# 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,
]

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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