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Let $x$ be a completely bounded map between operator spaces $W \subset \mathbf{B}(\mathcal{H})$ and $V \subset \mathbf{B}(\mathcal{K})$, where $\mathcal{H}$ and $\mathcal{K}$ are Hilbert spaces, and let $x_n$ be a sequence of completely bounded maps between the same operator spaces such that
$$ \| x - x_n\| \to 0,$$ where $\| x\| = \sup_{ h \in \mathcal{H} \colon\|h\| = 1} \|x(h)\|$. Do we get the convergence in the completely bounded norm $$ \| x - x_n\|_{\mathbf{cb}} \to 0$$ from the fact that both a limit and a sequence are completely bounded and they converge in standard operator norm? If not could you please give me a counterexample?

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The answer is no, the sequence need not converge in the completely bounded norm to $x$ even if $x$ and $x_n$ are completely bounded.

Let $W=V=B(\mathcal H)$, where $\mathcal H$ is a Hilbert space considered with some fixed orthonormal basis $\{ e_1, e_2, \dots \}$. Let $P_n$ be the orthogonal projection onto the space spanned by the first $n$ elements of this orthonormal basis.

Let $T_n$ be the transpose map on the $n \times n$ matrices $M_n(\mathbb C)$. Recall that $\Vert T_n \Vert =1$, but that $\Vert T_n \Vert_{cb}=n$.

Let $x_n = \frac{1}{n} P_nT_nP_n +I$ and $x=I$, where $I$ is the identity map on $\mathcal H$. Then each map $x_n$ is completely bounded (with $\Vert x_n \Vert_{cb} \leq 2$) and $$ \Vert x_n -x \Vert = \frac{1}{n} \Vert P_n T_n P_n \Vert =\frac{1}{n}, $$ which tends to $0$ as $n \to \infty$.

However, $$ \Vert x_n -x \Vert_{cb}= \frac{1}{n} \Vert P_n T_n P_n \Vert_{cb} =1, $$ and so this sequence does not converge in the completely bounded norm.

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