Is the family of compact sets $\{\lambda x \mid x \in B_{\ell^1}\} \subseteq \ell^1$, $\lambda \in c_0$, cofinal among all compact sets?

Question

For each $\lambda \in c_0$, the following set is a compact subset of $\ell^1$ with the norm topology: $$ B_\lambda = \{\lambda x \mid x \in B_{\ell^1}\} \subseteq \ell^1 $$ Equivalently, this is the image of $B_{\ell^1}$ under the map $T_\lambda: \ell^1 \to \ell^1, T_\lambda(x) = \lambda x$.

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Indeed, let's begin by showing that each $B_\lambda \subseteq \ell^1$, for $\lambda \in c_0$, is a compact subset of $\ell^1$.

By A relative compactness criterion in $\ell^p$ , the compactness of $B_\lambda$ is equivalent to all these three conditions:

1. $B_\lambda$ is bounded

2. $B_\lambda$ is closed

3. $$\underset{N \to \infty}{\operatorname{lim}} \underset{y \in B_\lambda}{\operatorname{sup}} \sum_{n = N}^\infty |y_n| = 0$$

Condition (1.) follows from $$ \|\lambda x\|_1 \leq \|\lambda\|_\infty \|x\|_1 \leq \|\lambda\|_\infty $$

Condition (2.) can be seen as follows: If $\lambda x_k \to y$, define $z$ by $$ z_n = \begin{cases} \frac{y_n}{\lambda_n}, & \lambda_n \neq 0 \\ 0, & \lambda_n = 0 \\ \end{cases} $$ Clearly $\lambda z = y$. It remains to show that $z \in B_{\ell^1}$. By Fatou's Lemma, we have $$ \|z\|_1 = \sum_{n = 1}^\infty |z_n| \leq \sum_{n = 1}^\infty \underset{k \to \infty}{\operatorname{lim inf}} |x_{k, n}| \leq \underset{k \to \infty}{\operatorname{lim inf}} \sum_{n = 1}^\infty |x_{k, n}| = \underset{k \to \infty}{\operatorname{lim inf}} \|x_k\|_1 \leq 1 $$

Condition (3.) is also easy to show: For $\varepsilon > 0$, there is $N_\varepsilon$ such that for all $n \geq N_\varepsilon$, $|\lambda_n| \leq \varepsilon$. Hence $$ \underset{y \in B_\lambda}{\operatorname{sup}} \sum_{n = N_\varepsilon}^\infty |y_n| = \underset{x \in B_{\ell^1}}{\operatorname{sup}} \sum_{n = N_\varepsilon}^\infty |\lambda_n x_n| \leq \varepsilon \underset{x \in B_{\ell^1}}{\operatorname{sup}} \sum_{n = N_\varepsilon}^\infty |x_n| \leq \varepsilon \underset{x \in B_{\ell^1}}{\operatorname{sup}} \|x\|_1 \leq \varepsilon $$ Hence $$\underset{N \to \infty}{\operatorname{lim}} \underset{y \in B_\lambda}{\operatorname{sup}} \sum_{n = N}^\infty |y_n| = 0$$

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Now, let's say that a family of compact sets $\mathcal{A}$ is cofinal in the set of all compact subsets of $\ell^1$ if for each compact $K \subseteq \ell^1$, there is some $K' \in \mathcal{A}$ with $K \subseteq K'$.

Is the family $\{B_\lambda\}_{\lambda \in c_0}$ cofinal in the set of all compact subsets of $\ell^1$?

Equivalently, is the following statement true? $$ \forall K \subseteq \ell^1 \text{ compact } \exists \lambda \in c_0 \ \forall y \in K \ \exists x \in \ell^1 \ \forall n \in \mathbb{N} : y_n = \lambda_n x_n $$

Answer 1

Yes, this family is cofinal in the set of compact subsets of $\ell^1$.

Consider - to avoid trivialities - a compact $K \subset \ell^1$ such that $$\mu_k = \sup_{x \in K} \sum_{n = k}^{\infty} \lvert x_n\rvert > 0$$ for all $k$. Evidently $(\mu_k)_{k \in \mathbb{N}}$ is monotonic (non-increasing), and by property 3, $\mu \in c_0$. Assume without loss of generality $\mu_0 = 1$.

Now, put $n_0 = 0$, for every $m > 0$ let $n_m = \min \{k > n_{m-1} : \mu_k \leqslant 4^{-m}\}$ and define $$\lambda_k = 2^{1-m} \quad \text{for } n_m \leqslant k < n_{m+1}\,.$$

Clearly $\lambda \in c_0$, and for every $x \in K$ we have \begin{align} \sum_{n = 0}^{\infty} \frac{\lvert x_n\rvert}{\lambda_n} &= \sum_{m = 0}^{\infty} \sum_{n = n_m}^{n_{m+1}-1} \frac{\lvert x_n\rvert}{\lambda_n} \\ &= \sum_{m = 0}^{\infty} 2^{m-1}\sum_{n = n_m}^{n_{m+1}-1} \lvert x_n\rvert \\ &\leqslant \sum_{m = 0}^{\infty} 2^{m-1} \mu_{n_m} \\ &\leqslant \sum_{m = 0}^{\infty} 2^{m-1}4^{-m} \\ &= \sum_{m = 0}^{\infty} 2^{-(m+1)} \\ &= 1\,, \end{align} that is, $K \subset B_\lambda$.