arXiv:1704.01643v1 [math.CA] 5 Apr 2017
Extending means to several variables
Attila Losonczi
16 March 2017
Abstract
We begin the study of how to extend few variable means to several
variable ones and how to shrink means of several variables to less
variables. With the help of one of the techniques we show that it is
enough to check an inequality between two quasi-arithmetic means in
2-variables and that simply implies the inequality in m-variables. The
technique has some relation to Markov chains. This method can be
applied to symmetrization and compounding means as well.
1
Introduction
In this paper we are going to study the ways of extensions of an n-variable
mean to an m-variable mean (n < m) and vica versa shrinking an m-variable
mean to an n-variable mean.
The origin of the problem was raised by M. Hajja in [6] Problem 14: is
there a natural way of deriving the definition of the n-variable arithmetic
mean from the definition of the 2-variable version. And what can one say
in general? Can we define when an n-variable mean is concordant to an mvariable mean i.e. they are the different variable versions of the same mean
(where ”mean” in this last context is just a variableless generic notion).
0
AMS (2010) Subject Classifications: 39B12, 39B22
Key Words and Phrases: generalized mean, iteration of mean, functional equation of
mean
1
Can we go that far? We start to answer these questions by presenting both
positive and negative results.
On basic facts on means the reader has to consult [4]. However we provide
some basic definitions.
A n-variable mean K is called stricly internal if min (a1 , . . . , an ) < K(a1 , . . . , an ) <
max (a1 , . . . , an ). An n-variable mean K is said to be monotone if ai ≤
bi (1 ≤ i ≤ n) implies K(a1 , . . . , an ) ≤ K(b1 , . . . , bn ). K is called continuous
(i)
(1)
(n)
if ak → c(i) (1 ≤ i ≤ n) then K(ak , . . . , ak ) → K(c(1) , . . . , c(n) ) i.e. K
is continuous as an n-variable function. K is symmetric if K(a1 , . . . , an ) =
K(ap(1) , . . . , ap(n) ) for all permutations p : {1, . . . n} → {1, . . . n}.
All means considered in this paper are symmetric, strictly internal, monotone and continuous if we do not say otherwise.
Sometimes we will denote a 2-variable mean by a ◦, so instead of K(a, b)
we will write a ◦ b.
Definition 1 A 2-variable mean ◦ is called round if it fulfils functional
equation (a ◦ k) ◦ (k ◦ b) = k for ∀a, b where k = a ◦ b.
2
Some basic observations
Unfortunately we cannot expect one generic, unique way to extend/shrink a
mean such that it keeps concordance. E.g. let us consider the following three
3-variable means.√
K1 (a, b, c) = qabc
K2 (a, b, c) =
ab+ac+bc
3
√
√
√
ab+ ac+ bc
3
K3 (a, b, c) =
√
For all 3 means we may expect K(a, b) = ab being the corresponding
2-variable mean. Hence when we extend K we cannot expect to get all three
3-varible means, or better to say there should be three extending methods at
least. And in the opposite when we reduce the means K1 , K2 , K3 to 2-variable
means, we cannot expect one generic way.
3
Extensions
We define the most natural way of extensions. We start with the simpliest
case and then gradually build the more complex ones.
2
3.1
Extending to 3-variables
Let a 2-variable mean ◦ be given and a, b, c ∈ R with a ≤ b ≤ c . Let us
define three sequences in the following way.
a0 = a, b0 = b, c0 = c
an+1 = an ◦ bn , bn+1 = an ◦ cn , cn+1 = bn ◦ cn
Obviously an ≤ bn ≤ cn since ◦ is monotone. Moreover an is increasing,
cn is decreasing and a ≤ an , cn ≤ c, hence both sequences are convergent. In
order to prove more we need some lemmas.
Lemma 1 Let ◦ be a strictly internal, continuous mean and (an ) be a
convergent sequence with an → a and (bn ) be a bounded sequence. If an ◦bn →
a then bn → a.
Proof: Let us suppose the opposite: there is a subsequence bnk of (bn )
converging to b 6= a (say a < b). Then ank ◦ bnk → a ◦ b. By strict internality
a < a ◦ b which is a contradiction.
With a slighly more assumption we can prove more:
Lemma 2 Let ◦ be a stricly monotone, continuous mean and (an ) be a
convergent sequence with an → a and (bn ) be a bounded sequence. If an ◦ bn
is convergent then bn converges as well.
