SUBADDITIVE FUNCTIONS AND THEIR (PSEUDO-)INVERSES
LARS PETER ØSTERDAL
Abstract. The paper considers non-negative increasing functions on intervals
with left endpoint closed at zero and investigates the duality between subadditivity and superadditivity via the inverse function and pseudo-inverses.
1. Introduction
This paper considers non-negative increasing functions f de…ned on intervals I;
where I = [0; !] for some ! > 0 or I = [0; 1[.
A function f is subadditive if f (x) + f (y) f (x + y) whenever x; y; x + y 2 I;
and f is superadditive if f (x) + f (y) f (x + y) whenever x; y; x + y 2 I: Subadditive functions have been studied by e.g. Hille (1948, chapter 7) and Rosenbaum
(1950) and Matkowski and Światkowski
¾
(1991, 1993), and non-negative superadditive functions have been treated by e.g. Bruckner (1960, 1962) and Bruckner and
Ostrow (1962).
Since f is subadditive if and only if f is superadditive (e.g. Rosenbaum, 1950)
results for non-negative superadditive functions translate directly to non-positive
subadditive functions as well. For many applications, however, the function class
of interest is the class of non-negative subadditive functions.1
There is an elementary connection between strictly increasing continuous subadditive and superadditive functions, which is likely to be folk knowledge.2 (Recall that if f is strictly increasing and continuous then J = f (I) is an interval
and there is a uniquely determined function f 1 on J, the inverse function, such
that f 1 (y) = fx j f (x) = yg; f 1 is continuous and strictly increasing and
(f 1 ) 1 = f:)
Proposition Let f : I ! R+ be strictly increasing and continuous. Then f is
subadditive if and only if f 1 is superadditive.
Proof: Let f be subadditive. We claim that if x; y; z 2 I, f (x) = x0 ; f (y) = y 0 and
f (z) = x0 + y 0 then x + y
z. For this, assume otherwise that x + y > z. Since
y > z x and since f is strictly increasing we have f (x) + f (y) > f (x) + f (z x)
f (z) = x0 + y 0 a contradiction proving the claim.
Date : Final version, May 2005.
2000 Mathematics Subject Classi…cation. Primary 26A48, 26A51; Secondary 91B02.
Key words and phrases. Subadditive functions, superadditive functions, pseudo-inverse.
1 In economic theory for instance, f can for example represent a non-linear tari¤ or long-run
production costs.
2
The following proposition is probably well-known, but we have been unable to …nd literature
where it has been stated and provide it here for ease of reference.
1
2
LARS PETER ØSTERDAL
Now, let x0 ; y 0 ; x0 + y 0 2 f (I) and x = f 1 (x0 ); y = f 1 (y 0 ). From our claim we
have x + y f 1 (x0 + y 0 ); i.e. f 1 (x0 ) + f 1 (y 0 ) f 1 (x0 + y 0 ); proving that f 1
is superadditive.
Conversely, let g f 1 be superadditive on J f (I). Let x0 ; y 0 ; x0 + y 0 2 g(J)
and x = g 1 (x0 ),y = g 1 (y 0 ). Since g is strictly increasing and superadditive we
have x + y
g 1 (x0 + y 0 ) since if otherwise (x + y < g 1 (x0 + y 0 )) we would
have x + y 2 J and g(x + y) < g(x) + g(y) contradiction superadditivity. Thus
g 1 (x0 + y 0 )
x + y = g 1 (x0 ) + g 1 (y 0 ) proving that g 1 = (f 1 ) 1 = f is
subadditive on g(J) = f 1 (f (I)) = I.
The assumption that f is strictly increasing and continuous is restrictive. However, when f is discontinuous or fails to be strictly increasing, the inverse f 1 may
not be a well-de…ned function. The classes of subadditive and superadditive functions may then also fail to share certain regularities. For instance, a non-negative
increasing superadditive function vanish at the origin, whereas this is not necessarily the case for non-negative increasing subadditive functions.
2. Results
Let R+ R+ [ f1g denote the extended non-negative reals. For an increasing
function g : I ! R+ , let
J
Then a function ge
1
[inf x2I g(x); supx2I g(x)], if g is bounded
[inf x2I g(x); 1[,
otherwise.
