Show that the set $$ B = \left\lbrace(x_n) \in \ell^1 : \sum_{n\geq 1} n|x_n|\leq 1\right\rbrace$$ is compact in $\ell^1$. Hint: You can use without proof the diagonalization process to conclude that every bounded sequence $(x_n)\in \ell^\infty$ has a subsequence $(x_{n_k})$ that converges in each component, that is $\lim_{k\rightarrow\infty} (x_{n_k}^{(i)})$ exists for all i. Moreover, sequences in $\ell^1$ are obviously closed by the $\ell^1$-norm.

My try: Every bounded sequence $(x_n) \in \ell^\infty$ has a subsequence $(x_{n_k})$ that converges in each component. That is $\lim_{k\rightarrow\infty} (x_{n_k}^{(i)})$ exists for all i. .And all sequences in $\ell^1$ are bounded in $\ell^1$-norm. I want to show that every sequence $(x_n) \in B$, has an Cauchy subsequence. Choose an N and M such that for $l,k > M$ such that $|x_{n_k}^{(i)} - x_{n_l}^{(i)}| < \frac{1}{N^2}$ Then $$\sum_i^N |x_{n_k}^{(i)} - x_{n_l}^{(i)}| + \sum_{i = N+1} ^\infty |x_{n_k}^{(i)} - x_{n_l}^{(i)}| \leqslant \frac{1}{N} + \frac{1}{N+1} \sum_{i = N+1} ^\infty i|x_{n_k}^{(i)} - x_{n_l}^{(i)}| \leqslant \frac{3}{N+1}$$ It feels wrong to compine $M,N$ like this, is it? what can I do instead?

  • $\begingroup$ I don't understand the first sentence: you mean that a bounded sequence of real numbers has a convergent subsequence, right? But it will help if you can show that you can extract, when you work with sequences whose terms are in $\ell^1$, that will work for all coordinates. $\endgroup$ – Davide Giraudo Dec 30 '12 at 11:38
  • $\begingroup$ Thanks, I corrected it now, can you please see if its wrong now? $\endgroup$ – Johan Dec 30 '12 at 12:09

We can use and show the following:

Let $K\subset \ell^1$. This set has a compact closure for the $\ell^1$ norm if and only if the following conditions are satisfied:

  • $\sup_{x\in K}\lVert x\rVert_{\ell^1}$ is finite, and
  • for all $\varepsilon>0$, we can find $N$ such that for all $x\in K$, $\sum_{k\geqslant N}|x_k|<\varepsilon$.

These conditions are equivalent to precompactness, that is, that for all $r>0$, we can find finitely many elements $x^1,\dots,x^N$ such that the balls centered at $x^j$ and for radius $r$ cover $K$.

  • $\begingroup$ Thanks! I have not seen this theorem, cant you prove it with the Hint? what is wrong with my try? $\endgroup$ – Johan Dec 30 '12 at 12:19

To prove compactness of $B$ you can use the following theorem:

A set $A \subseteq \ell_1$ is compact if and only if

  1. $A$ is closed
  2. $A$ is bounded
  3. $\sup_{x \in A} \sum_{j \geq n} |x_j| \to 0$ as $n \to \infty$

How to prove these properties of $B$?

  1. Let $X:= \{x \in \ell_1; \sum_j j \cdot |x_j|<\infty\}$. Define $$\|x\|_X := \sum_j j \cdot |x_j| \qquad (x \in X)$$ Then $(X,\|\cdot\|_X)$ is a normed space. Define $$T:X \to \ell_1, x=(x_j)_j \mapsto (j \cdot x_j)_j $$ Then $T$ is linear, surjective and isometric (in particular continuous) and therefore we conclude that $B=T^{-1}(B[0,1])$ is closed as a pre-image of a closed subset.
  2. Let $x=(x_n)_n \in B$, then $$\|x\|_1 = \sum_{n \geq 1} |x_n| \leq \sum_{n \geq 1} n \cdot |x_n| \leq 1$$ where we used the definition of $B$ in the last equation. Hence $$\sup_{x \in B} \|x\|_1 \leq 1$$ which means that $B$ is bounded.
  3. Use the following estimate: For all $x=(x_n)_n \in B$ we have $$\sum_{j \geq n} |x_j| = \frac{1}{n} \sum_{j \geq n} n \cdot |x_j| \leq \frac{1}{n} \cdot \sum_{j \geq n} j \cdot |x_j| \leq \frac{1}{n}$$

I think you mess things up. You wrote

Every convergent sequence $(x_n) \in \ell^\infty$ has a convergent subsequence $(x^{n_k})$. That converges component wise.

It doesn't make sense at all: If you consider a convergent sequence $(x_n)_{n} \in \ell_\infty$, then it's trivial that there exists a convergent subsequence (since the whole sequence is convergent). Probably you wanted to consider a sequence of sequences, i.e $(x^n)_n \subseteq \ell_\infty$ (i.e. $x^n \in \ell_\infty$). Similar for the next sentence:

And all sequences in $\ell^1$ are bounded in $\ell^1$-norm.

If you take one (!) sequence $x:=(x_n)_{n} \in \ell^1$, then (by definition) $\|x\|_1 < \infty$. But: If you consider a sequence of sequences, i.e. $(x^n)_n \subseteq \ell_1$, then it's not trivial that the sequence (of sequences) is bounded, i.e.

$$\sup_n \|x^n\|_1 < \infty$$

You have to differentiate between

  1. a sequence $x:=(x_n)_n$ which is an element of $\ell_1$, i.e. $\|x\|_1<\infty$
  2. a sequence in $\ell_1$, i.e. $(x^n)_{n \in \mathbb{N}} \subseteq \ell_1$ where $x^n \in \ell_1$ for all $n \in \mathbb{N}$.
  • $\begingroup$ Great Theorem, but I have not seen it before... Is it hard to prove? I rewrote the text, there was some typos, sorry $\endgroup$ – Johan Dec 30 '12 at 12:16
  • $\begingroup$ @Johan It's not that difficult to prove. When I attended functional analysis, it was an exercise left to the students. Probably you'll find this theorem (and a proof) in most books about functional analysis. (And I recognized that you rewrote the text, but I'm still the same oppinion.) $\endgroup$ – saz Dec 30 '12 at 12:38
  • $\begingroup$ I found one, but I cannot manage to prove 1. can u give me a hint? $\endgroup$ – Johan Dec 30 '12 at 13:46
  • 1
    $\begingroup$ @Johan I added one. $\endgroup$ – saz Dec 30 '12 at 14:15
  • $\begingroup$ @Johan Sorry, there was a typo in the definition of $T$. $\endgroup$ – saz Dec 30 '12 at 15:15

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