Proof: Let us suppose the opposite: there are two subsequences bnk , bmk
of (bn ) converging to c1 6= c2 (say c1 < c2 ). Then ank ◦ bnk → a ◦ c1 and amk ◦
bmk → a ◦ c2 . By strict monotonicity a ◦ c1 < a ◦ c2 which is a contradiction.
Remark 1 Strict monotonicity implies strict internality.
Proposition 1 For the sequences (an ), (bn ), (cn ) defined above, we get
that all three sequences converge to the same limit.
Proof: If an → α then by Lemma 1 we get bn → α. Similarly if cn → γ
then cn+1 = bn ◦ cn gives γ = α.
Definition 2 If K is a 2-varible mean then for a, b, c ∈ R let us denote
the common limit point of the three associated sequences by K (3) (a, b, c).
3
Remark 2 If n is fixed, an can be considered as a function of a, b, c. If
K is continuous then by induction one can derive that an (a, b, c) is continuous
as well. The same is true for bn , cn too.
Remark 3 If n is fixed, an can be considered as a 3-variable mean of
a, b, c. However we are not going to discuss such means because they are not
natural enough.
Theorem 1 K (3) is strictly internal, monotone, continuous.
Proof: a < a1 and (an ) being (strictly) increasing shows strict internality
(for c similarly).
If a ≤ a′ , b ≤ b′ , c ≤ c′ then clearly an ≤ a′n , bn ≤ b′n , cn ≤ c′n (∀n) therefore
α ≤ α′ (an → α, a′n → α′ ).
In order to prove that K (3) is continuous let a ≤ b ≤ c be given and
let k = K (3) (a, b, c). For ǫ > 0 we find N such that n > N implies that
k − ǫ ≤ an ≤ k ≤ cn ≤ k + ǫ. By Remark 2 aN (a, b, c), cN (a, b, c) are
continuous functions of a, b, c hence there is δ > 0 such that if a′ ∈ [a −
δ, a + δ], b′ ∈ [b − δ, b + δ], c′ ∈ [c − δ, c + δ] then k − ǫ ≤ aN (a′ , b′ , c′ ) ≤
k ≤ cN (a′ , b′ , c′ ) ≤ k + ǫ (the intervals are closed deliberately). If n > N
then k − ǫ ≤ aN (a − δ, b − δ, c − δ) ≤ an (a − δ, b − δ, c − δ) ≤ an (a′ , b′ , c′ ) ≤
cn (a′ , b′ , c′ ) ≤ cn (a + δ, b + δ, c + δ) ≤ cN (a + δ, b + δ, c + δ) ≤ k + ǫ where we
used that an (x, y, z) is increasing (x,y,z fixed) (cn is decreasing). Hence we
can conclude that K (3) is continuous.
Proposition 2 K (3) has the following properties:
(1) K (3) (a, a, a) = a
(2) a ≤ K(a, K(a, b)) ≤ K (3) (a, a, b) ≤ K(a, b) ≤ K(K(a, b), b) ≤
(3)
K (a, b, b) ≤ b (if K is strictly internal then ”≤” can be replaced by ”<”)
(3) K (3) (K(a, b), K(a, c), K(b, c)) = K (3) (a, b, c)
(4) If a < b then ∃x ∈ (a, b) such that K (3) (a, x, b) = x.
(5) If K (3) (a, x, b) = x then K (3) (a, a, b) ≤ x ≤ K (3) (a, b, b).
Proof: (1) an = a(∀n).
(2) The first inequality is obvious. For the second let us take the associated sequences for the 3-tuples a, a, b. We get a2 = K(a, K(a, b)) and
(an ) being increasing gives the second inequality. For the third consider
c1 = K(a, b) and cn is decreasing. The rest are similar.
4
(3) a1 = K(a, b), b1 = K(a, c), c1 = K(b, c) and if we create the three
sequences for a1 , b1 , c1 then we end up with the same sequences than for a, b, c
except that the indexes will be moved by 1. Hence K (3) (a1 , b1 , c1 )=K (3) (a, b, c).
Similarly we can get K (3) (an , bn , cn )=K (3) (a, b, c)∀n.
(4) Let us define f (x) = K (3) (a, x, b) − x. By Theorem 1 f is continuous.
By (2) f (a) ≥ 0, f (b) ≤ 0 implies the existence of x such that f (x) = 0.
(5) By monotonity K (3) (a, a, b) ≤ K (3) (a, x, b) = x.
