On Direct and Inverse Proportionality
Markku Halmetoja, Pentti Haukkanen, Teuvo Laurinolli,
Jorma K. Merikoski, Timo Tossavainen, and Ari Virtanen
1. INTRODUCTION. The motivation for this paper arises from a discussion of proportionality in a textbook for high schools. In the book, two quantities are said to be
directly proportional if, by whatever positive real number p one of them is multiplied,
the other one will change by the same ratio p. But is it necessary to go through all ps?
What more is required than insisting, say, that if one of the quantities doubles, then the
second doubles, too?
Let us state this problem more precisely. Let f : R → R be a continuous function
satisfying
f ( px) = p f (x)
(1)
for all x, where p is a given positive real number different from 1. Do we then have
f (x) = f (1)x
(2)
for all x? The answer: not necessarily, as the trivial counterexample f (x) = |x| shows.
However, this counterexample fails if we restrict the domain of f to the set R+ of
positive real numbers. On the other hand, a less trivial example, the function f : R+ →
R given by
f (x) = 2 p −2n x 3 − 3 p −n (1 + p)x 2 − p n+1 (1 + p)
(3)
whenever p n ≤ x < p n+1 (n ∈ Z) and where p > 1, shows that (1) does not imply
(2) even if f is continuously differentiable (see Figures 1–3). For the details, see the
appendix.
2
4
6
8
−20
−40
−60
−80
−100
Figure 1. The function f in (3) with p = 2.
At first glance, Figure 1 might suggest that the graph of f in (3) was, in an ambiguous way, a straight line. However, this illusion is partly due to the scaling of the
axes. As shown in Figure 2, f is in fact a cubic spline that also provides a good approximation to a linear function. Normally splines are used for the approximation of
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ON DIRECT AND INVERSE PROPORTIONALITY
871
more complicated functions (see, for example, [9]). Nevertheless, the graph of another
continuously differentiable function satisfying (1), namely, the one defined by
f (x) = 21−2n x 3 − 9 · 2−n x 2 + 14x − 6 · 2n
(4)
whenever 2n ≤ x < 2n+1 (n ∈ Z), plainly differs from a straight line (Figure 3).
2
4
6
8
−20
−40
−60
−80
−100
Figure 2. The function f in (3) with p = 2.
8
6
4
2
2
4
6
8
Figure 3. The function f in (4).
The foregoing observations lead us to state our problem in the following way:
Problem. Let p be a positive real number different from 1. Suppose that a function
f : R+ → R satisfies
f ( px) = p f (x)
for all x. Find relevant conditions on f to guarantee that
f (x) = f (1)x
for all x.
Having stated the problem, we observe immediately that we may without loss of
generality always assume that p > 1 (or 0 < p < 1). Namely, by replacing x with
x/ p in (1), we get
1
1
f
x = f (x).
p
p
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c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 114
Thus, if (1) holds for p, then it holds for 1/ p as well.
As (3) indicates, the continuity, or even continuous differentiability, of f and the
condition (1) are not sufficient for (2). However, in the next section we will see, for
example, that if f is both continuous and subadditive, then (1) implies (2). We also
give a sufficient condition on f to guarantee (2).
It is not difficult to see that f in (3) is not twice-differentiable. However, (2) does
not follow even if f is infinitely many times differentiable and satisfies (1). To outline
a counterexample, notice that the function
−1/x 2
if x = 0,
g(x) = e
0
if x = 0
has derivatives of all orders and g (0) = g (0) = · · · = 0, and define
h(x) = g(x − 1)g(x − p).
Then the function defined by
f (x) = p n h
x pn
(5)
whenever p n ≤ x < p n+1 (n ∈ Z) has derivatives of all orders and satisfies (1) but not
(2). For a monotonic counterexample, consider ϕ(x) = x + f (x).
In section 3, we solve our problem under the weaker assumption that f is continuous or monotone by adding another proportionality condition: f (qx) = q f (x) for all
x. Here q is a positive real number different from 1 such that p m = q n for all nonzero
integers m and n.
In the last section, we study inverse proportionality, or more precisely, functions
f : R+ → R+ satisfying
f ( px) =
1
f (x)
p
(6)
for all x. We seek relevant conditions on f to ensure that
f (x) =
f (1)
x
(7)
for all x. Not surprisingly, it turns out that the results in section 3 concerning direct
proportionality have inverse proportion counterparts. Also, we give a condition on f that suffices to imply (7).
