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A Hilbert algebra is an inner product space that is also a *-algebra where the various operations and structures interact according to some axioms. One of those axioms is that the linear operation given by left multiplication by an element must be continuous. That is $x \mapsto yx$ is bounded for all $y$. You can take the algebraic tensor product of hilbert algebras and turn it into something that might be a Hilbert algebra by defining all the structures in the obvious way. (Various arguments are required to show these definitions are consistent, as is always the case with a Tensor product, but I have that under control.) The problem is I don't see why left multiplication is continuous. Without the help of an orthonormal basis of $U$ the Hilbert Algebra, which need not exist, I just don't see how to work this out. Please only use purely algebraic concepts, or general analysis things, but no high-powered machinery from the subject of Hilbert Algebras that could possibly be logically dependent on the fact I'm trying to prove.

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Oh, the essential trick I missed was the recognization that we can get an ON set that spans all the vectors involved as factors in an arbitrary sum of fundamental tensors. – Jeff Aug 6 '12 at 7:55
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Let $U_i$ be the finite list of Hilbert algebras that I'm tensoring. WLOG I can assume the left multiplication $L$ is by a fundamental tensor $\otimes x_i$. Using the universal property of tensor products, we see that there is a linear map on $\otimes U_i$ to itself that, on fundamental tensors, multiplies by $x_i$ in the $i_{th}$ factor and then fixes the other factors. Since $L$ is a composition of such linear maps, it suffices to show that each such is continuous. Here's the trick: let $\sum_{j} \otimes_i x_{ij}$ be a general tensor, and in each $U_i$ find an ON set that spans all the $x_{ij}$ for that $i$. This shows that $\sum_{j} \otimes_i x_{ij}$ can be recast as a linear combination of mutually orthogonal fundamental tensors, from which the required boundedness estimates are clear.

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