Let I ⊆ ℝ be a nonempty open interval. The weighted quasi-arithmetic mean A φ : I 2 → I is defined by:

A φ ; λ ( x , y ) : = φ − 1 ( λ φ ( x ) + ( 1 − λ ) φ ( y ) ) ,   x , y ∈ I ,

where λ ∈ ( 0,1 ) and φ : I → ℝ is a continuous strictly monotone function.

Let t , s ∈ ℝ + , the weighted two-variable Bajraktarević mean B f , g : I 2 → I is defined by:

B f , g ; t , s ( x , y ) : = ( f g ) − 1 ( t f ( x ) + s f ( y ) t g ( x ) + s g ( y ) ) ,   x , y ∈ I ,

where f , g : I → ℝ are two continuous functions such that g is nowhere zero on I and the ratio function f g 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 n ≥ 3 , 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 n ≥ 2 , n ∈ ℕ . The case of fixed n = 2 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 ( f , g ) : I → ℝ 2 and ( h , k ) : I → ℝ 2 are equivalent if there exist constants a , b , c , d with a d ≠ c d such that:

h = a f + b g ,   k = c f + d g , (1)

and it can be written by ( f , g ) ~ ( h , k ) .

We will consider the equality problem of weighted Bajraktarević mean to weighted quasi-arithmetic means, that is:

B G , F ; t , s ( x , y ) = A h , λ ( x , y ) ,   x , y ∈ I , (2)

Applying G/H to the both sides of (2) and substituting f : = F ∘ h − 1 , g : = G ∘ h − 1 and φ : = g f , we get an equivalent formulation of (2) as follows:

( t f ( x ) + s f ( y ) ) ⋅ φ ( λ x + ( 1 − λ ) y ) = t f ( x ) φ ( x ) + s f ( y ) φ ( y ) ,   x , y ∈ I . (3)

For the case λ = 1 2 ,   t = s , this equation was considered and solved in [7] under strict monotonicity and continuity of φ and in [8] under continuity of φ , respectively. For the case λ = t = 1 − s , this equation was solved in [9] .

Lemma 1. Let φ : I → ℝ be a strictly monotone function, f : I → ℝ be an arbitrary function, and t , s ∈ ℝ + , λ ∈ ( 0,1 ) . Assume that the functional Equation (3) holds, then either f is identically zero, or f is nowhere zero, f and φ are infinitely many times differentiable and there exists a nonzero constant γ ∈ ℝ such that:

f μ φ ′ = γ , (4)

where

μ : = λ s + ( 1 − λ ) t λ ( 1 − λ ) ( t + s ) ∈ ( 0, + ∞ ) . (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 x ∈ I with x ≠ y 0 , the convex combination λ x + ( 1 − λ ) y 0 is strictly between the values x and y₀. Therefore, by the strict monotonicity of φ , we have that ( φ ( λ x + ( 1 − λ ) y 0 ) − φ ( x ) ) ⋅ ( φ ( y 0 ) − φ ( λ x + ( 1 − λ ) y 0 ) ) > 0 . Then, it follows from (3), that

f ( x ) = s t ⋅ f ( y 0 ) ⋅ φ ( y 0 ) − φ ( λ x + ( 1 − λ ) y 0 ) φ ( λ x + ( 1 − λ ) y 0 ) − φ ( x ) . (6)

This implies that f ( x ) is nonzero for all x ∈ I , furthermore, f ( x ) has the same sign as f ( y 0 ) , 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 φ is continuous. Denote by D φ the set of discontinuity point of φ . Then, the monotonicity of φ implies that D φ is countable.