The functional equations (1) and (6) are very special cases of the Schröder equation
f ϕ(x) = p f (x),
where ϕ is a function and p is a constant [1], [2], [5], [6]. For example, Theorem 6 can
be generalized to Theorem 9.5.1 in [6]. Since our exposition is elementary, we ignore
this general approach.
Our results are somewhat tied to the following well-known theorem (see [1, p. 34]):
Theorem 1. If f : R+ → R is additive (i.e., if
December 2007]
f (x + y) = f (x) + f (y)
(8)
ON DIRECT AND INVERSE PROPORTIONALITY
873
for all x and y) and if f is continuous at some point of R+ , then (2) holds for all x in
R+ .
This theorem remains valid if f is defined on the whole set of real numbers or
on the nonnegative real numbers. Theorem 1 also remains valid (see [7]) if, instead of
requiring the continuity of f at some point, f is assumed to be (Lebesgue) measurable.
2. THE FUNCTIONAL EQUATION f( px) = p f(x). We first provide a sufficient
condition for (2) in terms of the derivative of f .
Theorem 2. Let f : R+ → R be a differentiable function such that limx→0 f (x) exists. If f satisfies (1) for all x, then (2) also holds for all x.
Proof. We may assume that 0 < p < 1. Differentiating (1), we obtain f ( px) =
f (x), whence by induction
f ( p n x) = f (x)
for all x in R+ and for every natural number n. Hence, for each x in R+ ,
c = lim f (t) = lim f ( p n x) = lim f (x) = f (x).
t→0
n→∞
n→∞
Therefore f (x) = cx and (2) follows.
Notice that, by induction,
f (x) =
1
f ( p n x)
pn
for every n in N. Therefore, it is actually enough to assume in Theorem 2 that f is
differentiable on an interval (0, δ), where δ > 0. Similarly, if f is defined on the set of
nonnegative real numbers or on R, then we see by the same argument that (2) holds if
f is continuously differentiable at the origin.
The following two theorems show that there are several conditions on f that, together with continuity and (1), are sufficient for (2).
Theorem 3. Let f : R+ → R be continuous. If f is subadditive, (i.e., if
f (x + y) ≤ f (x) + f (y)
(9)
for all x and y) and satisfies (1) for all x, then (2) holds for all x. The same conclusion
holds if f is superadditive (i.e., if the inequality in (9) is reversed).
Proof. If x > 0, then by (1)
f ( p k x) = p k f (x)
for all integers k, and by (9)
f (mx) ≤ m f (x)
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c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 114
for all positive integers m. Thus
f (mp k x) ≤ mp k f (x)
(10)
for all integers k and positive integers m.
We claim that S = {mp k : k ∈ Z, m ∈ Z+ } is dense in the set of positive real numbers. We assume that p > 1. Let t > 0. Given > 0, we show that there exists s in S
such that |s − t| < . Since
t p n = t,
n→∞ p n
lim
where ·
is the floor function (i.e., x
is the largest integer that is less than or equal
to x), we find a positive integer n 0 satisfying
n
t p 0 < .
−
t
pn0
Thus s = mp k , where m = t p n0 and k = −n 0 , has the desired property, and the
claim follows.
By (10), the continuity of f , and the density of S,
f (t z) ≤ t f (z)
for all t and z in R+ . Hence, for all x and y in R+ we have
x
x
(x + y) ≤
f (x + y)
f (x) = f
x+y
x+y
and
f (y) = f
y
y
(x + y) ≤
f (x + y),
x+y
x+y
which implies that
f (x) + f (y) ≤ f (x + y).
In view of (9),
f (x) + f (y) = f (x + y).
Now (2) follows from Theorem 1. The superadditive case follows by considering − f
in place of f .
Recall that a function f : R+ → R is convex if
f t x + (1 − t)y ≤ t f (x) + (1 − t) f (y)
(11)
whenever 0 ≤ t ≤ 1. It is concave if (11) holds with the inequality reversed.
Theorem 4. Let f : R+ → R be a convex or concave function satisfying (1) for all x.
If limx→0 f (x) = 0, then (2) also holds for all x.
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ON DIRECT AND INVERSE PROPORTIONALITY
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Proof. We treat the convex case. If 0 ≤ t ≤ 1, then for z and w in R+
f t z + (1 − t)w ≤ t f (z) + (1 − t) f (w).
