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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.2012-12-11
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    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?2012-12-11
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    @confused I suggest you split the question in two pieces.2012-12-11
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    @confused: if you register one of your accounts, they can be merged.2012-12-12
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    @robjohn: I always thought that accounts can be merged even without being registered.2012-12-12

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