## products in the category of banach spaces

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Let `$\{X_{\alpha} \}_{\alpha \in A}$` be a collection of Banach spaces. It is easy to show that `$ P = \{(x_{\alpha}) : {\rm sup}_{\alpha} \|x_{\alpha} \| < \infty \} $` with `$\| (x_{\alpha} ) \| = {\rm sup}_{\alpha} \| x_{\alpha} \|$` is a banach space.

If the indexing set $A$ is finite, then it is easy to show that $P$ (with the natural projection maps) is the product of `$\{X_{\alpha} \}_{\alpha \in A}$` in the category of banach spaces.

Assume $Y$ is a banach space and $f_{\alpha} : Y \to X_{\alpha}$ is a linear continuous map for each $\alpha$. If `${\rm sup}_{\alpha} \|f_{\alpha} \| < \infty$` then it is easy to see that the induced linear map $Y \to \Pi X_{\alpha}$ has image a subset of $P$ and is continuous. If this condition is not satisfied then it is not clear to me that there exists a continuous linear map $g:Y \to P$ such that $\pi_{\alpha} \circ g = f_{\alpha}$ for all $\alpha$.

Is there any way to prove that $P$ is the categorical product of the collection `$\{X_{\alpha} \}_{\alpha \in A}$` when $A$ is infinite?

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In the category of Banach spaces and bounded linear maps, $P$ is NOT the product of $(X_\alpha)$ - that category does not have infinite products. – Yemon Choi May 19 2011 at 1:16
Just take each $X_\alpha$ and $Y$ to be a copy of the ground field and let $y_\alpha$ be multiplication by $\lambda(\alpha)$ where $\lambda: A \to {\bf R}$ is your favourite unbounded function. – Yemon Choi May 19 2011 at 1:47
(Sorry, in previous comment I meant "let $f_\alpha$ be multiplication...") – Yemon Choi May 19 2011 at 1:47
thanks Yemon. If you post this as an answer I will accept it – Daniel Barter May 19 2011 at 2:04

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@Zen Harper: the categorical product would be a Banach space $B$ with the property that a collection of maps $\{f_\alpha\colon Y\to X_\alpha\}$ was the same as a single map $f\colon Y\to B$. This is the "universal mapping property" of the product, hence the term "categorical". In the case of Banach spaces such a space $B$ cannot exist, and you've seen why: the condition on the $f_\alpha$ is that they be individually bounded, but their norms can be unbounded in $\alpha$, and it's not possible to bundle this up into a single space $B$ and bounded maps to it. – Tom Church May 19 2011 at 3:30
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## 1 Answer

[Promoted from comments, as requested]

The category of Banach spaces and bounded linear maps does not have arbitrary infinite products. Here is a simple example, using the notation of the question: take each $X_\alpha$ and $Y$ to be a copy of the ground field, fix an unbounded function $\lambda:A\to{\mathbb R}$, and let `$f_\alpha: Y \to X_\alpha$` be multiplication by `$\lambda(\alpha)$`. It is then easy to see that the cone `$(f_\alpha: Y\to X_\alpha)_{\alpha\in A}$` cannot factor through the cone `$(\pi_\alpha: P \to X_\alpha)_{\alpha\in A}$`. This shows that $P$ is not the product of the $X_\alpha$, and a little more work shows that no such product can exist (assume it does and then show it would have to be isomorphic to $P$).

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