Let x 0 ∈ I be fixed. Then, λ x 0 + ( 1 − λ ) I is a subinterval of I, hence, I \ D φ intersects λ x 0 + ( 1 − λ ) I . There, exists an element y 0 ∈ I such that λ x 0 + ( 1 − λ ) y 0 ∈ I \ D φ . Thus, φ is continuous at λ x 0 + ( 1 − λ ) y 0 . Therefore, (6) yields that f is continuous at x 0 . 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 0 < α < 1 2 | I | and I α : = ( I − α ) ∩ ( I + α ) . Then, I α is an nonempty interval and I α + [ − α , α ] ⊆ I . Let u ∈ I α , v ∈ [ − α , α ] and substituting x : = u − ( 1 − λ ) v and y : = u + λ v into (3), we obtain that:

( t f ( u − ( 1 − λ ) v ) + s f ( u + λ v ) ) φ ( u ) = t f ( u − ( 1 − λ ) v ) φ ( u − ( 1 − λ ) v ) + s f ( u + λ v ) φ ( u + λ v ) ,

holds for all u ∈ I α and for all v ∈ [ − α , α ] .

Integrating both sides of the above equation on v ∈ [ − α , α ] , it follows that:

φ ( u ) ∫ − α α ( t f ( u − ( 1 − λ ) v ) + s f ( u + λ v ) ) d v = t ∫ − α α     f ( u − ( 1 − λ ) v ) φ ( u − ( 1 − λ ) v ) d v + s ∫ − α α     f ( u + λ v ) φ ( u + λ v ) d v .

After simple change of the variable transformations, for all u ∈ I α , we get:

φ ( u ) ( t 1 − λ ∫ u − ( 1 − λ ) α u + ( 1 − λ ) α     f + s λ ∫ u − λ α u + λ α     f ) = t 1 − λ ∫ u − ( 1 − λ ) α u + ( 1 − λ ) α     f ⋅ φ + s λ ∫ u − λ α u + λ α     f ⋅ φ (7)

Hence, φ is continuously differentiable on I α . Since 0 < α < 1 2 | I | is arbitrary, it follows that φ is continuously differentiable and f is continuous on ∪ α > 0 I α = I . By (6), the continuous differentiability of φ implies that f is also continuously differentiable.

Now, we show that φ and f are twice continuously differentiable. Differentiating (3) with respect to x, we have:

λ φ ′ ( λ x + ( 1 − λ ) y ) ⋅ ( t f ( x ) + s f ( y ) ) + t φ ( λ x + ( 1 − λ ) y ) ⋅ f ′ ( x ) = t f ′ ( x ) φ ′ ( x ) ,   x , y ∈ I . (8)

Substituting x : = u − ( 1 − λ ) v and y : = u + λ v into the above equation and integrating both sides on v ∈ [ − α , α ] , we get:

λ φ ′ ( u ) ∫ − α α ( t f ( u − ( 1 − λ ) v ) + s f ( u + λ v ) ) d v = − t φ ( u ) ∫ − α α     f ′ ( u − ( 1 − λ ) v ) d v + t ∫ − α α ( f φ ) ′ ( u − ( 1 − λ ) v ) d v ,   u ∈ I α .

After similar change of the variable transformations as (7), for all u ∈ I α , we obtain:

λ φ ′ ( u ) ( t 1 − λ ∫ u − ( 1 − λ ) α u + ( 1 − λ ) α     f + s λ ∫ u − λ α u + λ α     f ) = − t 1 − λ φ ( u ) ∫ u − ( 1 − λ ) α u + ( 1 − λ ) α     f ′ + t 1 − λ ∫ u − ( 1 − λ ) α u + ( 1 − λ ) α ( f φ ) ′

Thus, φ is twice continuously differentiable on I α and hence on I. Then, by (6), this result implied that f is two times continuously differentiable on I.

To prove that φ and f are infinitely many times differentiable, differentiate (8) with respect to y, we get:

( 1 − λ ) λ φ ′ ′ ( λ x + ( 1 − λ ) y ) ⋅ ( t f ( x ) + s f ( y ) )   + φ ′ ( λ x + ( 1 − λ ) y ) ( λ s f ′ ( y ) + ( 1 − λ ) t f ′ ( x ) ) = 0. (9)

Substituting y : = x , we get:

λ ( 1 − λ ) ( t + s ) f φ ′ ′ + ( λ s + ( 1 − λ ) t ) f ′ φ ′ = 0, (10)

which is equivalent with

( f μ φ ′ ) ′ = 0 ,

where μ : = λ s + ( 1 − λ ) t λ ( 1 − λ ) ( t + s ) > 0 .

