There are two kinds of convergence for a sequence of functions described in the next definition, pointwise
convergence and uniform convergence. Of the two, uniform convergence is far better and tends to be the
kind of convergence most encountered in complex analysis. Pointwise convergence is more often encounted
in real analysis and necessitates much more difficult theorems.
Definition 2.2.30Let S ⊆ ℂ^{p}and let f_{n} : S → ℂ^{q}for n = 1,2,
⋅⋅⋅
. Then
{fn }
is said to convergepointwise to f on S if for all x ∈ S,
fn(x) → f (x)
for each x. The sequence is said to converge uniformly to f on S if
( )
lim sup|f (x)− f (x)| = 0
n→ ∞ x∈S n
sup_{x∈S}
|fn(x)− f (x)|
is denoted as
∥fn − f∥
_{∞}or just
∥fn − f∥
for short.
∥⋅∥
is called the uniform norm.
To illustrate the difference in the two types of convergence, here is a standard example.
Example 2.2.31Let
{
0 if x ∈ [0,1)
f (x) ≡ 1 if x = 1
Also let f_{n}
(x )
≡ x^{n}for x ∈
[0,1]
. Then f_{n}converges pointwise to f on
[0,1]
but does not convergeuniformly to f on
[0,1]
.
Note how the target function is not continuous although each function in the sequence is. The next
theorem shows that this kind of loss of continuity never occurs when you have uniform convergence. The
theorem holds generally when S ⊆ X a normed linear space and f,f_{n} have values in Y another normed
linear space. You should fill in the details to be sure you understand this. You simply replace
|⋅|
with
∥⋅∥
for an appropriate norm.
Theorem 2.2.32Let f_{n} : S → ℂ^{q}be continuous and let f_{n}converge uniformly to f on S. Then iff_{n}is continuous at x ∈ S, it follows that f is also continuous at x.
Proof: Let ε > 0 be given. Let N be such that if n ≥ N, then
sup|fn (y )− f (y)| ≡ ||fn − f||∞ < ε
y∈S 3
Pick such an n. Then by continuity of f_{n} at x, there exists δ > 0 such that if
Definition 2.2.33Let f_{n} : S → ℂ^{q}. Then f_{n}is said to be uniformly Cauchy if for every ε > 0, thereexists N such that if m,n > ε, then if m,n ≥ N, then
so, since x is arbitrary, it follows each f_{n} is bounded since
|fn(x)|
is bounded above by
max
{CN + ε∕2,Ck,k ≤ N }
. Also, for n ≥ N,
|fn(x)− f (x)| ≤ |fn (x)− fm (x)|+ |fm (x )− f (x)| < ε∕2 + |fm (x )− f (x)|
Now, take the limit as m →∞ on the right to obtain that for all x,
|fn(x)− f (x)| ≤ ε∕2
Therefore,
∥fn − f∥
< ε if n ≥ N so, since ε is arbitrary, this shows that lim_{n→∞}
∥f − fn∥
= 0.
⇒Conversely, if there exists bounded f : S → ℝ^{q} to which
{fn}
converges uniformly, why is
{fn}
uniformly Cauchy?
|fn(x)− fm(x)| ≤ |fn(x)− f (x)|+ |f (x)− fm (x)|
≤ ∥f − f∥+ ∥f − f ∥
n m
By assumption, there is N such that if n ≥ N, then
∥fn − f∥
< ε∕3. Then if m,n ≥ N, it follows that for
any x ∈ S,
|fn (x )− fm (x)| < ε∕3+ ε∕3 = 2ε∕3
Then
sup|fn (x) − fm (x)| ≡ ∥fn − fm ∥ ≤ 2ε∕3 < ε
x∈S
Hence
{fn}
is uniformly Cauchy. ■
Now here is an example of an infinite dimensional space which is also complete.
Example 2.2.36Denote by BC
(S;ℂq )
the bounded continuous functions defined on S with valuesin ℂ^{q}. Then this is a complex vector space with norm
∥⋅∥
defined above. It is also complete.
It is obvious that this is a complex vector space. Indeed, it is a subspace of the set of functions having
values in ℂ^{q} and it is clear that the given set of functions is closed with respect to the vector space
operations. It was explained above in Observation 2.2.34 that this uniform norm really is a norm. It
remains to verify completeness. Suppose then that
{fn}
is a Cauchy sequence. By Theorem 2.2.35, there
exists a bounded function f such that f_{n} converges to f uniformly. Then by Theorem 2.2.32 it follows that f
is also continuous. Since every Cauchy sequence converges, this says that BC
q
(S; ℂ )
is complete. Of course
this all works if S is a subset of a normed linear space and ℂ^{q} is replaced with Y a complete normed linear
space.
Example 2.2.37Let K be a compact set and consider C
(K; ℂq)
with the uniform norm. This isa complete normed linear space.
If you have a Cauchy sequence, then the functions are automatically bounded thanks to the extreme
value theorem. Thus, by the above example, the Cauchy sequence converges uniformly to a continuous
function.
One can consider convergence of infinite series the same way as done in calculus.
Definition 2.2.38The symbol∑_{k=1}^{∞}f_{k}
(x)
means lim_{n→∞}∑_{k=1}^{n}f_{k}
(x)
provided this limitexists. This is called pointwise convergence of the infinite sum.Thus the infinite sum means the limitof the sequence of partial sums. The infinite sum is said to converge uniformlyif the sequence ofparitial sums converges uniformly.
