Functional Equations and the Cauchy Mean Value Theorem

The aim of this note is to characterize all pairs of sufficiently smooth functions for which the mean value in the Cauchy Mean Value Theorem is taken at a point which has a well-determined position in the interval. As an application of this result, a partial answer is given to a question posed by Sahoo and Riedel.


Introduction
Given two differentiable functions F, G : R → R, the Cauchy Mean Value Theorem (MVT) states that for any interval [a, b] ⊂ R, where a < b, there exists a point c in (a, b) such that where f = F and g = G . A particular situation is the Lagrange MVT when G(x) = x is the identity function, in which case (1) reads as The problem to be investigated in this note can be formulated as follows.
For the case of the Lagrange MVT with c = a+b 2 , this problem was considered first by Haruki [5] and independently by Aczél [1], proving that the quadratic functions are the only solutions to (2). This problem can serve as a starting point for various functional equations [9]. More general functional equations have been considered even in the abstract setting of groups by several authors including Kannappan [6], Ebanks [3], Fechner-Gselmann [4]. On the other hand, the result of Aczél and Haruki has been generalized for higher order Taylor expansion by Sablik [8].
For the more general case of the Cauchy MVT much less is known. We mention Aumann [2] illustrating the geometrical significance of this equation and the recent contribution of Páles [7] providing the solution of a related equation under additional assumptions. In this note we provide a different approach to the Cauchy MVT. As it will turn out, the most challenging situation corresponds to c = a+b 2 in which case our main result is the following: Theorem 2. Assume that F, G : R → R are three times differentiable functions with derivatives F = f , G = g such that for all a, b ∈ R. Then one of the following possibilities holds: (a) {1, F, G} are linearly dependent on R; (c) there exists a non-zero real number µ such that F, G ∈ span{1, e µx , e −µx }, x ∈ R; (d) there exists a non-zero real number µ such that F, G ∈ span{1, sin(µx), cos(µx)}, x ∈ R.
The paper is organized as follows. In Section 2 we consider the problem first for the known case of the Lagrange mean value theorem as an illustration of our method. In Section 3 we provide a preliminary result that will allow to pass local information to a global one about the pairs of differentiable functions (F, G) satisfying (3). In Sections 4, 5 we consider the asymmetric (α β) and symmetric (α = β = 1/2) cases, respectively. Section 6 is for final remarks. Here we also provide a partial result to an open problem by Sahoo and Riedel which corresponds to a more general version of (3).

The Lagrange MVT with fixed mean value
Note that every c ∈ (a, b) can be written uniquely as c = αa + βb for some α, β ∈ (0, 1) with α + β = 1. It is easy to check that (2) holds for all a, b ∈ R with fixed α 1/2 if F is a linear function, and with α = 1/2 if F is a quadratic function. We claim that the converse of this statement is also true. As mentioned earlier, there are various proofs of the latter in the literature, see for example [1], [5] , [9]. Nevertheless, we give here a short and selfcontained argument mainly to illustrate our approach to the more general case of the Cauchy MVT. Proposition 3. Let α ∈ (0, 1) be fixed and β = 1 − α. Assume that F : R → R is a continuously differentiable function with F = f such that Then the following statements hold: 1. if α 1/2 then F is a linear function; 2. if α = 1/2 then F is a quadratic function.
Proof. Let us denote by αb + βa = x and b − a = h. Then (5) reads as From this equation it is apparent that f = F is differentiable as a linear combination of two differentiable functions and thus F is twice differentiable. By induction, it follows that F is infinitely differentiable. Differentiating (6) with respect to h, we obtain the relation Again, we differentiate (7) with respect to h and find that Since f is continuous, letting h 0, we obtain If α 1/2, this implies that f = 0 identically. Therefore f is constant and thus F is a linear function, proving the first statement. If α = 1/2, then the equation (7) reads as and twice differentiation with respect to h leads to Now letting h 0, we get f (x) = 0 for all x ∈ R, so f is linear and F is a quadratic function, proving the second statement.

