Let
$$ l^2 = \left\{ (x_n) : \sum_{n=1}^{\infty} x_n^2 < \infty \right\} $$
equipped with the norm
$$ \| (x_n) \| = \left( \sum_{n=1}^{\infty} x_n^2 \right)^{1/2}. $$
I am wondering how to prove that the set
- $\displaystyle \lbrace (x_n) \in l^2 : \sum_{n=1}^{k} x_n^2 \leq 1, x_n=0 $ for all $n>k$ $\rbrace$ where $k \in \mathbb{N}$ is fixed
is sequentially compact by using the definition that every sequence from the set has a subsequence which converges to an element of the set.
Can someone help me, please? Thanks so much.
EDIT: Here is an answer:
This is a well known proof technique.
1) For all $1 \leq j \leq k$, and for all $n \in \mathbb N$, $|(x_n)_j| \leq 1$.
2) Now , consider the sequence $(x_n)_1$, so we are taking the first component of each $x_n$ and creating a sequence . This is a bounded sequence, hence has a convergent subsequence $(x_{n_l})_1$. Furthermore, $|\lim (x_{n_l})_1|^2 \leq \lim |(x_{n_l})_1|^2$.
3) Look at $(x_{n_l})_2$ (the second components of the convergent subsequence), for the same reason, it has a convergent subsequence $(x_{n_{l_m}})_2$. Note that since $x_{n_{l_m}}$ is a subsequence of $x_{n_k}$, it's first component i.e. $(x_{n_{l_m}})_1$ will also be convergent, and the limits will have the same inequalities.
4) Now repeat with the third component etc. up till $k$, and let the rest of the entries be zero. This final subsequence will have the property that all it's components converge pointwise.
5)Let us call this subsequence $v_n$, then we see from pointwise inequality that $\sum |v_n|^2 \leq 1$, using the same limit logic, and that $v_n = 0$ for $n > k$. So $v_n$ is a limit point of the sequence in the given set, hence the given set is sequentially compact.
May I please ask why this answer assumed the fact that every bounded sequence has a convergent subsequence (bolzano weierstrass)? Is $l^2$ a subspace of $\Bbb R^k$? Or is it the case that the set is subset of $\Bbb R^k$?
Best Answer
A complete rewrite of my proof, so that you understand clearly what is going on. For simplicity, I'll assume we are talking about $l^2(\mathbb R)$, because the proof will easily work for $\mathbb R^k$ if you just change notation slightly.
We are given the following subset in $l^2(\mathbb R)$ : $$ A = \left\{ s \in l^2(\mathbb R) \ \Bigg|\ \sum_{i=1}^k (s_i)^2 \leq 1 , j > k \implies s_j = 0\right\} $$
We have to prove that this is compact. For this, we need to show that sequence in $A$ has a convergent subsequence. That is, given $x_n$, a sequence of sequences (that is, $x_1,x_2,x_3$ are all sequences now), we have to find a limit point of $x_n$ in $A$.
Notation : $x_1,x_2,x_3$ are sequences, so $x_i$ denotes a sequence. However, the $j$th component of $x_i$, is the $j$th component of a real sequence, so it's a real number. Thus, we will denote by $(x_i)_j$, the $j$th component of the sequence $x_i$, and this is a real number. (Note : If there's a pair of brackets in the expression, then it's a real number e.g. $(x_n)_j$ is a real number . If there is no pair of brackets in the expression then it's a sequence e.g. $x_{n_{k_l}}$ is a sequence).
To start , we will consider the sequence of first components of $x_n$, namely $(x_n)_1$. So the elements of this sequence are $(x_1)_1,(x_2)_1,(x_3)_1$ etc. Note that this is a sequence of real numbers.
Furthermore, this sequence is bounded. To see this, note that $|(x_n)_1|^2 \leq ||x_n||^2 \leq 1$ for all $n$. Now, by Bolzano-Weierstrass, this has a convergent subsequence (note : this convergent subsequence will converge to a real number). We will assume that the convergent subsequence is given by $(x_{n_k})_1$. So $(x_{n_k})_1 \to y_1$, where $y$ is a real number.
Now, suppose we have a sequence $a_n \to a$ in $\mathbb R^m$, then actually $||a_n|| \to ||a||$. So, $|(x_{n_k})_1| \to |y_1|$, and $|(x_{n_k})_1|^2 \to |y_1|^2$. Keep this in mind.
One more thing to keep in mind : If $(x_{n_{k_l}})$ is a subsequence of $x_{n_k}$, then since $(x_{n_k})_1 \to y_1$, it is also true that $(x_{n_{k_l}})_1 \to y_1$. (Because if a sequence converges, all it's subsequences also converge to the same point).
Now, consider $x_{n_k}$, which are sequences. The first components of these converge to $y$. Let us look at their second components, $(x_{n_k})_2$. This is a sequence of real numbers.
The same logic : These are bounded, and have a convergent subsequence $(x_{n_{k_l}})_2 \to y_2$, say. However, $(x_{n_{k_l}})$ is a subsequence of $x_{n_k}$, so $(x_{n_{k_l}})_1 \to y_1$ i also true.
Now, I want you to repeat this argument for the third component (I won't do it, it's costing me subscripts) and go up till $k$. At that point, we will have a sequence $z_n$ (which is a subsequence of $x_n$!), whose first, second, third, fourth ... kth components are all convergent, to $y_1,y_2, ... , y_k$. We will put $y_j = 0$ if $j>k$. (Here, note that $y_i$ are real numbers, the sequence itself is $y$).
We have to show that $y \in A$. To do this, note that $|y_i|^2 = \lim |(z_n)_i|^2$, hence summing from $i=1$ to $k$, $\sum |y_i|^2 = \lim \sum_{i=1}^n |(z_n)_i|^2 \leq 1$, since $z_n \in A$, so $\sum_{i=1}^n |(z_n)_i|^2 \leq 1$ for all $n$. Furthermore, after the $k$th component, $y$ is entirely zero. This shows that $y \in A$.
Finally, we have to show that $z_n \to y$ in $l^2$. This is done by showing that $||z_n - y|| \to 0$ as $n \to \infty$.
Let $\epsilon > 0$. Since $(z_n)_i \to y_i$ for $1 \leq i \leq k$, choose $N$ large enough such that for all $1 \leq i \leq k$, $n > N \implies |(z_{n})_i - y_i| < \frac{\epsilon }{\sqrt k}$.
Now, if $n > N$: $$ ||{z_n} - y||^2 = \sum_{i=1}^k ((z_{n})_i - y_i)^2 < \sum_{i=1}^k \frac {\epsilon^2} k < \epsilon $$
Hence, $z_n \to y$, hence $y$ is a limit point of $x_n$ lying in $A$. Hence, $A$ is sequentially compact.
Do ask if further doubts persist.