I have no idea how, where to start. I mean that we can show the compactness of the set via existence of convergent subsequence. But how can I take it? Please give a clue.
This is my problem
Show that set
(1)$K=\{a\in l^2(\Bbb N): |a(n)|\le \dfrac1n \ ,\ n=1,2,3...\}$ is a compact,in normed space $l^2(\Bbb N)$.
(2)How about $L=\{a\in l^2(\Bbb N): |a(n)|\le \dfrac{1}{\sqrt n} \ , n=1,2,3...\}$?
please help me.
$L$ is not compact even in the weak topology because it is not bounded: $$a_k(n)=\frac{1}{\sqrt{n}}\quad \text{if}\: n\leq k$$ $$a_k(n)=0\quad \text{if}\: n> k$$ this sequence does not admin a convergent subsequence because it escapes to infinity.
$K$ is closed and convex in the strong topology, and as such it is closed in the weak topology. In the weak topology it is compact because it is bounded and closed. Given any sequence $a_k\in l^2$ it admit a convergent subsequence in the weak topology, and thus converging pointwise. This subsequence is now converging in the strong topology from the dominated convergence theorem.