The Cauchy MVT with fixed mean value
Let us introduce the sets and also their complements Z f := R \ U g and Z g : Proposition 4. If U g ∅ but U f ∩ U g = ∅, then U f = ∅, i.e. f ≡ 0 on R and thus F is constant.
Proof. By assumption, there is a non-empty interval (p, q) ⊂ U g such that g(x) 0 on (p, q), but f (x) = 0 for all x ∈ [p, q]. Then with the change of variables h = b − a, x = αa + βb, (3) yields Denoting by y = x + αh for x ∈ [p, q] and h > 0, we get F(y) − F(y − h) = 0 if (h, y) lies within the semi-strip L := (h, y) : h > 0, p + αh < y < q + αh . Then, for y > p choosing h > 0 such that (h, y) ∈ L, we have ∂ ∂y for y > p. However, by (10), we have F(q+αh) = F(q−βh) and thus F(y) is the same constant for all y < q. Therefore, Proposition 4 shows that the condition U f ∩ U g = ∅ holds only if at least one of the sets U f and U g is empty. Then we have the simple cases described as in the beginning of the section.
and consider the representation (9). If {F, G, 1} are linearly dependent as functions on I σ for every σ ∈ Σ, then {F, G, 1} are linearly dependent on R.
Proof. For σ 1 , σ 2 ∈ Σ with σ 1 σ 2 , consider the intervals I σ 1 := (p 1 , q 1 ), I σ 2 := (p 2 , q 2 ) with and assume that {F, G, 1} are linearly dependent on I σ 1 and I σ 2 . Then it follows that there are constants A 1 , With the change of variables Inserting this value into (15), we obtain Put y = x + αh, then x − βh = y − h, and (17) means that (h, y) lies within the parallelogram Π := (h, y) : p 2 + αh < y < q 2 + αh, p 1 + h < y < q 1 + h . Since β ∈ (0, 1), (12) guarantees that Π ∅, and (16) implies Therefore, at any point of Π, we have But y − h ∈ I σ 1 by (17), so g(y − h) 0 and thus So far our analysis says nothing about B 1 , B 2 in (13), (14) but since σ 1 , σ 2 ∈ Σ were arbitrary, (18) together with (13), (14) imply On the other hand, by changing the roles of F and G in the above analysis, we come to the conclusion that By (11) there is a point x 0 ∈ U g ∩ U f so AK = 1 and these coefficients are not zero. But then (19) implies U g ⊂ U f and (20) implies U f ⊂ U g ; therefore, U g = U f and Z g = Z f . The latter means that by trivial reasons (all these values are zeros) so with (19) and (20) these identities are valid on the entire R = U f ∪ Z f = U g ∪ Z g . In particular, it follows that {F, G, 1} are linearly dependent on R.
The following proposition describes all pairs (F, G) of two times continuously differentiable functions satisfying (3) under the assumption (21) on α, β in the intervals where g = G does not vanish. Proposition 6. Let (F, G) be a solution of the Problem 1 with α, β satisfying (21) and I = (p, q), −∞ ≤ p < q ≤ +∞, be an interval where the derivative g(x) does not vanish. If F, G are two times continuously differentiable on I, then {F, G, 1} are linearly dependent on I.

Proof. With the change of variables
By differentiating both sides of (22) with respect to h twice, we obtain the following relation in T All the functions are continiuous so (24) holds on the closure T as well, in particular, on the interval {h = 0, p < x < q}. Therefore, with β 2 − α 2 = 1 − 2α  (21), we get f (x)g(x) = g (x) f (x) for all x ∈ I = (p, q). We can divide both sides by g 2 (x) and conclude that ( f /g) = 0 on I. This implies that f /g = A for some constant A ∈ R, and F ( The following theorem is the main result of this section.
Proof. Consider the following cases: In this case G is a constant on R and (3) holds for any differentiable function F. Hence (25) holds, for example, with A = 0, B = 1, C = −G and thus {F, G, 1} are linearly dependent on R.
In this case Proposition 4 yields that F is a constant on R and (3) holds for any differentiable function G. Hence (25) holds, for example, with A = 1, B = 0, C = −F and thus {F, G, 1} are again linearly dependent on R.
In this case Propositions 6 and 5 immediately imply that {F, G, 1} are linearly dependent on R.

