Extending a certain bound to the von Neumann tensor product I am wondering whether the following is true: Let $H_1, H_2$ denote two Hilbert spaces and let $$\phi_i\colon B(H_1)\overline{\otimes}B(H_2)\to B(H_2)$$ be ultraweakly continuous, completely positive linear maps for $i\in \{1,2\}$.
Assume now that there exists $\varepsilon>0$ such that $\Vert \phi_1(x)-\phi_2(x)\Vert\leq \Vert x\Vert \varepsilon$ for every $x$ in the algebraic tensor product $B(H_1)\odot B(H_2)$. Can we then conclude that $\Vert \phi_1(y)-\phi_2(y)\Vert\leq \Vert y\Vert \varepsilon$ for every $y\in B(H_1)\overline{\otimes}B(H_2)$?
My guess: by boundedness of $\phi_1-\phi_2$, we can extend the above inequality to hold true for elements in the minimal tensor product $B(H_1)\otimes B(H_2)$. Then, ultraweak continuity and a possible application of Kaplansky (?) takes us  home.
I am unsure about all of this and would love some help.
 A: Your reasoning is correct (or at least you mention the right ingredients).
More generally, if $\Phi\colon M\to N$ is an ultraweakly continuous linear map and $A\subset M$ is an ultraweakly dense $\ast$-subalgebra, then $\lVert \Phi\rVert=\lVert \Phi|_A\rVert$: Let $x\in M$ with $\lVert x\rVert\leq 1$. By Kaplansky's density theorem, there exists a net $(x_j)$ in $A$ such that $\lVert x_j\rVert\leq 1$ and $x_j\to x$ ultraweakly. Since $\Phi$ is ultraweakly continuous, it follows that $\Phi(x_j)\to \Phi(x)$ ultraweakly. Thus, if $\omega\in N_\ast$, then
$$
\lvert \omega(\Phi(x))\rvert=\lim_j \lvert \omega(\Phi(x_j))\rvert\leq \lVert \omega\rVert\liminf_j \lVert\Phi(x_j)\rVert\leq \lVert \omega\rVert\lVert \Phi|_A\rVert.
$$
Hence
$$
\lVert \Phi(x)\rVert=\sup_{\substack{\omega\in N_\ast\\\lVert \omega\rVert\leq 1}}\lvert \omega(\Phi(x))\rvert\leq \lVert \Phi|_A\rVert.
$$
Taking the supremum over all $x\in M$ with $\lVert x\rVert\leq 1$ gives $\lVert \Phi\rVert\leq \lVert \Phi|_A\rVert$, while the converse inequality is trivial.
