Periodic Functions On Non-Linear Temporal Models
By Alexia Puig
Over the course of this semester, I have been reading and working with two of Dr. Devito’s papers, “A Non-Linear Model For Time” and “Time Scapes.” These papers define time as a partially ordered set of instants. To define a partially ordered set, let I be an infinite set and let ≤ be a relation on the elements of I.[1] Then ≤ is a partial ordering of I, and the pair (I,≤) is a partially ordered set if:
a) for every i Î I we have i≤i
b) if i and j are in I and we have both i≤j and j≤i, then i = j
c) if i, j and k are in I and we have i≤j and j≤k, then i≤k
There is a duration function that assigns a non-negative real number for each pair of comparable instants.[2] This function is called duration or dur, for short and has the following properties:
1) dur(x,y) = p, where p is the “time” between x and y (two instants)
2) dur(x,y) = dur(y,x), and if x=y, then dur(x,y) = 0
3) if y is between x and z, meaning x≤y≤z or z≤y≤x, then dur(x,z) = dur(x,y) + dur(y,z).
I have been looking at functions defined on time tracks and studying their mathematical properties. A time track, T, is a non-empty subset of the infinite set of instants and has the following properties[3]:
a) Any two instants on T can be compared.
b) If x, y, and z are on T and y is between x and z, then dur(x,z) = dur(x,y)
+ dur(y,z).
c) Given y on T, and a positive, real number p, there are exactly 2 instants
x and z on T such that dur(x,y) = p and dur(y,z) = p.
I am trying to extend the idea of periodicity and integrability to functions defined on time tracks. Since instants cannot be added, this requires some new ideas. I was successful in extending periodicity using the translation function. This is defined as follows:
Definition 1: There is a translation function, tp, for any fixed number p, which maps T to T as follows:[4]
1) t0(x) = x for all x Î T
2) if p>0, tp(x) = y where y is the unique instant on T such that x<y and dur(x,y) = p
3) if p<0, tp(x) = z where z is the unique instant on T such that z<x and dur(z,x) = |p|
Then, a real number p is a period of the function f if f[tp(x)] = f(x) for all x. The set of all such numbers is denoted by P(f). We shall say that f is a periodic function if P(f) contains a non-zero number.
Lemma 1: If p and q are contained in the set P(f), then so are -p, p+q and p–q
Proof : Suppose that f[t-p(x)] = f(y) where y is the instant in the past such that dur(y,x) = p. Then tp(y) = x, so f[t-p(x)] = f(y) and f[tp(y)] = f(x). But by definition f[tp(y)] = f(y). Therefore f[t-p(x)] = f(x) for all x, which concludes that -p is also a period.
Now to prove that p+q and p–q are also included in the set of periods for f, when p and q are both periods:
f[tp(x)] = f(x) and f[tq(x)] = f(x) for all x
f[tp+q(x)] = f[tp[tq (x)]] = f[tq(x)] = f(x) for all x
\p+q Î P(f)
f[tp-q(x)] = f[tp[t-q (x)]] = f[t-q(x)] = f(x) for all x. Since the first proof shows that -p is a period contained in the set when p is one, so -q is a period when q is contained in the set.
\p-q Î P(f)
Corollary 1: If p is contained in the set P(f) then so is np, where n is an integer. When p is the smallest, positive member of P(f), these two sets coincide.
To prove that np is also a period contained in the set:
Since this is clearly true for n=1, I will need to prove it for n>1.
Suppose that (n-1)p Î P(f) and consider np. Since both 1p and (n-1)p are in P(f),
then 1p + np = np is in P(f) by Lemma 1
Now to prove that the sets nq Î P(f) and q Î P(f) coincide:
Def.: f[tp(x)] = f(x) for all x and let’s say that p is the smallest positive number for which this is true.
Suppose that f[tq(x)] = f(x) for all x and q>0
Then p<q, so divide q = np + r, where r is the remainder (0£r<p)
f[tq(x)] = f[tnp+r(x)] = f[tnp [tr(x)]]
Since p is a period so is np as shown above.
Then f[tnp [tr(x)]] = f[tr(x)] for all x, so r is a period.
