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If I have $ A = \{a \in \ell_2 : |a(n)| \leqslant c(n)\}$ for $c(n)\geqslant 0$ where $ n \in N $, and I want to show that is $A$ compact in $\ell_2$ iff $\sum{c(n)^2}<\infty$. How do I go about showing both directions?

If $f \in C(T)$ is the $1$-periodic continuous functions in $\Bbb R$, how to show $\lim \limits_{|n|\to\infty}\int_0^1 e^{-2\pi inx}f(x)=0 ?$ Also is this true if $f$ were in the closure of the set of 1-periodic step functions in $R$ Intuitively, I think the later is false since boundedness does not imply continuity.

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@martini That is probably the right question. – AD. Dec 11 '12 at 8:13
I'm confused about some notation. You say $\{a:\vert x(n)\vert\leq y(n)\}$. Do you mean $\{a:\vert a(n)\vert\leq y(n)\}$? Also, what is $c(n)$? Do you mean $y(n)$ there? – icurays1 Dec 11 '12 at 8:14
@confused I suggest you split the question in two pieces. – AD. Dec 11 '12 at 8:14
@confused: if you register one of your accounts, they can be merged. – robjohn Dec 12 '12 at 10:14
@robjohn: I always thought that accounts can be merged even without being registered. – Asaf Karagila Dec 12 '12 at 13:12

Hint for (1): If $\sum c_n^2 < \infty$, show the set is closed and totally bounded. If $\sum c_n^2 = \infty$, show that the set is not bounded.

Hint for (2): Riemann-Lebesgue lemma, but you probably want to prove it yourself for trigonometric polynomials and also for indicator functions of intervals.

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(1) Did you try to understand the hints? Don't work directly with Cauchy sequences. (2) Can you calculate the integral when $f(x) = \exp(2 \pi i k x)$ or the indicator function of an interval? – Robert Israel Dec 11 '12 at 20:01
Okay so for (1) I think I can figure it out but for (2) I am trying to apply Riemann-Lebesgue Lemma but it means that I need to show $f(x)$ is Lebesgue integrable. I'm not given anything except its 1-period continuous and so I dont know how to show $\int f^- <\infty$ and $\int f^+ <\infty$. Another approach was to show that its Riemann integrable which implies Lebesgue integrable. How can I show that the upper and lower integrals equal each other? $\overline{\int f} =\underline{\int f}$ – user52740 Dec 12 '12 at 7:06
A continuous function on a closed, bounded interval is bounded and measurable and hence Lebesgue integrable. – Robert Israel Dec 12 '12 at 10:57
Your proof by compactness is wrong. What you have to show is this: if $c(n) \ge 0$ and $\sum_n c(n)^2 = \infty$, then for any $B$ there is $a \in \ell_2$ with $|a(n)| \le c(n)$ for all $n$ and $\sum_n a(n)^2 \ge B$. – Robert Israel Dec 13 '12 at 6:34
As for the reverse direction, $\ell_2$ is not compact. Do you know what "totally bounded" means? – Robert Israel Dec 13 '12 at 6:35

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