Page 211 - DMTH401_REAL ANALYSIS

Unit 16: Uniform Convergence and Continuity

But then we have: Notes
|f(x ) – f(x)| |f(x ) – f (x )| + |f (x ) – f (x)| + |f (x) – f(x)| /3 + /3 + /3 =
0 0 N 0 N 0 N N
as long as |x – x| < . But that means that f is continuous at x .
0 0
Before we continue, we will introduce a new concept that will somewhat simplify our discussion
of uniform convergence, at least in terms of notation: we will use the supremum of a function to
define a ‘norm’ of f

16.2 Uniform Convergence and Supremum Norm

:
Definition 1: The supremum norm of a function f I ®  is
f = f = sup ( ) .
f
x
sup ¥
x I

Example: Let I =  and ( ) sin( ).f x = x Then f = 1.
sup
x
Example: Let I = [0,1] and ( )f x = - 2 . Then f = 2. The norm stays the same even if
sup
we change the interval [0, 1] to (0, 1).
Theorem 1: Let { }f ¥ be a sequence of real-valued functions on I. Then f ® f uniformly if and
n n= 1 n
only if f - f sup ® 0. Note that f - f sup is just a sequence of number.
n
n
x
x
n
n
x
Proof: f ® f uniformly Û " $ N "  N "  : I f n ( )- f ( )   Û " $ N "  N : sup x I
n

x
n
x
f ( )- f ( )   Û " $ N "  N : f - f  
n n
n
n
Example: Let f ( ) x on (0, 1). We can observe that f - f = sup x (0,1) x - 0 =
x
=
n
n sup
0
1 ® 0. As f ® uniformly.
n
ì 1, x  n
x
-
1
Example: Let f n ( ) = í on . Then f sup = sup x f n ( ) 0 = ® 0. So f ® 0
x
n
î 0, x < n
uniformly.
Using this proposition it can be easy to show uniform convergence of a function sequence,
especially if the sequence is bounded. Still, even with this idea of sup-norm uniform convergence
can not improve its properties: it preserves continuity but has a hard time with differentiability.
Example: Consider the sequence f (x) = 1/n sin(nx):
n
 Show that the sequence converges uniformly to a differentiable limit function for all x.
 Show that the sequence of derivatives f ’ does not converge to the derivative of the limit
n
function.
This example is ready-made for our sup-norm because |sin(x)| < 1 for all x. As for our proof: the
sequence converges uniformly to zero because:
||f – f|| = ||1/n sin(n x) – 0||  1/n ® 0
n D D
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