On the Equality of Weighted Bajratarević Means to Quasi-Arithmetic Means ()
1. Introduction
Let
be a nonempty open interval. The weighted quasi-arithmetic mean
is defined by:
where
and
is a continuous strictly monotone function.
Let
, the weighted two-variable Bajraktarević mean
is defined by:
where
are two continuous functions such that g is nowhere zero on I and the ratio function
is strictly monotone on I.
The research on the equality of Bajraktarević means has experienced a long history. As early as 1958, Bajraktarević [1] solved the equality of n-variable quasi-arithmetic means with weight function for a fixed
, and got the necessary and sufficient condition under twice differentiable assumption. Aczél and Daróczy [2] obtained the same result without differentiability conditions when the equality holds for all
,
. The case of fixed
is much more difficult and allows considerably more solutions. Losonczi [3] got 32 new families of solution under the six-time differentiable supposition. Several new characterizations of the equality of two-variable Bajraktarević means have been obtained by Losonczi, Páles and Zakaria in [4] under the same regularity assumptions. Recently, Páles and Zakaria [5] obtained the same conclusion under only first-order differentiability. Grünwald and Páles [6] considered the equality problem of generalized Bajraktarević means.
We say that two pairs of functions
and
are equivalent if there exist constants
with
such that:
(1)
and it can be written by
.
We will consider the equality problem of weighted Bajraktarević mean to weighted quasi-arithmetic means, that is:
(2)
Applying G/H to the both sides of (2) and substituting
,
and
, we get an equivalent formulation of (2) as follows:
(3)
For the case
, this equation was considered and solved in [7] under strict monotonicity and continuity of
and in [8] under continuity of
, respectively. For the case
, this equation was solved in [9] .
2. Some Necessary Conditions
Lemma 1. Let
be a strictly monotone function,
be an arbitrary function, and
,
. Assume that the functional Equation (3) holds, then either f is identically zero, or f is nowhere zero, f and
are infinitely many times differentiable and there exists a nonzero constant
such that:
(4)
where
(5)
Proof. If f is identically zero, then (3) holds. Now, we assume that there exists a point y0 such that f does not vanish at y0. Then, for
with
, the convex combination
is strictly between the values x and y0. Therefore, by the strict monotonicity of
, we have that
. Then, it follows from (3), that
(6)
This implies that
is nonzero for all
, furthermore,
has the same sign as
, i.e. the sign of f is constant.
In what follows, we prove that, at every point of I, the function f is continuous at every point where
is continuous. Denote by
the set of discontinuity point of
. Then, the monotonicity of
implies that
is countable.
Let
be fixed. Then,
is a subinterval of I, hence,
intersects
. There, exists an element
such that
. Thus,
is continuous at
. Therefore, (6) yields that f is continuous at
. Hence, f is continuous almost everywhere. On the other hand, f is bounded an every compact subinterval of I, it follows that f is Riemann integrable on every compact subinterval of I.
Let
and
. Then,
is an nonempty interval and
. Let
,
and substituting
and
into (3), we obtain that:
holds for all
and for all
.
Integrating both sides of the above equation on
, it follows that:
After simple change of the variable transformations, for all
, we get:
(7)
Hence,
is continuously differentiable on
. Since
is arbitrary, it follows that
is continuously differentiable and f is continuous on
. By (6), the continuous differentiability of
implies that f is also continuously differentiable.
Now, we show that
and f are twice continuously differentiable. Differentiating (3) with respect to x, we have:
(8)
Substituting
and
into the above equation and integrating both sides on
, we get:
After similar change of the variable transformations as (7), for all
, we obtain:
Thus,
is twice continuously differentiable on
and hence on I. Then, by (6), this result implied that f is two times continuously differentiable on I.
To prove that
and f are infinitely many times differentiable, differentiate (8) with respect to y, we get:
(9)
Substituting
, we get:
(10)
which is equivalent with
where
.
Hence, there exists a real constant
such that
. If
were zero, then this equation would imply that
is identically zero, which contradicts the strict monotonicity of
. As a consequence, (4) holds. Finally, using (4) and (6) repeatedly, we get that
and f are infinitely many times differentiable.
Lemma 2. Let
be a strictly monotone function,
be a non-identically-zero function, and
,
. If
solves (3), then we have
. And what’s more,
defined by (5) equals 2.
Proof. Differentiating (9) with respect to x, we obtain:
(11)
Inserting
, it follows that:
(12)
On the other hand, differentiating (9) with respect to x, we obtain:
(13)
Combing (12) and (13), we conclude that:
(14)
Firstly, we assume:
(15)
then Equation (14) can be rewritten by:
(16)
where
.
Integrating the above equation, we obtain that there exists a constant
such that:
(17)
By (17) and (4), letting
, we get:
Solving the above equation, we get there exist constants
with
such that:
(18)
where
.
Using (4) and (18), we obtain that there exist constants
such that:
(19)
where
,
.
For the case
, substituting (18) and (19) into (3), we obtain:
(20)
Comparing the coefficients of x after we make Taylor expansion of the above equation, we can get that:
which leads to contradictions with (15).
Similarly, for the case
, substituting (18) and (19) into (3) and comparing the coefficients, we can get:
that is
, which leads to contradictions with the assumption of
strictly monotone function.
Secondly, we assume:
(21)
then Equation (14) can be rewritten by:
(22)
Combing (4) and (25), it leads to
. Therefore, there exits
with
such that:
Substituting the above equations into (3) and comparing the coefficients of x, we obtain:
which leads to contradictions with (21).
Thirdly, we assume:
(23)
that is:
(24)
then Equation (14) can be rewritten by:
(25)
Using (5), (24) leads to
. Due to (4),
is nowhere zero. Therefore,
on I. Then, there exist
such that:
(26)
Using (4) and
, we get there exist
such that:
(27)
where
.
When
satisfy (26) and (27), respectively, it is easy to verify that (3) is valid.
Finally, for the case:
(28)
it holds that
and
which lead to
from (5).
3. Main Results
Using Lemma 2 and Theorem 5 in [9] , we get the following.
Theorem 1. Let
be a strictly monotone function,
be a non-identically-zero function, and
,
. Then, (3) holds if and only if
, f is nowhere zero and there exists
with
such that:
(29)
where
Corollary 1. Let
,
, and let
be two continuous functions such that g is nowhere zero on I and the ratio function
is strictly monotone on I,
be a continuous strictly monotone function. Then, (2) holds if and only if
and there exists
with
such that:
(30)