A Remark on the Uniform Convergence of Some Sequences of Functions ()
1. Introduction
Let X be a nonempty set,
be a function and
be a sequence of real-valued functions from X into
. Recall [1] - [3] that the sequence
is said to converge uniformly to f on X, if

Obviously, if
converges uniformly to f on X, then for each
fixed, the sequence
converges to
; that is,
converges pointwise to f. It is also obvious that when X is finite and
converges pointwise to f on X, then
converges uniformly to f on X. However this converse
doesn’t hold in general for an arbitrary (infinite) set X; i.e., the pointwise convergence may not imply the uniform convergence when X is an arbitrary (infinite) set.
One can observe that in the mathematical literature, there are very few known results that give conditions under which a pointwise convergence implies the uniform convergence. Concerning sequences of continuous functions defined on a compact set, we have the following facts:
Proposition A. (Dini’s Theorem) [4]
If K is a compact metric space,
a continuous function, and
a monotone sequence of continuous functions from K into
that converges pointwise to f on K, then
converges uniformly to f on K.
Proposition B. [5]
If E is a Banach space and
is a sequence of bounded linear operators of E that converges pointwise to a bounded linear operator T of E, then for every compact set
,
converges uniformly to T on K.
(For the sake of completeness, we give the proof of this proposition in the Appendix Section).
Therefore our aim is to highlight a new basic criterion that shows in some way how a sequence of real-valued functions can converge uniformly when it is more or less obvious that the sequence converges uniformly away from a finite number of points of the closure of its domain. In the case of sequences of functions of a real variable, our criterion avoids, unlike in most classical textbooks [3] [6] , the search of extrema (by differential calculus) of their general terms. Several examples that satisfy the criterion are given.
2. The Main Result (Remark)
2.1. Theorem
Let
be a metric space and
be a subset of E. Consider a sequence
of functions defined from
to
.
Suppose that there exists a function f from
to
, some points
, some positive real numbers
and a nonnegative constant M such that
(D)
Suppose furthermore that for each
,
converges uniformly to f on
; where
denotes the open ball of E centered at
and with radius
.
Then the sequence of functions
converges uniformly to f on
.
Proof
Let
be arbitrarily fixed (it may be sufficiently small in order to be meaningful). Then for every natural number n, we have
![]()
Thus
![]()
by the uniform convergence of
on
.
And so
![]()
i.e.,
![]()
2.2. Observation
The boundedness condition (D) of the above theorem can not be removed as shown by the sequence of functions defined from
into
as follows:
![]()
where
is equipped with its standard metric. Indeed,
converges uniformly to 0 on
for each
, but with
and
there is no positive number r for which the condition (D) is satisfied since
![]()
And we can see that
does not converge uniformly to 0 on
since
![]()
3. Examples
We give some examples that illustrate the theorem.
(1) Let
be an infinite metric space and let
be fixed. Denote by
the function defined from E into
by
![]()
Then the sequence of functions
defined by
![]()
converges uniformly to
on E.
(2) Given an infinite metric space
,
and
, we have that
i) the sequence of functions
defined by
![]()
converges uniformly to 0 on E,
ii) the sequence of functions
defined by
![]()
converges uniformly to 0 on E.
(3) Let
be an infinite metric space and
be a bounded and infinite subset of E, let a and b be two different points of
and let
and
be two fixed positive numbers.
i) Consider the sequence of functions
defined by
![]()
Then
converges uniformly to 0 on
.
ii) Consider the sequence of functions
defined by
![]()
Then
converges uniformly to 0 on
.
iii) Consider the sequence of functions
defined by
![]()
Then
converges uniformly to 0 on
.
(4) In real analysis, we can recover the facts that each of the following sequences converges uniformly to 0 on their respective domains:
![]()
Justifications (Proofs) of the examples
(1) For every
, we have
![]()
Therefore, on the one hand, for each
, we have
![]()
showing that
converges uniformly to
on
.
On the other hand, we have
![]()
fulfilling condition (D) of the above theorem.
Thus
converges uniformly to
on E.
(2) i) On the one hand, for each
, we have for all
and for all
with
:
![]()
and so
converges uniformly to 0 on
.
On the other hand, we have
![]()
fulfilling condition (D) of the above theorem.
Thus
converges uniformly to 0 on E.
ii) The uniform convergence of
, follows that of
since
![]()
Observe that the uniform convergence of
could also be proved using directly the above theorem.
(3) Note that for all natural number n, we have
![]()
because
![]()
following from
![]()
Therefore it suffices to prove that
converges uniformly to 0 on
, although each of these three sequences can be handled directly with the above theorem.
Let
be the diameter of
.
Then on the one hand, for each
, we have for all
and for all
:
![]()
and so
converges uniformly to 0 on
.
On the other hand, we have
![]()
showing condition (D) of the above theorem.
Thus
converges uniformly to 0 on
and we are done.
(4) i) Let us set
with
.
On the one hand, we have for every
:
![]()
On the other hand, we have for every
:
![]()
showing that
converges uniformly to 0 on
.
Therefore, by taking
,
,
,
,
and
, the above theorem implies that
converges uniformly to 0 on
.
ii) For
with
.
On the one hand, we have for every
:
![]()
On the other hand, we have for every
:
![]()
showing that
converges uniformly to 0 on
.
Therefore, by taking
,
,
,
,
and
, the above theorem implies that
converges uniformly to 0 on
.
iii) For
with
.
On the one hand, we have for every
:
![]()
On the other hand, we have for every
:
![]()
showing that
converges uniformly to 0 on
since
.
Therefore, by taking
,
,
,
,
and
, the above theorem implies that
converges uniformly to 0 on
.
iv) For
with
.
On the one hand, we have for every
:
![]()
On the other hand, we have for every
:
![]()
showing that
converges uniformly to 0 on
since
.
Therefore, by taking
,
,
,
,
and
, the above theorem implies that
converges uniformly to 0 on
.
v) The example of
with
, is a particular case of Example (2)-ii) above with
,
for all
,
and
.
Appendix
In this section, we prove Proposition B for the sake of completeness.
Proof of Proposition B
Let
be given. By the Uniform Boundedness Principle, we have that
. So let
. Then there exist
such that
.
Also,
. We have that
![]()
It follows that
and therefore
![]()
.