operator norm of completely bounded map Definition: An operator space $X$ is a closed subspace of a C*-algebra $A$. For each $n\in\mathbb{N}$, $\mathbb{M}_{n}(X)$ inherits a norm from $\mathbb{M}_{n}(A)$. Let $\phi$ be a linear map from an operator space $X\subset A$ into an operator space $Y\subset B$, $\phi$ is completely bounded (cb) if $$\|\phi\|_{cb}=sup_{n\in\mathbb N}\|\phi_{n}: \mathbb{M}_{n}(X)\to\mathbb{M}_{n}(Y)\|,$$ where $\phi_{n}$ is defined by $\phi_{n}([x_{i,j}])=[\phi(x_{i,j})]$ for $x=[x_{i,j}]\in\mathbb{M}_{n}(X)$.
My question is how to comprehend $\|\phi\|_{cb}=\|\phi\otimes id_{B(H)}\|$ for any infinite dimensional Hilbert space where $\phi:X\to Y$ is a c.b map. 
 A: You can trivially embed $Y\otimes M_n(\mathbb C)$ in $Y\otimes B(H)$, by considering $M_n(\mathbb C)$ as the $n\times n$ corner of $B(H)$; that is, identity $M_n(\mathbb C)$ with $PB(H)P$, where $P=E_{11}+\cdots+E_{nn}$. This embedding is isometric, because $PB(H)P$ acts on $PH\simeq \mathbb C^n$. Then
$$
\|\phi_n\|\leq\|\phi\otimes id_{B(H)}\|
$$
for all $n$. Conversely,
$$
\|\phi\otimes id_{B(H)}\|=\sup\{\|\phi\otimes id_{B(H)}(z)\|:\ z\in X\otimes B(H),\ \|z\|=1\}.
$$
Fix $\varepsilon>0$. Then there exists $z=\sum_{j=1}^m x_j\otimes T_j$, with norm one,  such that $$\tag{1}\|\phi\otimes id_{B(H)}\|<\|\phi\otimes id_{B(H)}(z)\|+\varepsilon.$$ Now, assuming we have $X\otimes B(H)\subset B(K\otimes H)$, we can choose $\xi=\sum_{j=1}^r\eta_j\otimes\xi_j\in K\otimes H$ with 
$$\tag{2}\|\phi\otimes id_{B(H)}(z)\|\leq\|\phi\otimes id_{B(H)}(z)\xi\|+\varepsilon.$$
Now, with $P$ the orthogonal projection onto the span of $\{T_j\xi_k:\ j,k\}\cup\{\xi_k: k\}$,
\begin{align}\tag{3}
\|\phi\otimes id_{B(H)}(z)\xi\|&=\|\sum_{j=1}^m\sum_{k=1}^r\phi(x_j)\eta_k\otimes T_j\xi_k\|=\|\sum_{j=1}^m\sum_{k=1}^r\phi(x_j)\eta_k\otimes PT_jP\xi_k\|\\ \ \\
&=\|(\phi\otimes I_{mr})((I\otimes P)z(I\otimes P))\xi\|\leq \|\phi\|_{\rm cb}\,\|z\|\,\|\xi\|=\|\phi\|_{\rm cb}.
\end{align}
Combining $(1)$, $(2)$, and $(3)$, we get 
$$
\|\phi\otimes id_{B(H)}\|\leq \|\phi\|_{\rm cb}+\varepsilon.
$$
As $\varepsilon$ was arbitrary, we are done. 

Note that in both approximations I used a sum of elementary tensors; these are not arbitrary elements of the tensor product, so another approximation is involved. 
