PAULO M. CARVALHO NETO1 AND PAULO A. LIBONI FILHO2
Abstract. We present a form of the Mean Value Theorem (MVT) for a continuous function f between metric spaces, connecting it with the possibility to choose the ε 7→ δ(ε) relation of f in a homeomorphic way. We also compare our formulation of the MVT with the classic one when the metric spaces are open subsets of Banach spaces. As a consequence, we derive a version of the Mean Value Propriety for measure spaces that also possesses a compatible metric structure.
Keywords metric space, mean value property, mean value theorem, homeomorphism, average of functions.
MSC (2010) Primary: 30L99, 54E40; Secondary: 26A15, 26E40.
1. Introduction
One of the most important results in Analysis is the Mean Value Theorem (MVT), which is used to prove many other significant results, from Several Complex Variables to Partial Differential Equations. It claims, in its original formulation (see Dieudonn´e, [5]) that if I = [a, b] ⊂ R is a closed real interval and B is a Banach space, then every pair of continuous functions f : I → B and φ : I → R with derivatives in (a, b) satisfies the following implication
(1) kf′(ξ)kB≤ φ′(ξ) for all ξ ∈ (a, b) =⇒ kf (b) − f (a)kB≤ φ(b) − φ(a).
The most known consequence of the previous result happens when we fix some constant M > 0 and set φ(ξ) = M (ξ − a). This guarantees that
(2) kf′(ξ)kB≤ M for all ξ ∈ (a, b) =⇒ kf (b) − f (a)kB≤ φ(b) − φ(a) = M (b − a).
There is also the classic form of the MVT for functions of several variables, which is directly obtained from the previous version. Consider B1, B2 Banach spaces and U ⊂ B1 an open subset.
If a, b ∈ U are such that the line segment [a, b] ⊂ U and f : U → B2 is a continuous function in [a, b] which is differentiable in (a, b), then we achieve a similar conclusion of (2) by considering f composed with the curve
t 7→ a + (b − a)t, where 0 ≤ t ≤ 1.
Inspired by Dieudonn´e, several authors generalized such theorems to more abstract spaces and/or to less regular functions. For instance, Clarke and Ledyaev in [1, 2] proposed the study of MVT in Hilbert spaces for lower semi-continuous functions in a new multidirectional sense. In a subsequent paper, Radulescu and Clarke (cf. [12]) proved that — in certain particular spaces — the locally Lipschitz functions also fulfill a variant of MVT.
Another distinct and celebrated approach was obtained by D. Preiss in [11]. He considered an Asplund space B and a Lipschitz continuous function f : B → R. By denoting as D the set of
1The author has been partially supported by FAPESP. Process 2013/00594-8.
2The author has been partially supported by CAPES. Process PNPD 2770/2011.
1
points where f is differentiable, D. Preiss proved that D is dense in B and then concluded that for any x, y ∈ B the following MVT holds
|f (x) − f (y)| ≤ Lkx − ykB, where L = sup
x∈D
kf′(x)kL(B,R).
An important formulation is due to M. Turinici, which proved (cf. [13]) a version resembling the original MVT in (1) to partially ordered metric spaces under some restrictive conditions on the metric. As we can see, there is a rich literature claiming different formulations of the MVT.
On the other hand, there is a lack of discussion for a version of the theorem which is suitable for metric spaces.
In what follows, inspired by Turinici’s paper and collected work about this topic, we propose a lot of results that discuss the viability of MVT for those general spaces. To do this, we must clarify what a suitable substitute of MVT means when no differential and/or vectorial framework is assumed.
To motivate our definition, let us examine the situation pictured in (2). Let (x−r, x+r) ⊂ (a, b) and consider Ψ : [0, r) ⊂ R+→ R+ given as
(3) Ψ(d) = sup
ξ∈[x−d,x+d]
kf′(ξ)kB.
If we write dR to indicate the Euclidian distance in R, then it follows that the MVT in (2) is equivalent to
(4) kf (x) − f (y)kB≤ Ψ(dR(x, y)) dR(x, y) for all y ∈ (x − r, x + r).
Note that the previous formulation is expressed in a metric fashion, with the aid of a certain function Ψ. Also, observe that extra proprieties of Ψ can be obtained from some regularity that f may possess. This discussion motivate us to consider the next definition.
Definition 1. Let (M1, d1) and (M2, d2) be metric spaces. Consider a function f : M1 → M2
and a point x ∈ M1. We say that f satisfies the Mean Value Inequality (MVI) in the open ball BM1(x, r) if there is a function Ψ : [0, r) ⊂ R+→ R+ such that
d2(f (x), f (y)) ≤ Ψ(d1(x, y)) d1(x, y) for all y ∈ BM1(x, r).
Also, the following properties must hold (i) Ψ is a continuous function;
(ii) The function I : [0, r) ⊂ R+→ R+ defined as I(d) = Ψ(d) d is a non decreasing homeomorphism over its image.
In this work we have two main objectives. First, establish a necessary and sufficient condition for the MVI in the open ball BM1(x, r). Note that our applications will be defined in abstract metric spaces, not necessarily partially ordered ones (cf. [13]). To deal with this problem, we introduce a subclass of continuous applications that, roughly speaking, are the ones which we can find a ε 7→ δ(ε) relation that possesses some regularity.
Our second goal is to show that the MVI in a Banach space is reduced to the supremum of the norm of the derivative in a open ball. We point out that in metric spaces there is no usual way to define directions, as opposed to what can be done in Banach spaces. Therefore, it is natural to expect that in our metric-only situation, the supremum is taken in the open ball, instead of the line segment.
To conclude, this paper is organized in the following sequence: in Section 2, we discuss the basic properties associated with sets that describes the regularity of a given function defined on a metric space. These sets are fundamental to the base structure of Sections 3 to 5, in which we investigate the ε 7→ δ(ε) regularity of continuous functions. In Section 6, we establish a necessary and sufficient condition for the MVI. Further, in Section 7, we derive a version of the Mean Value Propriety, which is about computing and estimating average of functions, and how it relates to its counterpart in Harmonic Analysis.
2. Preliminary Ideas and Initial Concepts
Throughout this paper, (M1, d1) and (M2, d2) denote metric spaces, unless stated otherwise.
For i = 1, 2, we write BMi(x, r) to represent the open ball centered at x ∈ Mi with radius r > 0.
