus, n! 1 limn f (x) = : Likewise, for x [; (k + 1) kx0] I; inequality (2.12) yields!1 2 x n < f n (x) < n (x ) ; k n and limn f (x) = :for x [x0; (k + 1) kx0] I: Since x0 < is any point in I; we have!1 proved convergence.2 3. Linear Functional Equations

There are di¤erent types of functional equations, but from now on we discuss only the linear case.

De…nition 19. A functional equation is linear if it can be written as

' (f (x)) = g (x) ' (x) + F (x) ; (3.1)

n where g; F [I] ; f T [I] ; g (x) = 0 in I;and ' is some unknown function in [I] : 2 2 6

The work in this chapter are mostly based on [1, p. 46-66]. Even though n many results are similar, [1, p. 46-66] is focused on the case when f R [I] and n 2 eventually f S [I]. The contribution of this work in this chapter is to generalize 2 n [1, p. 46-66] to T [I] which is a more general set of functions In most cases we establish almost the same results as in [1, p. 46-66].

3.1. Homogenous linear functional equations

De…nition 20. A homogenous linear functional equation is equation of the form

' (f (x)) = g (x) ' (x) : (3.2)

To be able to provide solution to (3.1), we recall the idea of a solution to a linear di¤erential equation where one starts by …nding a general solution for a homogenous equation. Thereafter, we search for a particular solution: As the following results show, some properties of linear functional equations are close those of linear di¤erential equations.

Proposition 21. Suppose that '1 and '2 are solutions to equation (3.1). Then '3 = ('1 '2) where R; is a solution to the corresponding homogenous linear functional equation (3.2).2

19 Proof. Let '1 and '2 be solutions to equation (3.1). Then

('1 (f (x)) '2 (f (x))) = (g (x) '1 (x) + F (x) (g (x) '2 (x) + F (x))) : That is, '3 (f (x)) = g (x) '3 (x) ; and '3 solves the homogenous equation.

Proposition 22. Suppose that '1 and '2 are solutions to a homogenous linear functional equation (3.2), then '3 = '1 + '2 where ; R solves that homogeneous equation. 2

Proof. The proof is immediate,

'3 (f (x)) = '1 (f (x)) + '2 (f (x)) =

g (x) '1 (x) + g (x) '2 (x) = g (x) '3 (x) ; and thus '3 solves (3.2). For a linear functional equation (3.2), properties of the functions g and f are very important.

De…nition 23. For a given linear functional equation, we de…ne Gn as

n 1 i Gn (x) = g f (x) ; n N; (3.3) i=0 2 Y and, provided the limit exists in I; we de…ne G (x) by

G (x) = lim Gn (x) : n !1 The behavior of this function is important for the appearance of the sets of solutions. We will consider three di¤erent cases and show that the behavior of solutions is quite di¤erent. Case (i): The limit of G (x) exists and is not 0 in I: Case (ii): There exists some interval J I such that G (x) = 0 in J and J: Finally, we have case (iii) which says simply that neither (i) or (ii) occurs. These2 cases will in the following part of this paper only be referred to as case (i) ; (ii) and (iii) ; due to their importance. 3.1.1. Case (i) Theorem 24. Suppose we have homogenous linear functional equation with the 0 case (i) satis…ed, g; F [I] and f T [I] : Then, for each initial value ; there exists exactly one2 solution for each2 value ' () = : 2 Proof. Since f : I I; we can use iterates to obtain, after p iterates, ! ' f p+1 (x) = g (f p (x)) ' (f p (x)) : (3.4) We will preform now the proof by induction, with the induction hypothesis ' (f n (x)) ' (x) = n N; (3.5) Gn (x) 2 and the basis n = 1: For n = 1; our claim is that ' (f 1 (x)) ' (f (x)) ' (x) = = : G1 (x) g (x) Since g (x) = 0; this is equivalent to (3.2), and thus is true. Suppose now that (3.5) holds for6 some n = p; ' (f p (x)) ' (x) = : Gp (x) It follows from (3.4) that ' (f p+1 (x)) = ' (f p (x)) : g (f p (x)) Using the induction hypothesis and (3.4), we obtain ' (f p (x)) ' (f p+1 (x)) ' (x) = = p ; (3.6) Gp (x) g (f (x)) Gp (x) which can be written as ' (f p+1 (x)) ' (x) = : Gp+1 (x) Finally, the induction says that this is true for all n 1: Since the limit of G (x) exists and f converges by Theorem 18, we can let n to obtain ! 1 ' () ' (x) = : G (x) For any value R; we …nd a unique solution as 2 ' (x) = : G (x) Since G (x) is supposed to be continuous, so is ' (x) : We also note that g () must be equal to 1 because otherwise, G () would not converge to a nonzero value and (i) would not occur.

3.1.2. Case (ii) An explicit solution is hard to …nd in this case. The reason is mainly because 0 the functions in our set T [I] are not necessarily strictly increasing. In case (ii) ; lack of this property makes it harder to …nd solutions. In Kuczma’s work [1] 0 corresponding theorems are established for the functions from R [I] ; which is a much more restrictive set. The mathematics of this will be left out here, but the details can be found in [1, pp. 49-50]. It is a result of the fact that, for functions in 0 R [I] ; monotonicity property is very important for deducing the following: Every continuous function '0 (x) ; de…ned in some subset [f (x0) ; x0] or [f (x0) ; x0] of I; can be uniquely extended to J I provided that '0 is a solution to (3.2) for x0: J is the part of I where x or x; respectively. This is not possible in our 0 case, but we provide a useful result tailored to the functions from T [I] :

Theorem 25. Suppose that Gn (x) converges uniformly to zero when n ; ! 1 in some interval J I; where J: Then, for each continuous function '0 that is a solution to the linear functional2 equation in some subinterval K = [c; k ( c) + ] ; where c < ; '0 (x) can be continuously extended to ' (x) on the entire interval I; and this extension is unique. Furthermore, ' () = 0: Proof. To prove that ' () = 0; we recall that identity (3.5) is equivalent to n ' (x) Gn (x) = ' (f (x)) : Let n and suppose that x J: Then we have, due to Theorem 18, ! 1 2 ' (x) 0 = ' () = 0: To prove the possible extension to I; observe that Lemma 16 ensures that f : K K: Since by Theorem 18 f converges to ; we conclude that, for each x I; ! 2 there exists an N N such that f N (x) K: Equation (3.5) yields 2 2 ' (f n (x)) ' (f n (x)) ' (x) = = 0 ; n N: Gn (x) Gn (x) This is a well de…ned function since '0 (x) is known, and it is unique since f : K K. Continuity follows easily from the fact that both f and g are continuos functions! applied a …nite number of times to obtain ' (x). 0 The fact that f T [I] makes it hard to …nd solutions to nonhomogeneous equations if we meet2 the case (ii); something we will experience in the following subsection as well.