Proposition 3 Let K be round (see Definition 1). Then ∀a ≤ b K (3) (a, k, b) =
k where k = K(a, b).
Proof: For the associated sequences a1 = K(a, k), b1 = k, c1 = K(k, b).
By roundness we get b2 = k. And b2 = K(a1 , c1 ). If we apply this for a1 , k, c1
we get b3 = k. By induction bn = k ∀n.
Proposition 4 If ∀a ≤ b K (3) (a, x, b) = x implies x = K(a, b) then K
is round.
Proof: Let k = K(a, b). With the associated sequences for a, k, b obviously
b1 = k. By Proposition 2 (3) K (3) (a1 , b1 , c1 ) = K (3) (a, k, b) = k = b1 . Because
b2 = K(a1 , c1 ) we have K (3) (a1 , b2 , c1 ) = b2 . By uniqueness b1 = b2 that is K
being round.
Proposition 5 Let ◦ be a strictly monotone, round mean, k = a ◦ b. If
(a ◦ x) ◦ (x ◦ b) = k then x = k.
Proof: Suppose x < k. Then a ◦ x < a ◦ k and x ◦ b < k ◦ b which implies
that k = (a ◦ x) ◦ (x ◦ b) < (a ◦ k) ◦ (k ◦ b) = k - a contradiction.
Proposition 6 If K1 , K2 are two 2-variable means and K1 (a, b) ≤
(3)
(3)
K2 (a, b)(∀a, b) then K1 (a, b, c) ≤ K2 (a, b, c)(∀a, b, c).
Proof: Obviously the associated sequences have the same property.
Proposition 7 K is a 2-variable quasi-arithmetic mean then K (3) is
the associated 3-variable quasi-arithmetic mean.
5
Proof: There is a sctricly monotone, continuous function f such that
(b)
K(a, b) = f −1 ( f (a)+f
). By induction one can show that
2
a2n = f −1 (
2n
2n
22n +2
f (a)+ 2 3−1 f (b)+ 2 3−1 f (c)
3
2n
2
22n −1
22n +2
22n −1
f (b)+ 3 f (c)
).
22n
22n −1
22n +2
22n −1
f (a)+ 3 f (b)+ 3 f (c)
c2n = f −1 ( 3
).
22n
22n+1 +1
22n+1 +1
22n+1 −2
f (a)+
f (b)+
f (c)
3
3
a2n+1 = f −1 ( 3
).
22n+1
2n+1
2n+1
2n+1
2
−2
2
+1
2
+1
f (a)+
f (b)+
f (c)
3
3
b2n+1 = f −1 ( 3
).
22n+1
22n+1 −2
22n+1 +1
22n+1 +1
f
(a)+
f
(b)+
f
(c)
3
3
c2n+1 = f −1 ( 3
).
22n+1
−1 f (a)+f (b)+f (c)
Clearly an → f (
) that is the associated
3
b2n = f −1 (
3
f (a)+
).
3
arithmetic mean.
3-variable quasi
In Theorem 5 we will provide a more general proof for this theorem.
3.2
Extension from 2-variable to n-variable
It is easy to generalize the previous method for extending a mean to m
variable (m ≥ 3). Let a 2-variable mean ◦ be given and a(1) , . . . , a(m) ∈ R
with a(i) ≤ a(j) (i < j) . Let us define m sequences in the following way.
(i)
a0 = a(i) (1 ≤ i ≤ m)
(1)
(1)
(2)
an+1 = an ◦ an .
(k)
(k−1)
(k+1)
an+1 = an
◦ an
(2 ≤ k ≤ m − 1).
(m)
(m−1)
(m)
an+1 = an
◦ an .
(i)
(j)
(1)
Obviously an ≤ an (i < j) since ◦ is monotone. Moreover an is increas(m)
(1)
(m)
ing, an is decreasing and a(1) ≤ an ≤ an ≤ a(m) , hence both sequences
are convergent.
We can prove all similar statements for m varible than we had for 3. We
enumerate those theorems mostly without proof as the proofs can be copied
from the 3-variable case. E.g. with the help of Lemma 1 one can show:
(i)
Proposition 8 For the sequences (an ) (1 ≤ i ≤ m) defined above, we
get that all m sequences converge to the same limit.
Definition 3 If K is a 2-varible mean then for a(1) , . . . , a(m) ∈ R let us
denote the common limit point of the m associated sequences by K (m) (a(1) , . . . , a(m) ).
6
Proposition 9 K (m) is strictly internal, monotone, continuous.