: J ! R+ is a pseudo-inverse to g if
supfx j g(x) < yg
ge
1
(y)
supfx j g(x)
yg;
1
whenever y 2 J: As noted by Denneberg (1994, p.5), ge is uniquely determined
except on an at most countable number of points on J. With respect to Lebesgue
integration (studied by Denneberg) there is no reason to distinguish between different pseudo-inverses, but in our context it is useful to consider two speci…c forms.
If g : I ! R+ is increasing and lower semi-continuous then we de…ne !
g :J !
R+ by !
g (x)
sup fy j g(y) xg : If g : I ! R+ is increasing and upper semicontinuous then g : J ! R+ is de…ned by g (x) min fy j x g(y)g :
Lemma 2.1. Let f : I ! R+ be increasing, then:
!
(1) If f is lower semi-continuous, then f is upper semi-continuous.
(2) If f is upper semi-continuous, then f is lower semi-continuous.
!
Proof: We prove (1). Assume that f is not upper semi-continuous, i.e. there exists
!
!
x and a sequence fxn g, such that limn!1 fxn g = x and lim sup f (xn ) > f (x): Let
!
!
!
" = lim supn!1 f (xn ) f (x) > 0: Since f is increasing there exists a decreasing
1
+
subsequence fx+
x+
x: By
n g of fxn g such that limn!1 fxn g = x and x + n
n
!
the de…nition of f and lower semi-continuity of f there exists y such that f (y) x
and f (y 0 ) > x for all y 0 > y: But this contradicts that for all y 00 2 [y; y + "[ we have
f (y 00 ) x+
x + n1 for all n, i.e. f (y 00 ) x.
n
The proof of (2) is similar.
For two functions g and h, we write g = h if they are de…ned on the same domain
I and f (x) = g(x) for all x 2 I.
SUBADDITIVE FUNCTIONS
3
Lemma 2.2. Let f : I ! R+ be increasing, then:
!
(1) If f is lower semi-continuous then f = f .
!
(2) If f is upper semi-continuous then f = f :
!
Proof: Suppose that f is lower semi-continuous. By Lemma 2.1, f is upper semi!
!
continuous, so f is well-de…ned. It is readily veri…ed that the domain of f is the
same as the domain of f . Now let x 2 I, then:
!
f (x) = minfy j x
!
f (y)g = minfy j x
supfz j f (z)
ygg:
!
Since x supfz j f (z) f (x)g, f (x) f (x): It remains to verify that if y 0 < f (x)
then y 0 2
= fy j x supfz j f (z) ygg: This follows from the fact that if y 0 < f (x)
!
then supfz j f (z) y 0 g < x since f is increasing. This means that f (x) f (x);
!
i.e. f (x) = f (x):
The proof of (2) is similar.
Notice that Lemma 2.2 would not be true for functions g : I ! R+ taking values
on the extended reals, since for the function g 1 on [0; 1] de…ned by
g 1 (x)
1
1 x
1
0 x<1
x = 1;
and the function g 2 on [0; 1[ de…ned by
1
1 x
0 x<1
1
x 1;
! !
we have g 1 = g 2 and the domain is [0; 1[, g 1 = g 2 and the domain is [0; 1]. Hence
!
g 2 6= g 2 .
g 2 (x)
Theorem 2.3. Let f : I ! R+ be increasing, then:
!
(1) If f is lower semi-continuous, then f is subadditive if and only if f is
superadditive.
(2) If f is upper semi-continuous, then f is superadditive if and only if f is
subadditive.
Proof: Suppose that f is lower semi-continuous.
!
First let f be subadditive. If f is not superadditive then there exist x ; y ; x +
y 2 J such that supfy 0 j f (y 0 ) x g + supfy 00 j f (y 00 ) y g > supfy 000 j f (y 000 )
x + y g: But f subadditive implies supfy 000 j f (y 000 )
x +y g
supfy 0 + y 00 j
0
00
0
0
00
00
f (y ) x ; f (y ) y g = supfy j f (y ) x g + supfy j f (y ) y g which is a
contradiction.
!
Now let f be superadditive. For arbitrary u; v 2 I such that f (u) = x and
f (v) = y; de…ne u = supfy 0 j f (y 0 )
xg and v = supfy 00 j f (y 00 )
yg..
!
0
0
00
00
xg + supfy j f (y )
yg =
Since f is superadditive we have supfy j f (y )
supfy 0 +y 00 j f (y 0 ) x; f (y 00 ) yg supfy 000 j f (y 000 ) x+yg for all x; y; x+y 2 J:
But then there exists y 000
u +v
u + v such that f (y 000 )
x + y: Then
000
f (u + v) f (u + v ) f (y ) x + y = f (u) + f (v); i.e. f is subadditive.
