Norm equality for operators on a Hilbert tensor product

Question

Ok, this could be tough. Suppose you are given bounded linear operators, $\{A_n\}_{n=1}^N\subset\mathcal{B}(\mathcal{H})$ on a Hilbert space $\mathcal{H}$ and $\{B_n\}_{n=1}^N\subset\mathcal{B}(\mathcal{K})$ on another Hilbert space $\mathcal{K}$, with $N\geq1$.

If $\{U_n\}_{n=1}^N\subset\mathcal{U}(\mathcal{K})$ is a family of unitary operators on $\mathcal{K}$, is it true that $$ \left\|\sum\limits_{n=1}^N A_n\otimes U_nB_n\right\|_{\mathcal{B}(\mathcal{H}\otimes\mathcal{K})}=\left\|\sum\limits_{n=1}^N A_n\otimes B_nU_n^*\right\|_{\mathcal{B}(\mathcal{H}\otimes\mathcal{K})}\,\,? $$ Notice that for $N=1$, the equality surely holds since $$ \|A\otimes UB\|_{\mathcal{B}(\mathcal{H}\otimes\mathcal{K})}=\|A\|_{\mathcal{B}(\mathcal{H})}\|UB\|_{\mathcal{B}(\mathcal{K})}=\|A\|_{\mathcal{B}(\mathcal{H})}\|BU^*\|_{\mathcal{B}(\mathcal{K})}=\|A\otimes BU^*\|_{\mathcal{B}(\mathcal{H}\otimes\mathcal{K})}. $$ From $N=2$, it seems that nothing can be said.

Answer 1

This isn’t true even when $H = \mathbb{C}$ and $N = 2$. Indeed, set $K = \mathbb{C}^2$, $B_1 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, $U_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, $B_2 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$, $U_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Then $U_1B_1 + U_2B_2 = \begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix}$ has norm $2$ but $B_1U_1^\ast + B_2U_2^\ast = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ has norm $1$.

Answer 2

It's even wrong for $H=K=\mathbb C$ and $N=2$. Take $A_1=A_2=1$ and $B_1=B_2^\ast=U_1^\ast=U_2=\frac 1{\sqrt 2}(1+i)$. Then $$ \lvert U_1B_1+U_2B_2\rvert=2\lvert B_1\rvert^2=2 $$ and $$ \lvert B_1 U_1^\ast+B_2 U_2^\ast\rvert=\lvert B_1^2+(B_1^\ast)^2\vert=0. $$