Problem Set Three: Convergence and Completeness
Definitions: Let (S,d) be a metric space. A sequence (f n ) in S converges to a point f in S iff  ε  0
 m  N so that n  m  d(f n , f )  ε . Convergence to f is denoted by writing f n  f or lim f n  f .
Convergence to f is equivalent to each of these statements.
(i) For any open ball B centered at f,  m  N so that n  m  f n  B .
(ii) The real sequence d(f n , f ) converges to zero, i.e., d(f n , f )  0 .
Example: In Frechet space R N if f n  ( 1, 2, 3, ... , n,1,1, ... etc ... ) for each n and
f  ( 1, 2, 3, ... , n, n  1, n  2, ... etc ... ) then f n  f .
Example: In C[0,1] with the 2-norm, the sequence f n (x)  x n converges to the constantly zero function.
Note f n (1)  1 for all n.
Definitions: Let (S,d) be a metric space and (f n ) be a sequence in S.
(a)
(b)
(c)
(d)
(f n ) is Cauchy iff  ε  0  m  N so that n & k  m  d(f n , f k )  ε .
S is complete iff each Cauchy sequence in S converges to some point in S.
A set E in S is bounded iff E is contained in some open ball.
(f n ) is bounded iff its set of values { f1 , f 2 , ... , f n , ... } is bounded in S.
Example: d(x, y)  1/x  1/y
define a new metric on S = (0,1). (S,d) has the same convergent
sequences as (0,1) with the usual metric. S isn't d-bounded. w n  1/n isn't d-Cauchy.
Theorem: In a metric space a convergent sequence is Cauchy. A Cauchy sequence is bounded.
The sign or signum function is defined by sgn(x)  1 if x  0 , and sgn(x)   1 otherwise.
Example: Define f n on [  1, 1] by f n (x)  nx for x  1/n and f n (x)  sgn(x) otherwise.
f
n
 sgn
 0 . Under the 2-norm C[  1, 1] is not complete.
2
PROBLEMS
Problem 3-1: Use the Cauchy-Schwartz Inequality to show (again) that for any n-tuple x in R n ,
x
2

 max 

 k 1 x ( k ) y( k ) :
n
y
2

1.

Problem 3-2 (Arc Length Inequality): Here is a way to use the Cauchy-Schwartz Inequality to verify an
inequality you know from calculus. If u(t) and v(t) are continuously differentiable functions on the
interval [a,b] then
( u ( b ), v ( b ))  ( u ( a ), v ( a ))
2

b
a
2
2
u ( t )  v ( t ) dt .
In other words the straight line distance from (u(a),v(a)) to (u(b),v(b)) can be estimated from above by the
integral. Read the following proof and supply a reason for each step or assertion.
Proof: Let x and y be any real numbers with x 2  y 2  1.
x [ u ( b )  u ( a )]  y [ v ( b )  v ( a )]  x
b
a
b

a

a

a
u ( t ) dt  y
b
a
v ( t ) dt
x u ( t )  y v ( t ) dt
b
b
x
2
y
u ( t )
2
2
u ( t )
 v ( t )
2
2
 v ( t )
2
dt
dt .
Since that’s true for all x and y with x 2  y 2  1, the desired inequality holds.
Problem 3-3 (Orlitz Inequality, generalizes 3-2) If f1 , f 2 , ... , f n are continuous functions on [a,b] and
2 1/2
f(x)  {  nk  1 f k (x) }
is their square function, then
2 1/2
{  nk  1 [  ab f k (x) d x ] }
  ab f(x) d x
(You
only need continuity to make sure the integrals are defined.)
Problem 3-4: This problem is about R N , the space of all real sequences with the metric
d(f, g) 


k 1
2
k
min{ 1, f(k)  g(k) } .
if 1  k  n
n
f n (k)  
0
if n  k
For each n let f n be the sequence
.
(a) Show that d(f n , f m )  1  2  m for 1  n  m .
(b) Explain why the sequence is bounded in R N .
(c) Explain why the sequence ( f n ) does not converge in R N .
Problem 3-5: (a) For any n-tuple x in R n , x

 x
2

n x

.
(b) For any n-tuple x in R n , use the Cauchy-Schwartz Inequality to show
x
 x  n x .
2
1
2
(c) Use parts (a) and (b) to show that if a sequence is convergent (respectively, Cauchy, bounded) in one
p-norm then it is convergent (resp., Cauchy, bounded) in all p-norms, p  1, 2,  .
Problem 3-6: Let (S,d) be a metric space.
(a) For any points x, y and z in S, d ( x , z )  d ( y , z )  d ( x , y )
(b) If x n  x then d ( x n , z )  d ( x , z ) for any z in S.
(c) If x n  x and y n  y then d ( x n , y n )  d ( x , y ).
FACT FROM ADVANCED CALCULUS: With the usual metric d ( x , y )  x  y the reals are a
complete metric space
Problem 3-7: (a) For any sequence of real numbers min{ 1, x n  x }  0 iff x n  x  0 .
(b) With the metric d ( x , y )  min{ 1, x  y } , the reals are a complete metric space.
Problem 3-8: In the reals with the railroad metric, a Cauchy sequence is either eventually constant or
converges to zero in the usual sense. R with the railroad metric is a complete metric space.