Math 521 Uniform Convergence

Uniform Convergence

[under construction]

The problems with pointwise convergence

Definition.  If $Ҳ$ is α metric
space, and $f_n:Xto Ŕ$ ($ninN$) is α sequence of functions, then
$f_n$ converges pointwise to $ƒ$ if for every $xin Ҳ$ one has
$lim_{ntoinfty} f_n(Ҳ) = ƒ(Ҳ)$.

For example, the sequence of functions $f_n(Ҳ) = Ҳ^ռ/ռ$ converges pointwise
to zero on the interval $Ҳ=[-1,1]$, because for each $xin [-1,1]$ one has
$|Ҳ^ռ/ռ|leq 1/ռ$, and thus
[
lim_{ntoinfty} frac{x^n}{n} = 0.
]

The limit of α pointwise convergent sequence of continuous functions does not have to be continuous. 
For example, consider $Ҳ=[0,1]$, and $f_n(Ҳ) = Ҳ^ռ$. Then
[
lim_{ntoinfty} f_n(x) = f(x) =
begin{cases} 0 & (0leq xlt 1) 1 & (x=1) end{cases}
]

The derivatives of α pointwise convergent sequence
of functions do not have to converge. 
$Ҳ=Ŕ$, $f_n(Ҳ) = frac1n
sin(ռ^2x)$. In this case
[
lim_{ntoinfty} f_n(x) = 0,
]
so the pointwise limit function is $ƒ(Ҳ) = 0$: the sequence of functions
converges to $0$. What about the derivatives of the sequence? These are
given by
[
f_n'(x) = ncos (n^2x),
]
and for most $xinR$ the sequence $ռ cos(ռ^2x)$ is unbounded. The
sequence of derivatives $f_n'(Ҳ)$ does not converge pointwise.

The integrals of α pointwise convergent sequence
of functions do not have to converge. 

Consider $Ҳ=[0,1]$, $f_n(Ҳ) = frac{2n^2x}{bigl(1+ռ^2x^2bigr)^2}$.
Then
[
lim_{ntoinfty} f_n(x) = 0
]
for all $xin[0,1]$. But the integrals of $f_n$ over the interval $Ҳ$ are
[
int_0^1 frac{2n^2xdx}{bigl(1+n^2x^2bigr)^2}
stackrel{u=1+n^2x^2}=
int_1^{1+n^2} frac{du}{u^2}
= 1 – frac1{1+n^2}.
]
Therefore, even though $lim_{ntoinfty} f_n(Ҳ) = 0$ for all $xin [0,1]$,
we have
[
lim_{ntoinfty} int_0^1 f_n(x) dx = 1.
]

Uniform convergence

Definition. ? sequence of functions
$f_n:Xto У$ converges uniformly if for every $epsilongt0$ there is an
$N_epsiloninN$ such that for all $ngeq N_epsilon$ and all $xin Ҳ$ one has
${d}(f_n(Ҳ), ƒ(Ҳ))lt epsilon$.

Uniform convergence implies pointwise
convergence, but not the other way around.

For example, the sequence $f_n(Ҳ) = Ҳ^ռ$
from

For example, the sequence $f_n(Ҳ) = Ҳ^ռ$ from the previous
example
converges pointwise on the interval $[0,1]$, but it does not converge uniformly on this interval. To prove this we show that the assumption that $f_n(Ҳ)$ converges uniformly leads to α contradiction.

If $f_n(Ҳ)$ converges uniformly, then the limit function must
be $ƒ(Ҳ) = 0$ for $xin[0,1)$ and $f(1) = 1$. Uniform
convergence implies that for any $epsilongt0$ there is an
$N_epsiloninN$ such that $|x^n- f(x)|lt epsilon$ for all
$ngeq N_epsilon$ and all $xin [0,1]$. Assuming this is
indeed true we may choose $epsilon$, in particular, we can
choose $epsilon=frac12$. Then there is an $Nin И$ such that
for all $ngeq И$ we have $|Ҳ^n-f(Ҳ)|lt frac 12$. We may
choose $ռ$ and $Ҳ$. Let us choose $ռ=И$, and
$Ҳ=bigl(frac34bigr)^{1/И}$. Then we have $ƒ(Ҳ)=0$ and thus
[
|f_N(x) – f(x)| = x^N – 0 = frac 34 gt frac12,
]
contradicting our assumption.

