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

## 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 fn 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.

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

Can you even have a Cauchy sequence?

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.

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.

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.

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.

Okay, thanks again for your assistance.

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?

Last edited:
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
$$\|f_n-f_m\|_2 = 4\pi -\int_{-\pi}^{\pi}e^{i(m-n)x}dx-\int_{-\pi}^{\pi}e^{i(n-m)x}dx$$
iomtt6076, I'm just double checking. You do know what $$\int_{-\pi}^{\pi}e^{i(m-n)x}\,dx$$ is, right?