5

We have the vector space $X=\{\vec{x}=(x_1, x_2,\cdots) | x_n\in\mathbb{R} (n\in\mathbb{N}), \sum_{n=1}^{\infty}\frac{1}{n}|x_n|<\infty \}$, and the norm $\|\vec{x}\|=\sum_{n=1}^{\infty}\frac{1}{n}|x_n|\ (\vec{x}=(x_1, x_2, \cdots))$ on it. This normed space is complete.

The question is about the compactness of the following subset $A\subset X$,

$A=\{\vec{x}\in X|\sum_{n=1}^{\infty} |x_n|^2\leq 1\}$.

I know this subset $A$ is bounded and closed (the boundedness comes from the fact $|x_n|\leq\frac{1}{\sqrt{n}}$ assuming $\{|x_n|\}$ is monotonically decreasing).

What I cannot tell is whether this subset $A$ is compact or not (in the Banach space $(X, \|\cdot\|)$).

Can you help me with this problem?

hoge
  • 53
  • In infinite dimension the conditions bounded and closed are not enough sufficient to prove the compactness. Maybe you can think about pre-compactness as $A$ is complete ! – Maman Oct 20 '21 at 14:06
  • Thanks, that's true. Total-boundedness should be checked to know whether this subset is compact or not. But I have no idea how we can check the total-boundedness of this subset. – hoge Oct 20 '21 at 14:12
  • Maybe you can use the fact that total-boundedness or "pre-compactness" is linked with Cauchy's subsequence ! – Maman Oct 20 '21 at 14:14
  • Thanks again. That is also true. the total-boundedness is linked with Cauchy subsequences. But still, I'm now sure how I can use Cauchy sequences or total-boundedness for this specific norm and subset. I'll welcome any comments and concrete strategies regarding this problem. – hoge Oct 20 '21 at 14:21

1 Answers1

3

Take any $\varepsilon >0$ then there exist a $n \in \mathbb{N}$ such that $\sum_{k=n+1}^{\infty}\frac{1}{k^2} <\varepsilon^2.$ The set $A^{n} =\{ (x_1, x_2, ..., x_n , 0,0 ...) : (x_1 , x_2 , ...,)\in A\}$ is compact since it is finite dimensional closed and bounded. Let $B^n =\{(0, 0, ..., x_{n+1} : (x_1 , x_2 , ...)\in A\}.$ If $u\in B$ then $$||u||=\sum_{j=1}^{\infty } \frac{|x_j |}{j}\leq \sqrt{\sum_{j=n+1}^{\infty } \frac{1}{j^2} }\cdot \sqrt{\sum_{j=n+1}^{\infty } |x_j |^2 }\leq \varepsilon .$$ Since $A^n $ is compact therefore exists $u_1 , u_2 , ..., u_m $ such that $$A^n \subset \bigcup_{i=1 }^m (u_j + B(0, \varepsilon ))$$ where $B(0,r)$ denotes the open ball of radius $r>0.$ Moreover $B^n \subset B(0, \varepsilon )$ hence $$A\subset A^n +B^n \subset B(0, \varepsilon) +\bigcup_{i=1 }^m (u_j + B(0, \varepsilon ))\subset \bigcup_{i=1 }^m (u_j + B(0, 3\varepsilon ))$$ so the set $A$ is totally bounded and hence compact.

  • Impressive. I should've noticed the Cauchy-Schwarz inequality. Thanks a million for your clear-cut answer. I really appreciate it. – hoge Oct 21 '21 at 01:17