# Uniform Convergence

The uniform convergence of a continuation of functions $\left(f_ \left\{N\right\}\right) _ \left\{N \ in \ mathbb \left\{NR\right\}\right\}$ is a form of convergence more demanding than the simple Convergence. This last request indeed only that, for each point $x$, the continuation $\left(f_ \left\{N\right\} \left(X\right)\right)_ \left\{N \ in \ mathbb \left\{NR\right\}\right\}$ has a limit. Convergence becomes uniform when all the continuations $\left(f_ \left\{N\right\} \left(X\right)\right)_ \left\{N \ in \ mathbb \left\{NR\right\}\right\}$ advances towards their respective limit with a kind of “overall movement”.

In the case of numerical functions of a variable, the concept takes a obviously geometrical form: the graph of the function $f_n$ “approaches” that of the limit.

## Definition

### Uniform convergence

• Is $X \, \!$ a topological Space, $\left(Y, d\right) \,$ a metric Space and $A \ subset X$ a subset of $X \,$.

That is to say $\left(f_ \left\{N\right\}\right) _ \left\{N\right\} \, \!$ a succession of functions defined on $X \, \!$ and with values in $Y \, \!$ and $f \, \!$ a function defined on $X \, \!$ with values in $Y \, \!$. It is said that the continuation $\left(f_ \left\{N\right\}\right) _ \left\{N\right\} \,$ converges uniformly towards $f \,$ on $A \,$ if:
$\left(1\right) \ quad \ forall \ varepsilon > 0, \ exists N_ \left\{\ varepsilon\right\} \ in \ NR, \ forall N \ in \ NR, N \ Ge N_ \left\{\ varepsilon\right\} \ Rightarrow \ forall X \ in has, D \left(f_ \left\{N\right\} \left(X\right), F \left(X\right)\right) < \ varepsilon$

Note:: the proposal $\left(1\right)$ is equivalent to:

$\ forall \ varepsilon > 0, \ exists N_ \left\{\ varepsilon\right\} \ in \ NR, \ forall N \ in \ NR, N \ Ge N_ \left\{\ varepsilon\right\} \ Rightarrow \ sup_ \left\{X \ in has\right\} \left(D \left(f_ \left\{N\right\} \left(X\right), F \left(X\right)\right) < \ varepsilon$

### Some explanations

One can ask for a posteriori which is the difference between the simple Convergence of a continuation in functions and the uniform convergence . Indeed, the continuation of functions $\left(f_ \left\{N\right\}\right) _ \left\{N\right\} \,$ converges simply towards $f \,$ on $A \,$ if:

$\ forall X \ in X, \ forall \ varepsilon > 0, \ exists N_ \left\{\ varepsilon, X\right\} \ in \ NR, \ forall N \ in \ NR, N \ Ge N_ \left\{\ varepsilon, X\right\} \ Rightarrow D \left(f_ \left\{N\right\} \left(X\right), F \left(X\right)\right) < \ varepsilon$
Here, the index $N_ \left\{\ varepsilon, X\right\} \,$ depends on $x \ in has \,$ whereas in the proposal $\left(1\right) \,$, the index $N_ \left\{\ varepsilon\right\} \,$ does not depend on it. This difference can appear alleviating to the uninitiated persons, but it is however essential:
• In the case of the simple convergence , for any element $x \ in has \,$, one can find a row from which the distance $D \left(f_ \left\{N\right\} \left(X\right), F \left(X\right)\right)\,$ becomes very small. A priori , if one chooses a $\ in has \ there,$ other than X then the row from which the distance $D \left(f_ \left\{N\right\} \left(there\right), F \left(there\right)\right)\,$ becomes very small will be different.

• In the case of the uniform convergence , one can find a row from which the distance $d \left(f_ \left\{N\right\} \left(X\right), F \left(X\right)\right) \,$ becomes very small for any $x \ in has \,$ at the same time. This condition is thus much stronger. In particular, a succession of functions which converges uniformly on a unit converges simply on this one. The reciprocal one is in general false except in very particular cases (see Théorèmes of Dini).

