# On Generalized Flett's Mean Value Theorem

Authors: Jana Molnárová
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## 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.


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## Body

## 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).
Iff:〈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 Theorem1.1 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 conditionf(a)=f(b)? The answer is well known as the Lagrange mean value theorem. For the sake of brevity put
(1.1)ab𝒦(f(n),g(n))=f(n)(b)-f(n)(a)g(n)(b)-g(n)(a),n∈ℕ∪{0},
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 ab𝒦(f(n)).Theorem 1.2 (Lagrange’s mean value theorem).
Iff:〈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)=ab𝒦(f).Clearly, Theorem1.2 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 Theorem1.1, the following question may arise: Are there changes if in Theorem1.1 the hypothesisf(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).
Iff:〈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
(1.2)f′(η)=𝒦(f)aη.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 assumptionf'(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).
Iff:〈a,b〉→ℝ is a differentiable function on 〈a,b〉, then there exist a number η∈(a,b) such that
(1.3)f′(η)=𝒦(f)aη+𝒦(f′)ab·η-a2.We point out that there are also other sufficient conditions guaranteeing the existence of a pointη∈(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).
Letf be n-times differentiable on 〈a,b〉 and f(n)(a)=f(n)(b). Then there exists η∈(a,b) such that
(1.4)f(η)-f(a)=∑i=1n‍(-1)i+1i!(η-a)if(i)(η).Observe that Pawlikowska’s theorem has a close relationship with thenth Taylor polynomial of f. Indeed, for
(1.5)Tn(f,x0)(x)=f(x0)+f′(x0)1!(x-x0)+⋯+f(n)(x0)n!(x-x0)n,
Pawlikowska’s theorem has the following very easy form f(a)=Tn(f,η)(a).Pawlikowska’s proof follows up the original idea of Flett; see [1], considering the auxiliary function
(1.6)Gf(x)={g(n-1)(x),x∈(a,b〉,1nf(n)(a),x=a,
where g(x)=ax𝒦(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(2.1)φk(x)=xf(n-k+1)(a)+∑i=0k‍(-1)i+1i!(k-i)(x-a)if(n-k+i)(x),k=1,2,…,n.
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, fork=1 we have φ1(x)=-f(n-1)(x)+xf(n)(a) and φ1′(x)=-f(n)(x)+f(n)(a). Clearly, φ1′(a)=0=φ1′(b), so applying Flett’s mean value theorem for φ1 on 〈a,b〉 there exists u1∈(a,b), such that φ1′(u1)(u1-a)=φ1(u1)-φ1(a); that is,
(2.2)f(n-1)(u1)-f(n-1)(a)=(u1-a)f(n)(u1).
Then for φ2(x)=-2f(n-2)(x)+(x-a)f(n-1)(x)+xf(n-1)(a), we get
(2.3)φ2′(x)=-f(n-1)(x)+(x-a)f(n)(x)+f(n-1)(a)
and φ2′(a)=0=φ2′(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 φ2′(u2)(u2-a)=φ2(u2)-φ2(a), which is equivalent to
(2.4)f(n-2)(u2)-f(n-2)(a)=(u2-a)f(n-1)(u2)-12(u2-a)2f(n)(u2).
Continuing this way after n-1 steps, n≥2, there exists un-1∈(a,b) such that