Hence, there exists a real constant γ such that f μ φ ′ = γ . If γ were zero, then this equation would imply that φ ′ is identically zero, which contradicts the strict monotonicity of φ . As a consequence, (4) holds. Finally, using (4) and (6) repeatedly, we get that φ and f are infinitely many times differentiable. □

Lemma 2. Let φ : I → ℝ be a strictly monotone function, f : I → ℝ be a non-identically-zero function, and t , s ∈ ℝ + , λ ∈ ( 0,1 ) . If ( φ , f ) solves (3), then we have λ = t t + s . And what’s more, μ defined by (5) equals 2.

Proof. Differentiating (9) with respect to x, we obtain:

λ ( λ s f ′ ( y ) + 2 ( 1 − λ ) t f ′ ( x ) ) ⋅ φ ′ ′ ( λ x + ( 1 − λ ) y )   + ( 1 − λ ) t f ′ ′ ( x ) ⋅ φ ′ ( λ x + ( 1 − λ ) y )   + λ 2 ( 1 − λ ) ( t f ( x ) + s f ( y ) ) ⋅ φ ′ ′ ′ ( λ x + ( 1 − λ ) y ) = 0 (11)

Inserting y : = x , it follows that:

λ ( 2 ( 1 − λ ) t + λ s ) f ′ φ ′ ′ + λ 2 ( 1 − λ ) ( t + s ) f φ ′ ′ ′ + ( 1 − λ ) t f ′ ′ φ ′ = 0. (12)

On the other hand, differentiating (9) with respect to x, we obtain:

( λ ( 1 − λ ) ( t + s ) + λ s + ( 1 − λ ) t ) f ′ φ ′ ′   + λ ( 1 − λ ) ( t + s ) f φ ′ ′ ′ + ( λ s + ( 1 − λ ) t ) f ′ ′ φ ′ = 0. (13)

Combing (12) and (13), we conclude that:

λ ( 1 − λ ) ( λ ( t + s ) − t ) f ′ φ ′ ′ + ( λ 2 s − ( 1 − λ ) 2 t ) f ′ ′ φ ′ = 0. (14)

Firstly, we assume:

λ ( 1 − λ ) ( λ ( t + s ) − t ) ≠ 0,   λ 2 s − ( 1 − λ ) 2 t ≠ 0, (15)

then Equation (14) can be rewritten by:

φ ′ ′ φ ′ = τ f ′ ′ f ′ , (16)

where τ : = ( 1 − λ ) 2 t − λ 2 s λ ( 1 − λ ) ( λ ( t + s ) − t ) ≠ 0 .

Integrating the above equation, we obtain that there exists a constant c 1 ∈ ℝ \ { 0 } such that:

φ ′ = c 1 f ′ τ . (17)

By (17) and (4), letting k = ( γ c 1 ) 1 / τ ≠ 0 , we get:

f ′ = k f − μ τ .

Solving the above equation, we get there exist constants k 1 , k 3 ∈ ℝ with k 1 ≠ 0 such that:

f ( x ) = { k 1 e k x , μ + τ = 0 , k 2 ( x + k 3 ) τ μ + τ ,   μ + τ ≠ 0 , (18)

where k 2 : = ( k ⋅ μ + τ τ ) τ μ + τ .

Using (4) and (18), we obtain that there exist constants b 1 , b 2 ∈ ℝ such that:

φ ( x ) = { a 1 e − μ k x + b 1 , μ + τ = 0 , a 2 ( x + k 3 ) 1 − μ τ μ + τ + b 2 , μ + τ ≠ 0 , (19)

where a 1 : = − γ / ( μ k k 1 μ ) , a 2 : = γ / ( k 2 μ ( 1 − μ τ μ + τ ) ) .