The Cauchy MVT with symmetric mean value
In this section we consider the problem of describing all pairs (F, G) of smooth functions for which the mean value in (3) is taken at the midpoint of the interval. Our first result gives a necessary (and also sufficient in case {1, F, G} are not linearly dependent) condition on such pairs in the intervals where g = G does not vanish. Proposition 8. Assume that F, G : R → R are three times differentiable functions with derivatives F = f , G = g. Let I ⊂ R be such an interval that g 0 for all x ∈ I and (4) holds for all a, b ∈ I. Then there exist constants A, K ∈ R and x 0 ∈ I such that Moreover, if (26) holds with K 0, then (4) holds if and only if x+h for all x, h ∈ R such that x, x + h, x − h ∈ I.
Proof. With the change of variables x = a+b 2 , h = b−a 2 , we can rewrite (4) as for all x, h ∈ R with the property that x, x + h, x − h ∈ I. By differentiating this equality three times with respect to h, we get Setting h = 0, we obtain for all x ∈ I, and thus f (x)g(x) − f (x)g (x) = K for some constant K. Then f g (x) = K g 2 (x) , x ∈ I, and integration over (x 0 , x) with any x 0 ∈ I yields (26). Now assume (26) holds with a nonzero constant K. Then we have and .
The following example illustrates that there are non-trivial functions satisfying (27) (and hence (4)) on R. A direct computation gives f (x) = sinh(x) = e x −e −x 2 , and consequently, We invite the interested reader to verify directly that the pair (F, G) in (31) satisfies the relation (4), giving a non-trivial example of such pairs. Now we assume that K 0 and analyze the property (27) for all x, h ∈ R such that x, x + h, x − h ∈ I. Differentiating it with respect to h, we obtain .
Differentiation two more times with respect to h gives for all x ∈ I and h ∈ R such that x, x + h, x − h ∈ I. Setting h = x − x 0 in these two equations, we obtain and for all x ∈ I with 2x − x 0 ∈ I. Since 2x − x 0 ∈ I and g has no zeros in I, both sides of (32) do not vanish. By comparing (33) and (32), we get for all x ∈ I such that 2x−x 0 ∈ I. Denoting by y(x) := g(2x−x 0 ) and λ := 4g (x 0 ) g(x 0 ) , (34) yields the second order differential equation y − λy = 0, whose general real-valued solution (depending on the sign of λ), has the following form where P, Q are real constants. Hence G has one of the following forms where A, B, C are real constants.
Remark 10. Altogether, we come to the following conclusion: on every interval I ⊂ R on which G 0, either {F, G, 1} are linearly dependent, or G and thus also F, cf. (26), has one of the forms described in (35)-(37).
In the sequel, we call a function G (resp. the pair (F, G)) to be of quadratic, exponential or trigonometric type on I if G has (resp. both of F and G have) the form (35), (36) or (37), respectively.
Consider the set U g and its representation, cf. (8), (9). The following lemma plays a crucial role in the analysis of the equation (4).
Proof. g(p) = 0 by Definition (9) so by Remark 10, it is sufficient to consider the following cases.
Case 1: G is of quadratic type on (p, q).
Case 2: G is of either exponential or trigonometric type on (p, q).
First suppose that G is of exponential type on (p, q). Then so is F and since the set of functions satisfying (4) is invariant with respect to the addition of constant functions, we can assume, without loss of generality, that F, G ∈ span e µ(x−p) , e −µ(x−p) for some µ 0. Hence there are real constants u, v such that F(x) = ue µ(x−p) + ve −µ(x−p) , x ∈ (p, q). Since F (p) = f (p) = 0, we get u = v and thus F(x) = 2u cosh(µ(x − p)). The same argument for G explains that G(x) = 2w cosh(µ(x − p)) for some real w, and consequently F and G are multiples of the same function cosh (µ(x − p)).
If G is of trigonometric type, then by the same way as above, we can conclude that F and G are multiples of the same function cos(µ(x − p)), implying that {F, G, 1} are linearly dependent on [p, q).
Proof of Theorem 2. Consider the set U g defined in (8). If U g = ∅, then g ≡ 0 on R, and thus G is identically constant on R. In this case F can be an arbitrary differentiable function on R and thus {1, F, G} are linearly dependent on R. If U g = R, then it follows (cf. Remark 10) that either {1, F, G} are linearly dependent or G has one of the forms (35)-(37) on the whole of R. Moreover, we get the same conclusion if U g ∩ U f = ∅ (cf. Proposition 4).
Next, let us assume that U g ∩ U f ∅ and U g is a proper subset of R. Consider the representation (9). It is clear (cf. Remark 10) that the index set Σ can be split into disjoint subsets as Σ = Σ lr ∪ Σ q ∪ Σ t ∪ Σ e , where Σ lr := σ ∈ Σ : {F, G, 1} are in linear relationship on I σ , Σ q := σ ∈ Σ : (F, G) are of quadratic type on I σ , Σ t := σ ∈ Σ : (F, G) are of trigonometric type on I σ , Σ e := σ ∈ Σ : (F, G) are of exponential type on I σ .
Claim 1. If Σ lr ∅, then Σ lr = Σ. Proof. Assume Σ lr is a proper subest of Σ. Then there exists σ 2 ∈ Σ such that σ 2 Σ lr . Since Σ lr ∅, there is σ 1 ∈ Σ lr and A 1 ∈ R such that f (x) = A 1 g(x) on x ∈ I σ 1 . Consider all x, h ∈ R such that x + h ∈ I σ 2 and x ∈ I σ 1 . Using (4) for a = x − h and b = x + h, and recalling that g 0 on I σ 1 , we get Therefore, From this it follows that F and G are in linear relationship on I σ 2 , that is, σ 2 ∈ Σ lr , which leads to a contradiction. Claim 2. If Σ lr = ∅, then only one of the index sets Σ q , Σ t , Σ e is non-empty. Proof. Let σ ∈ Σ and I σ = (p, q). Since U g is proper subset of R, one of p, q is finite. We can assume p > −∞. Then g(p) = 0, and Lemma 11 yields f (p) 0. Hence using (4) for a = p − h and b = p + h we get so the graph of G is symmetric with respect to the vertical line y = p.
If σ ∈ Σ q or σ ∈ Σ e , then q = +∞ since the functions of quadratic type has exactly one and the functions of exponential type has at most one critical point. Therefore, if σ ∈ Σ q , then G ∈ span{1, (x − p) 2 }, x ∈ R and Σ = Σ q . Similarly, it follows from (39) that if σ ∈ Σ e , then Σ = Σ e .
Next, assume Σ lr = Σ q = Σ e = ∅. Then Σ = Σ t and let σ ∈ Σ t . Since G is of trigonometric type on I σ = (p, q), we must have q < +∞. So g(p) = g(q) = 0 and it follows as in the proof of Lemma 11 that there are real constants u, v such that Using (39) we obtain that (40) holds on the whole of R.
Since U g ∅, at least one of Σ lr , Σ q , Σ t , Σ e is non-empty. If Σ lr ∅, then Claim 1 and Proposition 5 imply that {F, G, 1} are linearly dependent on R. If Σ lr = ∅, then Claim 2 yields that one of the possibilities (b) -(d) holds.

