1.4 Uniform Convergence 83
Althoughf(x)is continuous and has a continuous derivative in the interval
− 1 <x<1, the periodic extension offisnotcontinuous. The Fourier series
cannot converge uniformly in any interval containing 1 or−1becausethepe-
riodic extension offhas jumps there, but uniform convergence must produce
acontinuousfunction.
On the other hand, the functionf(x)=|sin(x)|, periodic with period 2π,is
continuous and has a sectionally continuous derivative. Therefore, its Fourier
series converges uniformly tof(x)everywhere.
Here is a restatement of Theorem 2 for a function given on the interval
−a<x<a. The condition at the endpoints replaces the condition of conti-
nuity of the periodic extension off.
Theorem 3.If f(x)is given on−a<x<a, if f is continuous and bounded and
has a sectionally continuous derivative, and if f(−a+)=f(a−), then the Fourier
series of f converges uniformly to f on the interval−a≤x≤a.(The series con-
verges to f(a−)=f(−a+)at x=±a.)
If an odd periodic function is to be continuous, it must have value 0 atx= 0
and at the endpoints of the symmetric period-interval. Thus, the odd periodic
extension of a function given in 0<x<amay have jump discontinuities even
though it is continuous where originally given. The even periodic extension
causes no such difficulty, however.
Theorem 4. If f(x)is given on 0 <x<a, if f is continuous and bounded and has
a sectionally continuous derivative, and if f( 0 +)=f(a−)= 0 , then the Fourier
sine series of f converges uniformly to f in the interval 0 ≤x≤a.(The series
converges to 0 at x= 0 and x=a.)
Theorem 5.If f(x)is given on 0 <x<a and if f is continuous and bounded
and has a sectionally continuous derivative, then the Fourier cosine series of f
convergesuniformlytof intheinterval 0 ≤x≤a.(The series converges to f( 0 +)
at x= 0 and to f(a−)at x=a.)
EXERCISES
- Determine whether the Fourier seriesof the following functions converge
uniformly or not. Sketch each function.
a. f(x)=ex, − 1 <x<1;
b. f(x)=sinh(x), −π<x<π;
c. f(x)=sin(x), −π<x<π;