## Uniform criterion of Cauchy

Now, one supposes in more than the metric Espace $\left(Y, d\right) \,$ is a complete Espace. It is the case of good number of metric spaces, such as for example of $\left(\ R,| \ cdot |\right) \,$ real line provided with its absolute Value or more generally of all Space of Banach.

Under these conditions, one shows that a succession of functions $\left(f_ \left\{N\right\}\right) _ \left\{N\right\} \,$ converges uniformly on $A \,$ if and only if it checks the uniform criterion of Cauchy , namely:

$\ forall \ varepsilon >0, \ exists N_ \left\{\ varepsilon\right\} \ in \ NR, \ forall p \ in \ NR, \ forall Q \ in \ NR, \left(p, Q \ Ge N_ \left\{\ varepsilon\right\}\right) \ Rightarrow \ forall X \ in has, D \left(f_ \left\{p\right\} \left(X\right), f_ \left\{Q\right\} \left(X\right)\right) < \ varepsilon$

As in the case of the series Cauchy, it is not necessary of exhiber the function towards which a succession of functions tends to show that convergence is uniform.

## Uniform convergence of functions to values in a normalized vector space

It is supposed now that $X \,$ is a metric Espace and that $\left(Y,||\ cdot||\right)$ is a vector Space normalized: it is a metric Espace from which topology is resulting from the distance $d \,$ such as:

$\ forall there \ in Y, \ forall y\text{'} \ in Y, D \left(there, y\text{'}\right) =||there there \text{'}||$.
The uniform convergence of a succession of functions $\left(f_ \left\{N\right\}\right) _ \left\{N\right\} \,$ on a part $A \,$ included in $X \,$ is thus written:
$\ forall \ varepsilon > 0, \ exists N_ \left\{\ varepsilon\right\} \ in \ NR, \ forall N \ in \ NR, N \ Ge N_ \left\{\ varepsilon\right\} \ Rightarrow \ forall X \ in has, ||f_ \left\{N\right\} \left(X\right) - F \left(X\right)|| < \ varepsilon$
What is still equivalent to:
$\ forall \ varepsilon > 0, \ exists N_ \left\{\ varepsilon\right\} \ in \ NR, \ forall N \ in \ NR, N \ Ge N_ \left\{\ varepsilon\right\} \ Rightarrow \ sup_ \left\{X \ in has\right\} \left( ||f_ \left\{N\right\} \left(X\right) - F \left(X\right)||\right) < \ varepsilon$

### Theorems

There is the following fundamental result:

If $\left(f_ \left\{N\right\}\right) _ \left\{N\right\} \,$ is a succession of continuous functions converging uniformly on $X \,$ towards a function $f \,$ then $f \,$ is continuous on $X \,$.

Proof. Is $\ epsilon >0 \,$ given. There exists an entirety $N \,$ such as, for all $x \ in X \,$, $d \ big \left(f_N \left(X\right), F \left(X\right) \ big\right) \ the \ epsilon$. The function $f_N \,$ is continuous in any point $a \ in X \,$. There exists thus open a $U \,$ container $a \,$ such as $d \ big \left(f_N \left(X\right), f_N \left(a\right) \ big\right) \ the \ epsilon$ for all $x \ in U \,$. Then, if $x \ in U \,$,

$d \ big \left(F \left(X\right), F \left(a\right) \ big\right) \ D \ big \left(F \left(X\right), f_N \left(X\right) \ big\right) +d \ big \left(f_N \left(X\right), f_N \left(a\right) \ big\right) + D \ big \left(f_N \left(a\right), F \left(a\right) \ big\right) \ 3 \ epsilon$

When $X \,$ is not compact, uniform convergence is a rare phenomenon. For example, $\left(1+ \ frac \left\{Z\right\} \left\{N\right\}\right) ^ \left\{N\right\}$ converges uniformly towards $e^z \,$ on all compact of $\ mathbb \left\{C\right\} \,$ when the entirety $n \,$ tends towards the infinite one, but not on $\ mathbb \left\{C\right\} \,$; one whole Series of ray of convergence $R \,$ converges uniformly on very compact open disc of center 0 and ray $R \,$, but one cannot say better in general.