3.1.3. Case (iii)

This condition on Gn is very restrictive, even though it might not seem like it.

Theorem 26. Assume that Gn (x) neither converges to a non-zero value, nor it converges uniformly to zero in a interval J I: Then, ' (x) = 0 is the only solution of (3.2) in I:

Proof. Suppose that ' (x) = 0 for some x I; then (3.5) yields 6 2 ' (f n (x)) G (x) = : n ' (x)

Since ' [f n (x)] does not diverge and ' (x) = 0; we can let n to obtain 6 ! 1 ' () lim Gn (x) = : n ' (x) !1 If ' () = 0; we have case (i) since G (x) would exist and be nonzero. Therefore, we must6 have that ' () = 0: We have an interval = J I such that ' (x) = 0; otherwise case (ii) would occur. Since ' (x) = 0 and2 '(x) is continuous, there6 exists a constant C such that 6

' (x) > C > 0; for x J: j j 2 Furthermore, for each given > 0; there exists a c < such that

' (x) < C; for x [c; (k + 1) kc] : j j 2 Due to Theorem 18 and Lemma 16, we can also …nd an N such that

f n (x) [c; (k + 1) kc] ; for x J and n N: 2 2 But then, f n (x) < C; and (3.5) gives us j j ' (f n (x)) Gn = j j < ; for x J and n N: j j ' (x) 2 j j Thus, we have uniform convergence to zero in the interval [c; (k + 1) kc] and the case (ii) : Therefore, the only possible solution for the homogenous equation is the trivial solution.

3.2. Criteria for Gn

Since the behavior of Gn is important, we will …nd now some criteria that help to determine which of the cases occurs.

0 Theorem 27. Suppose that g () < 1; g [I] ; and f T [I] : Then, Gn (x) converges uniformly to zero inj somej interval2J I as n 2 : ! 1 Proof. Since g () < 1 and g (x) is continuos, there exists a > 0 such that, for x J; J =j [ j; + k] ; we have that g (x) < C < 1 for some C: De…ne c = 2 ; then we have J = [c; (k + 1) kc]j : Sincej f converges to ; according to Theorem 18, f : J J and, due to Lemma 16, there exists an N such that ! f n (x) [c; (k + 1) kc] ; for x J and n N: 2 2

We evaluate now Gn in J;

n 1 N 1 n 1 i i i Gn (x) = g f (x) = g f (x) g f (x) j j i=0 i=0 i=N Y Y Y N 1 i n N < g f (x) C :

i=0 Y

The …rst product is bounded since it is a …nite product of …nite elements and the other approaches zero as n ; so case (ii) will occur. A similar proof is carried! out 1 for g () 1: j j Theorem 28. Suppose that there exists an interval J = [c; (k + 1) kc] such that g (x) 1; g () = 1; g (x) [I] and f T 0 [I] : Then, the case (iii) occurs. j j 6 2 2 Proof. We know that g () 1 in J: Furthermore, according to Theorem 18, f converges to and, byj Lemmaj 16, f : J J: Therefore, there exists an N such that ! f n (x) [c; (k + 1) kc] ; n N: 2 We evaluate Gn (x) to show that (ii) can not occur,

n 1 N 1 n 1 i i i Gn (x) = g f (x) = g f (x) g f (x) j j i=0 i=0 i=N Y Y Y N 1 g f i (x) > 0: i=0 Y

Hence, the case (ii) does not occur and, since g () = 1; neither can we have the case (i) : 6 There is, however, not an equally simple way to verify that (i) occurs. The reason is that the case where g () = 1 can generate all three cases. Loosely speaking, the case (ii) occurs when g (x) is much less than 1 for all x close to and the case (iii) occurs when g (x)j hasj much larger value than 1 for all x close to : Both conditions also dependj onj the convergence of f: We provide stronger conditions to ensure convergence however.

0 Theorem 29. Assume that f T [I] and g [I] : Suppose also that there exists a c < such that, for x [2c; (k + 1) kc2] ; we have f (x) x ; where 0 < < 1 is some constant.2 Then the case (i) occursj provided j thatj therej exist positive constants and M such that

g (x) 1 M x ; for x [c; k ( c) + ] (3.7) j j j j 2 and for some c < :

Proof. Since f (x) converges to ; according to Theorem 18, there exists an N such that, for a …xed value x I; 2 f n (x) [c; k ( c) + ] ; for n N: 2 But then, using conditions (3.7), we have

n n (n N) N g (f (x)) 1 M f (x) M f (x) : j j j j

Thus, 1 g f i (x) i=0 Y will converge uniformly in I and we have the case (i) :

3.3. Particular solution

The structure of solutions to nonhomogeneous linear functional equations is much like that in di¤erential equations; given the solution to a homogeneous equation, the particular solution to a nonhomogeneous equation is unique.

Proposition 30. Given the solution to the corresponding homogenous equation, a particular solution to (3.1) is unique.

Proof. In Proposition 21, we have shown that the di¤erence '3 of two functions '1 and '2 solving (3.1) is a solution to equation (3.2). Then, supposing that the homogenous part is the same in both equations, we have '3 = 0; and thus, '1 = '2: It is not always easy to …nd the general solution. Recall that the only solution of (3.2) in the case (iii) is the trivial one, and thus there exists only one solution to (3.1).

