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Let $H,K$ be non-zero Hilbert spaces. Then the algebraic tensor product $H \odot K$ is an inner product space for the unique inner product determined by $$\langle h \otimes k, h' \otimes k'\rangle = \langle h, h'\rangle_H \langle k, k' \rangle_K$$

I am trying to prove that $H \odot K$ is a Hilbert space (i.e. $H \odot K$ is complete for the norm induced by this inner product) implies that $\dim H <\infty$ or $\dim K < \infty$.

Let us prove the contrapositive, i.e. so suppose that $H$ and $K$ are both infinite dimensional. How can I show that there is a Cauchy sequence in $H \odot K$ that does not converge?

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    $\begingroup$ The statement is false unless $H \ne 0 \ne K$, so you probably want to assume that. \\ It is clear (when $K \ne 0$) that $H$ infinite dimensional implies $H \otimes K$ infinite dimensional (why $H \odot K$?). Surely you mean that you want to show that, if $H$ is infinite dimensional, then $H \otimes K$ is not complete? In that case, why not see if you can use the infinite dimensionality of $H$ to manufacture a non-convergent Cauchy sequence? $\endgroup$ – LSpice Oct 21 at 18:24
  • $\begingroup$ @LSpice Thanks for the comment. I made some edits. Finding this non-convergent Cauchy sequence is exactly what I can't. Do you have any hints for this? $\endgroup$ – user839372 Oct 21 at 18:35
  • $\begingroup$ Just out of curiosity: you rejected my edit that inserted the < missing from the title, and then inserted it yourself. Why? $\endgroup$ – LSpice Oct 21 at 22:04
  • $\begingroup$ I didn't have a specific sequence in mind, and, in fact, I think that the result is still not true. If $K$ is 1-dimensional, then $H \otimes K$ is isometric to $H$, so is complete even if $H$ is infinite dimensional; and, more generally, $H \otimes \mathbb C^n$ is isometric to $H^{\oplus n}$, which is complete for $n < \infty$. Are you sure that you don't mean to prove that $\dim(H)$ or $\dim(K)$ is finite? $\endgroup$ – LSpice Oct 21 at 22:07
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    $\begingroup$ As per en.wikipedia.org/wiki/… you just need an infinite matrix with finite sum of squares of the entries but not of finite rank; a diagonal one will do. $\endgroup$ – Max Oct 28 at 0:16
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It is very easy to exhibit a non-converging Cauchy sequence in $H\odot K$, but actually proving that it does not converge is a bit more difficult. The approach suggested by @Max, although somewhat sophisticated, is perhaps the easiest way to pin down all of the details. Here is a pedestrian way to describe this method.

For each pair of vectors $(x, y)\in H\times K$, consider the bounded linear operator $$ T_{x, y}:H\to K $$ given by $$ T_{x, y}(z) = \langle x,z\rangle y, \quad\forall z\in H. $$ (For simplicity I am assuming that we are working with real vector spaces but this can be fixed in the complex case by seeing $T_{x, y}$ as an operator defined on $\bar H$, namely the Hilbert space made out of $H$ by adopting the new scalar multiplication operation $\lambda \cdot x:= \bar \lambda x$).

It is easy to see that the map $$ (x, y)\in H\times K\mapsto T_{x, y}\in \mathscr {B}(H,K) $$ is bilinear so, by the universal property of (algebraic) tensor products, there exists a unique linear map $$ T :H\odot K\to \mathscr {B}(H,K), $$ such that $T (x\otimes y) = T_{x,y}$. We next claim that $$ \langle T (\xi )x,y\rangle = \langle \xi ,x\otimes y\rangle , \tag {1} $$ for every $\xi $ in $H\odot K$, $x\in H$, and $y\in K$. To prove this write $\xi =\sum_{i=1}^nx_i\otimes y_i$, and notice that $$ \langle T (\xi )x,y\rangle = \sum_{i=1}^n \langle T_{x_i,y_i}(x),y\rangle = $$ $$ = \sum_{i=1}^n \langle \langle x_i,x\rangle y_i,y\rangle = \sum_{i=1}^n \langle x_i,x\rangle \langle y_i,y\rangle = $$ $$ = \sum_{i=1}^n \langle x_i\otimes y_i,x\otimes y\rangle = \langle \xi ,x\otimes y\rangle, $$ proving the claim.

Note that the range of each $T_{x,y}$ is the one-dimensional space spanned by $y$, so $T_{x,y}$ has rank 1. Furthermore, every $\xi $ in $H\odot K$ may be writen as $\xi =\sum_{i=1}^nx_i\otimes y_i$, so $$ T (\xi ) = \sum_{i=1}^n T_{x_i,y_i}, $$ so we see that $T (\xi )$ has rank at most $n$, hence finite, for every $\xi $ in $H\odot K$.

The final contradiction will be achieved by proving that, if $H$ and $K$ are infinite dimensional, and $H\odot K$ is complete, there exists a vector $\xi $ in $H\odot K$ such that $T (\xi )$ has infinite rank.

Let us therefore assume from now on that $H$ and $K$ are both infinite dimensional, so we may find (necessarily infinite) orthonormal bases $\{e_i\}_{i\in I}$, and $\{f_j\}_{j\in J}$, for $H$ and $K$ respectively, and it is well known that $\{e_i\otimes f_j\}_{(i, j)\in I\times J}$ is an orthonormal basis for $H\otimes K$.

Assuming by contradiction that $H\odot K$ is complete, observe that $H\odot K$ coincides with its completion, so $H\odot K = H\otimes K$.

Choose infinite subsets $\{i_n:n\in {\mathbb N}\}\subseteq I$, and $\{j_n:n\in {\mathbb N}\}\subseteq J$, and let $$ \xi = \sum_{n=1}^\infty {1\over n^2} (e_{i_n}\otimes f_{j_n}). $$ This is well defined because the series is clearly absolutely convergent and we are working in a complete space!

Using (1) we have that $$ T (\xi )e_{i_n} = \sum_{j\in J} \langle T (\xi )e_{i_n}, f_j\rangle f_j = \sum_{j\in J} \langle \xi , e_{i_n}\otimes f_j\rangle f_j = \langle \xi , e_{i_n}\otimes f_{j_n}\rangle f_{j_n} = {1\over n^2} f_{j_n}. $$ This shows that ${1\over n^2} f_{j_n}$ lies in the range of $T (\xi )$, and hence that $T (\xi )$ has infinite rank, a contradiction.

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  • $\begingroup$ Definitely harder than I thought! Many thanks! I will go in detail through it soon! $\endgroup$ – user839372 Oct 28 at 19:32
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    $\begingroup$ You are welcome. And please fell free to ask for any clarification if needed! $\endgroup$ – Ruy Oct 28 at 19:47
  • $\begingroup$ Thanks! That's very kind of you! $\endgroup$ – user839372 Oct 28 at 19:48

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