4
LARS PETER ØSTERDAL
If f is upper semi-continuous, by Lemma 2:1 f is lower semi-continuous. By
!
(1) and Lemma 2.2 f is subadditive if and only if f = f is superadditive.
Let f : [0; !] ! R+ : For x 2 [0; 1[ we say that x1 ; :::; xn form a !-partition if
x + ::: + xn = x and 0 xi ! for all i. Let fb be a function on [0; !
b ], ! < !
b , or
[0; 1[ de…ned as
1
fb(x) = infff (x1 ) + ::: + f (xn ) j x1 ; :::; xn form a !-partition for xg:
One may show that fb is subadditive (see Bruckner, 1960, for a proof in the case
of superadditivity) and hence the maximal subadditive extension of f on [0; !
b ].3
A central problem is how to check whether a given function is subadditive. We
verify below that Bruckner’s test for superadditivity (Bruckner, 1962) has a counterpart for increasing, non-negative subadditive functions. For this, we make use
of the following lemma.
Lemma 2.4. Let f : [0; !] ! R+ be increasing, and !
b > !:
(1) If f is upper semi-continuous and superadditive, then the minimal superadditive extension fb on [0; !
b ] is upper semi-continuous.
(2) If f is lower semi-continuous and subadditive, then the maximal subadditive
extension fb on [0; !
b ] is lower semi-continuous.
Proof: By contradiction. Assume that fb is not upper semi-continuous at some point
x 2 [!; !
b [. Since fb is increasing there is an in…nite sequence fx1 ; x2 ; x3 ; :::g where
i
x > x for all i and fxi g ! x such that fb(x) < limi!1 fb(xi ). By Bruckner (1960)
fb(x) = supff (x1 ) + ::: + f (xn ) j x1 ; :::; xn forms a !-partition for xg; x 2 [0; !
b ].
By our assumption, for each xi there is a !-partition fx1i ; :::; xni i g such that
f (x1i ) + ::: + f (xni i ) f (x) + 12 (limi!1 fb(xi ) f (x)). Note that since f is superadditive on [0; !] we can assume that ni
N for all i and some positive integer
N . Since [0; !] is compact there is a subsequence fy 1 ; y 2 ; y 3 ; :::g fx1 ; x2 ; x3 ; :::g
such that fx1i ; :::; xni i g forms a !-partition for y i ; ni = K
N for all i, and
fx1i ; :::; xni i g converges pointwise to some !-partition fx1 ; :::; xK g for i ! 1.
We therefore have f (x1 ) + ::: + f (xK ) < limi!1 f (x1i ) + ::: + f (xni i ) contradicting
that f is upper semi-continuous hence right-continuous on [0; ![.
The proof of (2) is similar.
Let f : [0; !] ! R+ . The functions f1 , f2 ,...,fK on [0; ! 1 ], [0; ! 2 ]; :::; [0; ! K ]
respectively form a decomposition of f , if ! 1 + ::: + ! K = !, and:
8
f1 (x);
0 x !1
>
>
>
>
f
(x
!
)
+
f
(!
);
!1 < x !1 + !2
>
2
1
1
1
<
..
..
f (x) =
.
.
>
>
>
f
(x
!
:::
!
)
+
f
(!
)
>
1
K 1
1
1
>
! 1 + ::: + ! K 1 < x !:
: K
+::: + fK 1 (! K 1 );
Following Bruckner (1962), if f1 , f2 ,...,fK form a decomposition of f , we write
f = f1 ^ f2 ^
^ fK .
3 For a subadditive function f : I ! R , fb : Ib ! R is a maximal subadditive extension on
+
+
Ib I, if fb is subadditive and f = fb on I and if g is a subadditive function on Ib and g = f on I
then g fb.