3.3.1. Unique continuos solution Theorem 31. Suppose that g (x) 1 in J = [c; (k + 1) kc] for some c < : j j 0 Assume further that g () = 1 and f T [I] ; while g; F [I] : Then, the only continuos solution to (3.1)6 is given by2 2

1 F (x) 1 F (f i (x)) F (f i 1 (x)) ' (x) = + + : (3.8) 2 g (x) G (x) G (x) i=1 i+1 i ! X Proof. By Theorem, 28 case (iii) occurs and we thus have a unique continuous solution. Therefore, we only need to show that equation (3.8) de…nes a continuous solution. First we will show that ' (x) converges uniformly in I: Suppose that g (x) > 1; then, since g is continuous, there exists an interval K = [d; (k + 1)j kdj ] I such that 1 < C < g (x) ; for some constant C and j j x K: Theorem 18 and Lemma 16 claim that for some x I; there exists an N such2 that 2 f n (x) [d; (k + 1) kd] ; n N: 2 Now let D = maxK (F (x)) : We evaluate ' (x) to prove that ' (x) does not diverge. j j

1 F (x) 1 F (f i (x)) F (f i 1 (x)) ' (x) = + + : j j 2 g (x) Gi+1 (x) Gi (x) ! i=1 X By the triangle inequality,

1 F (x) 1 1 F (f i (x)) F (f i 1 (x)) ' (x) + + : j j 2 g (x) 2 Gi+1 (x) Gi (x) i=1 X Applying the triangle inequality once again, we obtain

1 F (x) 1 N F (f i (x)) F (f i 1 (x)) ' (x) + + + j j 2 g (x) 2 G (x) G (x) i=1 i+1 i X i i 1 1 1 F (f (x)) F (f (x)) + : 2 Gi+1 (x) Gi (x) i=N+1 X

The …rst two terms are clearly …nite, so we only have to prove that the last term is …nite and converges uniformly. Observe that

1 1 F (f i (x)) F (f i 1 (x)) + 2 Gi+1 (x) Gi (x) i=N+1 X 1 1 F (f i (x)) 1 1 F (f i 1 (x)) + : 2 Gi+1 (x) 2 Gi (x) i=N+1 i=N+1 X X D We use now the de…nition of to derive

1 1 F (f i (x)) F (f i 1 (x)) + 2 Gi+1 (x) Gi (x) i=N+1 X 1 1 D 1 1 D

i+1 N + i N ; 2 GN C 2 GN 1 C j j i=N+1 j j i=N+1 X X

which proves that the sum converges uniformly since C > 1 according to Weier- i 1 1 strass M-test. Furthermore since F (f (x)) Gi (x) is continuous for i N; the uniform limit theorem says that ' (x) is a continuous function. We will prove now convergence when g () = 1: In this case, according to our assumptions, g () = 1: Since g (x) j 1 inj J; we have that g (x) 1; due to continuity. For some x I; Theoremj 18j and Lemma 16 claim that there exists an N such that 2 f n (x) J; for n N: 2 We will evaluate ' (x) again to prove that it is …nite and it converges uniformly in I; j j 1 F (x) 1 F (f i (x)) F (f i 1 (x)) ' (x) = + + : j j 2 g (x) Gi+1 (x) Gi (x) ! i=1 X By the triangle inequality,

1 F (x) N F (f i (x)) F (f i 1 (x)) ' (x) + + + 2 g (x) G (x) G (x) j j i=1 i+1 i ! X i i 1 1 F (f (x)) F (f (x)) + : Gi+1 (x) Gi (x) i=N+1 X The …rst term is obviously …nite, so we only have to investigate the second term,

1 F (f i (x)) F (f i 1 (x)) + Gi+1 (x) Gi (x) i=N+1 X 1 g (f i (x)) F (f i (x)) + F (f i 1 (x)) = : Gi+1 (x) i=N+1 X i Since g (f (x)) 1; for i N; and Gi+1 (x) is alternating for i N;' (x) will converges uniformly. Thus, we have shown that the sum is a well-de…ned continuous function. It remains to show that (3.8) actually solves (3.1). Consider

1 F (f (x)) 1 F (f i (f (x))) F (f i 1 (f (x))) + + 2 g (f (x)) G (f (x)) G (f (x)) i=1 i+1 i ! X g (x) F (x) 1 F (f i (x)) F (f i 1 (x)) = + + + F (x) : 2 g (x) G (x) G (x) i=1 i+1 i ! X We simplify by evaluating functions and using the de…nition of Gn (x) ;

1 F (f (x)) 1 F (f i+1 (x)) g (x) F (f i (x)) g (x) + + 2 g (f (x)) G (f (x)) G (f (x)) i=1 i+2 i+1 ! X 1 1 F (f i (x)) F (f i 1 (x)) = F (x) + g (x) + + F (x) : 2 G (x) G (x) i=1 i+1 i ! X Transpose the terms in the equation,

1 F (f (x)) 1 F (f i+1 (x)) F (f i (x)) + g (x) + 2 g (f (x)) G (f (x)) G (f (x)) i=1 i+2 i+1 X 1 F (f i (x)) F (f i 1 (x)) 1 g (x) + = F (x) : G (x) G (x) 2 i=1 i+1 i !! X Multiplying by 2 and rearranging the sums, we have F (f (x)) 1 F (f i+1 (x)) F (f i (x)) + g (x) + g (f (x)) G (f (x)) G (f (x)) i=1 i+2 i+1 X 1 F (f i (x)) F (f i 1 (x)) F (f (x)) F (x) + + + = F (x) : G (x) G (x) G (x) G (x) i=2 i+1 i 2 1 !! X Rearranging the terms changing the index of the summation, we obtain, F (f (x)) F (f (x)) F (x) g (x) + + g (f (x)) G (x) G (x) 2 1 1 F (f i+1 (x)) F (f i (x)) g (x) + G (f (x)) G (f (x)) i=1 i+2 i+1 X 1 F (f i+1 (x)) F (f i (x)) + = F (x) : G (x) G (x) i=1 i+2 i+1 !! X By the de…nition of Gn (x) ; The latter equation can be simpli…ed as follows, F (f (x)) F (f (x)) + F (x) = F (x) : g (f (x)) g (f (x)) This is obviously true, and thus, equation (3.8) de…nes the only continuous solution to equation (3.1). 3.3.2. Arbitrary constant solution We will construct now a corresponding solution in the case where g (x) = 1 in some neighborhood of :

