How can we show that for $\lambda \in c_0$, $\{\lambda x \mid x \in B_{\ell^1}\} \subseteq \ell^1$ is closed?

Question

I would like to show that for $\lambda \in c_0$, 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 $T_\lambda: \ell^1 \to \ell^1, T_\lambda(x) = \lambda x$.

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

Question: How can we show that $B_\lambda$ fulfills condition (2.), i.e. how can we show that $B_\lambda$ is closed?

Answer 1

The problem can be rephrased as follows. Let $\lambda\in c_{0}$. Define $T\colon \ell^{1}\to \ell^{1}$ by $Tx := (\lambda (n) x(n))_{n\in\mathbb{N}}$. Note that $T$ is well-defined because $\lambda$ is a bounded sequence. We have \begin{equation} T(B_{\ell^{1}}) = B_{\lambda} = \{(\lambda (n) x(n) )_{n\in\mathbb{N}} : x\in B_{\ell^{1}} \}, \end{equation} where $B_{\ell^{1}}$ denotes the closed unit ball of $\ell^{1}$. We will show that $T(B_{\ell^{1}})$ is closed in $\ell^{1}$. Let $(x_{k})_{k\in\mathbb{N}}$ be a sequence in $B_{\ell^{1}}$ such that $(Tx_{k})_{k\in\mathbb{N}}$ converges to some $y\in \ell^{1}$. Define $x\colon \mathbb{N}\to\mathbb{K}$ by \begin{equation} x(n) := \begin{cases} (\lambda (n))^{-1} y(n) & \text{if } \lambda (n) \neq 0, \\ 0 & \text{if } \lambda (n) = 0 . \end{cases} \end{equation} Note that for each $n\in\mathbb{N}$ we have \begin{equation} \lambda (n) x(n) = \lim_{k\to\infty} \lambda (n) x_{k}(n) = y(n) . \tag{1} \end{equation} We claim $x\in B_{\ell^{1}}$ and $Tx = y$. Let $n\in\mathbb{N}$. As the sequence $(x_{k})_{k\in\mathbb{N}}$ belongs to $B_{\ell^{1}}$, we have $\sum_{j=1}^{n}|x_{k}(j)| \leq 1$ for each $k\in\mathbb{N}$. Since $|x(j)| \leq \liminf_{k\to\infty} |x_{k}(j)|$ for each $j\in\mathbb{N}$ by $(1)$, it follows that \begin{equation} \sum_{j=1}^{n}|x(j)| \leq \liminf_{k\to\infty} \sum_{j=1}^{n}|x_{k}(j)| \leq 1 . \end{equation} Hence $x\in B_{\ell^{1}}$. We now see from $(1)$ that $Tx = y$. This shows that $T(B_{\ell^{1}})$ is closed in $\ell^{1}$.