Problem Set Seven: Uniform Convergence
Definitions: Let S be a set, f : S  R be a function, and ( f n ) be a sequence of functions from S into R.
(a) ( f n ) converges pointwise on S to f iff, for each x in S, f n ( x )  f ( x ) in R. Equivalently,
 x  S  ε  0  m(x)  N, n  m(x)  f n (x)  f (x)
ε.
(b) ( f n ) converges uniformly on S to f iff  ε  0  m  N  x  S, n  m  f n (x)  f (x)  ε .
Note: (i) If ( f n ) converges uniformly on S to f then ( f n ) converges pointwise on S to f.
(ii) Pointwise convergence need not imply uniform convergence. For example f n ( x )  x n converges
0
if 0  x  1
1
if x  1
pointwise on [0,1] to f ( x )  
, but convergence is not uniform.
Definitions: Let S be a set.
(a) A function f : S  R is (uniformly) bounded iff there is a number b so that f ( x )  b for all x in S.
(b) The uniform norm of a bounded function f : S  R is
f
u
 min { b :  x  S f ( x )  b }  sup { f ( x ) : x  S } .
(c) B(S) denotes the set of all bounded functions f : S  R .
Theorem: (a) B(S) is a vector space under the pointwise operations
( f  g )( x )  f ( x )  g ( x ) and ( cf )( x )  c f ( x ).
(b) The uniform norm is a norm on B(S).
(c) ( f n ) converges to f in the metric space B(S) iff ( f n ) converges uniformly on S to f.
(d) B(S) is a complete under the uniform norm. A complete normed space is called a Banach space.
Theorem: If f : S  T is continuous and K  S is compact, then f ( K )  { f ( x ) : x  K } is a compact
subset of T.
Corollary: If f : S  R is continuous and S is compact then there are points p and q in S so that
f ( q )  f ( x )  f ( p ) for all x in S. In particular f is bounded on S.
Definition and Theorem: For S a compact metric space let C(S) denote the set of all continuous
functions f : S  R .
(a) C(S) is a vector space under the pointwise operations
(b) If ( f n ) is a sequence in C(S) and ( f n ) converges uniformly on S to f, then f is continuous.
(c) C(S) is a closed subset of B(S) and thus is complete under the uniform norm.
Lemma: If g is continuously differentiable on [a,b] then g
2
u
2 g
2
g
2
 (b  a )
1
Theorem: Let ( f n ) be a sequence of continuously differentiable functions on [a,b]. If
(i) ( f n ) is Cauchy in 2-norm, and
g
2
2
.
(ii) the derivatives ( f n ) are bounded in 2-norm,
then ( f n ) converges uniformly to a continuous function on [a,b].
Theorem: Let ( f n ) be a sequence of continuously differentiable functions on [a,b]. If
(i) lim f n (x 0 ) exists for some x 0 in [a,b], and
(ii) ( f n ) converges in 2-norm to some continuous g,
then there is a differentiable f so that f   g and ( f n ) converges uniformly to f on [a,b].
PROBLEMS
Problem 7-1: Answer these questions for each sequence ( f n ) . Does the sequence converge pointwise on
S? If so is the pointwise limit continuous? Is convergence uniform?
(a) f n ( x ) 
1
1 nx
, S  [ 0 , 1]
(b) f n ( x ) 
n
nx
, S  [ 0 , 1]
(c) f n ( x )  tan
1
( n x ), S  R
Problem 7-2: In B[0,1] let f n ( x )  max { 1  n x , 0 }.
(a) Calculate f n
u
fn fm u
and
for n  m .
(b) Is the set M  { f 1 , f 2 , f 3 , ... , f n , ... } totally bounded in B[0,1]?
(c) Does the sequence ( f n ) converge in B[0,1]? If so find its limit.
Definition: A function F : S  T between metric spaces is an isometry iff d(F(x), F(z))  d(x, z) for all x
and z in S. Note this doesn’t require that F is onto.
Problem 7-3: For (S,d) a metric space and x 0 a distinguished point in S, define F : S  B(S) by
F (x)(s)  d(x 0 , s)  d(s, x) Prove that F is an isometry with values in C(S).
Problem 7-4: Let ( f n ) be a sequence in C[a,b]. (a) Prove that if f n  0 uniformly then f n  0 in 2norm. Give (or review) an example showing the converse is false.
(b) Prove that if f n  0 in 2-norm then
b
a
f n (u) d u  0 and
b
 a f n (u) d u  0 .
Give (or review) an
example showing the converse is false.
Problem 7-5: If g is continuously differentiable on [a,b] and c 
g
u
 c{ g
2
2
 g
2 1/2
2
}
1  1/(b  a)
.
Problem 7-6: On [ 1, 1] let f n (x)  f( n x) for f(x)  x 2 sgn (x) /(1  x 2 ) .
(a)
f n  sgn
2
2

2
du
n
 0.
0
2 2
n
(1  u )
then
(b) ( f n ) can’t converge in 2-norm to a continuous function on [ 1, 1] .
(c) The sequence of derivatives ( f n ) converges pointwise to zero on [ 1, 1] .
(d)
f n
2
2
2
 8 n  0n
u du
2 4
(1  u )
 
.
Problem 7-7: On [ 1, 1] let f n (x) 
2
x  (1/n)
.
(a) ( f n ) converges uniformly to x on [ 1, 1] .
(b) ( f n ) converges to sgn(x) pointwise and in 2-norm on [ 1, 1] .
Definition (a Cantor Set): There is a recursively defined sequence (C n ) of subsets of [0,1] with the
following properties: (i) C 0  [0,1] ; (ii) C n is the union of a finite number of disjoint closed bounded
intervals, called the components of C n , and; (iii) C n  1 is obtained from C n by removing the open
middle third of each component of C n . For instance C 1  [0, 1/3]  [2/3, 1] ,
C 1  [0, 1/9]  [2/9, 1/3]  [2/3, 7/9]  [8/9, 1]
, and so on. The Middle Third Cantor Set is C   n 1 C n .
Problem 7-8: Write S  { 0, 1 } N and define f n : S  R by f n (x)  2 
n
k 1
3
k
x(k)
.
(a) Show that ( f n ) converges uniformly and in C(S) to a Lipschitz function f : S  R .
(b) Show that f is one-to-one and maps onto the Middle Third Cantor Set C.
(b) Show that the inverse function f  1 : C  S is continuous. A continuous, one-to-one and onto
function with continuous inverse is called a homeomorphism.