A MEAN VALUE THEOREM FOR METRIC SPACES
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 Dieudonné, [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 Dieudonné, 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
1 The
2 The
author has been partially supported by FAPESP. Process 2013/00594-8.
author has been partially supported by CAPES. Process PNPD 2770/2011.
1
2
P. M. CARVALHO NETO AND P. A. LIBONI FILHO
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 kf ′ (x)kL(B,R) .
x∈D
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
kf ′ (ξ)kB .
ξ∈[x−d,x+d]
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.
A MEAN VALUE THEOREM FOR METRIC SPACES
3
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,x is 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 δ2 is 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,x is 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;
4
P. M. CARVALHO NETO AND P. A. LIBONI FILHO
(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).
A MEAN VALUE THEOREM FOR METRIC SPACES
5
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

(0, |x|], ε ∈ (0, 1] and x 6= 0



 (0, ∞), ε > 1 and x 6= 0
∆f,x (ε) =

∅,
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 = d2 be the induced real Euclidean metric. If f : M1 → M2
is defined as f (x) = 1 − e−x , then
 i
1


 0, ln 1−ε , ε ∈ (0, 1)
∆f,0 (ε) =



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 → M2 and x ∈ M1 such 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 Πx is 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.
6
P. M. CARVALHO NETO AND P. A. LIBONI FILHO
(i) If {εn }∞
n=1 ⊂ Ef (x) is an increasing sequence that converges to ε, then
lim max ∆f,x (εn ) = max ∆f,x (ε);
n→∞
(ii) If {εn }∞
n=1 ⊂ Ef (x) is a decreasing sequence that converges to ε, then
lim max ∆f,x (εn ) = max ∆f,x (ε).
n→∞
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
δe = lim max ∆f,x (εn )
n→∞
and
δ = max ∆f,x (ε)
If δe 6= δ, then the non decreasing property above gives δe < δ. Consider the auxiliary sequence
{δn }∞
n=1 given by
#
"
δe + δ
δ − δe
δn =
+
2
2(n + 1)
and observe that it satisfies the following properties
(i) δn → (δe + δ)/2;
(ii) δe < δn < δ.
e we have that δn 6∈ ∆f,x (εn ).
Since for each n = {1, 2, . . .} the number δn is strictly bigger then δ,
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) ≤ (δe + 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 Πx (ε) = Πx (ε0 ) = lim+ Πx (ε).
t→ε0
ε→ε−
0
We shall begin proving the left limit. Let {e
εn } be sequence to the left of ε0 such that εen → ε0 .
It is not difficult to construct another increasing sequence {εn } with the following proprieties
(i) εn < εen , for each n ∈ N;
(ii) εn → ε0 , when n → ∞.
Then we know that ∆f,x (εn ) ⊂ ∆f,x (e
εn ) ⊂ ∆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 (e
εn ) ≤ max ∆f,x (ε0 ).
Finally, applying the limit when n → ∞ on both sides and using Lemma 10 we obtain
max ∆f,x (ε0 ) ≤ lim max ∆f,x (e
εn ) ≤ max ∆f,x (ε0 ).
n→∞
Since {e
εn } 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.
A MEAN VALUE THEOREM FOR METRIC SPACES
7
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 Πx function. 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
lim Πx (ε) = 0.
ε→0+
Proof. Since f is continuous and non constant, it follows easily (Theorem 6) that 0 is an accumulation 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
!
∞
\
lim Πx (εn ) = diam
∆f,x (εn ) .
n→∞
n=1
