On inifinite product of vector spaces

**bkarpuz** · March 24th 2011, 06:43 PM

Dear MHF members,

I have the following problem.

Problem. For a family of Banach spaces $\{A_{n}\}_{n\in\mathbb{N}}$ , let $A_{\Pi}:=\prod_{k\in\mathbb{N}}A_{k}=\big\{a=(a_{1 },a_{2},\ldots,a_{n},\ldots):\ \|a\|<\infty\big\}$ ,
where $\|a\|:=\sup_{n\in\mathbb{N}}\|a_{n}\|$ .

Show that $A_{\oplus}:=\oplus_{k\in\mathbb{N}}A_{k}=\{a=(a_{1 },a_{2},\ldots,a_{n},\ldots):\ \lim_{n\to\infty}\|a_{n}\|=0\}$ is the closure of the algebraic direct sum $A_{\Sigma}$ in the Banach space $A_{\Pi}$ .
If $\pi$ denotes the quotient map onto the quotient space $A_{\Pi}/A_{\oplus}$ , then show that the norm $\|\pi(a)\|=\limsup_{n\to\infty}\|a_{n}\|$ .

Thanks for being interested.
bkarpuz

**Drexel28** · March 24th 2011, 07:14 PM

Originally Posted by bkarpuz

Dear MHF members,

I have the following problem.

Problem. For a family of Banach spaces $\{A_{n}\}_{n\in\mathbb{N}}$ , let $A_{\Pi}:=\prod_{k\in\mathbb{N}}A_{k}=\big\{a=(a_{1 },a_{2},\ldots,a_{n},\ldots):\ \|a\|<\infty\big\}$ ,
where $\|a\|:=\sup_{n\in\mathbb{N}}\|a_{n}\|$ .

Show that $A_{\oplus}:=\oplus_{k\in\mathbb{N}}A_{k}=\{a=(a_{1 },a_{2},\ldots,a_{n},\ldots):\ \lim_{n\to\infty}\|a_{n}\|=0\}$ is the closure of the algebraic direct sum $A_{\Sigma}:=\sum_{k\in\mathbb{N}}A_{k}$ in the Banach space $A_{\Pi}$ .
If $\pi$ denotes the quotient map onto the quotient space $A_{\Pi}/A_{\oplus}$ , then show that the norm $\|\pi(a)\|=\limsup_{n\to\infty}\|a_{n}\|$ .

Thanks for being interested.
bkarpuz

Presumably you mean that you're considering $A_k$ embedded into $A_\Pi$ in the natural way (all coordinates zero, except the $k^{\text{th}}$ ) and by the 'algebraic direct sum' $A_\Sigma$ you mean $\displaystyle \left\{\sum_{k\in S}A_k:S\subseteq\mathbb{N}\text{ and }\#(S)<\infty\right\}$ . If so, let's show that $\overline{A_\Sigma}=A_\oplus$ . To see that $A_\oplus\subseteq\overline{A_\Sigma}$ let $(a_n)_{n\in\mathbb{N}}\in A_\oplus$ . Then, for any $\varepsilon>0$ we have, since $\|a_n\|_n\to 0$ there exists some $N\in\mathbb{N}$ such that $n\geqslant N\implies \|a_n\|_n<\varepsilon$ . Define then $(b_n)_{n\in\mathbb{N}}$ by

$b_n=\begin{cases}a_n & \mbox{if}\quad n\leqslant 0\\ 0 & \mbox{if}\quad n>N\end{cases}$

Evidently then $b_n\in A_\Sigma$ and moreover $\displaystyle \|a_n-b_n\|=\sup_{\substack{n\in\mathbb{N}\\ n>N}}\|a_n\|\leqslant \varepsilon$ . Since $\varepsilon$ was arbitrary the conclusion follows (technically you have to adjust the above argument so that $b_n\ne a_n$ but that's easy enough).

Conversely, suppose that $(a_n)_{n\in\mathbb{N}}\notin A_\oplus$ . Then, there exists some $\varepsilon>0$ such that for every $N\in\mathbb{N}$ there exists some $N'\geqslant N$ such that $\|a_{N'}\|_{N'}\geqslant \varepsilon$ . Evidently this implies that for every $N\in\mathbb{N}$ one has that $\displaystyle \sup_{\substack{n\in\mathbb{N}\\ n\geqslant N}}\|a_n\|_{n}\geqslant \varepsilon$ . So, let $(b_n)_{n\in\mathbb{N}}\in A_\Sigma$ be arbitrary. Then, by definition there exists some $N\in\mathbb{N}$ such that for every $n\geqslant N$ one has that $b_n=0$ . Consequently

$\displaystyle \begin{aligned}\|(a_n)-(b_n)\| &=\sup_{n\in\mathbb{N}}\|a_n-b_n\|_n\\ &\geqslant \sup_{\substack{n\in\mathbb{N}\\ n\geqslant N}}\|a_n-b_n\|_n\\ &= \sup_{\substack{n\in\mathbb{N}\\ n\geqslant N}}\|a_n\|_n\\ &\geqslant \varepsilon\end{aligned}$

Thus, $(b_n)\notin B_{\varepsilon}((a_n))$ . Since $(b_n)\in A_\Sigma$ was arbitrary it follows that $(a_n)\notin \overline{A_\Sigma}$ .

That should give you the basic idea...I might have skipped a minor detail in the case $(a_n)\in A_\Sigma$ but it's easily fixable.

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