On Generalized Flett's Mean Value Theorem

Abstract

We present a new proof of generalized Flett's mean value theorem due to Pawlikowska (from 1999) using only the original Flett's mean value theorem. Also, a Trahan-type condition is established in general case.

1. Introduction

Mean value theorems play an essential role in analysis. The simplest form of the mean value theorem due to Rolle is well known.

Theorem 1.1 (Rolle’s mean value theorem). If 𝑓∶⟨𝑎,𝑏⟩→ℝ is continuous on ⟨𝑎,𝑏⟩ and differentiable on (𝑎,𝑏) and 𝑓(𝑎)=𝑓(𝑏), then there exist a number 𝜂∈(𝑎,𝑏) such that 𝑓^′(𝜂)=0.

A geometric interpretation of Theorem 1.1 states that if the curve 𝑦=𝑓(𝑥) has a tangent at each point in (𝑎,𝑏) and 𝑓(𝑎)=𝑓(𝑏), then there exists a point 𝜂∈(𝑎,𝑏) such that the tangent at (𝜂,𝑓(𝜂)) is parallel to the 𝑥-axis. One may ask a natural question: What if we remove the boundary condition 𝑓(𝑎)=𝑓(𝑏)? The answer is well known as the Lagrange mean value theorem. For the sake of brevity put ^𝑏_𝑎𝒦𝑓^(𝑛),𝑔^(𝑛)=𝑓^(𝑛)(𝑏)−𝑓^(𝑛)(𝑎)𝑔^(𝑛)(𝑏)−𝑔^(𝑛)(𝑎),𝑛∈ℕ∪{0}, (1.1) for functions 𝑓,𝑔 defined on ⟨𝑎,𝑏⟩ (for which the expression has a sense). If 𝑔^(𝑛)(𝑏)−𝑔^(𝑛)(𝑎)=𝑏−𝑎, we simply write ^𝑏_𝑎𝒦(𝑓^(𝑛)).

Theorem 1.2 (Lagrange’s mean value theorem). If 𝑓∶⟨𝑎,𝑏⟩→ℝ is continuous on ⟨𝑎,𝑏⟩ and differentiable on (𝑎,𝑏), then there exist a number 𝜂∈(𝑎,𝑏) such that 𝑓^′(𝜂)=^𝑏_𝑎𝒦(𝑓).

Clearly, Theorem 1.2 reduces to Theorem 1.1 if 𝑓(𝑎)=𝑓(𝑏). Geometrically, Theorem 1.2 states that given a line ℓ  joining two points on the graph of a differentiable function 𝑓, namely, (𝑎,𝑓(𝑎)) and (𝑏,𝑓(𝑏)), then there exists a point 𝜂∈(𝑎,𝑏) such that the tangent at (𝜂,𝑓(𝜂)) is parallel to the given line ℓ.

In connection with Theorem 1.1, the following question may arise: Are there changes if in Theorem 1.1 the hypothesis 𝑓(𝑎)=𝑓(𝑏) refers to higher-order derivatives? Flett, see [1], first proved in 1958 the following answer to this question for 𝑛=1 which gives a variant of Lagrange’s mean value theorem with the Rolle-type condition.

Theorem 1.3 (Flett’s mean value theorem). If 𝑓∶⟨𝑎,𝑏⟩→ℝ is a differentiable function on ⟨𝑎,𝑏⟩ and 𝑓^′(𝑎)=𝑓^′(𝑏), then there exists a number 𝜂∈(𝑎,𝑏) such that 𝑓^′𝜂=^𝜂_𝑎𝒦𝑓. (1.2)

Flett’s original proof, see [1], uses Theorem 1.1. A slightly different proof which uses Fermat’s theorem instead of Rolle’s can be found in [2]. There is a nice geometric interpretation of Theorem 1.3: if the curve 𝑦=𝑓(𝑥) has a tangent at each point in ⟨𝑎,𝑏⟩ and if the tangents at (𝑎,𝑓(𝑎)) and (𝑏,𝑓(𝑏)) are parallel, then there exists a point 𝜂∈(𝑎,𝑏) such that the tangent at (𝜂,𝑓(𝜂)) passes through the point (𝑎,𝑓(𝑎)); see Figure 1.

