# On inifinite product of vector spaces

• Mar 24th 2011, 06:43 PM
bkarpuz
On inifinite product of vector spaces
Dear MHF members,

I have the following problem.

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

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

Thanks for being interested.
bkarpuz
• Mar 24th 2011, 07:14 PM
Drexel28
Quote:

Originally Posted by bkarpuz
Dear MHF members,

I have the following problem.

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

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

Thanks for being interested.
bkarpuz

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

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

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

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

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

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