Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(a_n)_{n}\in\ell^2:=\ell^2(\mathbb{R})$ be a fixed sequence. Consider the subspace $$C=\{(x_n)_{n}\in\ell^2 : |x_n|\le a_n\text{ for all }n\in\mathbb{N}\}.$$

According to the book [Dunford and Schwartz, Linear operators part I, page 453] $C$ is compact in the $\ell^2$-norm, but there is no proof. How can I show that $C$ is indeed compact in $\ell^2$ ?

share|improve this question
    
What díd you try? –  AD. Jul 25 '12 at 21:42
    
You should give a look to en.wikipedia.org/wiki/Hilbert_cube –  Romeo Jul 25 '12 at 21:43
    
Given a sequence $(x_n)$ in $C$, use a diagonalization argument to find a subsequence of $(x_n)$ that converges coordinatewise. Then use the given condition that elements in $C$ have uniformly small tails to show that in fact this subsequence is Cauchy in $\ell_2$ (and thus converges necessarily to its pointwise limit). –  David Mitra Jul 25 '12 at 22:02
    
@ AD : I've tried a diagonal argument. Take a sequence $(x(k))_{k}$ of elements in $C$, where $x(k)=\Big((x_n(1))_n, (x_n(2))_n,\ldots, (x_n(k))_n,\ldots\Big)$. Then I picked a subseqeunce for which the first coordinate converges, then I repeatedly chose a subseqeunce for which the 2and, 3rd, up to the k-th coordinate converges. I do not know how to prove that the sequence to which there coordinates converge is in $C$ and is a limit point of the $C$. But this is quite tedious. I would rather use some sort of approximation with finite rank operators. –  gifty Jul 25 '12 at 22:04
    
@DavidMitra I went for the diagonal argument. It's obnoxious, but it works. –  Alex Becker Jul 25 '12 at 22:14
add comment

5 Answers

up vote 14 down vote accepted

The set $C$ is the image of the compact set $X = [-1,1]^{\mathbb N}$ (with the product topology) under the map $F: X \to \ell^2$, where $F(x)_j = x_j a_j$. Since the image of a compact Hausdorff space under a continuous map is compact, we just need to show that $F$ is continuous. Note that $\Vert F(x) - F(y)\Vert^2 = \sum_{j=1}^\infty (x_j - y_j)^2 a_j^2$. Given $\varepsilon > 0$, there is $N\in\mathbb{N}$ such that $\sum_{j=N+1}^\infty a_j^2 < \varepsilon/8$. If $x, y \in X$ with $|x_j - y_j|^2 < \varepsilon/(2\|a\|^2)$ for $1 \le j \le N$, we have $\Vert F(x) - F(y)\Vert ^2 < \varepsilon$.

share|improve this answer
    
Nice and succinct. –  copper.hat Jul 25 '12 at 23:19
    
Thank you for your help :D –  gifty Jul 26 '12 at 8:44
add comment

We can use a standard, if lengthy, diagonal sequence argument. I use boldface to denote an element of $C$. Let $\{{\bf x}_n\}_{n=1}^\infty$ be a sequence in $C$. Keep in mind that each term $\bf x_n$ is itself a sequence in $\mathbb R$. We want to show that there is a convergent subsequence of this. To do so, let $$C_{ijk}=\left\{{\bf x}\in C : \frac{i-1}{j}a_m\leq |x_m|\leq \frac{i}{j}a_m,\forall m\leq k\right\}$$ and note that for $k,j$ fixed, varying $i$ from $1$ to $j$ gives all of $C$.

We first build a subsequence such that the first terms converge. Note that since $C=C_{121}\cup C_{221}$, infinitely many terms ${\bf x}_n$ must be in one of the sets $c_{i21}$. Let $\{{\bf x}_n^{(2)}\}_{n=1}^\infty$ be a subsequence of $\{{\bf x}_n\}_{n=1}^\infty$ such that each ${\bf x}_n^{(2)}$ is in the same set $C_{i21}$. We form a subsequence $\{{\bf x}_n^{(3)}\}_{n=1}^\infty$ of $\{{\bf x}_n^{(2)}\}_{n=1}^\infty$ in the same manner, so that each ${\bf x}_n^{(3)}$ is in the same set $C_{i31}$. Continuing in this manner, we get sequences $\{{\bf x}_n^{(j)}\}_{n=1}^\infty$ for all $j$. We then form the diagonal subsequence $\{{\bf y}_n\}_{n=1}^\infty=\{{\bf x}_n^{(n)}\}_{n=1}^\infty$, which has the property that the first terms of the sequences converge.