But since 0£r<p, and p is the smallest positive period, r must be 0.
If q<0, then -q Î P(f) and since -q is positive, -q = np for some integer n. But then
q = (-n)p , so r can be proven to be 0 for q<0 also.
I have also been interpreting the integrals of these functions and started
by studying the Riemann-Stieltjes Integral, its properties and applications.
The Riemann-Stieltjes Integral is defined as follows for two functions: Let
g(x) and f(x) be real functions of x defined on the interval [a,b], where a £x £b, (a=x0<x1<x2<…….<xn=b).
The limit of: 
, as max|xj – xj+1|
® 0, where xj-1≤xj*≤xj,
is denoted by: 
, and is called the integral of f
with respect to g.[5]
There is a Mean Value Theorem for the Riemann-Stieltjes Integral: = f(x)[g(b) –g(a)]
where a£x £b.[6]
This will be useful below. The advantage here is that f(x) and g(x) are
numbers while x is an instant.
Let g be a function such that dur(x,y) = |g(x)-g(y)|. Such a function is defined in Dr. Devito’s paper as: D(s(p1), s(p2)) = |p1-p2|.[7] Instead of “s”, I shall use “g”
Lemma 2: Let f be a
continuous and periodic function on the real line, and let a>0 be any
element contained in P(f), then for any real x and y, 
= 
.
Proof:
= 
– 
= 
+ 
–
Now =
First note that g[tα(u)] = g(u) because α is a constant and g(x) – g(y) = dur(x,y), so g[tα(x)]– g[tα(y)] = dur(x,y)
Then= 
So =
Therefore=
Lemma 3: If a continuous function has arbitrarily small periods, it is a constant
Proof: Consider
the integral: 
Suppose a and b are instants where a=x0<x1<x2<…….<xn=b.
Then the integral
can be represented by the sum 
where xj-1£ xj*£ xj
The limit of this
sum as max[g(xj) – g(xj-1)] goes to 0 is defined to be which is the
Riemann-Stieltjes Integral.
Then using the Mean Value Theorem: m[g(b)-g(a)] ££M[g(b)-g(a)] where
m stands for minimum and M stands for maximum.
So 
≤ 
≤
Now a = dur(tα(x),x) = g[tα(x)]-g(x) by the definition of g
Then 1/[ g(x+αn)
– g(x)] 
= f(x)/an [ g(x+αn)
– g(x)] =an f(x)/an = f(x)
where x£ x£x+αn
Since an Î P(f), then for
any x: 
1/anò
= f(x)
Then for any x and
y: 1/anò = 1/an
so f(x) = f(y)
Therefore, f is a constant.
Corollary 2: Any non-constant, continuous, periodic function has a smallest positive period
Proof: P(f[tα(x)]) = P(f(x)) for all x. In Lemma 1 it was proved that the set P+(f) = {a Î P(f)|a>0} is non-empty, where P+(f) is the set of positive periods for of P(f). Let a0 be the greatest lower bound for this set. If a0 is 0, then a sequence in P+(f) must converge to it, meaning that f would have arbitrarily small periods. But as proved above, if f has arbitrarily small periods, f must be a constant and this cannot be true since f is non-constant. Therefore a0 must be positive.
To prove that a0 Î P+(f),
I shall argue by contradiction. If a0 is not in this
set, then there must be some {an } Í P+(f)
such that an =a0. But since f is
continuous, 
=
for
any fixed x.
\a0 Î P+(f)
References
1. Devito, Carl L. “A Non-Linear Model For Time.” Astrophysics and Space Science
244 (1996): 357-369.
2. Devito, Carl L. “Time Scapes.” Chaos, Solitons & Fractals Vol. 9 No. 7 (1998):
1105-1114.
3. Olmstead, John M. H. Advanced Calculus. New York: Appleton-Century-Crofts, Inc.
1961.
4. Widder, David V. Advanced Calculus. Englewood Cliffs, N.J.: Prentice-Hall, Inc.
1947.
5. Widder, David Vernon. The Laplace Transform. London: Humphrey Milford Oxford
University Press. 1946.