Figure 1 

Geometric interpretation of Flett’s mean value theorem.

Similarly as in the case of Rolle’s theorem, we may ask about possibility to remove the boundary assumption 𝑓^′(𝑎)=𝑓^′(𝑏) in Theorem 1.3. As far as we know, the first result of that kind is given in the book [3].

Theorem 1.4 (Riedel-Sahoo). If 𝑓∶⟨𝑎,𝑏⟩→ℝ is a differentiable function on ⟨𝑎,𝑏⟩, then there exist a number 𝜂∈(𝑎,𝑏) such that 𝑓^′𝜂=^𝜂_𝑎𝒦𝑓+^𝑏_𝑎𝒦𝑓^′⋅𝜂−𝑎2. (1.3)

We point out that there are also other sufficient conditions guaranteeing the existence of a point 𝜂∈(𝑎,𝑏) satisfying (1.2). First such a condition was published in Trahan’s work [4]. An interesting idea is presented in paper [5] where the discrete and integral arithmetic mean is used. We suppose that this idea may be further generalized for the case of means studied, for example, in [6–8].

In recent years there has been renewed interest in Flett’s mean value theorem; see a survey in [9]. Among the many other extensions and generalizations of Theorem 1.3, see, for example, [10–13], we focus on that of Iwona Pawlikowska [14] solving the question of Zsolt Pales raised at the 35th International Symposium on Functional Equations held in Graz in 1997.

Theorem 1.5 (Pawlikowska). Let 𝑓 be 𝑛-times differentiable on ⟨𝑎,𝑏⟩ and 𝑓^(𝑛)(𝑎)=𝑓^(𝑛)(𝑏). Then there exists 𝜂∈(𝑎,𝑏) such that 𝑓𝜂−𝑓(𝑎)=^𝑛_𝑖=1(−1)^𝑖+1𝑖!𝜂−𝑎^𝑖𝑓^(𝑖)𝜂. (1.4)

Observe that Pawlikowska’s theorem has a close relationship with the 𝑛th Taylor polynomial of 𝑓. Indeed, for 𝑇_𝑛𝑓,𝑥₀(𝑥)=𝑓𝑥₀+𝑓^′𝑥₀1!𝑥−𝑥₀𝑓+⋯+^(𝑛)𝑥₀𝑛!𝑥−𝑥₀^𝑛, (1.5) Pawlikowska’s theorem has the following very easy form 𝑓(𝑎)=𝑇_𝑛(𝑓,𝜂)(𝑎).

Pawlikowska’s proof follows up the original idea of Flett; see [1], considering the auxiliary function 𝐺_𝑓(𝑥)=𝑔^(𝑛−1)(𝑥)(⟩,1,𝑥∈𝑎,𝑏𝑛𝑓^(𝑛)(𝑎),𝑥=𝑎, (1.6) where 𝑔(𝑥)=^𝑥_𝑎𝒦(𝑓) for 𝑥∈(𝑎,𝑏⟩ and using Theorem 1.1. In what follows we provide a different proof of Theorem 1.5 which uses only iterations of an appropriate auxiliary function and Theorem 1.3. In Section 3 we give a general version of Trahan condition, compare [4] under which Pawlikowska’s theorem holds.

2. New Proof of Pawlikowska’s Theorem

The key tool in our proof consists in using the auxiliary function 𝜑_𝑘(𝑥)=𝑥𝑓^{(𝑛−𝑘+1)}(𝑎)+^𝑘_𝑖=0(−1)^𝑖+1()𝑖!𝑘−𝑖)(𝑥−𝑎^𝑖𝑓^{(𝑛−𝑘+𝑖)}(𝑥),𝑘=1,2,…,𝑛. (2.1) Running through all indices 𝑘=1,2,…,𝑛, we show that its derivative fulfills assumptions of Flett’s mean value theorem and it implies the validity of Flett’s mean value theorem for the 𝑙th derivative of 𝑓, where 𝑙=𝑛−1,𝑛−2,…,1.