Theorem 32. Suppose that g (x) = 1 in J = [c; (k + 1) kc] ; for some c < 0 and f T [I] ; and let g; F [I] : Then, the continuos solution of (3.1) is, if it exists,2 given by 2 1 F [f i (x)] ' (x) = ; (3.10) G (x) G (x) i=0 i+1 X where : 2 Proof. First, we show that we have the case (i) : Theorem 18 and Lemma 16 yield that, for each x I; there exists an N N such that 2 2 f n (x) J; n N: 2 Thus, N 1 1 1 i i i lim Gn (x) = g f (x) = g f (x) g f (x) : n !1 i=0 i=0 i=N Y Y Y Since, for each i N; we have g (f i (x)) = 1; N 1 i lim Gn (x) = g f (x) = G (x) : n !1 i=0 Y 1 Thus, we have case (i) : For the homogenous equation, G (x) is the solution, so if we can show that 1 F [f i (x)] ' (x) = p G (x) i=0 i+1 X solves (3.1), so does ' (x) ; and this is the unique one-parameter family of solutions. Evaluating 'p (x) in (3.1), we have

1 F [f i+1 (x)] 1 F [f i (x)] = g (x) + F (x) : G (f (x)) G (x) i=0 n+1 i=0 i+1 X X Writing the sum in the form

1 F [f i+1 (x)] 1 F [f i (x)] = F (x) g (x) + F (x) G (f (x)) G (x) i=0 i+1 i=1 i+1 X X and simplifying it by using the de…nition of Gn (x) ; we obtain

1 F [f i+1 (x)] 1 F [f i (x)] = : G (f (x)) G (f (x)) i=0 i+1 n=1 i X X Therefore, if the series converges, it de…nes the only solution for that speci…c value ; otherwise, there exists no solution. We will investigate convergence of the function de…ned by (3.10) in I:

Theorem 33. Assume that assumptions of Theorem 32 are satis…ed for equation (3.1). Then, (3.10) is a well-de…ned continuous solution provided that F () = 0 and that we can …nd a function H (x) and C < 1 such that, for some c < ;

F (x) F () H (x) ; for x [c; k ( c) + ] ; j j 2 and H (f (x)) < CH (x) ; for = x [c; k ( c) + ] : 6 2 Proof. We will show that (3.10) converges uniformly if assumptions of the theorem are satis…ed. Since, by Theorem 18 and Lemma 16, f converges for each x; there exists an N such that

f n (x) [c; k ( c) + ] ; n N: 2 It follows from the assumptions that

n n n n 1 F (f (x)) = F (f (x)) F () H (f (x)) < CH f (x) : j j j j By induction, this yields

n n N N F (f (x)) < HC f (x) : (3.11) j j We evaluate (3.10) now

1 F [f i (x)] ' (x) = : j j G (x) Gi+1 (x) i=0 X

By the triangle inequality,

N 1 F [f i (x)] 1 F [f i (x)] ' (x) + : j j G (x) Gi+1 (x) Gi+1 (x) i=0 i=N X X

Factoring out GN ; we obtain

N 1 F [f i (x)] 1 1 F [f i (x)] ' (x) + : i j j G (x) Gi+1 (x) GN i=0 i=N g (f j (x)) X X j=N Y

Taking into account that f n (x) J; we have, by (3.11), 2 N 1 i 1 F [f (x)] 1 n N N ' (x) + C H f (x) : j j G (x) Gi+1 (x) GN i=0 i=N X X

Since C < 1;' (x) converges uniformly and thus, (3.10) is a continuous function, once again due to the Weierstrass M-test and the uniform limit theorem. Letting n ; we observe that, by Theorem 18 and Lemma 16, the identity (3.11) yields F !() 1 = 0: So if F () = 0; we cannot …nd a function H with desired properties. 6

3.3.3. Arbitrary functions solution The case not discussed so far is the one where g () < 1 and that case is not j j 0 easy to study. The reason is that we look for solutions when f T [I] ; and thus f is not necessarily a strictly increasing function. Hence, much2 of the theory in Kuczma’s work [1] is not applicable. We will, however, establish some results using the fact that Theorem 27 yields that (ii) occurs whenever g () < 1: j j Lemma 34. Suppose that g () < 1: Then, each solution ' to the equation (3.1), where g; F [I] andjf jT 0 [I] ; has the property 2 2 F () ' () = : 1 g () Proof. By (3.1), we get ' (f ()) = g () ' () + F () : Since is a …xed point of f; ' () = g () ' () + F () and, by simple algebra, we get the result. As in the homogeneous case, we suppose that we have a solution in the subinterval [c; k ( c) + ] I; where c < : Theorem 35. Assume that g () < 1, g; F [I] and f T 0 [I] : Suppose j j 2 2 also that we have a function '0 de…ning a continuous solution to (3.1) in K = [c; k ( c)] ; where c < and K I: Then, we can uniquely extend '0 to a solution to (3.1) in I:

Proof. It follows from Theorem 18 and Lemma 16 that, given an x = I; there 2 exists an n N such that 2 f n (x) K; n N: 2 We use the induction to show that n 1 n 1 n i j ' (f (x)) = Gn (x) ' (x) + F f (x) g f (x) : (3.12) i=0 j=i+1 X Y Namely, let (3.12) be our induction hypothesis with the basis n = 1: For the basis, we have 0 ' (f (x)) = G1 (x) ' (x) + F f (x) 1; which is exactly equation (3.1) and thus is satis…ed. Suppose now that our claim is true for p 1: By (3.1), ' f p+1 (x) = g (f p (x)) ' (f p (x)) + F (f p (x)) : By the induction assumption (3.12) is true for p; and thus we have

p 1 p 1 p+1 p i j ' f (x) = g (f (x)) Gp (x) ' (x) + F f (x) g f (x) i=0 j=i+1 ! X Y + F (f p (x)) : By the de…nition of Gn (x) ;

p+1 ' f (x) = Gp+1 (x) ' (x) p 1 p 1 + F f i (x) g (f p (x)) g f j (x) + F (f p (x)) i=0 j=i+1 X Y = Gp+1 (x) ' (x) p 1 p + F f i (x) g f j (x) + F (f p (x)) : i=0 j=i+1 X Y Finally, we add the last term to the sum to obtain

p p p+1 i j ' f (x) = Gp+1 (x) ' (x) + F f (x) g f (x) ; i=0 j=i+1 X Y and thus induction proves the formula (3.12). Since Gn (x) is non-zero as long as we do not let n ; we have by (3.12) ! 1 n 1 n 1 1 ' (x) = ' (f n (x)) F f i (x) g f j (x) : G (x) n i=0 j=i+1 ! X Y n n Since we know that '0 (f (x)) = ' (f (x)) ; the latter equation can be simpli…ed to n 1 ' (f n (x)) F (f i (x)) ' (x) = 0 : G (x) G (x) n i=0 i+1 X This a unique way to extend '0: It is continuous since there are only a …nite number of operations and '0 is assumed to be continuous in K: 4. Transformations

Since all our work in previous chapter was to …nd solutions to equations of the form (3.1), it might be interesting to …nd some cases where we can transform a given functional equation to a linear functional equation. These transformations can be rather complex, so we will only discuss simple transforms. We start by reducing a more general linear functional equation to (3.1).

