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Ukrainian Mathematical Journal, Vol. 65, No. 9, February, 2014 (Ukrainian Original Vol. 65, No. 9, September, 2013)

MEAN-VALUE THEOREM Yu. Yu. Trokhimchuk

UDC 517.5

We propose a new approach to the classical mean-value theorem in which two mean values are used instead of one. This approach is of especial importance for complex functions because there are no available theorems of this kind for these functions.

1. Real Functions The classical mean-value theorem for a real function f (x), x ∈ [a, b], states that if the function f is differentiable on the interval (a, b), then there exists a point c of this interval, a < c < b, at which the following equality is true: f (b) − f (a) = f 0 (c). b−a The following statement is proved in [1] for an arbitrary continuous function f : Mean-Value Theorem. Let f (x), x ∈ [a, b], be a continuous function. Then the following three (generally speaking, consistent) versions are true: (i) there exists a point c, a < c < b, for which the following equality is true: f (b) − f (a) = δ(c), b−a where δ(c) is a certain right derived number of the function f; (ii) the set E(x : f 0 (x) = ∞) is nonempty and its image f (E) has a positive measure: mes f (E) > 0; (iii) f is an ACG-function on [a, b] and 1 f (b) − f (a) = b−a b−a

Zb

0 fas (t)dt,

a

where, in the general case, the integral on the right-hand side is the Denjoy integral in a wide sense. Institute of Mathematics, Ukrainian National Academy of Sciences, Kyiv, Ukraine. Translated from Ukrains’kyi Matematychnyi Zhurnal, Vol. 65, No. 9, pp. 1276–1282, September, 2013. Original article submitted October 10, 2012. 1418

0041-5995/14/6509–1418

c 2014

Springer Science+Business Media New York

M EAN -VALUE T HEOREM

1419

Recall that a number A is called a derived number of the function f at the point x0 ∈ [a, b] if there exists a sequence hh → 0 such that f (x0 + hn ) − f (x0 ) = A; n→∞ hn lim

moreover, A is called a right derived number if all hn > 0 and a left derived number if hn < 0. We now present another formulation of the mean-value theorem that does not contain infinite derivatives but uses a couple of mean values. We prove the following theorem: Theorem 1. Let f (x), x ∈ [a, b], be an arbitrary finite function. If, at every point x, except at most a countable set of points, the right derived numbers can be equal only to infinity or to zero, then, in [a, b], one can find a dense set of the intervals of constancy of the function f. Proof. We introduce the following sets: A+ n

A− n

=

=

f (x + h) − f (x) ≥1 x: h

,

f (x + h) − f (x) 1 x: ≤ −1 for 0 < h ≤ h n

Bn =

1 for 0 < h ≤ n

f (x + h) − f (x) 1 x: ≤ 1 for 0 < h ≤ h n

,

and take an arbitrary segment [α, β] ⊂ [a, b]. By the condition of the theorem, we have [α, β] = A ∪ B, where A = E{x : f 0 (x) = ∞},

B = {x : f 0 (x) = 0}

and

− A = ∪ A+ ∪ ∪ A n n , n

n

B = ∪ Bn . n

0 0 If the set A is not of the first category, then one of the sets A± n is dense on the segment [α , β ] ⊂ [α, β] whose length can be smaller than 1/h. The definition of A± n immediately implies that f is also strictly monotone everywhere in this segment with derived numbers whose absolute values are greater than +1. However, this function is differentiable almost everywhere on [α0 , β 0 ] with derivatives different from ∞ and 0. Hence, everywhere on [α, β], the set B must be of the second category and, on the partial segment [α0 , β 0 ], f is a Lipschitz function without infinite derivatives. Thus, by the condition, there exist only zero derived numbers, which means that the function f is constant on [α0 , β 0 ]. Since the segment [α, β] ⊂ [a, b] is arbitrary, we arrive at the statement of the theorem. It can be proved that, under the conditions of Theorem 1, the complement to the obtained system of intervals of constancy is a countable nowhere dense set on [a, b]. However, this result is not used in what follows. For a function f strictly monotone on [a, b], this statement means that there exists a set of points dense on [a, b] at which finite right derived numbers exist and are different from zero.1 1

This statement is nontrivial because a strictly monotone function may have zero derived numbers both almost everywhere and on a set of the second category, infinite derivatives also on a set of the second category, etc.

