# comparing norms of tensor product of two Hilbert spaces

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?

1. 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.

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.