4.1. Transformation to simpler linear form

Suppose we have a linear functional equation,

g2 (x) ' (f (x)) = g1 (x) ' (h (x)) + F (x) ; (4.1) de…ned in a real interval I with the endpoints a and b; where a; b might be in…nite. 1 Furthermore, we suppose that h (x) has an inverse in I; namely, h (x) ; and that g2 (x) is not equal to zero. Even though (4.1) is not in the form (3.1), it is obvious that it is quite similar. We de…ne an interval J as h (I) = J. This will be an interval since h (x) is continuos and invertible, the end points of J will be h (a) and h (b) ; though the order might be reversed. De…ne now z = h (x); since h (x) 1 is invertible, we have x = h (z) : We apply now this de…nition of x to write equation (4.1) as

1 1 1 1 1 g2 h (z) ' f h (z) = g1 h (z) ' h h (z) + F h (z) ; which is equivalent to

1 1 1 g1 (h (z)) ' (z) F (h (z)) ' f h (z) = 1 + 1 : g2 (h (z)) g2 (h (z)) 1 Since f; h ; g1; g2 and F are known functions, we de…ne

1 ^ 1 1 1 f h (z) = f (z) ; g1 h (z) g2 h (z) = g (z) ; 35 and 1 1 1 F h (z) g2 h (z) = F^ (z) and write equation (4.1) as

' f^(z) = g (z) ' (z) + F^ (z) : (4.2) All functions in (4.2) are de…ned in our new interval J: Since (4.2) is in the form (3.1), all the theory developed for linear functional equations can be used. First, we check whether the functions de…ned in (4.2) satisfy our conditions for equation (3.1). Observe that even when the functional equation is in the desired form (3.1), we can obtain (4.2) as long as f (x) is invertible. Then, (4.2) yields

1 1 ' (z) F (f (z)) ' f (z) = 1 1 : g (f (z)) g (f (z)) Thus, another functional equation in the form (3.1) can be studied, provided that 1 0 f T [I] : 2 4.2. Transformation of a general equation

Another way to transform a functional equation to a linear functional equation is by de…ning some new function c ( (d (x))) = ' (x) ; where c and d are selected to satisfy certain conditions, while is a new unknown function. Sometimes this can lead to an equation of the form (3.1) or (4.1) with the unknown function ; these transforms are inspired by [2, p. 55-57]. We illustrate this idea with the example that follows. Let ' (f (x)) = eF (x) (' (x))g(x) ; where f (x) T n [I] ; g (x) ;F (x) [I] ; for I: 2 2 2 We de…ne as (x) = log (' (x)) a(x) = ' (x) ; a () where 1 = a > 0 is arbitrary. Using this transform, we have 6 g(x) eF (x) (' (x)) = eF (x)ag(x)(x) = ag(x)(x)+F (x) loga(e) = ' (f (x)) = a(f(x)); and the linear equation assumes the form

g (x) (x) + F (x) loga (e) = (f (x)) :

We can see now that we have found a way write a given equation in the form of a linear functional equation with the help of a one-to-one transformation. It is, however, not always possible to deduce a linear functional equation by these means and it is not always obvious which transformation should be attempted to reduce the equation to the form (3.1) or (4.1), even if this is possible. 5. Examples

To make the subject less abstract, we provide some examples related to the two major parts of this work, convergence of iterates and solutions to the linear functional equation. We start by introducing four di¤erent functions and show that 0 0 0 0 these functions belong to one the four sets R [I] ;S [I] ;P [I] ; and T [I] : Once we prove that these functions actually belong to a given set, we use them to construct solutions to linear functional equations.

5.1. Convergence

0 5.1.1. A function from R [I] We de…ne f and I as follows:

f (x) = log(x + 1);I = [0; 50]; = 0: (5.1) 0 To prove that this function is in the set R0 [I] ; we start by noting that it is continuous and also analytic in this interval. We can show that f is increasing by computing its derivative, 1 f 0 (x) = > 0 in I: x + 1 Thus, f is strictly increasing. We need to prove now that f satis…es (1.2) and (1.3) and that = 0 is a …xed point, f (0) = log(1) = 0: In order to show (1.2), note that, for x > 0; (log(x + 1) x)( x) > 0 log(x + 1) x < 0: () Similarly, to show (1.3), observe that, for x > 0; (log(x + 1)) ( x) < 0 log(x + 1) > 0: () Thus, f R0 [I] : 2 0 38 f x

n Figure 5.1: A function f R [I] ; as well as the boundaries y1 = x and y2 = . 2

0 5.1.2. A function from S [I] We de…ne f and I as follows:

cos(x) + 2x 1 f (x) = ;I = [ 5; 5]; = 0: (5.2) 3 0 We start by showing that this function belongs to S0 [I] : Note …rst that f is continuous. Since is an inner point satisfying (1.2) and (1.3), f () = due to continuity. First, we verify (1.2):

cos(x) + 2x 1 x ( x) > 0; x = 0; 3 6 or, in a simpli…ed form,

x + 1 cos(x) x > 0; x = 0: 3 6 We divide now the interval into three parts and use di¤erent methods to ensure that (1.2) is satis…ed. For x 2; it is easy to see that x + 1 cos(x) < 0 since 1 cos(x) 2: For x = 2; note that cos(2) = 1; and thus, x + 1 cos(x) < 0; for x 2: Since x < 0; we have 6 x + 1 cos(x) x > 0; x 2; 3 as we wanted to prove. For 2 < x < 0; we use …rst the identity and the fact that, x = 0 6 x 2 x2 1 cos (x) = 2 sin < : 2 2 Hence, get the inequality x2 x2 1 cos (x) < 1 cos (x) < 0: 2 () 2 Using this result, we have