Y U. Y U. T ROKHIMCHUK

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By m(x; f ) = m(x) we denote the set of all right derived numbers of the function f at the point x ∈ [a, b]. It is easy to see that all numerical values m(x0 ) appear as follows: the right neighborhood of the graph of B(f ) is projected from the point (x0 , f (x0 )) onto the vertical number axis R : x = x0 + 1 taking the point y = f (x0 ) on it as the zero point; for a continuous function f (x), the projection is connected. Shrinking this neighborhood into the point x0 , in the limit, we obtain a (closed) set m(x) on R = [−∞, +∞]. We prove the following theorem: Theorem 2. Let f (x), x ≥ 0, be strictly monotone and continuous at the point x = 0 and, in addition, f (x) f (0) = 0. Then limx→0 takes all intermediate values, i.e., the set m(0) is connected. x Note that this assertion does not follow solely from the continuity:

h(x) =

   0        x

for x ∈

∞ S

1 1 , , 2n 2n − 1

1 1 , , 2n − 1 2n

n=1

h(0) = 0. for x ∈

∞ S n=2

(1)

Proof. If f (x) has points of discontinuity (the theorem is nontrivial only in this case), then we close the graph of B(f ) by vertical segments at points of jumps to a continuous arc, which is an arc that is one-to-one projected onto the Oy -axis. Considering it as the graph of the inverse (and continuous) function ϕ(y), we obtain the connectedness of m(0; ϕ) and the equality m(0; f ) =

1 . m(0; ϕ)

Otherwise, it is an “empty” angle with vertex x = 0 that contains not points of B(f ) but only vertical segments complementing it [e.g., as in example (1)], which is not consistent with monotonicity of f (x). The theorem is proved. By this theorem, m(x, f ) is connected (except a countable set of discontinuity points) and if, e.g., f increases, then m(x, f ) ≥ 0. First, we recall the notion of sets of the first and second categories. A set A is called a set of the first category in R ≡ (−∞, +∞) if it is a union of a countable collection of sets nowhere dense in R: [ A= An . n

Thus, the set of rational points R is a set of the first category. Further, a set B ⊂ R is called residual (or a set of everywhere second category) if its complement is a set of the first category. According to the main theorem presented here, R (as any complete metric space) is a set of everywhere second category and cannot be represented in the form of a countable union of nowhere dense sets. We call a function f (x) nowhere constant if it does not have intervals of constancy. Theorem 3. If f (x), x ∈ [a, b], is a finite and nowhere constant function, then, on [a, b], there exists a dense set at every point of which this function has right finite derived numbers different from zero. Proof. First, we prove the theorem for a monotone (and, hence, strictly monotone) function f. Note that the graph of an arbitrary continuous monotone function f (x), x ∈ [a, b], is simultaneously the graph of the Lipschitz function k(ξ) with Lipschitz constant equal to 1 if the axes of the original Cartesian coordinate system xOy are


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π π for nondecreasing f or by − for nonincreasing f. As above, we complement the 4 4 graph of B(f ) by vertical segments to the continuous arc L. If the segment [α, β] ⊂ [a, b] does not contain points on L indicated in the lemma, then the derivative y of the changed function f (x) √ takes only values the ±∞ and 0 and the corresponding Lipschitz function k(ξ) on a segment of length (β − α) 2 has only derivatives equal to ±1 without intermediate values, which is impossible in view of the connectedness of m. In the general case, we consider the sets A± n and Bn . Again, if any segment [α, β] does not contain points mentioned in the theorem, then

rotated by an angle of +

[α, β] = A ∪ B, where A = E(x : f 0 (x) = ∞),

B = E(x : f 0 (x) = 0)

and ! A=

[

! [ [

A+ n

n

A− n

,

n

B=

[

Bn .

n

0 0 If A is a set not of the first category, then one of A± n is dense on the segment [α , β ] ⊂ [α, β] whose length can ± be smaller than 1/n. The definition of An readily implies that f is a strongly monotone function everywhere on this segment and, according the proved statement, this segment contains a set of points mentioned in the theorem, which contradicts our assumption. If B is a set not of the first category, then, on the segment [α0 , β 0 ], f is a Lipschitz function that does not have infinite derivatives. Thus, it has only zero derived numbers, which implies that the function f is constant on [α0 , β 0 ], which is impossible. The obtained contradictions prove Theorem 3. Earlier, we have introduced the sets m(x; f ) = m(x) of right derived numbers of the function f for x ∈ [a, b]). The following property of “continuity” of the family {m(x), x ∈ [a, b)} of these sets is proved in [1]: On [a, b], there exist a subset ε of everywhere second category for each point of which x0 and an arbitrary sequence xn → x0 , xn ∈ [a, b), the inclusion

lim sup m(xn ) ⊂ m(x) n→∞

is true; roughly speaking, m(x) “adsorbs” all close m(x). This statement can be regarded as an analog of the known Baire theorem on functions of the first class. In what follows, we need a series of propositions concerning the differential properties of strictly monotone functions. Let f (x), x ∈ [a, b] be a strictly increasing function. We introduce the sets A = {x : f 0 (x) = +∞},

B = {x : f 0 (x) = 0} and

C = {x : there exists δ(x; f ) : 0 < δ(x) < +∞}.

