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Let $\{H_i\}_{i\in I}$ be a family of Hilbert spaces, defined $$H = \{f\in \Pi_{i\in I} H_i, \sum_{i\in I}|f(i)|^2<\infty\}$$ and inner product $\langle f,g\rangle : = \sum_{i\in I} \langle f(i),g(j) \rangle$. If $I$ is uncountable, then $\sum_{i\in I}$ is defined to be the supremum over all finite sums.

My question: Is $H$ the product of $\{H_i\}_{i\in I}$ in the category of Hilbert spaces where morphisms are bounded linear maps (or unitary maps)?

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I guess no. Consider e.g. the linear maps $\varphi_i:\Bbb C\to H_i$ which select a unit vector $h_i\in H_i$ (i.e. $\varphi_i(1)=h_i$). Then we would need the map $\varphi:\Bbb C\to \prod_i H_i$ that maps $1\mapsto (h_i)_i$. – Berci Feb 12 '14 at 22:39
up vote 2 down vote accepted

No. But it has the following universal property: $H$ represents the contravariant functor which maps a Hilbert space $K$ to the set of bounded families of bounded linear maps $g_i : K \to H_i$ which are $l^2$-summable in the sense that $\sum_{i \in I} ||g_i(x)||^2<\infty$ for all $x \in K$. Because then we can define $g : K \to H$ by $g(k)=(g_i(x))$.

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