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Investigate the boundedness and compactness of the following subset of $( \ell^2 , \|\cdot\|_2 )$: $$S:=\{x=(x_n) \in \ell^2 : | x_n | \le 1/\sqrt{n} \ \forall n\in N\}.$$

Can the following be proof?

Boundedness:

If $S:=x=(x_n)$ is bounded in $\ell^2$ then there is a finite $M$ such that $|x_n|<M$ for all $n$.

Let's take $n=2^k$. Then we have $$\begin{align} 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n} &= 1+\left(\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right) %+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right) + \cdots %\\ &\phantom{=} \cdots + \left(\frac{1}{2^k+1}+\frac{1}{2^k+2}+\cdots+\frac{1}{2^k}\right) \\ &\ge 1+\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2} \\ &=1+\frac{k}{2} \end{align}$$

There cannot be any number $M$ which is greater than $1+\frac{k}{2}$ for all values of $k$ because if $k=2\lceil M\rceil$ is the smallest integer larger than $2M$, then $1+\frac{k}{2}\ge1+\frac{2M}{2}=1+M$, which is contradiction. So $S$ forms an unbounded set.

Compactness:

$S$ forms an unbounded set, hence it is not compact.

Thank you in advance. I am very poor in proving. I appreciate any comment.

user642796
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learner
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1 Answers1

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Indeed, $S$ is not bounded. We can use the idea you mention, but we can write it more simply. Define indeed for a fixed $N$ the sequence $x^{(N)}:=(1,\dots,j^{-1/2},\dots,N^{-1/2},0,\dots,0)$. Such a sequence belong to $\ell^2$ and its norm is $\sqrt{\sum_{j=1}^N\frac 1j}$.

Davide Giraudo
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