In the case where A or B are dense on the segment, they are sets of everywhere second category on this segment. Indeed, (for B ) we get 0

B = {f = 0} =

∞ [ \ n=1 x

1 1 f (x + h) − f (x) < for 0 < h < x: h n n

,


1422

where each term (x) contains an interval (x, x + ε(x)), ε(x) < 1/n, and their union is an open set Gn dense on [a, b]; hence, B contains a dense set Gδ . Thus, this is a set of the second category on this segment. For the set A, the reasoning is similar. Consider different versions for the function f. 1. The set A is dense on [α, β] ⊂ [a, b]. Then this is a set of the second category and we can assume that all m(x), x ∈ [α, β], are contained in narrow angles with vertical side: points of the set B are absent and the other points belong to C, which is a set of full measure (but of the first category). 2. The set B is dense on [α, β]. Then it is also a set of the second category. The proof implies that there exists a dense set on [α, β] and, in each interval, f is a Lipschitz function. Since f strictly increases, in each interval on [α, β], the set C is a set of positive measure of the first category. 3. The sets A and B are nowhere dense on [α, β]. Then C contains its dense open subset. We need the following theorem: Theorem 4. Let f and g be a strictly increasing functions on [a, b]. There exists a subset dense on [a, b] at every point of which both functions have right finite derived numbers different from zero. Proof. Parallel with f, we consider the sets for g : A1 = {x : g 0 (x) = ∞},

B1 = {x : g 0 (x) = 0},

and

C1 = {x : there exists δ(x; g) : 0 < δ(x) < ∞}

associated with sets introduced above. It is necessary to prove that the intersection C ∩ C1 is dense on [a, b]. Assume that [α, β] ⊂ [a, b] does not contain points C ∩ C1 , i.e., C1 ⊂ A

[

B

and C1 ⊂ A1

[

B1 .

In other words, C1 ⊂ B and C ⊂ B1 . The fact that C1 is dense implies that B is also dense and f ∈ Lip on [α0 , β 0 ] ⊂ [α, β], i.e., C1 is a set of everywhere positive measure (and of the first category). Similarly, on [α0 , β 0 ], we obtain a segment [α00 , β 00 ] on which C is a set of everywhere positive measure (and of the first category). In the plane xOy, we consider a set of points (“curve”) presented in the parametric form ξ = f (t),

η = g(t),

t ∈ [a, b].

Under our conditions, this curve is the graph of the strictly increasing function η = (ξ), by definition, generally speaking, on a disconnected set, namely, on the set f ([a, b]) of, e.g., zero dimension. According to our construction, the segment [α00 , β 00 ] is associated with a one-dimensional arc of the strictly increasing function η = η(ξ) generated by the Lipschitz functions f and g on this segment. Thus, on this segment, there exists an everywhere dense set of points with positive derivative, which contradicts the assumption of the absence of these points. The theorem is proved. By analogy, we prove all versions of the following statement: If f and g are arbitrary strictly monotone functions on the segment [a, b], then this segment contains a dense set at every point of which there exist finite and nonzero derived numbers and, furthermore, these numbers are right for one function and are left for the other. We now prove the main theorem of this section.


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Theorem 5. For any function f continuous on [a, b], there exist c1 , c2 ∈ (a, b) and a number λ, 0 < λ < 1, such that the following equality is true: f (b) − f (a) f (c2 ) − f (c1 ) , = λδ(c1 ) + (1 − λ)δ(c2 ) = b−a c2 − c1 where δ(c2 ) and δ(c2 ) are certain right finite derived numbers of the function f. Proof . By the well-known substitution f1 (x) = f (x) − f (a) −

f (b) − f (a) (x − a), b−a

we reduce the proof of the theorem to the case where the values of the function at the endpoints of the segment coincide. Let f (b) = f (a). It is necessary to find the corresponding triple (c1 , c2 , λ) with condition λδ(c1 ) + (1 − λ)δ(c2 ) = 0, i.e., δ(c2 ) 1 −1=− . λ δ(c1 )

(2)

Since the left-hand side for 0 < λ < 1 takes all positive values (0, +∞) and only these values, it suffices to find any two points c1 and c2 with finite (nonzero) right derived numbers of the function f = u of different signs with the required value f (c2 ) − f (c1 ) . c2 − c1 Let x = c be an extreme point of the function on [a, b]. If f (c) = f (a), then f = const on [a, b] and the theorem is trivial. Let f (c) > f (a). Consider levels of the function f for values y ∈ [f (a), f (c)]. By ϕ(y), ψ(y) : ϕ(y) < ψ(y) we denote the extreme values of the abscissas x of the level f (x) = y. The continuity of f implies that both these functions are semicontinuous and strictly monotone: ϕ increases, ψ decreases, the derived numbers of ϕ are nonnegative, and the derived numbers of ψ are nonpositive. It follows from Theorem 2 that, for dense subsets of the segment [f (a), f (c)], there exist nonzero values of these numbers. Any values of this kind give solutions of Eq. (2). The theorem is proved. We only note that λ in this theorem can be equal both to 0 and to 1, i.e., this theorem may have a unique mean point. Moreover, one can always find an infinite set of couples of mean points. We can “work” with these couples of mean points in the same way as with a single mean point in the classical case. 2. Complex Functions As for the mean-value theorem, the transition from real to complex (and analytic!) functions essentially changes the picture: In the general case, the theorem is not true. . . An attempt to get complete analogy with the classical case leads to the necessity of introducing additional geometric and analytic restrictions [2–4].