Proposition 10 If K1 , K2 are two 2-variable means and K1 (a, b) ≤
(m)
(m)
K2 (a, b)(∀a, b ∈ R) then K1 (a(1) , . . . , a(m) ) ≤ K2 (a(1) , . . . , a(m) ) (∀a(1) , . . . , a(m) ).
We will prove the following proposition later in Theorem 5 in a more
generic context. However one can also generalize the way in Proposition 7.
Proposition 11 K is a 2-variable quasi-arithmetic mean then K (m) is
the associated m-variable quasi-arithmetic mean.
Remark 4 Analysing this extension method one might think that why
do not we try to make all 2-means between all pairs of the points a(1) . . . , , a(m) .
Unfortunately that would not work as a simple case a(1) = a(2) = a, a(3) =
a(4) = b shows: the (first and last) sequences do not converge to the same
limit (another problem is that the number of sequences would tend to infinity).
3.3
Extension from n-variable to m-variable
We are going to follow similar way of extension that we have already shown
previously for simpler cases.
We are going to descibe the method when there are m points and we
create m sequences from them in a way that a new element of a sequence is
based on the n-mean of the previous step sequence elements and always from
the same ones.
In order to describe such generic method we need some definitions first.
Definition 4 Im = {1, . . . , m} (m ∈ N).
n
Im
= {(j1 , . . . , jn ) : ji ∈ Im } (n ∈ N, n < m).
n
A partial order on Im
is defined by (j1 , . . . , jn ) ≤ (k1 , . . . , kn ) ⇐⇒ j1 ≤
k1 , . . . , jn ≤ kn .
Let a n-variable mean K be given and a(1) , . . . , a(m) ∈ R with a(i) ≤
a(j) (i < j) . If we have m defining sequences, how can we decribe such
defining relations? We have definitions like this:
(1)
(m)
a0 = a(1) , . . . , a0 = a(m) .
(j )
(j )
(i)
ak+1 = K(ak i,1 , . . . , ak i,n ) (1 ≤ i ≤ m, k ∈ N) where ji,1 , . . . , ji,n ∈ Im
depend on i only.
7
n
So we can associate a ti ∈ Im
∀i to that, namely ti = (ji,1 , . . . , ji,n ).
n
And conversely if we have a T = {t1 , . . . , tm } (ti ∈ Im
∀i) then we can
(j )
(ji,1 )
(i)
create sequences with the help of it: ak+1 (T ) = K(ak , . . . , ak i,n ) where
ti = (ji,1 , . . . , ji,n ).
Therefore such associated sequences can be definied by a system T =
n
{t1 , . . . , tm } where ti ∈ Im
(1 ≤ i ≤ m).
We also prefer the following four properties of T :
(1) t1 ≤ t2 ≤ · · · ≤ tm
(2) ∀k ∈ Im , |{i : k ∈ ti }| = n
(3) ∀i min ti ≤ i ≤ max ti ; if 2 ≤ i ≤ m − 1 then min ti < i < max ti
(4) ∀i ≥ 2 ∃j < i such that i ∈ tj .
Definition 5 T = {t1 , . . . , tm } is called admissible if it satisfies (1),(2),(3),(4).
Theorem 2 If T is admissible then all sequences converges to the same
limit.
Proof: Obvious consequence is that 1 ∈ t1 , m ∈ tm and (1) implies that if
(i)
(j)
i ≤ j then ak ≤ ak ∀k.
(1)
(2)
(m)
(1)
(m)
a(1) ≤ ak ≤ ak · · · ≤ ak ≤ a(m) ∀k and ak is inceasing and ak is
decreasing hence both converges. We show that all converges to the same
(i)
(1)
limit. Let ak → c. Suppose there is i ≥ 2 such that (ak ) does not converge
to c. Let i denote the least such index. Then by (4) there is j < i such that
(p)
i ∈ tj = (u1 , . . . , un ). All sequences (ak ) are bounded (p ∈ tj ) hence we can
(p)
find a subsequence of (k) say (kq ) such that all sequences (akq ) are converges,
(p)
(i)
say (akq ) → wp . Let us choose (kq ) such that (akq ) → wi 6= c. By assumption
(j)
(j)
(u )
(un )
(akq ) → c. By definition (ak+1 ) = K(ak 1 , . . . , ak
). K being stricly inter-
K(wu1 ,...,wun )−c
.