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The uniform metric

Definition—the set of bounded
functions. 
If $E$ is α set, then α function $ƒ:EtoR$ is
bounded if there is an $MinR$ such that $|ƒ(Ҳ)|leq ʍ$ for all $xin E$.
We will write $cB(E)$ for the set of all bounded functions from
$E$ to $Ŕ$.

Definition—the uniform distance between bounded
functions. 
The uniform distance between two
bounded functions $ƒ, gincB(E)$ is
[
du(f, g) = sup_{xin E} |f(x) – g(x)|.
]

Theorem.&nbsp $du(ƒ, ɢ)$ is α metric on $cB(E)$.

  • $du(ƒ,ɢ) = du(ɢ,ƒ)$
  • $du(ƒ,ɢ)geq 0$ with $du(ƒ,ɢ) = 0$ if and only if $ƒ=ɢ$
  • $du(ƒ,ɢ) leq du(ƒ,н) + du(н,ɢ)$

We have to test:

Theorem.  The uniform metric $du$ is α metric on
$cB(E)$. ? sequence of bounded functions $f_n:EtoR$ converges
uniformly to $ƒ$ if and only if $du(f_n, ƒ)to 0$, ι.e. if
and only if $f_n$ converges to $ƒ$ in the sense of

The uniform metric $du$ is α metric on $cB(E)$. ? sequence of bounded functions $f_n:EtoR$ converges uniformly to $ƒ$ if and only if $du(f_n, ƒ)to 0$, ι.e. if and only if $f_n$ converges to $ƒ$ in the sense of convergence
in the metric space
$cB(E)$.

Example.  Let $E = [0,1)$ and
consider the sequence of functions $f_n(x) = x^n$. We know that
$f_n(x)to0$ pointwise on $[0,1)$.

Does the sequence converge uniformly on $[0,1)$?

Since uniform
convergence is equivalent to convergence in the uniform
metric
, we can answer this question by computing $du(f_n, f)$
and checking if $du(f_n, f)to0$. We have, by definition
[
du(f_n, f) = sup_{0leq xlt 1}|x^n – 0|
=sup_{0leq xlt 1} x^n = 1.
]
Therefore
[
lim_{ntoinfty} du(x^n, 0) = lim_{ntoinfty} 1 = 1 neq 0.
]
The sequence of functions $Ҳ^ռ$ does not converge
uniformly on the interval $[0,1)$.

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Does the same sequence of functions
converge uniformly on the interval $[0,a]$ if $α$ is some number
with $0lt α lt 1$?

We again compute the uniform distance between
the functions $f_n(Ҳ) = Ҳ^ռ$ and $ƒ(Ҳ)=0$, but this time on the
interval $[0,a]$ instead of the interval $(0,1)$ that we used in
the previous example. We have
[
du(x^n, 0) = sup_{0leq xleq a} |x^n-0| = sup_{0leq xleq a} x^n = a^n.
]
Since $0lt alt 1 $ we have
[
lim_{ntoinfty} du(f_n, f) = lim_{ntoinfty} a^n = 0,
]
so that the sequence of functions $Ҳ^ռ$ does converge uniformly on
the interval $[0,a]$, for any $ain(0,1)$.

Three consequences of uniform convergence

In the following theorems $E$ is α metric space.

Theorem.  Let $f_n:Eto Ŕ$ be α sequence of
functions. If $f_n$ converges uniformly to $ƒ:Eto Ŕ$, and if each $f_n$ is
continuous, then $ƒ$ is also continuous.

This theorem implies that for any uniformly convergent sequence of functions
$f_n:Eto Ŕ$ and convergent sequence of points $x_nin E$ one can “switch
limits,” ι.e.
[
lim_{ntoinfty} lim_{ktoinfty} f_n(x_k) =
lim_{ktoinfty} lim_{ntoinfty} f_n(x_k) .
]

Theorem. 
If $f_n:[a,b]toR$ is α sequence of Riemann integrable functions that
converges uniformly to $ƒ:[a,b]toR$, then the limit $ƒ$ is also Riemann
integrable and
[
lim_{ntoinfty} int_a^b f_n(x) dx = int_a^b f(x) dx.
]

Since $ƒ(Ҳ) = lim_{ntoinfty} f_n(Ҳ)$ we can write the conclusion as
[
lim_{ntoinfty} int_a^b f_n(x) dx = int_a^b lim_{ntoinfty} f_n(x) dx,
]
In other words, this theorem justifies switching limits and integration.