Note how this theorem includes the case of ∑_{k=1}^{∞}a_{k} as a special case. Here the a_{k} don’t depend on
x.
The following theorem is very useful. It tells how to recognize that an infinite sum is converging or
converging uniformly. First is a little lemma which reviews standard calculus.
Lemma 2.2.39Suppose M_{k}≥ 0 and∑_{k=1}^{∞}M_{k}converges. Then
∞∑
lim Mk = 0
m→ ∞ k=m
Proof: By assumption, there is N such that if m ≥ N, then if n > m,
Then letting n →∞, one can pass to a limit and conclude that
∞
∑
Mk < ε
k=m+1
It follows that for m > N,∑_{k=m}^{∞}M_{k}< ε. The part about passing to a limit follows from the fact that
n →∑_{k=m+1}^{n}M_{k} is an increasing sequence which is bounded above by ∑_{k=1}^{∞}M_{k}. Therefore, it
converges by completeness of ℝ. ■
Theorem 2.2.40For x ∈ S, if∑_{k=1}^{∞}
|fk(x)|
< ∞, then∑_{k=1}^{∞}f_{k}
(x)
converges pointwise. Ifthere exists M_{k}such that M_{k}≥
|fk(x)|
for all x ∈ S, then∑_{k=1}^{∞}f_{k}
(x)
converges uniformly.
Proof: Let m < n. Then
||∑n m∑ || ∞∑
|| fk (x )− fk (x )||≤ |fk (x )| < ε∕2
|k=1 k=1 | k=m
whenever m is large enough due to the assumption that ∑_{k=1}^{∞}
|fk(x )|
< ∞. Thus the partial sums are a
Cauchy sequence and so the series converges pointwise.
If M_{k}≥
|fk(x)|
for all x ∈ S, then for M large enough,
||n m || ∞ ∞
||∑ f (x)− ∑ f (x)||≤ ∑ |f (x)| ≤ ∑ M < ε∕2
|k=1 k k=1 k | k=m k k=m k
and so the partial sums are uniformly Cauchy sequence. Hence they converge uniformly to what is defined
as ∑_{k=1}^{∞}f_{k}
(x)
for x ∈ S. ■
The latter part of this theorem is called the Weierstrass M test. As a very interesting application,
consider the question of nowhere differentiable functions.
Now consider the following description of a function. The following is the graph of the function on
[0,1]
.
PICT
The height of the function is 1/2 and the slope of the rising line is 1 while the slope of the falling line is −1.
Now extend this function to the whole real line to make it periodic of period 1. This means
f
(x + n)
= f
(x )
for all x ∈ ℝ and n ∈ ℤ, the integers. In other words to find the graph of f on
[1,2]
you
simply slide the graph of f on
[0,1]
a distance of 1 to get the same tent shaped thing on
[1,2]
. Continue
this way. The following picture illustrates what a piece of the graph of this function looks like. Some might
call it an infinite sawtooth.
PICT
Now define
∑∞ ( )k ( )
g(x) ≡ 3 f 4kx .
k=0 4
Letting M_{k} =
(3∕4)
^{−k}, an application of the Weierstrass M test, Theorem 2.2.40 shows g is everywhere
continuous. This is because each function in the sum is continuous and the series converges uniformly on ℝ.
However, this function is nowhere differentiable. This is shown next.
Let δ_{m} = ±
14
(4−m )
where we assume m > 2. That of interest will be m →∞.
( ( ))
f (4k(x+ δm )) − f (4kx) = f 4k x ± 1 (4− m) − f (4kx)
4
( ◜int◞e◟ger◝)
| k 1 k−m| ( k )
= f|(4 x ± 44 |) − f 4x = 0
Therefore,
( )
g(x-+-δm-)−-g(x) 1--m∑ 3 k( ( k ) ( k ))
δm = δm 4 f 4 (x + δm) − f 4 x
k=0
The absolute value of the last term in the sum is
||( 3)m m m ||
|| 4 (f (4 (x+ δm))− f (4 x))||
and we choose the sign of δ_{m} such that both 4^{m}
(x + δm)
and 4^{m}x are in some interval [k∕2,
(k+ 1)
∕2)
which is certainly possible because the distance between these two points is 1/4 and such half open
intervals include all of ℝ. Thus, since f has slope ±1 on the interval just mentioned,
|( )m | ( )m
|| 3 (f (4m(x + δ ))− f (4mx))||= 3 4m |δ | = 3m|δ |
| 4 m | 4 m m
As to the other terms, 0 ≤ f
(x)
≤ 1∕2 and so
| ( ) | ( ) ( )
||m∑−1 3 k ( ( k ) ( k ))|| m∑−1 3 k 1−-(3∕4)m- 3 m
|| 4 f 4 (x+ δm) − f 4 x || ≤ 4 = 1∕4 = 4− 4 4
k=0 k=0
Thus
||g (x+ δm)− g (x)|| m ( ( 3)m ) m
||------δm------|| ≥ 3 − 4− 4 4 ≥ 3 − 4
Since δ_{m}→ 0 as m →∞, g^{′}
(x)
does not exist because the difference quotients are not bounded.
■
This proves the following theorem.
Theorem 2.2.41There exists afunction defined on ℝ which is continuous and bounded but failsto have a derivative at any point.
Proof: It only remains to verify that the function just constructed is bounded. However,
∞ ( )k
0 ≤ g(x) ≤ 1 ∑ 3 = 2 ■
2 k=0 4
Note that you could consider
(ε∕2)
g
(x)
to get a function which is continuous, has values between 0 and
ε which has no derivative.