(12)
Since f is convex on an open interval, it is continuous (see [9, p. 5]). Letting w → 0,
we infer that the left-hand side of (12) tends to f (t z). At the same time, the right-hand
side tends to t f (z), and thus we have
f (t z) ≤ t f (z).
As in the proof of Theorem 3, we conclude that
f (x) + f (y) ≤ f (x + y).
Thus f is superadditive, and (2) follows from Theorem 3.
3. THE TWO-EQUATION CASE. We mentioned in the introduction that (1) does
not imply (2) even if f is continuously differentiable. When f satisfies a second equation of this type, however, the situation already changes dramatically for continuous f .
Theorem 5. Let f : R+ → R be continuous. If p and q are positive real numbers
different from 1 such that p m = q n for all nonzero integers m and n, and if
f ( px) = p f (x)
and
f (qx) = q f (x)
for all x, then
f (x) = f (1)x
for all x.
Proof. Let S = { p m q n : m, n ∈ Z}. If x belongs to S, then the assertion clearly holds.
Since f is continuous, the theorem follows provided S is dense in the set of positive
real numbers. Taking logarithms of the elements of S, we obtain an equivalent claim
stating that T = {m ln p + n ln q : m, n ∈ Z} is dense in the set of real numbers. Dividing the elements of T by ln p, we obtain another equivalent claim, namely, that the set
U = {m + n ln q/ ln p : m, n ∈ Z} is dense in the set of real numbers. Since ln q/ ln p
is irrational (if it were a rational m/n, then m ln p = n ln q, so p m = q n contrary to
our assumption), the claim follows immediately from a theorem of Dirichlet: if α is an
irrational number, then { m + nα : m ∈ Z, n ∈ N } is a dense subset of R (see [3, p.
16]).
This theorem remains valid if, instead of continuity, we assume monotonicity.
Theorem 6. Let f : R+ → R be a monotone function. If f satisfies the proportionality conditions in Theorem 5, then f (x) = f (1)x for all x in R+ .
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c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 114
Proof. Write c = f (1). If c = 0, then f (x) = 0 in the dense set S described in the
proof of Theorem 5. Thus, by monotonicity, f (x) = 0 for every x in R+ , and the
desired conclusion follows.
Now assume that c > 0. As in the proof of Theorem 5, f (x) = cx holds for x
in S. This fact and the density of S imply that f is increasing. Suppose that there
exists x0 in R+ such that f (x0 ) = cx0 + d, where d = 0. First consider the case where
d > 0. By the density of S, there exist integers m and n such that x1 = p m q n > x0 and
x1 − x0 < d/c. But then
f (x1 ) = cx1 = cx0 + c(x1 − x0 ) < cx0 + d = f (x0 ),
contradicting the fact that f is increasing. Similar reasoning rules out the possibility
that d < 0. Thus f (x) = cx must hold for each x in R+ .
If c < 0, then f is decreasing, and a similar argument works.
Finally, we remark that the assumption on the continuity of f in Theorem 5 or
on the monotonicity of f in Theorem 6 cannot be weakened, say, to measurability
of f . For a counterexample, define f (x) = x if x is rational, and f√
(x) = −x if√x is
irrational. Then, f (2x) = 2x and f (3x) = 3x for all x in R+ yet f ( 2) = f (1) 2.
4. THE FUNCTIONAL EQUATION f( px) = f(x)/ p. We now switch our attention to inverse proportionality and discuss the possibilities of inverse proportional
counterparts to the results in the previous sections. We start by observing that, for
a function f : R+ → R+ and a positive constant p different from 1, the condition (6)
is equivalent to the condition
f (x) =
1
g(x)
where g : R+ → R+ is a function such that
g( px) = pg(x)
(13)
for every x in R+ . As in the introduction, we may also assume that 0 < p < 1 or that
p > 1, whichever we prefer.
As in the directly proportional case, the continuity (or differentiability) of f and (6)
by themselves are not sufficient to imply (7) for all x in R+ . For example, let p > 1,
and for each n in Z and t in [0, 1) set
f (x) = t p −n−1 + (1 − t) p −n
(14)
whenever
x = t p n+1 + (1 − t) p n .
Then f is positive and continuous on R+ and satisfies (6) but not (7).
