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

$$2\pi -2\text{Re}\int_{-\pi}^{\pi}g(x)e^{-inx}dx+\int_{-\pi}^{\pi}|g(x)|^2 dx$$

So

$$g\sim\sum\limits_{-\infty}^{\infty}c_n e^{inx}$$

and by Parseval's Theorem,

$$\sum\limits_{-\infty}^{\infty}|c_n|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}|g|^2 dx$$

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?

$$\|f_n-f_m\|_2 = 4\pi -\int_{-\pi}^{\pi}e^{i(m-n)x}dx-\int_{-\pi}^{\pi}e^{i(n-m)x}dx$$

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 $$f\in S$$

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

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, $$\|f-g\|_2<M$$

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]

$$\|f_n-f_m\|_2 = 4\pi -\int_{-\pi}^{\pi}e^{i(m-n)x}dx-\int_{-\pi}^{\pi}e^{i(n-m)x}dx$$

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