We can do the same thing to get a subsequence $\{{\bf y}_n^{(2)}\}_{n=1}^\infty$ of $\{{\bf y}_n\}_{n=1}^\infty$ such that the first and second terms converge, by using the sets $C_{ij2}$ instead of $C_{ij1}$. Continuing in this manner, we get sequences $\{{\bf y}_n^{(k)}\}_{n=1}^\infty$ for all $k$, which are such that the first $k$ terms converge. Again we form the diagonal sequence $\{{\bf z}_n\}_{n=1}^\infty=\{{\bf y}_n^{(n)}\}_{n=1}^\infty$. Note that the $k^{th}$ term of $\{{\bf z}_n\}_{n=1}^\infty$ converges for all $k$, say to a limit $L_k$.

Putting this together, let ${\bf L}=\{L_k\}_{k=1}^\infty$. We have $$\left\|{\bf z}_n-{\bf L}\right\|_2^2=\sum\limits_{k=1}^\infty \left|({\bf z}_n)_k-L_k\right|^2\leq \sum\limits_{k=1}^t \left|({\bf z}_n)_k-L_k\right|^2+\sum\limits_{k=t+1}^\infty |a_k|^2$$ and since $\{a_k\}_{k=1}^\infty\in \ell^2$, for any $\epsilon>0$ we can choose $t$ such that $\sum\limits_{k=t+1}^\infty |a_k|^2<\epsilon/2$. Since $({\bf z}_n)_k\to L_k$ for all $k$, for $n$ sufficiently large we have $\left|({\bf z}_n)_k-L_k\right|^2<\epsilon/2t$. Thus ${\bf z}_n\to \bf L$, hence every sequence in $C$ has a convergent subsequence. It follows that $C$ is compact.

share|improve this answer
    
@t.b. Thanks for the formatting advice. –  Alex Becker Jul 25 '12 at 23:53
    
Okay this looks more palatable :) But why do you need those sets $C_{ijk}$ in the first place? To get a subsequence such that the first coordinate converges just observe that $\{\mathbf{x}_n(1) \,:\,n \in \mathbb{N}\} \subset \{z \in \mathbb{C}\,:\,|z| \leq a_1\}$ and the latter set is compact in $\mathbb{C}$ (or in your case in $\mathbb{R}$, it doesn't matter), so you don't need to chop $C$ up into further pieces. –  t.b. Jul 26 '12 at 6:25
add comment

Show that $C$ is complete and totally bounded.

Completeness is straightforward, suppose $x_n \in C$ is Cauchy. Then each component is also Cauchy since $|[x_n]_k - [x_m]_k | \leq \|x_n -x_m\|$, hence each component converges to some $\alpha_k$. Furthermore, it is clear that $|\alpha_k| \leq a_n$, so define $\hat{x}$ by $[\hat{x}]_n = \alpha_n$. Clearly, $\hat{x} \in C$, and we just need to show that $x_n \to \hat{x}$. Let $\epsilon > 0$, and choose $N'$ such that $\sum_{k>N'} a_k^2 < \frac{\epsilon}{2}$. Furthermore, we can choose $N>N'$ such that if $n>N$, then $|[x_n]_k - [\hat{x}]_k|^2< \frac{\epsilon}{2N'}$, for all $k \leq N'$. Then, if $n>N$, we have $\|x_n-\hat{x}\|^2 < \sum_{k\leq N'} |[x_n]_k - [\hat{x}]_k|^2 + \frac{\epsilon}{2} \leq \epsilon$. Hence $x_n \to \hat{x}$.

To show total boundedness, we need to find a finite $\epsilon$-net for all $\epsilon>0$. Choose $N'$ such that $\sum_{k>N'} a_k^2 < \frac{\epsilon^2}{2}$. It is clear that since $A=[-a_1,a_1]\times \cdots \times [-a_{N'},a_{N'}]$ is compact (as a subset of $\mathbb{R}^{N'}$), we can find an $\frac{\epsilon^2}{2}$-net for $A$. Let $E \subset A$ be a $\frac{\epsilon^2}{2}$-net for $A$, then define the set (with slight abuse of notation) $E' = E \times \{0\} \times \{0\} \times \cdots$. The set $E' \subset C$ is finite, and is an $\epsilon$-net for $C$. To see this, choose $x \in C$. Then let $\tilde{x} \in \mathbb{R}^{N'}$ be the first $N'$ components of $x$. Then there exists $\tilde{y} \in E$ such that $\|\tilde{y}-\tilde{x} \| < \frac{\epsilon^2}{2}$. Let $y \in E'$ be the element (again abusing notation slightly) $\tilde{y} \times 0 \times 0 \times \cdots$. Then we have the estimate $\|y-x\|^2 = \sum_{k\leq N'} |[\tilde{y}]_k-[x]_k|^2 + \sum_{k\leq N'} |[x]_k|^2 \leq \frac{\epsilon^2}{2}+ \sum_{k\leq N'} a_n^2 \leq \epsilon^2$.

share|improve this answer
add comment

I just want to mention that the "diagonal argument" mentioned in comments is nothing other than the compactness of $\prod_{i \in \mathbb{N}} [a_i, b_i]$ (or equivalently $[0,1]^\mathbb{N}$), an important (and easier to prove) special case of Tychonoff's theorem. Essentially all compactness results in infinite-dimensional spaces flow from this fact, so it would be well to become comfortable with it.