The function f in (14) is not differentiable. To provide a differentiable counterexample, let
December 2007]
f (x) = 3 · 2−3n x 2 − 7 · 21−2n x + 21 · 2−n
(15)
ON DIRECT AND INVERSE PROPORTIONALITY
877
0.5
0.4
0.3
0.2
0.1
4
8
12
16
Figure 4. The function f in (14) with p = 2.
5
4
3
2
1
4
8
12
16
Figure 5. The function f in (15).
whenever 2n ≤ x < 2n+1 . In fact, let g : R+ → R+ be a differentiable function such
that the condition (13) holds but g(x) = g(1)x does not. Such functions exists—for
instance, the one in (3). Then f = 1/g is differentiable and satisfies (6) but not (7).
Observe that the functions f in (14) and (15) are convex. Therefore, convexity and
(6) do not guarantee (7), despite the fact that limx→0 f (x) = ∞ (cf. the limit condition
in Theorem 4). Trivially, a concave function cannot satisfy (7) for all x in R+ .
Finally, the functions in (14) and (15) reveal that subadditivity and (6) do not imply
(7). Obviously, no superadditive f : R+ → R+ satisfies (7) for all x in R+ .
From the foregoing observations, it now seems clear that Theorems 3 and 4 have
no inverse proportional counterparts. Nevertheless, we complete our study by showing
that Theorems 2, 5, and 6 do admit such analogues.
To that end, notice first that if f in (6) is continuous or monotone, then also g =
1/ f is continuous or monotone, respectively. Second, if the constant p is replaced
with q = 1 in (6), the same can be done in (13). And lastly, a function f : R+ → R+
satisfying (7) cannot be increasing. Theorem 7 thus follows from Theorems 5 and 6.
Theorem 7. Let f : R+ → R+ be continuous or decreasing. If p and q are positive
real numbers different from 1 such that p m = q n for all nonzero integers m and n and
if
f ( px) =
878
1
f (x)
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c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 114
and
f (qx) =
1
f (x)
q
for all x in R+ , then
f (x) =
f (1)
x
for all x in R+ .
The counterpart for Theorem 2 follows easily if the limit condition is modified
slightly:
Theorem 8. Let p be a positive real number different from 1. If f : R+ → R+ is a
differentiable function such that
lim
x→0
f (x)
f (x)2
exists and if
f ( px) =
1
f (x)
p
holds for all x in R+ , then
f (x) =
f (1)
x
for all x in R+ .
Proof. If g = 1/ f , then g is differentiable and both
g( px) = pg(x)
and
g (x) =
− f (x)
f (x)2
hold in R+ . By the assumption, limx→0 g (x) exists. Therefore, it follows from Theorem 2 that
g(x) = g(1)x
and thus,
f (x) =
f (1)
x
for all x in R+ .
Observe that, similar to the situation in Theorem 2, it would now be enough to
require only that f be differentiable on some interval (0, δ). Also, Theorems 7 and 8
hold for f : R+ → R− with minor modifications.
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5. APPENDIX. We prove here that f in (3) is continuously differentiable and satisfies (1). The first claim is trivial unless x = p n for some n in Z. In the case of x = p n ,
we see easily that
lim
h→0+
f ( p n + h) − f ( p n )
= −6 p,
h
whence
f + ( p n ) = −6 p.
Replacing n with n − 1 in (3), we obtain
f (x) = 2 p −2(n−1) x 3 − 3 p −(n−1) (1 + p)x 2 − p n−1 ( p + p 2 ),
(16)
when p n−1 ≤ x < p n . A straightforward calculation based on (3) and (16) then reveals
that
lim
h→0−
f ( p n + h) − f ( p n )
= −6 p
h
as well, so
f − ( p n ) = −6 p.
Thus f is differentiable at p n and f ( p n ) = −6 p.
To show that f is continuous at p n , note that
f (x) = 6 p −2n x 2 − 6 p −n (1 + p)x
when p n < x < p n+1 . Therefore
lim
x→( p n )+
f (x) = 6 p −2n ( p n )2 − 6 p −n (1 + p) p n = 6 − 6(1 + p) = −6 p = f ( p n ).
Similarly,
f (x) = 6 p −2(n−1) x 2 − 6 p −(n−1) (1 + p)x
when p n−1 < x < p n , which leads to
lim
x→( p n )−
f (x) = 6 p −2(n−1) ( p n )2 − 6 p −(n−1) (1 + p) p n
= 6 p 2 − 6 p(1 + p) = −6 p = f ( p n ).
