• Support PF! Buy your school textbooks, materials and every day products Here!

Sequence of functions in L^2[-pi,pi]

  • Thread starter iomtt6076
  • Start date
  • #1
42
0

Homework Statement


Consider the functions [tex]f_n(x)=e^{inx},n\in\mathbb{Z},-\pi\leq x\leq\pi[/tex] viewed as points in [tex]\mathscr{L}^2[-\pi,\pi][/tex]. Prove that this set of functions is closed and bounded, but not compact.


2. The attempt at a solution
I'm first trying to prove that the set of functions is closed. Let g be a limit point of the set, so given any positive number epsilon, there exists fn such that [tex]\|f_n-g\|_2<\sqrt{\epsilon}[/tex]. Squaring both sides, I get on the left hand side

[tex]
2\pi -2\text{Re}\int_{-\pi}^{\pi}g(x)e^{-inx}dx+\int_{-\pi}^{\pi}|g(x)|^2 dx
[/tex]

So

[tex]
g\sim\sum\limits_{-\infty}^{\infty}c_n e^{inx}
[/tex]

and by Parseval's Theorem,

[tex]
\sum\limits_{-\infty}^{\infty}|c_n|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}|g|^2 dx
[/tex]

I don't know where to go from here.
 

Answers and Replies

  • #2
392
0
What would [tex]\|f_n-f_m\|_2[/tex] be?

Can you even have a Cauchy sequence?
 
  • #3
42
0
What would [tex]\|f_n-f_m\|_2[/tex] be?

Can you even have a Cauchy sequence?
[tex]
\|f_n-f_m\|_2 = 4\pi -\int_{-\pi}^{\pi}e^{i(m-n)x}dx-\int_{-\pi}^{\pi}e^{i(n-m)x}dx
[/tex]

so when m and n are different, there exists a positive number less than the expression above. So I would say that no, you can't have a Cauchy sequence and so this shows the set of functions is not compact.

But I'm still having trouble with proving the set is closed.
 
  • #4
392
0
Could show the set is complete as a metric space, hence closed.

Note: I guess technically there are Cauchy sequences, but only the sequences that eventually become constant.
 
  • #5
42
0
Thanks for your help; I have to say though that I don't see how there are sequences that eventually become constant. I can't seem to find a sequence {fn(x)} that converges to a constant value given any x.
 
  • #6
392
0
I mean a sequence [tex](a_1, a_2, a_3,\dots)[/tex] in X with the property that there exists x in X and an integer N such that for all n>N, a_n=x.

For example, the sequence [tex](f_2, f_8, f_8, f_8, f_8,\dots)[/tex] where all the rest of the "points" (i.e. functions) are f_8

Such a sequence is trivially Cauchy.
 
  • #7
42
0
Okay, thanks again for your assistance.
 
  • #8
42
0
Sorry, just one last question: isn't any subset S of [tex]{\cal L}^2[-\pi,\pi][/tex] bounded by the zero function g(x)=0, since for any [tex]f\in S[/tex]

[tex]
\|f-g\|_2 = \left(\int_{-\pi}^{\pi}|f-g|^2 d\mu\right)^{1/2} = \left(\int_{-\pi}^{\pi}|f|^2 d\mu\right)^{1/2}
[/tex]

which by definition is finite?
 
Last edited:
  • #9
392
0
No. Using g=0 is OK, but you would need one constant M such that for all f in S, [tex]\|f-g\|_2<M[/tex]

Fortunately, in this case it is easy to find such an M, with your g.
 
  • #10
Dick
Science Advisor
Homework Helper
26,258
618
All of your functions are orthogonal in L^2 and they are all of constant L^2 norm. Aren't you trying to prove the sequence is a closed set? It's a discrete set.
 
  • #11
392
0
[tex]
\|f_n-f_m\|_2 = 4\pi -\int_{-\pi}^{\pi}e^{i(m-n)x}dx-\int_{-\pi}^{\pi}e^{i(n-m)x}dx
[/tex]
iomtt6076, I'm just double checking. You do know what [tex]\int_{-\pi}^{\pi}e^{i(m-n)x}\,dx[/tex] is, right?
 

Related Threads for: Sequence of functions in L^2[-pi,pi]

Replies
2
Views
941
  • Last Post
Replies
6
Views
800
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
12
Views
2K
Replies
5
Views
6K
  • Last Post
Replies
3
Views
2K
Top