SUBADDITIVE FUNCTIONS
5
Theorem 2.5. Let f1 and f2 be non-negative subadditive functions de…ned on
[0; ! 1 ] and [0; ! 2 ] respectively, and f = f1 ^ f2 : Let fb1 be the maximal subadditive
extension of f1 : Then f is subadditive on [0; !] if and only if f fb1 on [0; !]:
Proof: If g1 and g2 are non-negative superadditive upper semi-continuous functions
de…ned on [0; ! 1 ] and [0; ! 2 ] respectively, and g = g1 ^g2 ; then g is superadditive on
[0; !] if and only if gb1 g on [0; !]; where gb1 is the minimal superadditive extension
of g1 (Theorem 1, Bruckner 1962):
By Lemma 2.4 gb1 is upper semi-continuous, hence by Lemma 2.1 and Theorem
2.3 gb1 is lower semi-continuous and subadditive.
We claim that gb1 is the maximal subadditive extension of g1 : For this, assume
that the maximal subadditive extension of g1 ; c
g1 , does not coincide with gb1 . Then
c
we must have g1 (x)
gb1 (x) for all x 2 [!; !
b ], with at least one strict inequality.
!
!
c
However, then (by Theorem 2.3) g is a superadditive extension of g and c
g (x)
1
1
1
gb1 (x) with at least one strict inequality - a contradiction proving the claim.
!
!
c
We conclude that f
fb1 if and only if f
f1 and the result follows from
Bruckner’s Theorem 1.
By induction, it follows from Theorem 2.5 that:
Theorem 2.6. Let f1 ; :::; fK be non-negative and subadditive on [0; ! 1 ]; :::; [0; ! K ]
respectively and let f = f1 ^ ::: ^ fK : Let fbk be the maximal subadditive extension
of fk , k = 1; :::; K: Then f is subadditive on [0; !] if fk ^ ::: ^ fK fbk for every k:
Proof: K = 1 is trivial, and K = 2 is Theorem 2.5. For K 3; it remains to verify
that if fk ^ ::: ^ fK is subadditive then fk 1 ^ ::: ^ fK fbk 1 is subadditive. But
that follows from Theorem 2.5.
If ffn g is a pointwise convergent sequence4 of strictly increasing continuous functions, then ffn 1 g is not necessarily pointwise convergent as the following example
shows.
Example 2.1. Let f : [0; 100] ! R+ be the piecewise a¢ ne (subadditive) function
spanned by the points f(0; 0); (1; 10); (3; 10); (100; 12)g. Let ffn g be the sequence of
(subadditive) functions on [0; 100] where fn is the function spanned by the points
f(0; 0); (1; 10 2 n ); (3; 10 + 2 n ); (100; 12)g; where n = 12 n 1 if n is odd, n =
1) and n = 12 n 1 if n is even n = n 1 otherwise ( 0
n 1 otherwise ( 0
1). Then ffn g converges pointwise to f , but ffn 1 g is not pointwise convergent
since fn 1 (10) = 2 for n even and fn 1 (10) = 53 for n odd.
Theorem 2.7 below shows that any increasing subadditive (superadditive) function can be approximated by a bijective subadditive (superadditive) function pointwise. The class of strictly increasing continuous subadditive (superadditive) functions is in this particular sense dense in the class of increasing subadditive (superadditive) functions, and one may note that Theorem 2.7 together with the Proposition
constitute an alternative proof of Theorem 2.3.
Theorem 2.7. Let f : I ! R+ be increasing:
4 An in…nite sequence of functions {f g converges pointwise to f if for all x 2 I and all " > 0
n
there exists N such that jfn (x) f (x)j " for all n N:
6
LARS PETER ØSTERDAL
(1) f is subadditive if and only if there exists a pointwise convergent sequence of
strictly increasing continuous subadditive functions ffn g where limn!1 ffn g =
!
f and limn!1 ffn 1 g = f .
(2) f is superadditive if and only if there exists a pointwise convergent sequence of strictly increasing continuous superadditive functions ffn g where
limn!1 ffn g = f and limn!1 ffn 1 g = f .
Proof: We prove (1). It is readily veri…ed that if ffn g is a pointwise convergent
sequence of subadditive (superadditive) functions, limn!1 ffn g = f , then f is
subadditive (superadditive) (see e.g. Rosenbaum 1950, Theorem 1.3.2; Bruckner
and Ostrow, 1962, Theorem 2) and the “if” part of (1) follows.
For the “only if” part of (1), let f be increasing and subadditive. We construct
a function fn in a two step procedure. Roughly speaking, …rst the discontinuities
are smoothened out (step 1) and then ‡at segments are eliminated (step 2).
