Are inner products always dominated by the norm of tensor products?

Question

Prove or disprove the following possible generalization of Cauchy-Schwarz inequality: Let $H$ be a Hilbert space and let $A_1, \dots A_n, B_1, \dots B_n \in H$. Then

$|\sum_{i=1}^{n} \langle A_i, B_i \rangle| \leq || \sum_{i=1}^{n} A_i \otimes B_i ||$

where the norm is in the tensor Hilbert space $H \hat{\otimes} H$. In essence we are asking whether a bunch of inner products are always dominated by the norm of the corresponding tensor products.

I say that it is a generalization of Cauchy-Schwarz inequality because if one takes $n=1$ above, one gets

$|\langle A_1, B_1\rangle| \leq ||A_1 \otimes B_1||=||A_1||||B_1||$

which is the familiar Cauchy-Schwarz inequality. I have tried applying standard tricks used to prove Cauchy-Schwarz, such as using the fact that the norm of $A \wedge B = A\otimes B - B \otimes A$ is non-negative but to no avail.

Answer 1

Given a Banach space $X$, the quadratic form $q: X \times X^\ast \to \mathbb{C}$ given by $q(x,x^\ast) = \langle x, x^\ast \rangle$ is bounded. Indeed, it is contractive. Therefore it lifts to the projective tensor product $q: X \hat\otimes X^\ast \to \mathbb{C}$. Now, use that, when $X$ is a Hilbert space you have self-duality.

Another approach is that you can identify $H \hat\otimes H$ with the trace class operators $S^1[H] \subset B(H)$. It is easy to check that $$ \mathrm{Tr} ( A \otimes B ) = \langle A, B \rangle. $$ So you only need to see that the trace is bounded in $S^1[H]$, which follows by definition.

Answer 2

Certainly false. Here's a counterexample: Let $A_1 = B_1 = e_1$ and $A_2 = B_2 = e_2$ where $e_1, e_2 \in H$ and are orthonormal. Then $$ \sum_{i=1}^2 \langle A_i, B_i\rangle = 2$$ but, \begin{align} \Big\| \sum_{i=1}^2 A_i \otimes B_i\Big\| &= \sqrt{\langle e_1, e_1 \rangle\langle e_1, e_1 \rangle + \langle e_1, e_2 \rangle\langle e_1, e_2 \rangle + \langle e_2, e_1 \rangle\langle e_2, e_1 \rangle + \langle e_2, e_2 \rangle\langle e_2, e_2 \rangle} \\ &= \sqrt 2 \end{align} which is a contradiction.