Also, let F(M1, M2) (or simply F) be the set of all functions between M1 and M2. An element of F × M1× R+\ {0} is called a triplet. We say that a positive real number δ is suitable for the (f, x, ε) triplet if f (BM1(x, δ)) ⊂ BM2(f (x), ε). In other words, δ > 0 is suitable for the (f, x, ε) triplet if the following implication holds
y ∈ M1 and d1(x, y) < δ =⇒ d2(f (x), f (y)) < ε.
With the previous definition, it is clear that f : M1 → M2 is continuous at x ∈ M1 if, and only if, for every ε > 0 there exists a suitable number for the (f, x, ε) triplet. Based in the notion introduced above, we now discuss properties of some new structures.
Definition 2. Given the (f, x, ε) triplet, consider the set
∆f,x(ε) = {δ > 0 : δ is suitable for the (f, x, ε) triplet}.
The idea behind the previous set, is to capture all possible choices of δ > 0 one can find while trying to prove that a certain function is continuous. Note that we are not excluding the possibility of ∆f,x(ε) = ∅, which would means that f is discontinuous at x. It is important to understand every possible scenario for the set of suitable numbers for a triplet.
Theorem 3. Let (f, x, ε) be a triplet. One, and only one of the following alternatives occurs.
(i) ∆f,x(ε) = ∅;
(ii) ∆f,x(ε) = (0, ∞);
(iii) There exists a certain δ > 0 such that ∆f,x(ε) = (0, δ].
Proof. Let us first prove that ∆f,xis connected. We claim that if δ2∈ ∆f,x(ε) and if 0 < δ1≤ δ2, then δ1∈ ∆f,x(ε). Indeed, if y ∈ M1 and d1(x, y) < δ1, then d1(x, y) < δ2. Since δ2is suitable for the (f, x, ε) triplet, we conclude that d2(f (x), f (y)) < ε. This guarantees that δ1 is also suitable for the (f, x, ε) triplet.
Now let us check that ∆f,xis closed from the right side. More precisely, if {δn}∞n=1⊂ ∆f,x(ε) is an increasing sequence that converges to δ, then δ ∈ ∆f,x(ε). Fix y ∈ M1 such that d1(x, y) < δ.
Choose a natural number N such that |δ − δN| < δ − d1(x, y). Since the sequence is increasing, we conclude that d1(x, y) < δN. But we already know that δN is suitable for the (f, x, ε) triplet, therefore we conclude that d2(f (x), f (y)) < ε. Once y ∈ M1 was an arbitrary choice, δ is also suitable for the (f, x, ε) triplet.
Now, we are able to conclude the argument. Three alternatives occur.
(i) ∆f,x(ε) = ∅. Then, we are done;
(ii) ∆f,x(ε) is nonempty unbounded. Then, by the hereditary property above, ∆f,x(ε) = (0, ∞);
(iii) ∆f,x(ε) is nonempty bounded. Then, by the sequential argument we just developed,
∆f,x(ε) = (0, δ], where δ = sup ∆f,x(ε) = (0, δ].
To study more properties of ∆f,x, fix x ∈ M1 and consider the family {∆f,x(ε) : ε > 0}.
Roughly speaking, we are going to show that the set of suitable numbers for the (f, x, ε) triplet does not became smaller as ε gets bigger.
Proposition 4. Consider any function f : M1→ M2. For each element x ∈ M1, if 0 < ε1≤ ε2, then ∆f,x(ε1) ⊂ ∆f,x(ε2).
Proof. If ∆f,x(ε1) = ∅, there is nothing to be done. On the other hand, suppose that there exists a value δ ∈ ∆f,x(ε1). Hence, for y ∈ M1 with d1(x, y) < δ, we have that d2(f (x), f (y)) < ε1. But ε1 ≤ ε2, therefore for any y ∈ M1 with d1(x, y) < δ, we have that d2(f (x), f (y)) < ε2. In other
words, δ ∈ ∆f,x(ε2).
Definition 5. Given x ∈ M1 and a function f : M1→ M2, define
Ef(x) = {ε > 0 : ∆f,x(ε) is a non empty bounded set}.
Shortly, we may say that Ef(x) captures all the possible values of ε > 0 such that there exists a maximum real number δ > 0 that verifies the sentence
y ∈ M1 and d1(x, y) < δ =⇒ d2(f (x), f (y)) < ε.
Working in the same way as we did in Theorem 3, we examine the full range of topological possibilities about the set Ef(x). Note that, from now on, the continuity of the function f at x plays an important role. This allow us to state and prove the following result.
Theorem 6. Let f : M1 → M2 be a given function which is continuous at x ∈ M1. Then one, and only one of the following alternatives occurs.
(i) If f is an unbounded, then Ef(x) = (0, ∞);
(ii) If f is a constant function, then Ef(x) = ∅;
(iii) In the other cases, there exists some real number ε > 0 such that Ef(x) = (0, ε] or Ef(x) = (0, ε).
Proof. Note that since f is continuous at x ∈ M1, Proposition 4 guarantees that Ef(x) is an interval. Now let us prove that f is a constant function if, and only if, Ef(x) = ∅. Indeed, if there exists x ∈ M1 such that Ef(x) = ∅, then for any ε > 0 we have that ∆f,x(ε) = (0, ∞). Thus, for any ε, δ > 0 we must have the following
y ∈ M1 and d1(x, y) < δ =⇒ d2(f (x), f (y)) < ε.
Therefore, for any y ∈ M1 and any ε > 0 we have that d2(f (x), f (y)) < ε; which implies that for any y ∈ M1, d2(f (x), f (y)) = 0. The other part of the claim is trivial.
To conclude, let us prove that Ef(x) is a proper subset of (0, ∞) if, and only if, f is bounded.
Indeed, let f be bounded. In this case, there exists ε > 0 such that f (M1) ⊂ BM2(f (x), ε). In other words, for any δ > 0
f (BM1(x, δ)) ⊂ BM2(f (x), ε).
This last statement implies that any δ is suitable for the (f, x, ε) triplet and therefore ∆f,x(ε) = (0, ∞), which means that ε /∈ Ef(x). If there exists ε ∈ (0, ∞) \ Ef(x), by definition ∆f,x(ε) = (0, ∞). Hence, for any y ∈ M1 we have that d2(f (x), f (y)) < ε, which implies that f is bounded.