2 x + 1 cos(x) x + 1 cos(x) x + 1 cos(x) x > x 2 x 3 3 3 x x2 x2 x3 = 2 x = 2 > 0; 2 < x < 0; 3 3 and we have the desired inequality. For the last case, x > 0; we note that 0 1 cos (x) and x + 1 cos(x) x + 1 cos(x) 1 cos(x) x2 x x x = > 0; 3 3 3 3 so (1.2) is true for all x: To check (1.3), we have to show that cos(x) + 2x 1 ( x) < 0; x = 0: 3 6 For x < 0; since 0 cos(x) 1; we have cos(x) + 2x 1 cos(x) + 2x 1 ( x) ( x) 3 3 cos(x) 1 2x2 ( x) = < 0: 3 3 For 0 < x < 2; cos(x) + 2x 1 cos(x) + 2x 1 ( x) < ( x) 3 3 x2 1 + cos(x) 2x x2 2 ( x) = 2 x < 0: 3 3 For x 2; it is easy to see that cos(x) + 2x 1 > 0; and thus, cos(x) + 2x 1 ( x) < 0: 3 Thus, the function f satis…es (1.3) as well, hence f S0 [I] : 2 0 f x

n Figure 5.2: A function f S [I] ; as well as the boundaries y1 = x and y2 = . 2

0 5.1.3. A function from P [I] We de…ne f and I as follows:

f (x) = sin(x);I = [ 5; 5]; = 0: (5.3) 0 To prove that f P0 [I] ; we need to show (1.2), (1.4) and that f : I I. Due to Lemma 16, f :2I I;. To verify (1.2), suppose that x < 0: Then, ! ! (sin(x) x)( x) > 0 (sin(x) x) > 0; () and the inequality is true for all x < 0: Suppose now that x > 0; then

(sin(x) x)( x) > 0 (sin(x) x) < 0; () which is true for all x > 0: It remains to verify (1.4). Suppose …rst that x < 0; then

(sin(x) + x)( x) < 0 (sin(x) + x) < 0; () which is true for all x < 0: Suppose now that x > 0; then

(sin(x) + x)( x) < 0 (sin(x) + x) > 0; () 0 and the inequality is true for all x > 0: Thus, f P0 [I] since f (x) is obviously continuous. 2 f x

n Figure 5.3: A function f P [I] ; as well as the boundaries y1 = x and y2 = 2 x. 2

0 5.1.4. A function from T [I] We de…ne f and I as follows:

x 2 1; for x 0; x2 15 f (x) = 2 ; for 0 < x 1; I = 3; ; = 0: (5.4) 8 x 2 1 ; for x > 1; < 2

It is not obvious: that this function is even continuous, but limx 0 f (x) = 0 and 1 ! limx 1 f (x) = 2 ; and at any other point of I the function is clearly continuous. ! 0 The …xed point is easy to …nd as 0: To prove that f belongs the set T0 [I] ; we must show that (1.2) and (1.5) hold and f : I I. We start by demonstrating (1.2), which will be done in three steps, one for! each of the expressions de…ning f: Suppose that x < 0; then (1.2) gives

x x 2 1 x ( x) > 0 2 1 x > 0: () x Since x < 0; properties of the exponential function yield 1 < 2 ; and we have

x 2 1 x > x > 0; so the inequality is true. Now, for the interval 0 < x 1; we must verify x2 x2 x ( x) > 0 x < 0; 2 () 2 which can be easily done. For x > 1; we need to show that x 3x 1 x ( x) > 0 1 < 0: 2 () 2 Since x > 1; it is easy to see that the inequality is true. Since we have shown it for all subintervals, it is true in I as well. We have to show now (1.5). This is, however, more tricky since we must show the existence of a k > 0 such that both inequalities in (1.5) are veri…ed. Let us check whether k = 5=2 veri…es the inequalities (1.5). We start with the …rst inequality which only a¤ects the interval x where x < 0; so f (x) = 2 1 :

x 5 5 x 5 2 1 + 1 0 x ( x) < 0 2 1 + x < 0: 2 2 () 2 x 5 Since 2 1 2 x is a convex function, it is enough to prove that it is negative at the endpoints of the interval [3; 0) ; it might be equal to 0 at 0 since this is not a part of the interval. For x = 3; 5 1 23 1 3 = < 0; 2 2 and for x = 0; 0 5 2 1 0 = 0; 2 so the …rst inequality is satis…ed. We proceed with the part where 0 < x 1; then 5 x2 5 5x2 + 1 0 x ( x) < 0 + x > 0: 2 2 2 () 4 The latter is obviously true since x is positive. Finally, for x > 1; we have by (1.5)

5 x 5 10 x 1 + 1 0 x ( x) < 0 > 0; 2 2 2 () 4 which is true since x < 10 in I: Hence, we have found a k such that both inequal- 5 ities hold. Because I = 3; 2 3 ; Lemma 16, ensures that this function takes on values in I as well and thus, the function de…ned by (5.4) is in T 0 [I] : 0 f x

n Figure 5.4: A function f T [I] ; as well as the boundaries y1 = x and y2 de…ned by (1.5) 2

5.2. Solutions to linear functional equations

We will show now some explicit solutions of linear functional equations. Since we have de…ned in Section 5.1 four converging functions, we use them in the examples. To make it easier to follow, the examples are constructed in such a way that case (iii) is the only case that occurs, so there is always at most one solution to a given linear functional equation. We do this mainly for the sake of visualization. That is also why = 0 in all the examples. For other values of ; one can simply move the origin with a very simple transformation, z = x :

0 5.2.1. A function from R [I] For the function (5.1), we have chosen the linear functional equation

' (log(x + 1)) + ' (x) = sin (x) : (5.5)

We have g (x) = 1 in I and F (x) = sin (x) [I] : According to Theorem 31, equation (5.5) has a unique solution described2 by

1 F (x) 1 F (f i (x)) F (f i 1 (x)) ' (x) = + + : 2 g (x) G (x) G (x) i=1 i+1 i ! X x

2.0

1.5

1.0

0.5

x 10 20 30 40 50 0.5

1.0

1.5

Figure 5.5: Solution for equation (5.5) in I

Since g (x) = 1 and F (x) = sin (x) ; (5.5) becomes

1 1 i i i 1 ' (x) = sin (x) ( 1) sin log (x + 1) sin log (x + 1) ; 2 i=1 ! X and can be also written as

1 1 i i i 1 ' (x) = sin (x) + ( 1) sin log (x + 1) sin log (x + 1) : 2 i=1 ! X We let Mathematica provide the solution and also show how iterates approach the real solution.