Final Remarks
As a consequence of our main result we can give a partial answer to following still open question of Sahoo and Riedel (cf. [9, Section 2.7] for an equivalent formulation). Problem. Find all functions F, G, φ, ψ : R → R satisfying for all x, y ∈ R.
We provide a partial result to this problem under certain assumptions on the unknown functions. First let us make the change of variables s = x+y 2 , t = x−y 2 and write (41) equivalently as Theorem 12. Let F, G : R → R be three times differentiable and φ, ψ : R → R be an arbitrary functions satisfying (42) on R. If either φ 0 or ψ 0 on R, then one of the following possibilities holds: (a) there exist constants A 0 , A 1 , A 2 ∈ R such that for all s ∈ R, we have Proof. Let f, g be the derivatives of F, G, respectively and the sets U g , U f (resp. Z g , Z f ) be defined as in Section 3. Without loss of generality, assume that φ does not vanish on R. By differentiating (42) with respect to t and setting t = 0 in the resulting equation, we get f (s)φ(s) = g(s)ψ(s), s ∈ R.
For any s ∈ U g and t ∈ R, by (42) and (43), we have On the other hand, observe that we have Z g ⊂ Z f by (43) since φ 0 on R. So (44) holds for all s ∈ U g ∪ Z g = R. Therefore, Theorem (2) can be applied to (44) and the four characterizations follows immediately.
It is likely that the methods of this paper work for related equations when we replace the linear mean αa + (1 − α)b by the p-mean M The essence of our approach is to reduce a functional equation to an ODE. For this strategy we need certain smoothness assumptions. It would be interesting to provide an alternative way that will not require this additional smoothness assumptions.