Example 7. Note that all the situations listed in the last case can happen, as we can see in the following couple examples.
(i) Consider M1 = M2 = R and d1 = d2 the real Euclidean metric. If f : M1 → M2 is the characteristic function of [0, ∞), then
∆f,x(ε) =
(0, |x|], ε ∈ (0, 1] and x 6= 0 (0, ∞), ε > 1 and x 6= 0
∅, ε ∈ (0, 1] and x = 0 (0, ∞), ε > 1 and x = 0 which implies that Ef(0) = ∅ and Ef(x) = (0, 1] for x 6= 0;
(ii) Let M1= M2= [0, ∞) and d1= d2be the induced real Euclidean metric. If f : M1→ M2
is defined as f (x) = 1 − e−x, then
∆f,0(ε) =
0, ln1−ε1 i
, ε ∈ (0, 1)
0, ∞), ε ≥ 1.
Therefore Ef(0) = (0, 1).
Until this moment, the constructions that we made do not seem to have a direct connection with the original function f . However, now that we have discussed these prerequisites, we are ready to define the function that is related to the continuity of f .
Definition 8. Given f : M1→ M2and x ∈ M1such that Ef(x) 6= ∅, define Πf,x: Ef(x) → (0, ∞) as
Πf,x(ε) = max ∆f,x(ε).
Remark 9. A few remarks are in order.
(i) Since the set ∆f,x(ε) admits a maximum for every ε ∈ Ef(x), the function Πx is well defined;
(ii) Is a direct consequence, from the previous section, that Πxis a non decreasing function that for every ε ∈ Ef(x) provided the largest possible number such that f (BM1(x, Πf,x(ε))) ⊂ BM2(f (x), ε);
(iii) Even if, during the construction of the function Πf,x, the original function f plays an important role, whenever there is no possibility of confusion, we omit f as an index, writing instead just Πx.
3. The Continuity of Πx
We now focus our attention to the continuity properties of the function Πx.
Lemma 10. Let f : M1→ M2 be a non constant and continuous function. Choose x ∈ M1 and suppose that for any r > 0 the closure of BM1(x, r) is compact in M1. Under these conditions, for all ε ∈ Ef(x) the following holds.
(i) If {εn}∞n=1⊂ Ef(x) is an increasing sequence that converges to ε, then
n→∞lim max ∆f,x(εn) = max ∆f,x(ε);
(ii) If {εn}∞n=1⊂ Ef(x) is a decreasing sequence that converges to ε, then
n→∞lim max ∆f,x(εn) = max ∆f,x(ε).
Proof. We only verify the first part; the second one follows in a similar way. By Proposition 4, the map ε 7→ max ∆f,x(ε) is non decreasing. Denote, for simplicity
eδ = lim
n→∞max ∆f,x(εn) and δ = max ∆f,x(ε)
If eδ 6= δ, then the non decreasing property above gives eδ < δ. Consider the auxiliary sequence {δn}∞n=1 given by
δn= eδ + δ
2 +
"
δ − eδ 2(n + 1)
#
and observe that it satisfies the following properties (i) δn→ (eδ + δ)/2;
(ii) eδ < δn< δ.
Since for each n = {1, 2, . . .} the number δnis strictly bigger then eδ, we have that δn 6∈ ∆f,x(εn).
Therefore, for each n, there exists yn ∈ M1 such that
d1(yn, x) < δn and d2(f (yn), f (x)) ≥ εn.
Since yn ∈ BM1(x, δ), choosing a subsequence if necessary, we can assume that there exists y ∈ BM1(x, δ) such that yn→ y. Then, when n → ∞, we obtain
d1(y, x) ≤ (eδ + δ)/2 and d2(f (y), f (x)) ≥ ε,
and conclude that δ 6∈ ∆f,x(ε), which is a contradiction.
Theorem 11. Let f : M1→ M2 be a non constant and continuous function. Choose x ∈ M1 and suppose that for any r > 0 the closure of BM1(x, r) is compact in M1. Under these conditions Πx
is a continuous function.
Proof. Choose and fix ε0∈ Ef(x). Lets prove the equality lim
ε→ε−0 Πx(ε) = Πx(ε0) = lim
t→ε+0 Πx(ε).
We shall begin proving the left limit. Let {eεn} be sequence to the left of ε0 such that eεn → ε0. It is not difficult to construct another increasing sequence {εn} with the following proprieties
(i) εn< eεn, for each n ∈ N;
(ii) εn→ ε0, when n → ∞.
Then we know that ∆f,x(εn) ⊂ ∆f,x(eεn) ⊂ ∆f,x(ε0) for all natural number n. Using the maximum function in this sequence of inclusions, we deduce
max ∆f,x(εn) ≤ max ∆f,x(eεn) ≤ max ∆f,x(ε0).
Finally, applying the limit when n → ∞ on both sides and using Lemma 10 we obtain max ∆f,x(ε0) ≤ lim
n→∞max ∆f,x(eεn) ≤ max ∆f,x(ε0).
Since {eεn} was an arbitrary sequence, we conclude the proof of the left limit. The right one is
obtained in a similar way and therefore we are done.
The next step in our work is to extend the domain of Πx. To this end, we recall a useful definition.
Definition 12. A function f : M1→ M2 is called locally constant at a point x ∈ M1, if there is exists r > 0 such that f |BM1(x,r)is a constant function.
Observe that locally constant functions are almost similar to the constant functions, when we compute its Πxfunction. Hence, we shall exclude this class of functions in our future results.
Theorem 13. Let f : M1→ M2 be a continuous function. Choose x ∈ M1 and suppose that f is not locally constant at x. In this case, 0 is an accumulation point of Ef(x) and
ε→0lim+Πx(ε) = 0.
Proof. Since f is continuous and non constant, it follows easily (Theorem 6) that 0 is an accumu- lation point of Ef(x). In fact, Ef(x) turns out to be an interval in this case.
Now, consider {εn} ⊂ Ef(x) a decreasing sequence that converges to 0. By monotonicity, we have that ∆f,x(εn+1) ⊂ ∆f,x(εn). If diam A is the diameter of the set A, then diam ∆f,x(εn+1) ≤ diam ∆f,x(εn). Since Πx(ε) = max ∆f,x(ε) = diam ∆f,x(ε), we observe that
n→∞lim Πx(εn) = diam
\∞ n=1
∆f,x(εn)
! .
