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Exercise:

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My proposal

Recall that

\begin{equation} \|u\|_p := \left( \sum\limits_{j \in \mathbb{Z}} | u_{j} |^{p} \right)^{1/p}, \,\,\, j \in \mathbb{Z}, \forall u = \left( u_{j} \right) \in l_{p}, \, u_{j} \neq 0 \, \text{ if } 1 \leq p < \infty, \end{equation}

Let $u \in l^{p}(\mathbb{Z})$, $V \subset l^{p}(\mathbb{Z})$, and let $\{v_{j}\}_{j}$ be a sequence in $V$ which converges to $u$. The countable union of closed set is closed in $\mathbb{Z}$ when $u_{j} \neq 0$. (unsure about this step!) Because $V$ has finite dimension, we have a basis $\{ u_{1}, ..., u_{k} \}$ of $V$. Also, $u \in$ Span$(u_{1}, ..., u_{k}, u)$. But, we have that $V$ is closed in Span$(u_{1}, ..., u_{k}, u)$ with $v_{n} \to u$, and then $u \in V. \square$

Svetoslav's answer rewritten

Recall that $$ V=\left\{u\in l^p(\mathbb Z): \left\{j\in\mathbb Z:u_j\neq 0\right\} \text{is finite}\right\}\subset l^p(\mathbb Z)$$ where $$\|u\|_p := \left( \sum\limits_{j \in \mathbb{Z}} | u_{j} |^{p} \right)^{1/p}, \,\,\, \forall \,u = \left\{ u_{j} \right\}_{j\in\mathbb Z} \in l_{p},\,\, 1 \leq p < \infty. $$

Let an element be $u \in l^{p}(\mathbb{Z})$, and a sequence of elements $(v^{j})_{j=1}^{\infty}\subset V$. The series \begin{equation*} ||u||_{p}^{p} = \sum\limits_{j \in \mathbb{Z}} |u_{j}|^{p} < \infty, \end{equation*} is (absolutely) convergent which implies that its tail tends to zero. Indeed, let $S_n := \sum\limits_{j\in\mathbb Z,j\leq |n|} |u_{j}|^{p}$ be the partial sums of the series and let $A=\sum\limits_{j \in \mathbb{Z}} |u_{j}|^{p} < \infty$ be its sum. By the definition for convergent series, given $\epsilon >0$, there exists $N\in\mathbb N$ such that for all $n>N$ we have \begin{equation*} |\sum_{ |j| > n } |u_{j}|^{p} |=|A-S_n| < \epsilon. \end{equation*} implying $\sum_{ |j| > n } |u_{j}|^{p} \to 0$, when $n \to \infty$. Take for approximating sequence $\{ v^{j} \}_{j=1}^{\infty} \subset V$: \begin{equation*} v^{j} = (0,0, ..., 0, u_{-j}, u_{-j+1}, ..., u_{-1}, u_{0}, u_{1}, ..., u_{j-1}, u_{j}, 0, 0, ...) \in V \end{equation*} and see that this sequence converges to $u$, i.e \begin{equation*} || v^{j} - u ||_{p}=\sum_{ |k| > j } |u_{k}|^{p} \to 0 \, \text{ when } \, j \to \infty. \end{equation*}

Therefore $V$ is a dense subspace of $l^{p}(\mathbb{Z})$ for $1 \leq p < \infty. \square$

How can you show better that V is a dense subspace of $l^{p}(\mathbb{Z})$?

Svetoslav
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Léo Léopold Hertz 준영
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    $V$ is not finite dimensional . – Svetoslav Sep 14 '15 at 13:10
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    And your proposal is completely messed up :) You should show that for an arbitrary $u\in l^p(\mathbb Z)$ there is a sequence ${v_j}{j=1}^{\infty}\subset V$ such that $|v_j-u|{l^p}\rightarrow 0$ when $j\rightarrow \infty$ – Svetoslav Sep 14 '15 at 13:14
  • How can you say that? Is the following right? The countable union of closed set is closed in $\mathbb{Z}$ when $u_{j} \neq 0$. There is no limit point of 0. – Léo Léopold Hertz 준영 Sep 14 '15 at 13:23

1 Answers1

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Hint:

Notation: An element $u\in l^p(\mathbb Z)$ is a sequence $\{u_j\}_{j\in\mathbb Z}$ (with subscript) and a sequence of elements in $l^p(\mathbb Z)$ we denote with superscript: $\left\{v^k\right\}_{k=1}^{\infty}$

Because $\|u\|_p^p=\sum\limits_{j\in\mathbb Z}{|u_j|^p}<\infty$ (absolutely convergent series) then the tail of this series tends to $0$, i.e $\sum\limits_{|j|>n}{|u_j|^p}\rightarrow 0$ when $n\rightarrow \infty$. Then take for approximating sequence $\{v^j\}\subset V$ : $$v^j=(0,0,...,0,u_{-j},u_{-j+1},..,u_{-1},u_0,u_1,..,u_{j-1},u_{j},0,0,.. )\in V$$ and see that $\|v^j-u\|_p\rightarrow 0$ when $j\rightarrow \infty$

Svetoslav
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