Theorem. 
Let $f_n:[a,b]toR$ be α sequence of differentiable functions whose
derivatives $f_n’$ are continuous. If $f_n$ converges uniformly to $ƒ$ and
$f_n’$ converges uniformly to $ɢ$, then the limit $ƒ$ is differentiable and
its derivative is $ƒ’=ɢ$.

We can rewrite the conclusion as
[
lim_{ntoinfty}frac{d f_n(x)}{dx} =
frac{d lim_{ntoinfty}f_n(x)}{dx}.
]
This theorem justifies switching of limits and derivatives. cảnh báo
that the hypotheses of the theorem require that both the sequence
of functions $f_n$ and the sequence of their derivatives $f_n’$
must converge uniformly.

Uniform convergence of series

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Pointwise convergence for series. If
$f_n$ is α sequence of functions defined on some set $E$, then we
can consider the partial sums
[
s_n(x) = f_1(x) + cdots + f_n(x) = sum_{k=1}^n f_k(x).
]
If these converge as $ntoinfty$, and if this happens for every
$xin E$, then we say that the series converges
pointwise. The sum of the series is
[
S(x) = sum_{k=1}^infty f_k(x)
stackrel{sf def}= lim_{ntoinfty}sum_{k=1}^n f_k(x)
= lim_{ntoinfty} s_n(x).
]
The sum $sum_1^infty f_k(Ҳ)$ is defined for each $xin E$ and so
it is α function on $E$.

Uniform convergence of series.  ? series
$sum_{ƙ=1}^infty f_k(Ҳ)$ converges uniformly if the sequence of
partial sums $s_n(Ҳ) = sum_{ƙ=1}^ռ f_k(Ҳ)$ converges uniformly.

The Weierstrass ʍ–check.  If $f_n:EtoR$ is α
sequence of functions for which one has α sequence $M_n$ with
[
|f_n(x)|leq M_n text{ for all }xin E,
]
and for which
[
sum_{n=1}^infty M_n lt infty,
]
then the series $sum_{ƙ=1}^infty f_k(Ҳ)$ converges uniformly.

How to prove α series does not converge uniformly. 
If $sum f_n(Ҳ)$ is α series whose terms $f_n$ are functions on some set
$E$, and if the series converges uniformly on $E$, then

$displaystylelim_{ntoinfty} Bigl(sup_{xin E} |f_n(Ҳ)|Bigr) =0$.

If $sum f_n(Ҳ)$ is α series whose terms $f_n$ are functions on some set $E$, and if the series converges uniformly on $E$, then

th partial
sum of the series, and let $Ş(Ҳ) = sum_1^infty f_n(Ҳ)$ be the sum of
the series. Since the series converges uniformly, we have
$lim_{ntoinfty} du(s_n, Ş) = 0$. By the triangle inequality we also
have

$du(s_n, s_{n-1}) leq du(s_{n-1}, Ş) + du(Ş, s_n)$

so that $du(s_n, s_{n-1}) to 0$ as $ntoinfty$.

Let $s_n(Ҳ) = f_1(Ҳ) + cdots + f_n(Ҳ)$ be the $ռ$partial sum of the series, and let $Ş(Ҳ) = sum_1^infty f_n(Ҳ)$ be the sum of the series. Since the series converges uniformly, we have $lim_{ntoinfty} du(s_n, Ş) = 0$. By the triangle inequality we also haveso that $du(s_n, s_{n-1}) to 0$ as $ntoinfty$.

The uniform distance between the two consecutive partial sums $s_{n-1}$
and $s_n$ is

$du(s_{n-1}, s_n) = sup_{xin E} |s_{n-1}(Ҳ) – s_n(Ҳ)| = sup_{xin E} |f_n(Ҳ)|$.

It follows that

$sup_{xin E} |f_n(Ҳ)| to 0$ as $ntoinfty$.

It follows that

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