2
nal gives that c < K(wu1 , . . . , wun ). Let ǫ =
K is continuous
′
therefore there exists δ > 0 such that wur ∈ (wur − δ, wur + δ) (1 ≤ r ≤ n)
implies that K(wu′ 1 , . . . , wu′ n ) ∈ (K(wu1 , . . . , wun ) − ǫ, K(wu1 , . . . , wun ) + ǫ).
(j)
There is N ∈ N such that k ≥ N implies ak ∈ (c−ǫ, c+ǫ) and q ≥ N implies
(j)
(u )
(u )
(p)
akq ∈ (wur − δ, wur + δ) (1 ≤ r ≤ n). We get akN +1 = K(akN1 , . . . , akNn ) ∈
(K(wu1 , . . . , wun ) − ǫ, K(wu1 , . . . , wun ) + ǫ) and ∈ (c − ǫ, c + ǫ) at the same
time which is a contradiction.
Definition 6 If we can construct an admissible T = Tn,m from nvariable to m-variable (n < m) then the associated sequences will have all
8
properties to define the extension of the n-varible mean K to a m-variable
mean. Let us denote it by K (T ) .
Theorem 3 If T is admissible then K (T ) will be strictly internal, monotone, continuous.
Proof: The proof of Theorem 1 can be copied for n points (instead of 3)
where we use the corresponding version of Remark 2 for n points.
We now show a way to construct such T . We go by recursion. In 3.2 we
had such T from 2-variables to m-variables, namely
t1 = (1, 2), tk = (k −1, k +1) (2 ≤ k ≤ m−1), tm = (m−1, m). Obviously
it has properties (1),(2),(3),(4).
Let us suppose we have such T = {t1 , . . . , tm−1 } for extension from (n−1)variables to (m − 1)-variables and we construct T ′ for n to m. Let ti =
(ji,1 , . . . ,ji,n−1 ) (∀i). Let us define t′i ∈ T ′ (i ∈ Im )

if 1 ≤ i ≤ n − 1
(1, ji,1 + 1, . . . , ji,n−1 + 1)
′
ti = (i − (n − 1), ji,1 + 1, . . . , ji,n−1 + 1) if n ≤ i ≤ m − 1


(m − n + 1, m − n + 2, . . . , m)
if i = m
Theorem 4 T ′ is admissible.
n
Proof: Obviously t′i ∈ Im
(∀i).
n
′
in the first line type
In the definition of ti let us call the elements of Im
1, second line type 2, third line type 3 elements.
(1): If T satisfies (1) then so does T ′ using also that t′m is the greatest
n
element in Im
.
(2): If i = 1 it is clear.
If 2 ≤ i ≤ m − n then we know that |{h : i − 1 ∈ th ∈ T }| = n − 1. Take
the type 2 element that starts with i. With that element we get |{h : i ∈
t′h ∈ T ′ }| = n.
If m − n + 1 ≤ i then the type 3 element will provide the missing point.
(3): It is easy to check for type 1, 2 and 3 elements.
(4): If i = 2 then t′1 will satisfies the condition. If 3 ≤ i ≤ m is given,
we know that there is j ∈ Im−1 , j < i−1 such that (i−1) ∈ tj . Hence i ∈ t′j . The following theorem gives that the n-variable quasi-arithmetic means
are concordant in this way.
9
Theorem 5 For a quasi-arithmetic n-variable mean K this extension
give the expected result i.e. K (Tn,m ) will be the associated m-variable quasiarithmetic mean.
Proof: There is a sctricly monotone, continuous function f such that
(1)
(a(n) )
K(a(1) , . . . , a(n) ) = f −1 ( f (a )+···+f
). By induction one can prove that
n
(i)
(1)
i,k
(n)
−1 i,k
ak = f (s1 f (a ) + · · ·+ sn f (a )) and in the bracket there is a weighted
i,k
average of f (a(1) ), . . . , f (a(n) ) (in other words si,k
1 + · · · + sn = 1). The first
question is how can one determine those factors. The second is what are
their limit. Our aim is to prove that limk→∞ si,k
→ m1 (∀i∀j) that would
j
prove the theorem completely.
Let us start
( with the first question. Let us define an m × m matrix M:
1
if j ∈ ti
Mi,j = n
.
0 otherwise
k
One can easily show by induction that si,k
j = (M )i,j .
Now we handle the second question that is the limit. In the theory of
Markov chains there is a theorem that states that for an irreducible, aperiodic, positive recurrent and doubly stohastic Markov chain with m states it
holds that limn→∞ P (X(k) = s) = m1 where X is the associated probability
variable, X(k) denotes the state that X has in the k th step and s is any state.