This guarantees there is a subsequence $x_{n_k}$ which converges pointwise (i.e. coordinate-wise) to some $x$. But $|(x_{n_k})_i| \le |a_i|$ with $a_i \in \ell^2$, so by dominated convergence, the convergence also takes place in $\ell^2$.

share|improve this answer
add comment

Theorem. Let $p\in[1,+\infty]$ and $M\subset\ell_p$ is bounded subset such that $$ \lim\limits_{N\to\infty}\sup\{\Vert (0,0,\ldots,0,x_N,x_{N+1},\ldots) \Vert_p:x\in M\}=0 $$ then $M$ is totally bounded.

Proof. Take arbitrary $\varepsilon>0$ then there exist $N\in\mathbb{N}$ such that for all $x\in M$ we have $$ \Vert (0,0,\ldots,0,x_N,x_{N+1},\ldots) \Vert_p<\varepsilon/2\tag{1} $$ Since $M$ is bounded, the set $$ M_{reduced}=\{(x_1,x_2,\ldots,x_N,0,0,\ldots):x\in M\} $$ is bounded. But $M_{reduced}$ is a bounded subset of $\mathbb{R}^N$ so we can find finite $\varepsilon/2$-net $N\subset \ell_p$. This is possible because all norms on finite dimensional spaces are equivalent to euclidean norm, and sets bounded in euclidean norm are always totally bounded. We claim that $N$ is a finite $\varepsilon$-net for $M$. Take arbitrary $x\in M$ and consider $x_{reduced}=(x_1,x_2,\ldots,x_N,0,0\ldots,)\in M_{reduced}$. Then there exist $y\in N$ such that $\Vert x_{reduced}-y\Vert_p<\varepsilon/2$. Then using $(1)$ we get $$ \Vert x-y\Vert_p\leq\Vert x-x_{reduced}\Vert_p+\Vert x_{reduced}-y\Vert_p< \Vert (0,0,\ldots,0,x_N,x_{N+1},\ldots) \Vert_p+\varepsilon/2<\varepsilon $$ Since $x\in M$ is arbitrary, then $N$ is an $\varepsilon$-net for $M$. Since $\varepsilon>0$ is arbitrary and by construction $N$ is finite, then $M$ is totally bounded.

Theorem. Let $1\leq p\leq +\infty$ and $a\in \ell_p \cap c_0$, then the set $$ C=\{x\in\ell_p:\;\forall n\in\mathbb{N}\quad|x_n|\leq |a_n|\} $$ is compact.

Proof. We need to show that $C$ is complete and totally bounded

1) Completeness. Since $C$ is a subset of complete space $\ell_p$, it is enough to show that $C$ is closed. Consider continuous linear functionals $f_n:\ell_p\to\mathbb{R}:x\mapsto x_n$. Since $|f_n(x)|=|x_n|\leq\Vert x\Vert_p$ for all $x\in\ell_p$, then $f_n$ is continuous and obviously linear. Hence $f_n^{-1}([-a_n,a_n])$ is a closed set as preimage of closed set. Note that $$ C=\bigcap\limits_{n\in\mathbb{N}}f_n^{-1}([-a_n,a_n]) $$ so $C$ is closed as intersection of closed sets.

2) Total boundedness. From the formula of the norm in $\ell_p$ it follows that for all $x\in C$ we have $\Vert x\Vert_p\leq\Vert a\Vert_p$, so $C$ is bounded. By the same reasonong for all $x\in C$ we have $$ \Vert (0,0,\ldots,0,x_N,x_{N+1},\ldots) \Vert_p\leq \Vert (0,0,\ldots,0,a_N,a_{N+1},\ldots) \Vert_p $$ Hence $$ \lim\limits_{N\to\infty}\sup\left\{\Vert (0,0,\ldots,0,x_N,x_{N+1},\ldots) \Vert_p:x\in C\right\}\leq \lim\limits_{N\to\infty}\Vert (0,0,\ldots,0,a_N,a_{N+1},\ldots) \Vert_p=0 $$ Since $a\in\ell_p\cap c_0$ then the last limit is $0$, so $$ \lim\limits_{N\to\infty}\sup\left\{\Vert (0,0,\ldots,0,x_N,x_{N+1},\ldots) \Vert_p:x\in C\right\}=0 $$ Thus from previous theorem we see that $C$ is totally bounded.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.