Indeed, for 𝑘=1 we have 𝜑₁(𝑥)=−𝑓^(𝑛−1)(𝑥)+𝑥𝑓^(𝑛)(𝑎) and 𝜑^′₁(𝑥)=−𝑓^(𝑛)(𝑥)+𝑓^(𝑛)(𝑎). Clearly, 𝜑^′₁(𝑎)=0=𝜑^′₁(𝑏), so applying Flett’s mean value theorem for 𝜑₁ on ⟨𝑎,𝑏⟩ there exists 𝑢₁∈(𝑎,𝑏), such that 𝜑^′₁(𝑢₁)(𝑢₁−𝑎)=𝜑₁(𝑢₁)−𝜑₁(𝑎); that is, 𝑓^(𝑛−1)𝑢₁−𝑓^(𝑛−1)(𝑎)=𝑢₁−𝑎𝑓^(𝑛)𝑢₁. (2.2) Then for 𝜑₂(𝑥)=−2𝑓^(𝑛−2)(𝑥)+(𝑥−𝑎)𝑓^(𝑛−1)(𝑥)+𝑥𝑓^(𝑛−1)(𝑎), we get 𝜑^′₂(𝑥)=−𝑓^(𝑛−1)(𝑥)+(𝑥−𝑎)𝑓^(𝑛)(𝑥)+𝑓^(𝑛−1)(𝑎) (2.3) and 𝜑^′₂(𝑎)=0=𝜑^′₂(𝑢₁) by (2.2). So, by Flett’s mean value theorem for 𝜑₂ on ⟨𝑎,𝑢₁⟩, there exists 𝑢₂∈(𝑎,𝑢₁)⊂(𝑎,𝑏) such that 𝜑^′₂(𝑢₂)(𝑢₂−𝑎)=𝜑₂(𝑢₂)−𝜑₂(𝑎), which is equivalent to 𝑓^(𝑛−2)𝑢₂−𝑓^(𝑛−2)(𝑎)=𝑢₂−𝑎𝑓^(𝑛−1)𝑢₂−12𝑢₂−𝑎²𝑓^(𝑛)𝑢₂. (2.4) Continuing this way after 𝑛−1 steps, 𝑛≥2, there exists 𝑢_𝑛−1∈(𝑎,𝑏) such that 𝑓^′𝑢_𝑛−1−𝑓^′(𝑎)=^𝑛−1_𝑖=1(−1)^𝑖+1𝑖!(𝑢_𝑛−1−𝑎)^𝑖𝑓^(𝑖+1)𝑢_𝑛−1. (2.5) Considering the function 𝜑_𝑛, we get 𝜑^′_𝑛(𝑥)=−𝑓^′(𝑥)+𝑓^′(𝑎)+^𝑛−1_𝑖=1(−1)^𝑖+1()𝑖!𝑥−𝑎^𝑖𝑓^(𝑖)(𝑥)=𝑓^′(𝑎)+^𝑛−1_𝑖=0(−1)^𝑖+1()𝑖!𝑥−𝑎^𝑖𝑓^(𝑖+1)(𝑥). (2.6) Clearly, 𝜑^′_𝑛(𝑎)=0=𝜑^′_𝑛(𝑢_𝑛−1) by (2.5). Then by Flett’s mean value theorem for 𝜑_𝑛 on ⟨𝑎,𝑢_𝑛−1⟩, there exists 𝜂∈(𝑎,𝑢_𝑛−1)⊂(𝑎,𝑏) such that 𝜑^′_𝑛𝜂𝜂−𝑎=𝜑_𝑛𝜂−𝜑_𝑛(𝑎). (2.7) Since 𝜑^′_𝑛𝜂𝜂−𝑎=𝜂−𝑎𝑓^′(𝑎)+^𝑛_𝑖=1(−1)^𝑖(𝑖−1)!𝜂−𝑎^𝑖𝑓^(𝑖)𝜂,𝜑_𝑛𝜂−𝜑_𝑛(𝑎)=𝜂−𝑎𝑓^′(𝑎)−𝑛𝑓𝜂−𝑓(𝑎)+^𝑛_𝑖=1(−1)^𝑖+1()𝑖!𝑛−𝑖𝜂−𝑎^𝑖𝑓^(𝑖)𝜂, (2.8) the equality (2.7) yields −𝑛𝑓𝜂−𝑓(𝑎)=^𝑛_𝑖=1(−1)^𝑖(𝑖−1)!𝜂−𝑎^𝑖𝑓^(𝑖)𝜂1+𝑛−𝑖𝑖=𝑛^𝑛_𝑖=1(−1)^𝑖𝑖!𝜂−𝑎^𝑖𝑓^(𝑖)𝜂, (2.9) which corresponds to (1.4).