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 continuous, 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 considered 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
8
P. M. CARVALHO NETO AND P. A. LIBONI FILHO
(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 ye ∈ M1 and
d1 (x, ye) < d1 (x, y) + 1, then d2 (f (x), f (e
y )) < ε′ .
Once we can suppose that ye = 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.
A MEAN VALUE THEOREM FOR METRIC SPACES
9
Without loss of generality, assume that there exists yeε such that limn→∞ yn,ε = yeε . Using a
limit argument, we have that for all ε ∈ [ε1 , ε2 ], there is some yeε ∈ M1 such that d1 (x, yeε ) ≤ δ
and d2 (f (x), f (yeε )) ≥ ε. If we suppose, for an instant, that d1 (x, yeε ) < δ = Πx (ε), a directly
consequence would be that d2 (f (x), f (yeε )) < ε, which is a contradiction. Because of that, for all
ε ∈ [ε1 , ε2 ], there is some yeε ∈ M1 such that d1 (x, yeε ) = δ and d2 (f (x), f (yeε )) ≥ ε. Since f is
adherent at x, we conclude that
∀ ε ∈ [ε1 , ε2 ], ∃ yeε ∈ M1 such that d1 (x, yeε ) = Πx (ε) and d2 (f (x), f (yeε )) = ε.
In other words, there exists yf
ε2 )) = ε2 .
ε2 ) = Πx (ε2 ) and d2 (f (x), f (yf
ε2 ∈ M1 such that d1 (x, yf
Since f is adherent at x and d1 (x, yf
ε2 )) ≤ ε1
ε2 ) = Πx (ε1 ), we conclude that ε2 = d2 (f (x), f (yf
what implies that ε1 = ε2 . Therefore, Πx is 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 ∈ M1 such 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)
(ii)
(iii)
(iv)
f is continuous;
f is adherent at x;
f is not locally constant at x;
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)
(ii)
(iii)
(iv)
0 ∈ Ix ⊂ R+ ;
Πx is a monotonic increasing homeomorphism;
f (BM1 (x, δ)) ⊂ BM2 (f (x), ε) ⇐⇒ δ ≤ Πx (ε), for all ε ∈ Ix ;
Πx (0) = 0.
10
P. M. CARVALHO NETO AND P. A. LIBONI FILHO
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 enclosed 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 C 1 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
d2 (f (x), f (yr ))
d2 (f (x), f (y))
=
.
d1 (x, y)
d1 (x, yr )
y∈BM1 [x,r]\{x}
sup
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
|1 − e−y |
1 − er
d2 (f (0), f (−r))
Γ(r) =
sup
=
=
.
|y|
−r
d1 (0, −r)
y∈BR [0,r]\{0}
In other words, f satisfies the Lagrange Propriety at 0 with Γ(r) =
er −1
r .
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 C k diffeomorphism, provided that Γ is a C k application.
Proof. Let ∆ : (−ζ, ζ) → R be given by ∆(t) = t Γ(t). Since ∆ is a C k application and ∆′ (0) is an
isomorphism, we can use the Inverse Function Theorem and assume that ∆|(−ζ1 ,ζ1 ) : (−ζ1 , ζ1 ) →
(−ζ2 , ζ2 ) is a C k diffeomorphism.
A MEAN VALUE THEOREM FOR METRIC SPACES
11
Let ∆−1 be the inverse map of ∆|(−ζ1 ,ζ1 ) . Then, ∆−1 (ε)Γ(∆−1 (ε)) = ε, which implies that
∆−1 (ε) =
ε
d2 (f (x), f (y))
d1 (x, y)
−1
y∈BM1 [x,∆ (ε)]\{x}
.
sup
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)) −1
∆ (ε) = ε.
d1 (x, y)
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 C 1 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 C k diffeomorphism, the proof is finished.
12
P. M. CARVALHO NETO AND P. A. LIBONI FILHO
Corollary 25. Assuming the last theorem hypothesis, we also have that
ε
ε
,
=
Πx (ε) =
d2 (f (x), f (y))
Γ(ξ(ε))
sup
d1 (x, y)
y∈BM1 [x,ξ(ε)]\{x}
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
1
ε2
≈ε+
+ O(ε3 ).
1−ε
2
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).
r
We also know that f satisfies the Lagrange Propriety with Γ(r) = 1−e
−r . Under these conditions,
ξ satisfies the following system