For the case μ + τ = 0 , substituting (18) and (19) into (3), we obtain:

e − μ k ( λ x + ( 1 − λ ) y ) ( t e k x + s e k y ) = t e k x ( 1 − μ ) + s e k y ( 1 − μ ) . (20)

Comparing the coefficients of x after we make Taylor expansion of the above equation, we can get that:

t = λ ( t + s ) ,

which leads to contradictions with (15).

Similarly, for the case μ + τ ≠ 0 , substituting (18) and (19) into (3) and comparing the coefficients, we can get:

μ τ μ + τ = 1,

that is φ ( x ) = a 2 + b 2 , which leads to contradictions with the assumption of φ strictly monotone function.

Secondly, we assume:

λ ( 1 − λ ) ( λ ( t + s ) − t ) ≠ 0,   λ 2 s − ( 1 − λ ) 2 t = 0, (21)

then Equation (14) can be rewritten by:

f ′ φ ′ ′ = 0. (22)

Combing (4) and (25), it leads to f ′ = φ ′ ′ = 0 . Therefore, there exits a , b , c ∈ ℝ with a ≠ 0 such that:

f ( x ) = c ,   φ ( x ) = a x + b ,   x ∈ I .

Substituting the above equations into (3) and comparing the coefficients of x, we obtain:

t = λ ( t + s ) ,

which leads to contradictions with (21).

Thirdly, we assume:

λ ( 1 − λ ) ( λ ( t + s ) − t ) = 0,   λ 2 s − ( 1 − λ ) 2 t ≠ 0, (23)

that is:

λ = t t + s ,   t ≠ s , (24)

then Equation (14) can be rewritten by:

f ′ ′ φ ′ = 0. (25)

Using (5), (24) leads to μ = 2 . Due to (4), φ ′ is nowhere zero. Therefore, f ′ ′ = 0 on I. Then, there exist a , b ∈ ℝ such that:

f ( x ) = a ( x + b ) ,   x ∈ I . (26)

Using (4) and μ = 2 , we get there exist c 1 ∈ ℝ such that:

φ ( x ) = a 1 ( x + b ) − 1 + c 1 , (27)

where a 1 : = − γ a − 2 .

When f , φ satisfy (26) and (27), respectively, it is easy to verify that (3) is valid.

Finally, for the case:

λ ( 1 − λ ) ( λ ( t + s ) − t ) = 0,   λ 2 s − ( 1 − λ ) 2 t = 0, (28)

it holds that λ = 1 2 and s = t which lead to μ = 2 from (5). □

Using Lemma 2 and Theorem 5 in [9] , we get the following.

Theorem 1. Let φ : I → ℝ be a strictly monotone function, f : I → ℝ be a non-identically-zero function, and t , s ∈ ℝ + , λ ∈ ( 0,1 ) . Then, (3) holds if and only if λ = t t + s , f is nowhere zero and there exists p ∈ ℝ with ( λ − 1 2 ) p = 0 such that:

( f , f ⋅ φ ) ~ ( S p , C p ) , (29)

where

S p ( x ) : = { sin ( − p x ) , p < 0 , x , p = 0 , sinh ( p x ) , p > 0 ,     and     C p ( x ) : = { cos ( − p x ) , p < 0 , 1 , p = 0 , cosh ( p x ) , p > 0.

Corollary 1. Let t , s ∈ ℝ + , λ ∈ ( 0,1 ) , and let G , F : I → ℝ be two continuous functions such that g is nowhere zero on I and the ratio function f g is strictly monotone on I, h : I → ℝ be a continuous strictly monotone function. Then, (2) holds if and only if λ = t t + s and there exists p ∈ ℝ with ( λ − 1 2 ) p = 0 such that:

( F , G ) ~ ( S p ∘ h , C p ∘ h ) . (30)

The author declares no conflicts of interest regarding the publication of this paper.

Yang, Y.X. (2024) On the Equality of Weighted Bajratarević Means to Quasi-Arithmetic Means. Journal of Applied Mathematics and Physics, 12, 1126-1133. https://doi.org/10.4236/jamp.2024.124070