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

[iomtt6076]

Homework Statement

Consider the functions f_n(x)=e^{inx},n\in\mathbb{Z},-\pi\leq x\leq\pi viewed as points in \mathscr{L}^2[-\pi,\pi]. 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 f_n such that \|f_n-g\|_2<\sqrt{\epsilon}. Squaring both sides, I get on the left hand side

<br /> 2\pi -2\text{Re}\int_{-\pi}^{\pi}g(x)e^{-inx}dx+\int_{-\pi}^{\pi}|g(x)|^2 dx<br />

So

<br /> g\sim\sum\limits_{-\infty}^{\infty}c_n e^{inx}<br />

and by Parseval's Theorem,

<br /> \sum\limits_{-\infty}^{\infty}|c_n|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}|g|^2 dx<br />

I don't know where to go from here.

 

[Billy Bob]

What would \|f_n-f_m\|_2 be?

Can you even have a Cauchy sequence?

 

[iomtt6076]

What would \|f_n-f_m\|_2 be?

Can you even have a Cauchy sequence?

<br /> \|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<br />

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.

 

[Billy Bob]

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.

 

[iomtt6076]

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 {f_n(x)} that converges to a constant value given any x.

 

[Billy Bob]

I mean a sequence (a_1, a_2, a_3,\dots) 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 (f_2, f_8, f_8, f_8, f_8,\dots) where all the rest of the "points" (i.e. functions) are f_8

Such a sequence is trivially Cauchy.

 

[iomtt6076]

Okay, thanks again for your assistance.

 

[iomtt6076]

Sorry, just one last question: isn't any subset S of {\cal L}^2[-\pi,\pi] bounded by the zero function g(x)=0, since for any f\in S

<br /> \|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}<br />

which by definition is finite?

 

[Billy Bob]

No. Using g=0 is OK, but you would need one constant M such that for all f in S, \|f-g\|_2&lt;M

Fortunately, in this case it is easy to find such an M, with your g.

 

[Dick]

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.

 

[Billy Bob]

<br /> \|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<br />

iomtt6076, I'm just double checking. You do know what \int_{-\pi}^{\pi}e^{i(m-n)x}\,dx is, right?