If δ ∈ ∩∞n=1∆f,x(εn), then f (B(x, δ)) ⊂ B(f (x), εn) for any n ∈ N. Since εn → 0, we conclude that f is locally constant at x. Since by hypothesis this situation cannot happen, we obtain that ∩∞n=1∆f,x(εn) = ∅. Therefore limn→∞Πx(εn) = 0. Following the same final steps done in
Theorem 11, we conclude the proof of this theorem.
This last theorem allow us to continuously extend our definition of Πx to ε = 0 if f is not locally constant at x. This extension will be useful in the next result, which connects the image of Πx with the set of suitable values.
Theorem 14. Let f : M1 → M2 be a continuous function. Choose x ∈ M1 such that f is not locally constant at x. Suppose that for any r > 0 the closure of BM1(x, r) is compact in M1. Under these conditions, for all ε ∈ Ef(x) we have that ∆f,x(ε) = Πx(0, ε].
Proof. If δ > 0 is suitable for the (f, x, ε) triplet, then Πx(0) < δ ≤ Πx(ε). Since Πx is con- tinuous, it must exist some ε′ ∈ (0, ε] such that Πx(ε′) = δ. Conversely, if ε′ ∈ (0, ε], then f (BM1(x, Πx(ε′))) ⊂ BM2(f (x), ε′) what guarantees that f (BM1(x, Πx(ε′))) ⊂ BM2(f (x), ε). This proves that Πx(ε′) is suitable for the (f, x, ε) triplet.
4. Πx as a Homeomorphism
There is no special reason to expect that Πx is a homeomorphism. For example, if R is con- sidered with the discrete metric, than any given f : R → R is a continuous function. In the case where f (x) = x, it is quite easy to see that 1 is the maximal suitable value for the (f, x, ε) triplet.
More precisely, we have that Πx(ε) = 1 for all ε ∈ (0, 1). Inspired by this example, we shall now investigate under which circumstances we can guarantee that Πx is a homeomorphism. We start with some basic definition.
Definition 15. Given a function f : M1 → M2 and a point x ∈ M1, we say that f is adherent at x if the following holds
(i) f is continuous at x;
(ii) If y ∈ M1 and ε ∈ Ef(x) are such that d1(x, y) = Πx(ε), then d2(f (x), f (y)) ≤ ε.
If f : M1→ M2 is adherent at any x ∈ M1, we simply say that f is adherent.
This last technical definition has the purpose of categorize a critical behavior of the function f . Recall that if f is continuous at x and d1(x, y) < Πx(ε), then d2(f (x), f (y)) < ε. Then, it is natural to ask what happens with f when d1(x, y) = Πx(ε). In other words, suppose that there exists some point y such that d1(x, y) = Πx(ε) and d2(f (x), f (y) ≥ ε. Intuitively speaking, if f is adherent at x, then we are requesting that the number d2(f (x), f (y)) stays, at the worst possible scenario, equal to ε.
Example 16. As outlined previously, consider R with the discrete metric and f : R → R given by f (x) = x. Then we can verify that f is not adherent at any x ∈ R. Indeed, if ε = 1/2 and y is such that d1(x, y) = Π(ε) = 1, then y can be any real number, except x itself. Under these conditions, it is false that d2(x, y) = d2(f (x), f (y)) ≤ 1/2 for all y such that d1(x, y) = 1.
The next theorem investigates the relation between Πx and the definition of adherence at a point x.
Theorem 17. Consider f : M1 → M2 and x ∈ M1. Assume that for any r > 0 the closure of BM1(x, r) is compact in M1. If f is continuous at x, then the following are equivalent
(i) f is adherent at x.
(ii) Πx is injective.
Proof. Suppose that f is not adherent at x. Under these conditions, there exist y ∈ M1 and ε ∈ Ef(x) such that d1(x, y) = Πx(ε) and d2(f (x), f (y)) > ε > 0. We observe that ε′ = d2(f (x), f (y)) ∈ Ef(x). Indeed, since f is continuous at x and ε′ > 0 we already know that
∆f,x(ε′) is a non empty set. Now, assume that ∆f,x(ε′) is not a bounded set. Using Theorem 3, we conclude that ∆f,x(ε′) = (0, ∞). Therewith, 0 < d1(x, y) + 1 ∈ ∆f,x(ε′), which implies that d1(x, y) + 1 is suitable for the triplet (f, x, ε′). Therefore, if ey ∈ M1and
d1(x, ey) < d1(x, y) + 1, then d2(f (x), f (ey)) < ε′.
Once we can suppose that ey = y, in the previous statement, we obtain that ε′ < ε′; contradiction.
In other words, ∆f,x(ε′) is a bounded set and ε′ = d2(f (x), f (y)) ∈ Ef(x).
Now, since Πx is a non decreasing function, we obtain that Πx(ε) ≤ Πx(ε′). Assume, for an instant, that Πx(ε) < Πx(ε′). Using the fact that d1(x, y) = Πx(ε), we deduce that d1(x, y) <
Πx(ε′). Therefore d2(f (x), f (y)) < ε′, which is a contradiction. Hence, Πx(ε) = Πx(ε′), which implies that Πx is not injective.
Conversely, suppose that f is adherent at x. Consider ε1 and ε2 in Ef(x) such that Πx(ε1) = Πx(ε2). Without loss of generality, assume that ε1 ≤ ε2. By Lemma 6, we already know that [ε1, ε2] ⊂ (0, ε2] ⊂ Ef(x).
Let δ = Πx(ε1) = Πx(ε2). Since Πx is a non decreasing function and [ε1, ε2] ⊂ Ef(x), we have that Πx(ε) = δ for all ε ∈ [ε1, ε2]. But δ is the largest suitable number for the (f, x, ε) triplet, whenever ε lies in [ε1, ε2]. Writing it in another way: for all ε ∈ [ε1, ε2] and for any number δ + 1/n, we have that δ + 1/n is not suitable for the (f, x, ε) triplet. Hence, for all ε ∈ [ε1, ε2] and for any natural number n, there is some yn,ε ∈ M1 such that d1(x, yn,ε) < δ + 1/n and d2(f (x), f (yn,ε)) ≥ ε.