It is also possible to state the result which no longer requires any endpoint conditions. If we consider the auxiliary function 𝜓_𝑘(𝑥)=𝜑_𝑘(𝑥)+(−1)^𝑘+1()!()𝑘+1𝑥−𝑎^𝑘+1⋅^𝑏_𝑎𝒦𝑓^(𝑛), (2.10) then the analogous way as in the proof of Theorem 1.5 yields the following result also given in [14] including Riedel-Sahoo’s Theorem 1.4 as a special case (𝑛=1).

Theorem 2.1. If 𝑓∶⟨𝑎,𝑏⟩→ℝ is 𝑛-times differentiable on ⟨𝑎,𝑏⟩, then there exists 𝜂∈(𝑎,𝑏) such that 𝑓(𝑎)=𝑇_𝑛𝑓,𝜂(𝑎)+𝑎−𝜂^𝑛+1()!⋅𝑛+1^𝑏𝑎𝒦𝑓^(𝑛). (2.11)

Note that the case 𝑛=1 is used to extend Flett’s mean value theorem for holomorphic functions; see [10]. An easy generalization of Pawlikowska’s theorem involving two functions is the following one.

Theorem 2.2. Let 𝑓, 𝑔 be 𝑛-times differentiable on ⟨𝑎,𝑏⟩ and 𝑔^(𝑛)(𝑎)≠𝑔^(𝑛)(𝑏). Then there exists 𝜂∈(𝑎,𝑏⟩ such that 𝑓(𝑎)−𝑇_𝑛𝑓,𝜂(𝑎)=^𝑏_𝑎𝒦𝑓^(𝑛),𝑔^(𝑛)⋅𝑔(𝑎)−𝑇_𝑛𝑔,𝜂(𝑎). (2.12)

This may be verified applying Pawlikowska’s theorem to the auxiliary function ℎ(𝑥)=𝑓(𝑥)−^𝑏_𝑎𝒦𝑓^(𝑛),𝑔^(𝑛)⋅𝑔(𝑥),𝑥∈⟨𝑎,𝑏⟩. (2.13) A different proof will be presented in the following section.

3. A Trahan-Type Condition

In [4] Trahan gave a sufficient condition for the existence of a point 𝜂∈(𝑎,𝑏⟩ satisfying (1.2) under the assumptions of differentiability of 𝑓 on ⟨𝑎,𝑏⟩ and inequality 𝑓^′(𝑏)−^𝑏_𝑎𝒦𝑓⋅𝑓^′(𝑎)−^𝑏_𝑎𝒦𝑓≥0. (3.1)

Modifying Trahan’s original proof using Pawlikowska’s auxiliary function 𝐺_𝑓, we are able to state the following condition for validity (1.4).

