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? 



products in the category of banach spaces
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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 $P$).


