Uniform Convergence

The uniform convergence of a continuation of functions $(f\_\; \{N\})\; \_\; \{N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\}$ is a form of convergence more demanding than the simple Convergence. This last request indeed only that, for each point $x$, the continuation $(f\_\; \{N\}\; (X))\_\; \{N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\}$ has a limit. Convergence becomes uniform when all the continuations $(f\_\; \{N\}\; (X))\_\; \{N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\}$ 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 $X\; \backslash ,\; \backslash !$ a topological Space, $(Y,\; d)\; \backslash ,$ a metric Space and $A\; \backslash \; subset\; X$ a subset of $X\; \backslash ,$.

That is to say $(f\_\; \{N\})\; \_\; \{N\}\; \backslash ,\; \backslash !$ a succession of functions defined on $X\; \backslash ,\; \backslash !$ and with values in $Y\; \backslash ,\; \backslash !$ and $f\; \backslash ,\; \backslash !$ a function defined on $X\; \backslash ,\; \backslash !$ with values in $Y\; \backslash ,\; \backslash !$. It is said that the continuation $(f\_\; \{N\})\; \_\; \{N\}\; \backslash ,$ converges uniformly towards $f\; \backslash ,$ on $A\; \backslash ,$ if:

Note:: the proposal $(1)$ is equivalent to:

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 $(f\_\; \{N\})\; \_\; \{N\}\; \backslash ,$ converges simply towards $f\; \backslash ,$ on $A\; \backslash ,$ if:

Here, the index $N\_\; \{\backslash \; varepsilon,\; X\}\; \backslash ,$ depends on $x\; \backslash \; in\; has\; \backslash ,$ whereas in the proposal $(1)\; \backslash ,$, the index $N\_\; \{\backslash \; varepsilon\}\; \backslash ,$ 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 $x\; \backslash \; in\; has\; \backslash ,$, one can find a row from which the distance $D\; (f\_\; \{N\}\; (X),\; F\; (X))\backslash ,$ becomes very small. A priori , if one chooses a $\backslash \; in\; has\; \backslash \; there,$ other than X then the row from which the distance $D\; (f\_\; \{N\}\; (there),\; F\; (there))\backslash ,$ becomes very small will be different.

• In the case of the uniform convergence , one can find a row from which the distance $d\; (f\_\; \{N\}\; (X),\; F\; (X))\; \backslash ,$ becomes very small for any $x\; \backslash \; in\; has\; \backslash ,$ 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 $(Y,\; d)\; \backslash ,$ is a complete Espace. It is the case of good number of metric spaces, such as for example of $(\backslash \; R,|\; \backslash \; cdot\; |)\; \backslash ,$ 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 $(f\_\; \{N\})\; \_\; \{N\}\; \backslash ,$ converges uniformly on $A\; \backslash ,$ if and only if it checks the uniform criterion of Cauchy , namely:

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

$\backslash \; forall\; there\; \backslash \; in\; Y,\; \backslash \; forall\; y\text{'}\; \backslash \; in\; Y,\; D\; (there,\; y\text{'})\; =||there\; there\; \text{'}||$.

The uniform convergence of a succession of functions $(f\_\; \{N\})\; \_\; \{N\}\; \backslash ,$ on a part $A\; \backslash ,$ included in $X\; \backslash ,$ is thus written:

What is still equivalent to:

Theorems

There is the following fundamental result:

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

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

$d\; \backslash \; big\; (F\; (X),\; F\; (a)\; \backslash \; big)\; \backslash \; D\; \backslash \; big\; (F\; (X),\; f\_N\; (X)\; \backslash \; big)\; +d\; \backslash \; big\; (f\_N\; (X),\; f\_N\; (a)\; \backslash \; big)\; +\; D\; \backslash \; big\; (f\_N\; (a),\; F\; (a)\; \backslash \; big)\; \backslash \; 3\; \backslash \; epsilon$

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

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

Examples

• When $X\; \backslash ,$ is Localement compact, or when its topology is defined by metric.

Under these conditions, if a continuation $(f\_n)\; \_\; \{,\; N\; \backslash \; Ge\; 0\}$ of continuous functions converges uniformly on very compact towards a function $f\; \backslash ,$, then $f\; \backslash ,$ are continuous.

• One with the same conclusion when $X\; \backslash ,$ is a space of Banach, if uniform convergence takes place

on any closed ball of center $0\; \backslash ,$. 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 $X=\; \backslash ,$ is a Intervalle of $\backslash \; R$, if $Y=\; \backslash \; R$ or $Y=\; \backslash \; mathbb\; \{C\}$, then if a succession of functions $(f\_\; \{N\})\; \_\; \{N\}\; \backslash ,$ integrable converges uniformly towards a function $f\; \backslash ,$ integrable then: $\backslash \; lim\_\; \{N\; \backslash \; rightarrow\; +\; \backslash \; infty\}\; \backslash \; int\_\; \{has\}\; ^\; \{B\}\; f\_\; \{N\}\; (X)\; .dx\; =\; \backslash \; int\_\; \{has\}\; ^\; \{B\}\; F\; (X)\; .dx$.

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

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

Notation

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

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

$\backslash \; lim\_\; \{N\; \backslash \; rightarrow\; +\; \backslash \; infty\}\; ||f\_\; \{N\}\; -\; F||\_\; \{\backslash \; infty,\; has\}\; =0$

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

Case where X is compact

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

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

Space continuous numerical functions on

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

Theorem of Weierstrass

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

$\backslash \; forall\; \backslash \; varepsilon>0,\; \backslash \; exists\; P\_\; \{\backslash \; varepsilon\}\; \backslash \; in\; \backslash \; R,\; ||f-P\_\{\backslash varepsilon\}||\_\; \{\backslash \; infty,\}\; \backslash \; Leq\; \backslash \; varepsilon$.

where $\backslash \; R$ indicates the whole of the polynomials with real coefficients.

See too

• simple Convergence
• Theorem of Stone-Weierstrass

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