And that is exactly that we need. Let us show that M has all required
properties:
By (2) M is doubly stohastic.
M is irreducible since there is only one communication class because state
”1” and state ”j” communicate (∀j > 1) by (4).
(1)
Aperiodic: p11 > 0 hence for state ”1” the period is 1 and all states in a
communication class have the same period.
Positive recurrent: An irreducible finite-state Markov chain is always positive recurrent.
We can also answer one of the questions of Hajja, namely: is there a
natural way to derive the n-variable arithmetic mean from the 2-variable
arithmetic mean? Our method just provides that (use Theorem 5 with f (x) =
x).
Remark 5 Analysing this extension method property (2) cannot be abandoned if we want to keep Theorem 5 valid as a simple 4-variable case would
show.
10
Proposition 12 If K1 , K2 are two n-variable means and K1 ≤ K2 then
(T
)
≤ K2 n,m .
(T
)
K1 n,m
Now we can formulate one of our main results namely that an inequality
between quasi-arithmetic means is enough to check in 2 variables only.
Theorem 6 If K1 , K2 are n-variable quasi-arithmetic means and K1 ≤
(m)
(m)
(m)
K2 holds in n variables then K1 ≤ K2 holds as well where Ki denotes
the associated m-variable quasi-arithmetic mean (n < m).
Proof: Theorem 5 and Proposition 12.
We state a theorem regarding equivalent means. We recall the classic
definition.
Definition 7 Two means K and L are equivalent if there is a homeomorphism f of R (or between the domains of K, L) such that L = Kf where
Kf (a1 , . . . , an ) = f −1 (K(f (a1 ), . . . , f (an ))).
Theorem 7 Let T be an admissible system for n < m. Let two nvariable means K, L are equivalent by function f . Then K (T ) , L(T ) are equivalent means as well and the same function f shows that.
Proof: Let a(1) , . . . , a(m) ∈ R be given. Let us create the associated sequences to L.
(1)
(m)
a0 = a(1) , . . . , a0 = a(m) .
(j )
(j )
(i)
ak+1 = f −1 (K(f (ak i,1 ), . . . , f (ak i,n ))) (1 ≤ i ≤ m, k ∈ N) where ji,1 , . . . , ji,n ∈
Im depend on i only.
Let us investigate these sequences: (b(i) ) where
(1)
(m)
b0 = f (a(1) ), . . . , b0 = f (a(m) ),
(i)
(i)
bk = f (ak ) (1 ≤ i ≤ m).
(1)
(m)
If we run the same process for K and b0 , . . . , b0 then we end up with
(1)
(m)
(i)
K (T ) (b0 , . . . , b0 ) that equals to K (T ) (f (a(1) ), . . . , f (a(m) )). I.e. limk→∞ f (ak ) =
(i)
K (T ) (f (a(1) ), . . . , f (a(m) )) or limk→∞ ak = f −1 (K (T ) (f (a(1) ), . . . , f (a(m) )))
but this limit gives L(T ) (a(1) , . . . , a(m) ) as well.
11
4
Shrinking
We descibe a generic way of reducing the number of variables of a mean that
is similar the technique that we had in the previous section.
Let K be a stricly internal, monotone, continuous m-variable mean. Let
n < m and a(1) ≤ · · · ≤ a(n) ∈ R be given. We create sequences in the
following way:
(1)
(n)
a0 = a(1) , . . . , a0 = a(n) .
(i)
(1)
(i−1) (i)
(i) (i+1)
(n)
ak+1 = K(ak , . . . , ak , ak , . . . , ak , ak , . . . , ak ) (1 ≤ i ≤ n) where
(i)
in the middle there are (m − n + 1) pieces of ak .
So the associated defining system T = Tm,n is the following: T = {t1 , . . . , tn }
where ti ∈ Inm (1 ≤ i ≤ n) namely ti = (1, . . . , i − 1, i, . . . , i, i + 1, . . . , n).
For these we can prove all usual things:
Proposition 13 T is admissible.
(1)
(n)
(1)
Corollary 1 a(1) ≤ ak ≤ · · · ≤ ak ≤ a(n) , (ak ) converges and all
other sequences converge to the same limit.
Definition 8 Let us denote the common limit point by K (T ) .
Corollary 2 K (T ) is stricly internal, monotone, continuous n-variable
mean.