 ξ(0) = 0
1
 ξ ′ (ε Γ(ε)) =
Γ(ε) + εΓ′ (ε)
Π0 (ε) = ln
for sufficiently small ε. Therefore ξ(ε) = ln(1 + ε) and
Π0 (ε) = ln(1 + ε) ≈ ε −
ε2
+ O(ε3 ).
2
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 hypothesis, suppose that M1 is an open set of B1 and 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
ε
≤ Πx (ε) ≤
M
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−1
x is differentiable at 0 ;
(ii) f satisfies the MVI (see Definition 1) for metric spaces in some BM1 (x, R).
A MEAN VALUE THEOREM FOR METRIC SPACES
13
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

Π−1

x (d)

, if d ∈ Jx \ {0}
d
Ψ(d) =

 (Π−1 )′ (0+ ), if d = 0,
x
where Jx = Πx (Ix ).
Observe that Ψ is continuous and that the function d 7→ Ψ(d)d is a non decreasing homeomorphism. 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−1
x on both sides and taking the limit as ε → 0, we achieve
d2 (f (x), f (y)) ≤ Π−1
x (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−1
x (d1 (x, y)) .
−1
Since d1 (x, y) < Πx (ε) and Πx is strictly increasing, we obtain that Π−1
x (d1 (x, y)) < Πx (Πx (ε)) =
ε, i.e., d2 (f (x), f (y)) < ε.
Finally, to prove the last statement, assume that M1 and M2 are 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, ye) < d1 (x, y),
we have that
d2 (f (x), f (e
y )) ≤ d1 (x, ye) supBM1 [x,d1 (x,ey)] kf ′ (t)kL(B1 ,B2 )
≤
d1 (x, ye) 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−1
x (d1 (x, y)) ≤ d1 (x, y)
sup
BM1 [x,d1 (x,y)]
kf ′ (t)kL(B1 ,B2 ) ,
14
P. M. CARVALHO NETO AND P. A. LIBONI FILHO
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−1
x at 0 is the natural substitute for f being differentiable on its domain. Also note that, since Ψ is a continuous function and supBM1 [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 B1 and 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 Ix and Jx where 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) supBM1 [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 )
which implies
Πx
d1 (x, y)
sup
BM1 [x,d1 (x,y)]
′
kf (t)kL(B1 ,B2 )
< d1 (x, y) + ε,
!
≤ 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−1
x (d1 (x, y)) ,
which implies that Ψ(d1 (x, y)) ≥ supBM1 [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→ supBM1 [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−1
L(B1 ,B2 ) supBM1 [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).
A MEAN VALUE THEOREM FOR METRIC SPACES
15
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
ε
< d1 (x, y) + ε.
ye = x + h. Then d1 (x, ye) = λy < d1 (x, y) + M
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) −
y ) − f (x)kB2 ≥ d1 (x, y) supBM1 [x,d1 (x,y)] kf ′ (t)kL(B1 ,B2 )
f (x) − f ′ (x) · hkB2 is small enough, so kf (e
even if d1 (x, ye) < 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 ye are equidistant points from x, then Σ(y) = Σ(e
y );
(ii) If Σ(y) = Σ(e
y ), then y and ye 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 Mλ fA to indicate the shifted average of f over A. In other words,
Z
Z
1
1
Mλ fA =
f − λ dµ =
f − λ dµ.
µ(A) A
|A| A
16
P. M. CARVALHO NETO AND P. A. LIBONI FILHO
Observe that if MfA denotes the standard average over A, then we have that
Mλ fA = 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
−1
function at the same point x, then |Mf (x) fB(x,r0 ) | ≤ Π−1
x (r0 ) for all r0 in the domain of the Πx
function.
Proof. Since r0 is in the domain of the Π−1
x function, we have that if d(x, t) < r0 then |f (x)−f (t)| <
−1
Πx (r0 ). Therefore,
Z
Z
Z
f (t) − f (x) dµ(t) ≤
|f (t) − f (x)| dµ(t) ≤
Π−1
x (r0 ) dµ(t).
B(x,r0 )
B(x,r0 )
B(x,r0 )
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−1
x (r0 ).
It is now clear that Π−1
x 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−1
x (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 statement;
(iii) If r0 ∈ Πx (0, ∞), then r ≤ r0 .
In addition to that, if f ∈ L1 (X), then r0 7−→ 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 r0 7−→ MfB(x,r0 ) is a continuous application.
A MEAN VALUE THEOREM FOR METRIC SPACES
17
We can also state the following version of the theorem, which is a trivial consequence of the
previous one.
Theorem 34 (The Mean-Value Property for Enclosed Functions). Let f : X −→ C be a enclosed
function at some point x and take a number r0 > 0. Additionally, suppose 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) − f (x)| < |MfB(x,r0 ) − f (x)|;
(ii) If f is strongly enclosed at x, then r is the maximum radius satisfying the previous statement;
(iii) If r0 ∈ Πx (0, ∞), then r ≤ r0 .
In addition to that, if f ∈ L1 (X), then r0 7−→ r(r0 ) is a continuous application.
Some remarks are in order. First of all, the last theorem can be seen as a version of the Mean
Value Property. In harmonic functions, one can expect that the image of the center of the ball is
given by the average value of the function in the interior of the ball. In this version, for enclosed
functions, we have shown that the image of any point which lies in a certain set is strictly small
than the average taken on some larger ball.
Also note that we call it a version — not a generalization — because of the following: suppose that f is a harmonic function. Under the assumptions of the last theorem, we have that
|MfB(x,r0 ) − f (x)| = 0 and r = Πx (0) = 0. This is a natural and expected phenomenon, since
harmonic functions behave much more nicely with respect to averages than enclosed functions.
To conclude, observe that the unboundedness of the function should not pose a threat to the
usability of the last theorem. Let us consider the following two cases.
First, suppose that we have Πx (0, ∞) = (0, ∞) and r0 ≤ R, for some R > 0. In this case,
if there exists an unbounded function F such that F |B(x,R) = f , then we can obtain the same
conclusions.
On the other hand, if we do not have any control over Πx (0, ∞), but R is small enough, then
we can still claim the same conclusions.
Corollary 35. 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 |MfB(x,r) | ≤ M|f |B(x,r) ≤ |MfB(x,r0 ) |.
Acknowledgement The authors would like to thank the anonymous referee for his/her time and
valuable comments. Moreover, the first author would like to thank the Federal University of São
Carlos (UFSCar) for the hospitality and support during a short term visit in São Carlos.
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18
P. M. CARVALHO NETO AND P. A. LIBONI FILHO
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(P. M. Carvalho Neto) Instituto de Matemática, Estatı́stica e Computação Cientı́fica, Universidade
Estadual de Campinas, Rua Sérgio Buarque de Holanda 651, Campinas, Brazil.
E-mail address: pmat@ime.unicamp.br
(P. A. Liboni Filho) Departamento de Matemática, Universidade Federal de São Carlos, Rod. Washington Luı́s, Km 235, São Carlos, Brazil.
E-mail address: liboni@dm.ufscar.br