Theorem 3.1. Let 𝑓 be 𝑛-times differentiable on ⟨𝑎,𝑏⟩ and 𝑓^(𝑛)(𝑎)(𝑎−𝑏)^𝑛𝑛!+𝑀_𝑓𝑓^(𝑛)(𝑏)(𝑎−𝑏)^𝑛𝑛!+𝑀_𝑓≥0, (3.2) where 𝑀_𝑓=𝑇_𝑛−1(𝑓,𝑏)(𝑎)−𝑓(𝑎). Then there exists 𝜂∈(𝑎,𝑏⟩ satisfying (1.4).

Proof. Since 𝐺_𝑓 is continuous on ⟨𝑎,𝑏⟩ and differentiable on (𝑎,𝑏⟩ with 𝐺^′_𝑓(𝑥)=𝑔^(𝑛)(𝑥)=(−1)^𝑛𝑛!(𝑥−𝑎)^𝑛+1𝑓(𝑥)−𝑓(𝑎)+^𝑛_𝑖=1(−1)^𝑖()𝑖!𝑥−𝑎^𝑖𝑓^(𝑖)(𝑥), (3.3) for 𝑥∈(𝑎,𝑏⟩, [14; page 282], then 𝐺_𝑓(𝑏)−𝐺_𝑓(𝑎)𝐺^′_𝑓(𝑏)=𝑔^(𝑛−1)(𝑏)−1𝑛𝑓^(𝑛)(𝑎)𝑔^(𝑛)(𝑏)()!=−𝑛!𝑛−1(𝑏−𝑎)^2𝑛+1𝑓^(𝑛)(𝑎)(𝑎−𝑏)^𝑛𝑛!+𝑇_𝑛−1(𝑎)(𝑎)⋅𝑓𝑓,𝑏−𝑓_(𝑛)(𝑏)(𝑎−𝑏)^𝑛𝑛!+𝑇_𝑛−1𝑓,𝑏(𝑎)−𝑓(𝑎)≤0. (3.4) According to [4, Lemma  1], there exists 𝜂∈(𝑎,𝑏⟩ such that 𝐺^′_𝑓(𝜂)=0 which corresponds to (1.4).

Now we provide an alternative proof of Theorem 2.2 which does not use, original Pawlikowska’s theorem.

Proof of Theorem 2.2. For 𝑥∈(𝑎,𝑏⟩ put 𝜑(𝑥)=^𝑥_𝑎𝒦(𝑓) and 𝜓(𝑥)=^𝑥_𝑎𝒦(𝑔). Define the auxiliary function 𝐹 as follows: 𝐹(𝑥)=𝜑^(𝑛−1)(𝑥)−^𝑏_𝑎𝒦𝑓^(𝑛),𝑔^(𝑛)⋅𝜓^(𝑛−1)(𝑥)(⟩,1,𝑥∈𝑎,𝑏𝑛𝑓^(𝑛)(𝑎)−^𝑏_𝑎𝒦𝑓^(𝑛),𝑔^(𝑛)⋅𝑔^(𝑛)(𝑎),𝑥=𝑎. (3.5) Clearly, 𝐹 is continuous on ⟨𝑎,𝑏⟩, differentiable on (𝑎,𝑏⟩, and for 𝑥∈(𝑎,𝑏⟩ there holds 𝐹^′(𝑥)=𝜑^(𝑛)(𝑥)−^𝑏_𝑎𝒦𝑓^(𝑛),𝑔^(𝑛)⋅𝜓^(𝑛)(𝑥). (3.6) Then 𝐹^′(𝑏𝐹(𝑏)(𝑎𝑛)[−𝐹)]=−()(𝐹(𝑏)(𝑎𝑏−𝑎−𝐹))²≤0, (3.7) and by [[4]; Lemma  1] there exists 𝜂∈(𝑎,𝑏⟩ such that 𝐹^′(𝜂)=0; that is, 𝑓(𝑎)−𝑇_𝑛𝑓,𝜂(𝑎)=^𝑏_𝑎𝒦𝑓^(𝑛),𝑔^(𝑛)⋅𝑔(𝑎)−𝑇_𝑛𝑔,𝜂(𝑎), (3.8) which completes the proof.

Acknowledgment

The work was partially supported by the Internal Grant VVGS-PF-2012-36.