(2.5)f′(un-1)-f′(a)=∑i=1n-1‍(-1)i+1i!(un-1-a)if(i+1)(un-1).
Considering the function φn, we get
(2.6)φn′(x)=-f′(x)+f′(a)+∑i=1n-1‍(-1)i+1i!(x-a)if(i)(x)=f′(a)+∑i=0n-1‍(-1)i+1i!(x-a)if(i+1)(x).
Clearly, φn′(a)=0=φn′(un-1) by (2.5). Then by Flett’s mean value theorem for φn on 〈a,un-1〉, there exists η∈(a,un-1)⊂(a,b) such that
(2.7)φn′(η)(η-a)=φn(η)-φn(a).
Since
(2.8)φn′(η)(η-a)=(η-a)f′(a)+∑i=1n‍(-1)i(i-1)!(η-a)if(i)(η),φn(η)-φn(a)=(η-a)f′(a)-n(f(η)-f(a))+∑i=1n‍(-1)i+1i!(n-i)(η-a)if(i)(η),
the equality (2.7) yields
(2.9)-n(f(η)-f(a))=∑i=1n‍(-1)i(i-1)!(η-a)if(i)(η)(1+n-ii)=n∑i=1n‍(-1)ii!(η-a)if(i)(η),
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(2.10)ψk(x)=φk(x)+(-1)k+1(k+1)!(x-a)k+1·ab𝒦(f(n)),
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.
Iff:〈a,b〉→ℝ is n-times differentiable on 〈a,b〉, then there exists η∈(a,b) such that
(2.11)f(a)=Tn(f,η)(a)+(a-η)n+1(n+1)!·𝒦(f(n))ab.Note that the casen=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.
Letf, g be n-times differentiable on 〈a,b〉 and g(n)(a)≠g(n)(b). Then there exists η∈(a,b〉 such that
(2.12)f(a)-Tn(f,η)(a)=𝒦(f(n),g(n))ab·[g(a)-Tn(g,η)(a)].This may be verified applying Pawlikowska’s theorem to the auxiliary function(2.13)h(x)=f(x)-𝒦(f(n),g(n))ab·g(x),x∈〈a,b〉.
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
(3.1)(f′(b)-𝒦(f)ab)·(f′(a)-𝒦(f)ab)≥0.Modifying Trahan’s original proof using Pawlikowska’s auxiliary functionGf, we are able to state the following condition for validity (1.4).Theorem 3.1.
Letf be n-times differentiable on 〈a,b〉 and
(3.2)(f(n)(a)(a-b)nn!+Mf)(f(n)(b)(a-b)nn!+Mf)≥0,
where Mf=Tn-1(f,b)(a)-f(a). Then there exists η∈(a,b〉 satisfying (1.4).Proof.
SinceGf is continuous on 〈a,b〉 and differentiable on (a,b〉 with
(3.3)Gf′(x)=g(n)(x)=(-1)nn!(x-a)n+1(f(x)-f(a)+∑i=1n‍(-1)ii!(x-a)if(i)(x)),
for x∈(a,b〉, [14; page 282], then
(3.4)(Gf(b)-Gf(a))Gf′(b)=(g(n-1)(b)-1nf(n)(a))g(n)(b)=-n!(n-1)!(b-a)2n+1(f(n)(a)(a-b)nn!+Tn-1(f,b)(a)-f(a))·(f(n)(b)(a-b)nn!+Tn-1(f,b)(a)-f(a))≤0.
According to [4, Lemma  1], there exists η∈(a,b〉 such that Gf′(η)=0 which corresponds to (1.4).Now we provide an alternative proof of Theorem2.2 which does not use, original Pawlikowska’s theorem.Proof of Theorem2.2.
Forx∈(a,b〉 put φ(x)=ax𝒦(f) and ψ(x)=ax𝒦(g). Define the auxiliary function F as follows:
(3.5)F(x)={φ(n-1)(x)-ab𝒦(f(n),g(n))·ψ(n-1)(x),x∈(a,b〉,1n[f(n)(a)-ab𝒦(f(n),g(n))·g(n)(a)],x=a.
Clearly, F is continuous on 〈a,b〉, differentiable on (a,b〉, and for x∈(a,b〉 there holds
(3.6)F′(x)=φ(n)(x)-𝒦(f(n),g(n))ab·ψ(n)(x).
Then
(3.7)F′(b)[F(b)-F(a)]=-n(b-a)(F(b)-F(a))2≤0,
and by [[4]; Lemma  1] there exists η∈(a,b〉 such that F′(η)=0; that is,
(3.8)f(a)-Tn(f,η)(a)=𝒦(f(n),g(n))ab·(g(a)-Tn(g,η)(a)),
which completes the proof.



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