1424

Nevertheless, the mean-value theorem for an analytic function f (z), z ∈ D ⊂ C is locally true, i.e., for every point z0 ∈ D and any ε > 0, there exists δ ≤ ε such that, for any couple of points a, b from the δ -neighborhood Uδ (z0 ), the following equality is true: f (b) − f (a) = f 0 (c), b−a where the point c belongs to an ε-neighborhood of the point z0 . The result obtained above for real functions shows that a similar result may be true for complex functions. We now show that this result remains completely true in the complex case. Assume that an analytic function f (z) and a rectilinear segment l with endpoints a and b are given in the domain D ⊂ C. From the viewpoint of mechanics, the relation f (b) − f (a) τ= = b−a

Z1

f 0 (a + t(b − a))dt

0

gives the center of gravity of the f 0 -image L = f 0 (l) regarded as a uniform smooth curve, which, as known, belongs to a convex shell of this curve. According to this relation, if, e.g., Im f 0 (z) 6= 0 at some points of l, then this center of gravity does not belong to the real axis. If Im f 0 = 0 everywhere on l, then this center belongs to the rectilinear segment L. This directly implies that, in the general case where Im f 0 6≡ 0 on l, τ is an interior point of the convex shell O(L) of the curve L. We prove that there exists a straight line S passing through the point τ such that both rays of the straight line cross the curve L at the points ζ1 and ζ2 , so that the segments τ ζ1 and τ ζ2 do not contain other points of intersection with L. We draw all rectilinear rays of τ ζ up to the first points of their intersection with the curve L. If the curve L itself passes through the point τ, then we fix a value on the segment l for which f 0 (ζ) = τ. If the extreme rays of τ ζ in our construction form an angle smaller that π, then τ is not even a boundary point of the shell O(L). This yields our statement on the existence of the required straight line. Denoting the points of the segment l corresponding to the points ζ1 and ζ2 by c1 and c2 , we can formulate the established result in the form of the following theorem: Theorem 6. Let an analytic function f (z) and a rectilinear segment l with endpoints a and b be given in the domain D ⊂ C. Then there exist a number λ ∈ [0, 1] and points c1 and c2 of this segment such that the equality f (c2 ) − f (c1 ) f (b) − f (a) = λf 0 (c1 ) + (1 − λ)f 0 (c2 ) = b−a c2 − c1 is true. Note that λ can be equal to 0 (or 1, which is the same) and, instead of two mean points of the segment l, we can get a single point. Nevertheless, recalling our proof and taking into account the openness of the mapping w = f 0 (z), we can state that, on l, one can find two intervals all points of which play the roles of c1 and c2 . This means that choosing c1 in one of the intervals, we can find c2 in the second interval such that, for the proper λ, the previous equality remains true (moreover, with the same left-hand side!)


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As in the case of real functions, by using this theorem, we obtain the mean-value theorem for any holomorphic function in the domain D ⊂ Cn : f (b) − f (a) = ∆f = df +

d2 f dn−1 f 1 n + ... + + λd f c1 +(1 − λ)dn f c2 , 2! (n − 1)! n!

where c1 and c2 are points of the segment ab ⊂ D. These points fill certain intervals of the indicated segment. REFERENCES 1. Yu. Yu. Trokhimchuk, Differentiation, Inner Mappings, and Criteria of Analyticity [in Russian], Institute of Mathematics, Ukrainian National Academy of Sciences, Kiev (2008). 2. J. Ievirtz, “On the mean-value theorem for analytic functions,” Mich. Math. J., 33, 365–375 (1986). 3. E. I. Radzievskaya and G. V. Radzievskii, “The remainder of the Taylor formula for a function holomorphic in a domain admits the representation in the Lagrange form,” Sib. Mat. Zh., 44, No. 2, 402–414 (2003). 4. A. M. Gomilko, E. I. Radzievskaya, and G. V. Radzievskii, “Local mean-value theorem for divided differences of holomorphic functions,” in: Proc. of the Institute of Mathematics, Ukrainian National Academy of Sciences [in Russian], 6, No. 1 (2009), pp. 82–95.