K
Theorem 8 For a quasi-arithmetic m-mean K this shrinking gives that
will be the associated n-variable quasi-arithmetic n-mean (n < m).
(Tm,n )
Proof: We follow the proof of Theorem 5 i.e. we use the theory of Markov
chains.
With the mean defining function f we get
(i)
(1)
i,k
(n)
ak = f −1 (si,k
)).
1 f (a ) + · · · + sn f (a
In this (
case the associated stohastic n × n matrix M is
1
if i 6= j
Mi,j = m
.
m−n+1
if i = j
m
k
For the weights we get si,k
j = (M )i,j as in Theorem 5.
For M it can be shown similarly that it is irreducible, aperiodic, positive
recurrent and doubly stohastic hence provides a uniform limit distribution
1
i.e. limk→∞ si,k
j = n.
We can formulate a similar statement to Theorem 6.
12
Theorem 9 If K1 , K2 are m-variable quasi-arithmetic means and K1 ≤
K2 holds in m variables then K1 ≤ K2 holds in n variables as well (n < m).
We just formulate the corresponding theorem on shrinking of equivalent
means since the proof is the same (see Theorem 7).
Theorem 10 Let two m-variable means K, L are equivalent by function
f . Then K (Tm,n ) , L(Tm,n ) are equivalent means as well and the same function
f shows that (n < m).
For shrinking means there are many other ways as well, we provide two
more.
Proposition 14 Let K be a n-variable strictly internal and continuous
mean (n ≥ 3). If a < b are given then there exists x ∈ (a, b) such that
K(a, x, . . . , x, b) = x.
Proof: We can follow the proof of Proposition 2 for f (x) = K(a, x, . . . , x, b)−
x where we only used the mentioned two properties of the mean.
Definition 9 If K is a n-variable strictly internal and continuous mean,
a < b then let K (s1 ) (a, b) = inf{x : K(a, x, . . . , x, b) = x}.
Remark 6 Actually the infimum is a minimum because of continuity
of K.
Proposition 15 The definition of K (s1 ) provides a strictly internal,
lower semi continuous mean.
Proof: Let an → a, bn → b, K (s1 ) (an , bn ) = ln . If ln → l then K(an , ln , . . . , ln , bn ) →
K(a, l, . . . , l, b) implies that K(a, l, . . . , l, b) = l that gives K (s1 ) (a, b) ≤ l. Proposition 16 Suppose K is a 2-variable, round mean. Let k =
K(a, b). Then K (T2,4 ) (a, k, k, b) = k.
Proof: By the definition of the usual sequences (a0 = a, b0 = c0 = k, d0 =
b) we get that a1 = b1 = K(a, k), c1 = d1 = K(k, b) and a2 = K(a, k), b2 =
c2 = k, d2 = K(k, b) by roundness. From this point we can go by induction
and get that b2n = k ∀n hence bn → k.
13
Proposition 17 If K is a n-variable quasi-arithmetic mean then K (s1 )
is the corresponding 2-variable quasi-arithmetic mean.
We provide one more way of shrinking.
Proposition 18 (1) If a < b are given, K is a 2n-variable strictly internal, monotone and continuous mean then K (s2 ) (a, b) = K(a, . . . , a, b, . . . , b)
is strictly internal, monotone, continuous where there are n pieces of a and
n-pieces of b inside.
(2) Similarly K (s2 ) (a(1) , . . . , a(n) ) = K(a(1) , . . . , a(n) , a(1) , . . . , a(n) ) is strictly
internal, monotone, continuous.
Proposition 19 Suppose K is a 2-variable, round mean. If we construct K (T2,4 ) as in Subsection 3.2 then K (T2,4 ) (a, a, b, b) = K(a, b).
Proof: Let k = K(a, b). By the definition of the usual sequences (a0 =
b0 = a, c0 = d0 = b) we get that a1 = a, b1 = c1 = k, d1 = b and a2 = b2 =
K(a, k), c2 = d2 = K(k, b). By roundness we have c3 = d3 = k. From this
point we can go by induction and get that c2n+1 = k ∀n hence cn → k. Proposition 20 If K is a 2n-variable quasi-arithmetic mean then K (s2 )
is the corresponding 2-variable quasi-arithmetic mean.
We close this section with some counterexamples.
q
ab+ac+bd+cd
Example 1 Let K(a, b, c, d) =
. Then K is strictly in4
ternal, monotone and continuous.
