CB-norm of mappings from a matrix space

Question

The following result of Roger Smith is well known to operator algebraists:

$$\| \phi: E \rightarrow M_n\|_{cb} =\| \phi^{(n)} \otimes id_{M_n}: E \otimes_{min}M_n \rightarrow M_n \otimes M_n\|,$$ where $\phi^{(n)}$ is the $n$-amplification of $\phi$, i.e., $$\phi^{(n)}: = \phi \otimes id_{M_n}: E \otimes_{min}M_n \rightarrow M_n \otimes M_n. $$

My question is: for any $m$, does there exist $N = N(m)$, such that for any bounded linear map $u : M_m \rightarrow \mathscr{K}(\ell_2)$, we have $$ \| u \|_{cb} \le 2 \| u^{(N)} \|? $$

I suspect the answer is no, but I can not provide myself an argument. Does anyone know the answer and reference for it ?

Answer 1

Please allow me to give a short argument here, probably I should not have asked the question here, anyway, for me, it seems to be interesting and worthwhile to know why it is not so.

1. It is easy to see we can replace $\mathscr{K}(\ell_2)$ by $B = B(\ell_2)$ without changing the answer for the original problem.

2. Now assume that the answer is YES. Then by identifying $u : M_m \rightarrow B$ with an element in $S_1^m \otimes_{min}B$, the ''YES'' answer would imply that $$S_1^m \otimes_{min} (X_i)_{\mathscr{U}} \xrightarrow{4-isomorphic} (S_1^m \otimes_{min} X_i)_{\mathscr{U}},$$ for any ultroproduct $(X_i)_\mathscr{U}$. This implies that the finite dimensional trace class $S_1^m$ viewed as an operator space is exact with a constant independent of $m$, which is known to be false.

3. Hence the answer is NO.