0 5.2.2. A function from S [I] For the next example, we use the function de…ned by (5.2) and we have chosen the linear functional equation cos(x) + 2x 1 ' + x2 + 1 ' (x) = ex: (5.6) 3 We see that g (x) = (x + 1)2 1; g () = 1; and F (x) = ex [I] ; so by Theorem 31,j therej exists a unique solution to6 (5.6) de…ned by 2

f i(x) f i 1(x) 1 ex 1 e( ) e( ) ' (x) = + + : 2 g (x) G (x) G (x) i=1 i+1 i !! X x

x 4 2 2 4

Figure 5.6: Solution for equation (5.6) in I

We use Mathematica to visualize the solution.

0 5.2.3. A function from P [I] In this case, we want to solve an equation with the function de…ned by (5.3). We de…ne the linear functional equation as follows:

' (sin (x)) 2ex' (x) = cos (x) : (5.7) Observe that in the case (5.7), we have g (x) = 2ex; which means that g () = 2 > 1 and F (x) = cos (x) [I] ; so Theorem 31 provides the unique solution.j j 2

0 5.2.4. A function from T [I] The function (5.4) in T is the most complicated, but it is also continuous and de…ned in I: We use it for a linear functional equation

2 x 2 e 4 (2x + 1) ' (f (x)) + ' (x) = x : (5.8) 3 ! j j p We have 2 x 2 g (x) = e 4 2x + 1 ; x

x 4 2 2 4

Figure 5.7: Solution for equation (5.7) in I

1.5

1.0

0.5

x 1.0 0.5 0.5 1.0 1.5 2.0 2.5

Figure 5.8: Solution of equation (5.8) in 1; 5 2 In order to show that the values of g (x) for all x close to x = 0 will be less then negative one,.we investigate the second derivative of g (x) ;

2 0 4 2 e 4 (2 0 + 19 x + 14) 7 g00 (0) = = : 4 2 Since the value of the derivative is negative at 0; we have a maximum at 0 and g (0) = 1: There must therefore exist a neighborhood of 0 where g (x) 1; so, according to Theorem 31, we have the case (iii) : Taking into account that F [I] ; we know that there is a unique solution, and we use Mathematica to generate2 the solution. 6. Conclusions

Initially, the purpose of this work was to analyze formulas describing solutions to linear functional equations in di¤erent cases. After reading the work of Kuczma [1], I understood that the theoretical basis of the subject is much larger than I …rst presumed. Furthermore, a good starting point for the project was to de…ne functions that converge to a …xed point. Therefore, this paper is divided into two major parts, the …rst, where the purpose is to expand the sets with the convergence property, and the second, where the work of Kuczma [1] is partly adjusted to be used with these new sets. Since the theory of functional equations is not discussed in mathematical courses at a bachelor level, I had no notable understanding for the subject before this work. However, much of the theory in [1] is presented in a way that a student with some experience in analysis, di¤erential equations and set theory is able to comprehend. Personally, I believe that I have got a much wider perspective not only on the subject of functional equations, but also on a more general mathematical …eld of analysis, especially on the ways of solving mathematical problems with a non-direct approach through the development of the corresponding theory around the subject. Despite being largely based on Marek Kuczma’s book [1], this work provides also examples and historical facts related to functional equations along with the material related to iterative solutions from the book by Small [2] and websites [3]. A very important source of information is the book [2]. It provided an idea on how to transform some general functional equations to linear, as well as a formula to work with exponents. As a …nal statement, I can say that even though applications of functional equations is another interesting part of the subject, it has not been presented here due to a mostly theoretical standpoint of this report.

49 Bibliography

[1] Marek Kuczma, Functional Equations in a Single Variable, Polish Scienti…c Publishers, Warszawa, 1968

[2] Christopher G. Small, Functional Equations and How to Solve Them, Springer, New York, 2007

[3] http://eqworld.ipmnet.ru/en/solutions/fe/fe1208.pdf

[4] http://en.wikipedia.org/wiki/Gr%C3%A9goire_de_Saint-Vincent

[5] http://commons.wikimedia.org/wiki/File:CharlesBabbage.jpg

[6] http://ste…zu.wordpress.com/augustin-louis-cauchy-3

50 7. Appendix

See separate document: Appendix

51 Exit

@D R j[Log[x+1]] +j[x] =Sin[x], Convergens of Log[x+1] in I=[0,50],

In[1]:= fr x_ := Log x + 1

In[2]:= Manipulate Plot x, Nest fr, x, n , x, 0, 50 , AxesLabel ® x, f^n x , n, 1, 15, 1 @ D @ D

@ @8 @ D< 8 < 8 H L@ D

30 Out[2]=

x 10 20 30 40 50

j[Log[x+1]] +j[x] =Sin[x], solving for by j (4,7)

In[3]:= Manipulate Plot Sin x + Sum -1 ^i Sin Nest fr, x, i - Sin Nest fr, x, i - 1 , i, 1, n 2, x, 0, 50@ , AxesLabel ® x, j x , n, 0, 20, 1 @H @ D @H L H @ @ DD @ @ DDL 8

n j x

10 2.0H L

1.5

1.0 Out[3]= 0.5

x 10 20 30 40 50 -0.5

-1.0

-1.5

j[Log[x+1]] +j[x] =Sin[x], vertify solution for by j (4,7) 2 Examples.nb

In[4]:= Manipulate Plot Sin x , Sin fr x + Sum@ -1@8^i Sin Nest fr, fr x , i - Sin Nest fr, fr x , i - 1 , i, 1, n 2 + Sin@ Dx H+ Sum@ @-1DD^i Sin Nest fr, x, i - Sin Nest fr, x, i - 1 , i, 1, n 2 , x, 0, 50 @,H AxesLabelL H ®@ x,@F x @, Dn, 0D,D20, 1@ @ @ D DDL 8