Accordingly, f is continuously differentiable.
To prove that f satisfies (1), let p n ≤ x < p n+1 . Then p n+1 ≤ px < p (n+1)+1 ,
hence, by (3),
f ( px) = 2 p −2(n+1) ( px)3 − 3 p −(n+1) (1 + p)( px)2 − p n+2 (1 + p)
= 2 p −2n+1 x 3 − 3 p −n+1 (1 + p)x 2 − p n+2 (1 + p)
= p[2 p −2n x 3 − 3 p −n (1 + p)x 2 − p n+1 (1 + p)] = p f (x).
ACKNOWLEDGMENTS. In preparing the first version of the manuscript, we were not aware of the Schröder
equation. We thank one referee for introducing us to it and for giving relevant references. We also thank the
other referee for noting an error in our original function f in (3). Finally, we thank Dan Velleman for suggesting
the counterexample (5).
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6.
J. Aczél, Lectures on Functional Equations and Their Applications, Academic Press, New York, 1966.
, A Short Course of Functional Equations, D. Reidel, Dordrecht, 1987.
A. Browder, Mathematical Analysis: An Introduction, Springer-Verlag, New York, 1996.
E. Kreyszig, Introductory Functional Analysis with Applications, John Wiley, New York, 1978.
M. Kuczma, Functional Equations in a Single Variable, Polish Scientific Publishers, Warsaw, 1968.
M. Kuczma, B. Choczewski, and R. Ger, Iterative Functional Equations, Cambridge University Press,
Cambridge, 1990.
7. G. Letac, Cauchy functional equation again, this M ONTHLY 85 (1978) 663–664.
8. J. van Tiel, Convex Analysis, John Wiley, New York, 1984.
9. E. W. Weisstein, Cubic Spline, from MathWorld—A Wolfram Web Resource, http://mathworld.
wolfram.com/CubicSpline.html.
MARKKU HALMETOJA received his M.Sc. degree from Åbo Akademi University in 1977. He is a teacher
at Mänttä senior high school. He has coauthored textbooks for both senior high schools and universities. In his
spare time he plays saxophone in an amateur jazz quartet.
Mäntän lukio, P.O. Box 87, FI-35801 Mänttä, Finland
[email protected]
PENTTI HAUKKANEN received his Ph.D. degree in 1988 from the University of Tampere. He is a docent
and senior assistant at the University of Tampere. He has published research papers on number theory. In his
spare time, he enjoys sports and literature.
Department of Mathematics, Statistics and Philosophy, FI-33014 University of Tampere, Finland
[email protected]
TEUVO LAURINOLLI received his Ph.Lic. degree from the University of Oulu in 1977. He is a teacher at
the senior high school Oulun Lyseon Lukio. He has coauthored textbooks for both comprehensive schools and
senior high schools. He has also published one research paper on automata theory. His leisure time activities
include swimming and attending concerts of classical music.
Oulun Lyseon Lukio, Kajaaninkatu 3, FI-90100 Oulu, Finland
[email protected]
JORMA K. MERIKOSKI received his Ph.D. degree from the University of Tampere in 1976. He is retired
from the University of Tampere but continues working as a docent. He has coauthored textbooks for comprehensive schools, senior high schools, and universities. He has also published research papers, mostly on linear
algebra. In his spare time, he enjoys jogging and literature.
Department of Mathematics, Statistics and Philosophy, FI-33014 University of Tampere, Finland
[email protected]
TIMO TOSSAVAINEN received his Ph.D. degree from the University of Jyväskylä in 2000. He is a lecturer
at the University of Joensuu, and a docent at the University of Tampere. He has coauthored textbooks for both
senior high schools and universities. He has also published research papers on analysis, linear algebra, and
mathematics education. Since he is father to three small children, he has no leisure time.
Savonlinna Department of Teacher Education, University of Joensuu,
P.O.Box 86, FI-57101 Savonlinna, Finland
[email protected]
ARI VIRTANEN received his Ph.D. degree from the University of Tampere in 2006. He is a departmental
coordinator at the University of Tampere. He has coauthored textbooks for universities and has also published
research papers on linear algebra and modal logic. His leisure activities involve ice-water swimming.
Department of Mathematics, Statistics and Philosophy, FI-33014 University of Tampere, Finland
[email protected]
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