In fact, continuity being a local property, uniform convergence on " suffisamment" parts of $X \,$ is enough to ensure the continuity of the function limite.

Examples

• When $X \,$ is Localement compact, or when its topology is defined by metric.
Under these conditions, if a continuation $\left(f_n\right) _ \left\{, N \ Ge 0\right\}$ of continuous functions converges uniformly on very compact towards a function $f \,$, then $f \,$ are continuous.
• One with the same conclusion when $X \,$ is a space of Banach, if uniform convergence takes place
on any closed ball of center $0 \,$. Thus one shows for example the continuity of the function Exponentielle in a Algebra of Banach.

The following result, less extremely than the Theorem of dominated convergence, is also much less difficult with montrer.

If $X= \,$ is a Intervalle of $\ R$, if $Y= \ R$ or $Y= \ mathbb \left\{C\right\}$, then if a succession of functions $\left(f_ \left\{N\right\}\right) _ \left\{N\right\} \,$ integrable converges uniformly towards a function $f \,$ integrable then: $\ lim_ \left\{N \ rightarrow + \ infty\right\} \ int_ \left\{has\right\} ^ \left\{B\right\} f_ \left\{N\right\} \left(X\right) .dx = \ int_ \left\{has\right\} ^ \left\{B\right\} F \left(X\right) .dx$.

Its use is at the base of the following result of Analyze complexes.

Is $\left(f_n\right) _ \left\{N \ Ge 0\right\} \,$ a succession of holomorphic functions on open of $U \ subset \ mathbb \left\{C\right\} \,$, converging uniformly on very compact of $U \,$ towards a function $f \,$. Then $f \,$ is holomorphic.

### Notation

The following notation is introduced: $\ forall has \ subset X, \ forall F: X \ rightarrow Y, ||F||_ \left\{\ infty, has\right\} = \ sup_ \left\{X \ in has\right\} \left( ||F \left(X\right)|| \right)$

It follows directly that a succession of functions $\left(f_ \left\{N\right\}\right) _ \left\{N\right\} \,$ converges uniformly towards a function $f \,$ if and only if:

$\ lim_ \left\{N \ rightarrow + \ infty\right\} ||f_ \left\{N\right\} - F||_ \left\{\ infty, has\right\} =0$

$\ triangle$: $||\ cdot||_ \left\{\ infty, has\right\}$ is in general not a standard on the vector Space of the functions of $A \,$ with values in $Y \,$.

### Case where X is compact

One supposes from now on that X is a metric Espace compact, $\left(Y,||\ cdot||\right)$ being always a vector Space normalized. One notes $\ mathcal \left\{C\right\} \left(X, Y\right)$ the whole of the continuous functions definite on $X \,$ and with values in $Y \,$.

Then: $\left(\ mathcal \left\{C\right\} \left(X, Y\right),||\ cdot||_ \left\{\ infty, X\right\}\right)$ is a vector Space normalized. So moreover, $Y \,$ are complete then $\ mathcal \left\{C\right\} \left(X, Y\right)$ is him also complete.

## Space continuous numerical functions on

One chooses in this section $X= \,$ a compact interval of $\ R$ and $Y= \ R$. Since $\ R$ provided with the absolute value is complete, it results from it that the vector Space normalized $\ mathcal \left\{C\right\} \left(, \ R\right)$ provided with the standard $||\ cdot||_ \left\{\ infty,\right\}$ is complete.

### Theorem of Weierstrass

The theorem of Weierstrass affirms that one can approach in a uniform way any function numerical continues on $\,$ by a succession of very regular functions to knowing by polynomials. More precisely, if $f \,$ is a continuous function on $\,$ then:

$\ forall \ varepsilon>0, \ exists P_ \left\{\ varepsilon\right\} \ in \ R, ||f-P_\left\{\varepsilon\right\}||_ \left\{\ infty,\right\} \ Leq \ varepsilon$.
where $\ R$ indicates the whole of the polynomials with real coefficients.

## See too

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