Step 1 : It is readily veri…ed that if f is discontinuous then limx!0+ ff (x)g > 0: If
f is discontinuous de…ne
fen (x) =
nxf ( n1 ); 0
1
f (x),
n
x
x;
1
n
and notice that ffen g converges pointwise to f: If f is continuous de…ne fen = f: Let
fbn be the function on I de…ned by
fbn (x) = infffen (x1 ) + ::: + fen (xsn )g;
where 0 xi x; x1 + ::: + xsn , sn 1. It is easily veri…ed that fbn is subadditive,
and notice that fbn is continuous. Further, fbn converges pointwise to f since for all
x 2 I then fen (x) = fbn (x) for all n su¢ ciently large.
Step 2 : We say that s is a ‡at segment for fbn if fbn (x0 ) = fbn (x00 ) = s for some
x0 6= x00 : Let S = f:::; s 1 ; s0 ; s1 ; :::g denote the increasingly ordered set of ‡at
segments. De…ne x(si ) = minfx j fbn (x) = si g and x(si ) = supfx j fbn (x) = si g:
Now, since fbn is strictly increasing on [0; x(s0 )] the maximal subadditive extension of fbn ’s restriction to [0; x(s0 )] is strictly increasing. Hence by Theorem 2.5
there is some
1
zn0 2 [x(s0 ); minfx(s0 ) + 2 ; x(s1 )g];
n
with the convention zn0 = 1 if x(s0 ) = 1 and if s0 is the largest ‡at segment then
replace minfx(s0 ) + n12 ; x(s1 g with x(s1 ) + n12 , and there is a subadditive strictly
increasing function fbn0 where
(
fbn (x(s0 )) + c0n x; x(s0 ) x zn0
0
fbn (x) =
fbn (x);
otherwise,
for some c0n > 0. Repeating this procedure for each ‡at segment in a way such
that each ‡at segment will selected at some point (for example continuing with
s = s1 ; s 1 ; s2 ; s 2 ; :::), in the limit we obtain a strictly increasing continuous subadditive function fn on I. Furthermore, it is readily veri…ed that ffn g converges
pointwise to fbn hence also to f .
!
It remains to verify that limn!1 ffn g 1 = f (pointwise). If z is a ‡at segment
!
of f , then limn!1 (fn ) 1 (z) = supfy j f (y) = zg = supfy j f (y) zg = f (z): If
z 0 is not a ‡at segment, and furthermore [f (x) < z 0 or f (x) > z 0 ] for all x 2 I then
SUBADDITIVE FUNCTIONS
7
!
limn!1 (fn ) 1 (z 0 ) = supfy j f (y) z 0 g = f (z 0 ): Finally if z 00 is a continuous and
strictly increasing point for a pseudo-inverse of f , then clearly (fn ) 1 (z 00 ) = fy j
!
f (y) = z 00 g = supfy j f (y) z 00 g = f (z 00 ):
The proof of (2) is similar.
3. Acknowledgements
I would like to thank Andrew Bruckner, Jens Leth Hougaard, Lars ThorlundPetersen and an anonymous referee for helpful comments and advice. The usual
disclaimer applies.
References
[1] Bruckner, A., 1960, Minimal superadditive extensions of superadditive functions, Paci…c J
Math 10, 1155-1162.
[2] Bruckner, A., 1962, Tests for the superadditivity of functions, Proc Amer Math Soc 13, 126130.
[3] Bruckner, A. and E. Ostrow, 1962, Some function classes related to the class of convex functions, Paci…c J Math 12, 1203-1215.
[4] Denneberg, D., 1994, Non-additive measure and integral, Kluwer Academic Publishers.
[5] Hille, E. and R. Phillips, 1957, Functional analysis and semi-groups, Amer Math Soc Colloq
Publ, Vol. 31, Revised edition.
[6] Matkowski, J. and T. Światkowski,
¾
1991, Quasi-monotonicity, subadditive bijections of R+ ,
and characterization of LP -norm. J Math Anal Appl 154, 493-506.
[7] Matkowski, J. and T. Światkowski,
¾
1993, On subadditive functions. Proc Amer Math Soc 119,
187-197.
[8] Rosenbaum, R. A., 1950, Sub-additive functions. Duke Math J 17, 227-247.
Lars Peter Østerdal, Institute of Economics, University of Copenhagen, Studiestraede 6, DK-1455 Copenhagen K. Denmark.
E-mail address : [email protected].
URL: http://www.econ.ku.dk/lpo