Without loss of generality, assume that there exists eyε such that limn→∞yn,ε = eyε. Using a limit argument, we have that for all ε ∈ [ε1, ε2], there is some eyε ∈ M1 such that d1(x, eyε) ≤ δ and d2(f (x), f ( eyε)) ≥ ε. If we suppose, for an instant, that d1(x, eyε) < δ = Πx(ε), a directly consequence would be that d2(f (x), f ( eyε)) < ε, which is a contradiction. Because of that, for all ε ∈ [ε1, ε2], there is some eyε ∈ M1 such that d1(x, eyε) = δ and d2(f (x), f ( eyε)) ≥ ε. Since f is adherent at x, we conclude that
∀ ε ∈ [ε1, ε2], ∃ eyε∈ M1 such that d1(x, eyε) = Πx(ε) and d2(f (x), f ( eyε)) = ε.
In other words, there exists fyε2 ∈ M1 such that d1(x, fyε2) = Πx(ε2) and d2(f (x), f ( fyε2)) = ε2. Since f is adherent at x and d1(x, fyε2) = Πx(ε1), we conclude that ε2 = d2(f (x), f ( fyε2)) ≤ ε1
what implies that ε1= ε2. Therefore, Πxis an injective function. Remark 18. Since any continuous and injective function g : I → R, defined on any interval I ⊂ R, is a homeomorphism (from I to g(I)), this last theorem guarantees that function Πx is a homeomorphism. In other words, it guarantees that the dependence ε 7→ δ(ε) from the continuity of a function in a metric space can be described by a homeomorphism.
Next theorem specifies more clearly which functions can be expected to be adherent. From now on, if (M1, d1) denotes a metric space, we write BM1[x, r] to denote the closed ball BM1[x, r] = {y ∈ M1: d1(x, y) ≤ r}.
Theorem 19. Let f : M1→ M2 be continuous at x ∈ M1. Suppose that M1 and M2 are metric spaces such that the closure of the ball BMi(x, r) is the closed ball BMi[x, r]. Then f is adherent at x.
Proof. Let ε ∈ Ef(x). Notice that f (BM1(x, Πx(ε)) ⊂ BM2(f (x), ε). Indeed, let p ∈ f (BM1(x, Πx(ε)).
Then, there exists z ∈ BM1(x, Πx(ε)) such that f (z) = p. On the other hand, z is the limit of some sequence (zn) of elements in BM1(x, Πx(ε)). Since f is a continuous function, we must have that p = lim f (zn). In other words, p ∈ f (BM1(x, Πx(ε)). But f is a continuous map, therefore f (BM1(x, Πx(ε)) ⊂ BM2(f (x), ε) what guarantees that p ∈ BM2(f (x), ε).
Now, using the hypothesis about the closure of the balls, we conclude that f (BM1[x, Πx(ε)]) ⊂ BM2[f (x), ε]. At last, suppose that there exists y ∈ M1such that d1(x, y) = Πx(ε). It follows that y ∈ BM1[x, Πx(ε)], what guarantees that f (y) ∈ BM2[f (x), ε]. This concludes the proof. We finish this section with an important theorem, that put together some previous results and discussions.
Theorem 20 (Πx as a Homeomorphism). Suppose that f : M1 → M2 is any given function, x ∈ M1 and assume the following
(i) f is continuous;
(ii) f is adherent at x;
(iii) f is not locally constant at x;
(iv) For any r > 0 the closure of BM1(x, r) is compact in M1.
Then there exists an interval Ix and a function Πx: Ix→ Jx⊂ R+ such that (i) 0 ∈ Ix⊂ R+;
(ii) Πx is a monotonic increasing homeomorphism;
(iii) f (BM1(x, δ)) ⊂ BM2(f (x), ε) ⇐⇒ δ ≤ Πx(ε), for all ε ∈ Ix; (iv) Πx(0) = 0.
5. Enclosed Functions
At this point, we observe that for a certain class of functions (see Theorem 20) we have already managed to construct a homeomorphism that is related to the continuity properties of f .
Definition 21. Given a function f : M1 → M2 and a point x ∈ M1, we say that f is strongly enclosed at x if f satisfies the conclusions (i), (ii), (iii) and (iv) of Theorem 20. Furthermore, the Πx map is called the strong continuity function of f at x. We can also say that f is en- closed at x if f satisfies all the previous conclusions, with the possible exception of the implication f (BM1(x, δ)) ⊂ BM2(f (x), ε) =⇒ δ ≤ Πx(ε). In this case, the Πx map is just called a continuity function of f at x. If the function is strongly enclosed (or enclosed) at all points of the domain, we say that the function is strongly enclosed (or enclosed).
Roughly speaking, we may not have a maximum suitable number for enclosed functions, while the opposite occurs with the strongly enclosed functions. In general, to guarantee that a function is strongly enclosed, we require some compactness hypothesis (see Theorem 20). On the other hand, all locally Lipschitz functions are enclosed.
It is interesting to note that the strong continuity function associated with a strong enclosed function f is unique, where the uniqueness here is considered in the sense of germs at 0. Note that this does not happen in the enclosed case. We are now interested in finding other classes of function which are enclosed or strongly enclosed.
Definition 22. Given a function f : M1→ M2 and a point x ∈ M1, we are going to say that f satisfies the Lagrange Propriety at x if the following two conditions hold
(i) There exists a C1 function Γ : (−ζ, ζ) ⊂ R → R such that Γ(r) = sup
y∈BM1[x,r]\{x}
d2(f (x), f (y)) d1(x, y) for all r ∈ (0, ζ);
(ii) For all r ∈ (0, ζ), there is an element yr∈ BM1[x, r] \ {x} such that sup
y∈BM1[x,r]\{x}
d2(f (x), f (y))
d1(x, y) = d2(f (x), f (yr)) d1(x, yr) .
Example 23. Consider M1 = M2 = R and d1 = d2 the real Euclidean metric. Assume that f : M1→ M2 is given by f (x) = 1 − e−x. Since (1 − e−y)/y is a positive decreasing function, we have that
Γ(r) = sup
y∈BR[0,r]\{0}
|1 − e−y|
|y| = 1 − er
−r =d2(f (0), f (−r)) d1(0, −r) . In other words, f satisfies the Lagrange Propriety at 0 with Γ(r) = err−1.
The truly importance of the Lagrange class is uncovered by the following result, which connects the Γ function with the strong enclosed definition.
