Question

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?

Answer 1

[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 $P$).