Fourier tranforms

  • #1
maria clara
58
0
If a function belongs to G(R) but has points that are jump discontinuities, it's Fourier transform will not belong to G(R).
But would it be correct to claim that if a function in G(R) is continuous than its Fourier transform also belongs to G(R)? I guess it's not true, but can someone suggest a counterexample?
Thanks.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
43,021
971
Frst, what do YOU mean by "G(R)"?
 
  • #3
maria clara
58
0
Sorry, I wasn't sure whether this sign is well known.
G(R) is the space of functions that might have points of discontinuity only of first kind, and which are absolutely integrable.
 
  • #4
maria clara
58
0
ideas? anyone?...:frown:
 

Suggested for: Fourier tranforms

  • Last Post
Replies
0
Views
312
Replies
1
Views
299
  • Last Post
Replies
3
Views
245
Replies
2
Views
101
Replies
2
Views
375
  • Last Post
Replies
1
Views
285
  • Last Post
Replies
23
Views
522
Replies
2
Views
436
  • Last Post
Replies
6
Views
616
Replies
24
Views
291
Top