Theorem 24. Suppose that f : M1 → M2 is any given function that satisfies the Lagrange Propriety at x ∈ M1. If Γ(0) 6= 0 and Γ′(0) > 0, then f is strongly enclosed at x. Furthermore, the strong continuity function of f is a Ck diffeomorphism, provided that Γ is a Ck application.
Proof. Let ∆ : (−ζ, ζ) → R be given by ∆(t) = t Γ(t). Since ∆ is a Ck application and ∆′(0) is an isomorphism, we can use the Inverse Function Theorem and assume that ∆|(−ζ1,ζ1): (−ζ1, ζ1) → (−ζ2, ζ2) is a Ck diffeomorphism.
Let ∆−1 be the inverse map of ∆|(−ζ1,ζ1). Then, ∆−1(ε)Γ(∆−1(ε)) = ε, which implies that
∆−1(ε) = ε
sup
y∈BM1[x,∆−1(ε)]\{x}
d2(f (x), f (y)) d1(x, y)
.
We claim that ∆−1(ε) is suitable for (f, x, ε) triplet. Indeed, if d1(x, y) < ∆−1(ε), then d2(f (x), f (y)) ≤ sup
y∈BM1[x,∆−1(ε)]\{x}
d2(f (x), f (y))
d1(x, y) d1(x, y).
This implies that
d2(f (x), f (y)) < sup
y∈BM1[x,∆−1(ε)]\{x}
d2(f (x), f (y))
d1(x, y) ∆−1(ε) = ε.
If ε is small enough, let us prove that δ = ∆−1(ε) is the maximum possible suitable number for (f, x, ε) triplet. Indeed, if δ = ∆−1(ε), then ∆(δ) = ε, which implies that δ Γ(δ) = ε. Remember that for each δ, there exists an element yδ such that
Γ(δ) = d2(f (x), f (yδ)) d1(x, yδ) .
Hence, δ d2(f (x), f (yδ)) = ε d1(x, yδ). It is now clear that d2(f (x), f (yδ)) = ε if, and only if, d(x, yδ) = δ. Since yδ ∈ BM1[x, δ] \ {x}, we also conclude that d2(f (x), f (yδ)) < ε if, and only if, d(x, yδ) < δ.
Now suppose that δ is not the maximum possible suitable number for (f, x, ε) triplet. Therefore, we can find some E > 0 such that
d1(x, y) < E + δ =⇒ d2(f (x), f (x)) < ε.
Since yδ ∈ BM1[x, δ] \ {x}, we have that
d1(x, yδ) < E + δ =⇒ d2(f (x), f (yδ)) < ε.
Making E → 0, we then conclude that
d1(x, yδ) ≤ δ =⇒ d2(f (x), f (yδ)) < ε.
Therefore, as proved above,
d1(x, yδ) ≤ δ =⇒ d2(f (x), f (yδ)) < ε =⇒ d(x, yδ) < δ.
Because of the last inequality, we have that yδ ∈ BM1[x, δ] \ {x} actually satisfies d(x, yδ) < δ.
Since Γ′(0) > 0 and Γ is a C1 function, we know that Γ′ is non negative on a certain interval starting at the origin. Considering that ε is small enough, we have that
Γ(d1(x, yδ)) < Γ(δ) = d2(f (x), f (yδ) d1(x, yδ) . Note that the last inequality is a contradiction, since by definition
Γ(d1(x, yδ)) ≥ d2(f (x), f (yδ) d1(x, yδ) .
Therefore δ = ∆−1(ε) is the maximum suitable number for the (f, x, ε) triplet. In other words, Πx = ∆−1 is the strongly enclosed function associated to f . Since the function ∆|(−ζ1,ζ1) : (−ζ1, ζ1) → (−ζ2, ζ2) is a Ck diffeomorphism, the proof is finished.
Corollary 25. Assuming the last theorem hypothesis, we also have that
Πx(ε) = ε
sup
y∈BM1[x,ξ(ε)]\{x}
d2(f (x), f (y)) d1(x, y)
= ε
Γ(ξ(ε)),
where ξ satisfies the following conditions
ξ(0) = 0
ξ′(ε Γ(ε)) = 1 Γ(ε) + εΓ′(ε)
Proof. We just have to note that ξ = Πx = ∆−1. The rest of the conclusion follows from the
Chain Rule.
Example 26. If M1= M2 = [0, ∞) and d1 = d2 are the real Euclidean metric, we know that if function f : M1→ M2 is given by f (x) = 1 − e−x, then
Π0(ε) = ln 1
1 − ε≈ ε +ε2
2 + O(ε3).
Now let us consider a slightly different situation. Suppose that M1 = M2 = R and d1 = d2
are the real Euclidean metric. Suppose that f : M1 → M2 is given by f (x) = 1 − e−x. Since our domain is an open set, we can use the last theorem to find an expression for Π0(ε).
We also know that f satisfies the Lagrange Propriety with Γ(r) = 1−e−rr. Under these conditions, ξ satisfies the following system
ξ(0) = 0
ξ′(ε Γ(ε)) = 1 Γ(ε) + εΓ′(ε) for sufficiently small ε. Therefore ξ(ε) = ln(1 + ε) and
Π0(ε) = ln(1 + ε) ≈ ε −ε2
2 + O(ε3).
It is noteworthy that a simple domain change can alter the Π continuity function expression.
This phenomenon should not be surprising, since there is a connection between the ε 7→ δ(ε) relation of f and its domain.
Corollary 27. Suppose that B1 and B2 are Banach spaces. In addition to the last theorem hy- pothesis, suppose that M1is an open set of B1and also assume that M2⊂ B2. If f is differentiable and M is the maximum value of function t 7→ kf′(t)kL(B1,B2), then
ε
M ≤ Πx(ε) ≤ ε Γ(0)
for all sufficiently small ε. If t 7→ kf′(t)kL(B1,B2) does not reach a maximum value, then the first inequality is reduced to 0 ≤ Πx(ε).
6. The Mean Value Inequality Now we are ready to present our main theorem.
Theorem 28 (Mean Value Inequality for Metric Spaces). Given a function f : M1 → M2 and a point x ∈ M1, the following are equivalent
(i) f is enclosed in x and Π−1x is differentiable at 0+;
(ii) f satisfies the MVI (see Definition 1) for metric spaces in some BM1(x, R).
