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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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    @robjohn: I always thought that accounts can be merged even without being registered.2012-12-12

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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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    OK, now try a nice finite collection of $x_i$'s, say those where each $x_i(n)$ is a multiple of $\delta$ and $x_i(n) = 0$ for n > N, for some $\delta$ and $N$.2012-12-13