comparing norms of tensor product of two Hilbert spaces

Question

Suppose $H_1$ and $H_2$ are two Hilbert spaces with dimension $n$ and $m$, for $ x \in H_1 \otimes H_2$

consider $$\|x\|_\pi = \inf \left\{ \sum_{i=1}^n \|a_i\| \|b_i\| : x = \sum_{i} a_i \otimes b_i \right\}$$ where $x \in A \otimes B$.

$ \|a\otimes b\|_{2}^2 =\langle a \otimes b,a\otimes b\rangle = \langle a,a\rangle_1 \, \langle b,b\rangle_2 \quad \mbox{for all } a \in H_1 \mbox{ and } b \in H_2 $ respectively.

Questions: 1. We know norms on finite dimensional spaces are equivalent. Hence there are $c_1$ and $c_2$ such that $$ c_1\|x\|_2 \leq \|x\|_\pi \leq c_2\|x\|_2 $$ since $\|x\|_\pi$ is largest cross norm so $c_1$ can be taken equal to $1$. what about $c_2$

2.What about when $H_1$ and $H_2$ are arbitrary Hilbert spaces, are these two norm equivalent?

3. Where can I know more about such comparing norms on tensor of Hilbert spaces. (any book or article)

thanks, and pardon me for poor English.

Answer 1

The norms are equivalent if and only if $\min(\dim(H_1),\dim(H_2))<\infty$.

One way to think about this is to dualize. The space of bounded linear functionals on $(H_1 \otimes H_2, \Vert\cdot\Vert_\pi)$ can be identified naturally with the space of bounded linear maps from $H_1^*$ to $H_2$. On the other hand, the space of bounded linear functionals on $(H_1 \otimes H_2, \Vert\cdot\Vert_2)$ can be identified naturally with the space of Hilbert-Schmidt operators $H_1^*\to H_2$. If both $H_1$ and $H_2$ are infinite-dimensional then there are bounded linear maps $H_1^*\to H_2$ that are not Hilbert-Schmidt, e.g. the "identity map on $\ell_2$" if both $H_1$ and $H_2$ are separable.

A good source to learn about tensor norms on Banach spaces is Ryan's book Introduction to Tensor Products of Banach Spaces, or if you want something more condensed and more advanced, the classic book of Defant and Floret.