In addition, suppose that M1 and M2 are open sets contained in Banach spaces B1 and B2, respectively. If f is differentiable and strongly enclosed at x, then there exists R > 0 such that
Ψ(d1(x, y)) ≤ sup
BM1[x,d1(x,y)]
kf′(t)kL(B1,B2)
for any y ∈ BM1(x, R).
Proof. Suppose that f is enclosed at x and consider Ix= D(Πx). Define Ψ : Jx→ R+ by
Ψ(d) =
Π−1x (d)
d , if d ∈ Jx\ {0}
(Π−1x )′(0+), if d = 0, where Jx= Πx(Ix).
Observe that Ψ is continuous and that the function d 7→ Ψ(d)d is a non decreasing homeomor- phism. Now choose R > 0 with the following two properties holding.
(i) R < (1/2) sup Jx;
(ii) For any y ∈ BM1(x, R), we have d2(f (x), f (y) < sup Ix.
Let y ∈ BM1(x, R). If ε ∈ (0, R), then the value d1(x, y)+ε is not suitable for the (f, x, d2(f (x), f (y))) triplet. Therefore we obtain the following inequality
Πx(d2(f (x), f (y))) < d1(x, y) + ε.
Applying Π−1x on both sides and taking the limit as ε → 0, we achieve d2(f (x), f (y)) ≤ Π−1x (d1(x, y)) = Ψ(d1(x, y)) d1(x, y).
Conversely, let us suppose that f satisfies the MVI. Then, there exists a function Ψ : [0, r) ⊂ R+ → R+ with the properties described on Definition 1. Set Πx as the inverse of the function d 7→ Ψ(d)d. Hence, if d1(x, y) < δ and δ ≤ Πx(ε), MVI guarantees that
d2(f (x), f (y)) ≤ Ψ(d1(x, y)) d1(x, y) = Π−1x (d1(x, y)) .
Since d1(x, y) < Πx(ε) and Πxis strictly increasing, we obtain that Πx−1(d1(x, y)) < Π−1x (Πx(ε)) = ε, i.e., d2(f (x), f (y)) < ε.
Finally, to prove the last statement, assume that M1and M2are open sets contained in Banach spaces B1 and B2, respectively and f is differentiable. Choose R > 0 such that for any y ∈ BM1(x, R)
d1(x, y) sup
BM1[x,d1(x,y)]
kf′(t)kL(B1,B2)< sup Ix/2.
Now let us prove that for any y ∈ BM1(x, R), the value d1(x, y) is suitable for the (f, x, r) triplet, where the value r = d1(x, y) supBM1[x,d1(x,y)]kf′(t)kL(B1,B2). Indeed, if d1(x, ey) < d1(x, y), we have that
d2(f (x), f (ey)) ≤ d1(x, ey) supB
M1[x,d1(x,ey)]kf′(t)kL(B1,B2)
≤ d1(x, ey) supBM1[x,d1(x,y)]kf′(t)kL(B1,B2)
< d1(x, y) supBM1[x,d1(x,y)]kf′(t)kL(B1,B2). Therefore, since f is strongly enclosed at x, d1(x, y) ≤ Πx
d1(x, y) supBM1[x,d1(x,y)]kf′(t)kL(B1,B2)
what guarantees
Π−1x (d1(x, y)) ≤ d1(x, y) sup
BM1[x,d1(x,y)]
kf′(t)kL(B1,B2),
which finally implies that
Ψ(d1(x, y)) ≤ sup
BM1[x,d1(x,y)]
kf′(t)kL(B1,B2),
and the proof is complete.
Notice that the differentiability requirement of Π−1x at 0 is the natural substitute for f being dif- ferentiable on its domain. Also note that, since Ψ is a continuous function and supB
M1[x,d1(x,y)]kf′(t)kL(B1,B2)
may not be continuous, there is no hope to expect anything better than the inequality provided by the last theorem.
On the other hand, if f is continuously differentiable we have that y 7→ supBM1[x,d1(x,y)]kf′(t)kL(B1,B2)
defines a continuous function. Moreover, it provide us more precise information about the function Ψ, as we can verify in the following result.
Theorem 29. In addition to the last theorem hypothesis, suppose that M1 and M2 are open sets contained in Banach spaces B1and B2, respectively. If f is continuously differentiable and if there is a direction v ∈ B1 such that kf′(x) · vkB2 = kf′(x)kL(B1,B2) 6= 0, then there exists R > 0 such that
Ψ(d1(x, y)) = sup
BM1[x,d1(x,y)]
kf′(t)kL(B1,B2)
for all y ∈ BM1(x, R).
Proof. Choose R > 0 with the following two properties
(i) d1(x, y) supBM1[x,d1(x,y)]kf′(t)kL(B1,B2)< sup Ix, for any y ∈ BM1(x, R);
(ii) R < (1/2) sup Jx.
where Ixand Jxwhere defined on Theorem 28.
Let y ∈ BM1(x, R). Suppose that for any ε ∈ (0, R), the positive value d1(x, y) + ε is not suitable for the (f, x, r) triplet, where r = d1(x, y) supB
M1[x,d1(x,y)]kf′(t)kL(B1,B2). Under these circumstances, we obtain
Πx d1(x, y) sup
BM1[x,d1(x,y)]
kf′(t)kL(B1,B2)
!
< d1(x, y) + ε,
which implies
Πx d1(x, y) sup
BM1[x,d1(x,y)]
kf′(t)kL(B1,B2)
!
≤ d1(x, y).
Applying the inverse of Πx, we have d1(x, y) sup
BM1[x,d1(x,y)]
kf′(t)kL(B1,B2)≤ Π−1x (d1(x, y)) , which implies that Ψ(d1(x, y)) ≥ supB
M1[x,d1(x,y)]kf′(t)kL(B1,B2).
To complete this proof we show that for any ε > 0, the value d1(x, y) + ε is not suitable for the triplet (f, x, r). Since y 7→ supB
M1[x,d1(x,y)]kf′(t)kL(B1,B2)is a continuous application, than for all ξ > 0 we can find ζ > 0 such that if d1(x, y) < ζ, then kf′(x)k−1L(B
1,B2) supB
M1[x,d1(x,y)]kf′(t)kL(B1,B2)<
ξ + 1.