0.5

Out[4]= x 10 20 30 40 50

-0.5

-1.0

¢ | £ Examples.nb 3

S j[(Cos[x]+2x-1)/3] +(x^2+1)j[x] =Exp[x], Convergens of (Cos[x]+2x-1)/3 in I=[-5,5],

In[5]:= fs x_ := Cos x + 2 x - 1 3

Manipulate Plot x, Nest fs, x, n , x, -5, 5 , AxesLabel ® x, f^n x , n, 1, 20, 1 @ D H @ D L

@ @8 @ D< 8 < 8 H L@ D

H L 4

x -4 -2 2 4

-2

-4

j[(Cos[x]+2x-1)/3] +(x^2+1)j[x] =Exp[x], solving for by j (4,7)

In[6]:= gs x_ := - x^2 + 1 Manipulate Plot@ D- ExpH x gsLx + Sum Exp Nest fs, x, i Product gs Nest fs, x, j , j, 0, i + Exp@ Nest fs, x, i - 1 Product gs Nest fs, x, j , j, 0, i - 1 , i, 1, n 2,@ xH, -5@, 5D , AxesLabel@ D @®H x,@ j x @ , n, 0D,D20 H, 1 @ @ @ DD 8

10 H L 5

Out[7]= 3

x -4 -2 2 4

j[(Cos[x]+2x-1)/3] +(x^2+1)j[x] =Exp[x], vertify solution for by j (4,7) 4 Examples.nb

In[8]:= Manipulate Plot Exp x , - Exp fs x gs fs x + Sum Exp Nest fs, fs x , i @ Product@8 @ Dgs Nest fs, fs x , j , j, 0, i + Exp Nest fs, fs x , i - 1 H @ @ DProductD @ gs@ DNestD fs@H, fs @x , j@ , j,@0,D i -D1D , i, 1, n 2 + gs x Exp x Hgs x + Sum@ @ Exp@Nest fs@ ,Dx, iDD 8 Product

n F x

H L 40

30 Out[8]=

x -4 -2 2 4

¢ | £ Examples.nb 5

P j[Sin[x]] -2Exp[x]j[x] =Cos[x], Convergens of Sin[x] in I=[-5,5],

In[9]:= fp x_ := Sin x Manipulate Plot@ D x, -x,@ NestD fp, x, n , x, -5, 5 , AxesLabel ® x, f^n x , n, 1, 20, 1 @ @8 @ D< 8 < 8 H L@ D

H L 4

Out[10]= x -4 -2 2 4

-2

-4

j[Sin[x]] -2Exp[x]j[x] =Cos[x], solving for by j (4,7)

In[11]:= gp x_ := 2 Exp x Manipulate Plot@ D- Cos x @ Dgp x + Sum Cos Nest fp, x, i Product gp Nest fp, x, j , j, 0, i + Cos@ Nest fp, x, i - 1 Product gp Nest fp, x, j , j, 0, i - 1 , i, 1, n 2,@ xH, -5@, 5D , AxesLabel@ D @®H x,@ j x @ , n, 0D,D20 H, 1 @ @ @ DD 8

n j x

10 10 H L

x -4 -2 2 4

Out[12]= -10

-20

-30

j[Sin[x]] -2Exp[x]j[x] =Cos[x], vertify solution for by j (4,7) 6 Examples.nb

In[13]:= Manipulate Plot Cos x , - Cos fp x gp fp x + Sum Cos Nest fp, fp x , i @ Product@8 @ Dgp Nest fp, fp x , j , j, 0, i + Cos Nest fp, fp x , i - 1 H @ @ DProductD @ gp@ DNestD fp@H, fp @x , j@ , j,@0,D i -D1D , i, 1, n 2 + gp x Cos x Hgp x + Sum@ @ Cos@Nest fp@ ,Dx, iDD 8 Product

n F x 1.0 H L

0.5

Out[13]= x -4 -2 2 4

-0.5

-1.0

¢ | £ Examples.nb 7

T j[ft[x]] +((Exp[-(x^2)/4](2x^2+1))/3)j[x] =Sqrt[Abs[x]], Convergens of Sin[x] in I=[-3,7.5],

In[14]:= p x_ := If x < 0, 2^ -x - 1, x^2 2 ft x_ := If x < 1, p x , 1 - x 2 r@x_D := If@x < 0, -2.5H xL, -2 xH 5 L D Manipulate@ D @ @ D D @PlotD x, r@ x , Nest ft, x, n ,D x, -3, 7.5 , AxesLabel ® x, f^n x , n, 1, 20, 1 @ @8 @ D @ D< 8 < 8 H L@ D

H L 6

4 Out[17]=

x -2 2 4 6

-2

j[ft[x]] +((Exp[-(x^2)/4](2x^2+1))/3)j[x] =Sqrt[Abs[x]], solving for by j (4,7)

In[18]:= gt x_ := - Exp - x^2 4 2 x^2 + 1 3 Manipulate Plot - Sqrt Abs x gt x + @ D Sum HHSqrt@AbsH NestL ftD H, x, i LL ProductL gt Nest ft, x, j , j, 0, i + Sqrt@ Abs@ H Nest@ ft,@ xD,Di- 1@ D Product gt Nest ft, x, j , j, 0, i - 1 , i@,H1, n @ 2@, x,@-1, 2.5 D,DDAxesLabel H ®@ x@, j x@ , n, 0D,D 208, 1

n j x

10 1.5H L

1.0 Out[19]=

0.5

x -1.0 -0.5 0.5 1.0 1.5 2.0 2.5 8 Examples.nb

j[ft[x]] +((Exp[-(x^2)/4](2x^2+1))/3)j[x] =Sqrt[Abs[x]], vertify solution for by j (4,7)

In[20]:= Manipulate Plot Sqrt Abs x , - Sqrt Abs ft x gt ft x + Sum@ Sqrt@8 Abs@ Nest@ DftD , ft x , i Product gt Nest ft, ft x , j , j, 0, i + H @ Sqrt@ @AbsDDDNest @ft,@ ftDD x , i - 1 @H Product@ @gt Nest@ ft, @ftD x ,DDjD ,H j, 0, i@- 1@ ,@ i, 1, @n D D2D+ 8

n F x 4 H L

Out[20]= 2

x -2 2 4 6

¢ | £