We may expect that the corresponding 2
√
-variable mean would be ab. Let us see what our three methods gives.
.
(1) K(a, a, b, b) = a+b
2
(2) K(a, x, x, b) = x implies that x = a+b
.
2
(3) When we use the general shrinking method K (T4,2 ) for e.g. a = 1, b = 3
then we get b4 < 2 = a+b
hence the limit will be less than a+b
because (bn )
2
2
is decreasing. (We conjecture that it gives something between the geometric
and arithmetic mean.)
q
ab+ac+bc
Example 2 Let K(a, b, c) =
. Then K is strictly internal,
3
monotone and continuous. Let us see what our
√ applicable two methods gives.
(1) K(a, x, b) = x implies that x =
a+b+
14
(a+b)2 +12ab
.
6
(2) When we use the general shrinking method K (T3,2 ) it gives a different
result, for e.g. a = 0.1, b = 2 then we get b3 < 0.781 hence the limit will be
less than that because (bn ) is decreasing. And for the same values the method
in (1) gives approx. 0.784.
Example 3 Let L(a, b, c) = min{a,b,c}+max{a,b,c}
or simply L(a, b, c) =
2
a+c
if
a
≤
b
≤
c.
Then
there
does
not
exist
a
2-variable
strictly internal,
2
(T2,3 )
monotone, continuous mean such that L = K
because K (T2,3 ) (a, a, b) <
(T2,3 )
K
(a, b, b) (a < b) which obviously does not hold for L. Clearly L(T3,2 ) (a, c) =
a+c
and then (L(T3,2 ) )(T2,3 ) 6= L.
2
5
On copounding
We can simply generalize our extension method from n-variable to m-variable
by interchanging K to m pieces of means K1 ≤ · · · ≤ Km i.e. the defining
sequences would be:
(1)
(m)
a0 = a(1) , . . . , a0 = a(m)
(j )
(j )
(i)
ak+1 = Ki (ak i,1 , . . . , ak i,n ) (1 ≤ i ≤ m, k ∈ N) where ji,1 , . . . , ji,n ∈ Im
depend on i only.
In the usual way one can prove the existence of a common limit, therefore
we can derive a new kind of mean of points a(1) ≤ · · · ≤ a(m) and means
K1 ≤ · · · ≤ Km .
If n = m and ji,h = h ∀i then it is a generalization of compounding of
two means.
6
Symmetrization
Using similar technique we can symmetrize a non-symmetric 2-variable mean.
Let ◦ be a non-symmetric, strictly internal, monotone, continuous mean. Let
a < b ∈ R be given. Let us define two sequences:
a0 = a, b0 = b.
an+1 = min {an ◦ bn , bn ◦ an }, bn+1 = max {an ◦ bn , bn ◦ an }.
Obviously a ≤ an is increasing while bn ≤ b is decreasing, therefore both
converges. By continuity they must converge to the same limit point. Let us
denote it by K (sym) (a, b).
Proposition 21 K (sym) is symmetric.
15
Proof: The associated sequences for (a, b) and (b, a) are the same.
References
[1] R. Abu-Saris, M. Hajja, On Gauss compounding of symmetric
weighted arithmetic means, Journal of Mathematical Analysis and
Applications 322 (2006), 729–734.
[2] J. Aczél, On mean values, Bull. Amer. Math. Soc. 54 (1948), 392–400.
[3] J. M. Borwein, P. B. Borwein, The way of all means, Amer. Math.
Monthly 94 (1987), 519-522.
[4] P. S. Bullen, Handbook of means and their inequalities, vol. 260
Kluwer Academic Publisher, Dordrecht, The Netherlands (2003).
[5] Z. Daróczy and Zs. Páles, On functional equations involving means,
Publ. Math. Debrecen 62 no. 3–4 (2003), 363–377.
[6] M. Hajja, Some elementary aspects of means, International Journal of
Mathematics and Mathematical Sciences, Means and Their Inequalities, Volume 2013, Article ID 698906, 1–9.
[7] Z. Makó, Zs. Páles, The invariance of the arithmetic mean with respect to generalized quasi-arithmetic means, Journal of Mathematical
Analysis and Applications 353 Issue 1 (2009), 8–23.
[8] Z. Makó, Zs. Páles, On the equality of generalized quasi-arithmetic
means, Publ. Math. Debrecen 72/3-4 (2008), 407–440.
Attila Losonczi,
Hungary 2120 Dunakeszi Bojtorjan u. 3/1
E-mail: [email protected]
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