Added tensor gymnastics.
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@@ -53,6 +53,32 @@
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In constructing the \hyperref[projective tensor product]{definition:projective-tensor-product}, it may be more natural to obtain its topology as a projective topology using its universal property. However, doing so requires taking a least upper bound across \textit{all continuous linear maps defined on} $E \times F$, a collection too big to be a set. As such, constructing it as a projective topology is logically dubious, or at the very least beyond my abilities.
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\end{remark}
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\begin{proposition}
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\label{proposition:projective-tensor-product-dual}
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Let $E, F$ be locally convex space over $K \in \RC$, then
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\[
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(E \wh \otimes_\pi F)^* = (E \otimes_\pi F)^* \iso L^2(E, F; K) \iso L(E; F^*) \iso L(F; E^*)
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\]
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where:
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\begin{enumerate}
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\item The dual pairing between $E \otimes_\pi F$ and $L^2(E, F; K)$ is given by
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\[
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\angles{\sum_{k = 1}^n x_k \otimes y_k, \lambda}_{E \otimes_\pi F} = \sum_{k = 1}^n \lambda(x_k, y_k)
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\]
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\item The dual pairing between $E \otimes_\pi F$ and $L(E; F^*)$ is given by
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\[
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\angles{\sum_{k = 1}^n x_k \otimes y_k, T}_{E \otimes_\pi F} = \sum_{k = 1}^n \dpn{y_k, Tx_k}{F}
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\]
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\item The dual pairing between $E \otimes_\pi F$ and $L(F; E^*)$ is given by
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\[
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\angles{\sum_{k = 1}^n x_k \otimes y_k, T}_{E \otimes_\pi F} = \sum_{k = 1}^n \dpn{x_k, Ty_k}{E}
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\]
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\end{enumerate}
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\end{proposition}
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\begin{definition}[Cross Seminorm]
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