In other words, if d1(x, y) < ζ then
d1(x, y) supBM1[x,d1(x,y)]kf′(t)kL(B1,B2)
kf′(x)kL(B1,B2)
< d1(x, y) ξ + d1(x, y) < ζ ξ + d1(x, y).
Let M > 1 and set ξ = ε/M . Without loss of generality, we assume that ζ < 1. Hence, if d1(x, y) < ζ, then
d1(x, y) supBM1[x,d1(x,y)]kf′(t)kL(B1,B2)
kf′(x)kL(B1,B2)
< d1(x, y) + ε M. Choose λy such that
d1(x, y) supBM1[x,d1(x,y)]kf′(t)kL(B1,B2)
kf′(x)kL(B1,B2) < λy< d1(x, y) + ε M.
Since there is a direction v ∈ B1 such that kf′(x) · vkB2 = kf′(x)kL(B1,B2), define h = λv and e
y = x + h. Then d1(x, ey) = λy< d1(x, y) +Mε < d1(x, y) + ε.
On the other hand,
kf′(x) · hkB2 = λkf′(x)kL(B1,B2)> d1(x, y) sup
BM1[x,d1(x,y)]
kf′(t)kL(B1,B2).
Choosing a suitable M and a sufficiently small value R > 0, if y ∈ BM1(x, R) then kf (x + h) − f (x) − f′(x) · hkB2 is small enough, so kf (ey) − f (x)kB2 ≥ d1(x, y) supBM1[x,d1(x,y)]kf′(t)kL(B1,B2)
even if d1(x, ey) < d1(x, y) + ε. Therewith, d1(x, y) + ε is not suitable for the desired triplet if
y ∈ BM1(x, R), which concludes the proof.
Corollary 30. Suppose that M1 and M2 are open sets contained in Banach spaces B1 and B2, respectively, f : M1 → M2 is continuously differentiable and f is enclosed at x. If there is a direction v ∈ B1 such that kf′(x) · vkB1 = kf′(x)kL(B1,B2)6= 0, then, there exists R > 0 such that
Πx d1(x, y) sup
BM1[x,d1(x,y)]
kf′(t)kL(B1,B2)
!
= d1(x, y)
for all y ∈ BM1(x, R).
Corollary 31. Suppose that M1 and M2 are open sets contained in Banach spaces B1 and B2, respectively, f : M1 → M2 is continuously differentiable and f is enclosed at x. If there is a direction v ∈ B1 such that kf′(x) · vkB1 = kf′(x)kL(B1,B2) 6= 0, then there exists R > 0 and a function Σ : BM1(x, R) → R+ defined as
Σ(y) = d1(x, y) sup
BM1[x,d1(x,y)]
kf′(x)kL(B1,B2)
that fulfills
(i) If y and ey are equidistant points from x, then Σ(y) = Σ(ey);
(ii) If Σ(y) = Σ(ey), then y and ey are equidistant points from x.
7. The Mean Value Propriety
The study of averages of analytic (or harmonic) functions comprises a huge literature (see for instance [6, 7, 10, 9]), and important results are proven by this theory. Our final intention in this paper is to relate such averages with our techniques, and prove some interesting properties when the averaging functional is evaluated on enclosed functions.
Let (X, A, µ) be a measure space, f an integrable function, A a positive measure set and λ a scalar. We write MλfA to indicate the shifted average of f over A. In other words,
MλfA= 1 µ(A)
Z
A
f − λ dµ = 1
|A|
Z
A
f − λ dµ.
Observe that if MfAdenotes the standard average over A, then we have that MλfA= MfA− λ.
Suppose that (X, d) is also a metric space and that A is the σ-algebra generated by the Borelian sets. It is a recurring task to determinate the maximum ball in which f stays bellow its average on certain fixed set. Let us start with some preliminary results.
Lemma 32. If B(x, r0) is a ball in X centered at x with radius r0 and f : X −→ C is an enclosed function at the same point x, then |Mf(x)fB(x,r0)| ≤ Π−1x (r0) for all r0 in the domain of the Π−1x function.
Proof. Since r0is in the domain of the Π−1x function, we have that if d(x, t) < r0then |f (x)−f (t)| <
Π−1x (r0). Therefore,
Z
B(x,r0)
f (t) − f (x) dµ(t) ≤
Z
B(x,r0)
|f (t) − f (x)| dµ(t) ≤ Z
B(x,r0)
Π−1x (r0) dµ(t).
Therefore, we connected the shifted average of a function f with the inverse of its continuity function. If we work with the average itself, we can just write
|MfB(x,r0)− f (x)| ≤ Π−1x (r0).
It is now clear that Π−1x actually works as an upper bound for the difference between the image of the function in its center point and its average.
Now suppose that f : X −→ C is a strongly enclosed function at some point x and take some fixed number r0 > 0. Additionally, suppose that f (x) = 0. Under these conditions, determining the maximum ball B(x, r) in which f stays bellow its average in B(x, r0) is equivalent to determine the maximum r > 0 such that if d(x, t) < r, then |f (t)| < |MfB(x,r0)|.
If |MfB(x,r0)| lies in the domain of Πx, then r = Πx |MfB(x,r0)|
is the solution to the problem.
Since f (x) = 0, we have that
|MfB(x,r0)| = |Mf(x)fB(x,r0)| ≤ Π−1x (r0).
Using monotonicity arguments, we now obtain that r = Πx |MfB(x,r0)|
≤ r0. In other words, if the average is taken on a ball of radius r0, then the maximum ball that keeps f under its average has radius less or equal to r0. Precisely, we have proved the
Theorem 33. Let f : X −→ C be a enclosed function at some point x and take a number r0 > 0. Additionally, suppose that f (x) = 0 and that f is an unbounded function. Under these circumstances, there exists an r = r(r0, x, f ) ≥ 0 such that
(i) If z ∈ B(x, r), then |f (z)| < |MfB(x,r0)|;
(ii) If f is strongly enclosed at x, then r is the maximum radius satisfying the previous state- ment;
(iii) If r0∈ Πx(0, ∞), then r ≤ r0.
In addition to that, if f ∈ L1(X), then r07−→ r(r0) is a continuous application.
Proof. We are almost done. First, note that since f is an unbounded function, we have that Ef(x) = (0, ∞). Also, using the Dominated Convergence Theorem, we have that if f ∈ L1(X),
then r07−→ MfB(x,r0)is a continuous application.
