For $\pi$ complete contraction, show that $ \|(\pi(x_1),...,\pi(x_n)) \| \leq \|(x_1,…x_n) \|$ Let $X \subset B(H)$ be an operator space and $ \pi : X \to B(K)$ be a complete contraction.

For $x=(x_1,...,x_n) \in M_{1,n}(X)$, is it obvious that $ \|(\pi(x_1),...,\pi(x_n)) \| \leq \|x \|$. Here all the notations have natural meaning.

I think it should be obvious but I ain't able to see it clearly.
 A: The "obvious" part is, I guess, about being familiar with how things work with matrices of operators.
Let $\tilde x$ be the $n\times n$ matrix $\begin{bmatrix} x\\ 0\end{bmatrix}$. We have, writing $\tilde\xi=\begin{bmatrix} \xi_1&\cdots&\xi_n\end{bmatrix}^T$,
\begin{align}
\|\tilde x\|^2
&=\sup\big\{\|\tilde x\,\tilde\xi\|^2:\ \tilde\xi\in H^n\big\}\\[0.3cm]
&=\sup\big\{\sum_k\Big\|\sum_j(\tilde x)_{kj}\xi_j\Big\|^2:\ \xi_1,\ldots,\xi_n\in H,\ \sum_j\|\xi_j\|^2=1\big\}\\[0.3cm]
&=\sup\big\{\Big\|\sum_jx_j\xi_j\Big\|^2:\ \xi_1,\ldots,\xi_n\in H,\ \sum_j\|\xi_j\|^2=1\big\}\\[0.3cm]
&=\|x\|^2.
\end{align}
Then
\begin{align}
\|\begin{bmatrix}\pi(x_1)&\cdots&\pi(x_n)\end{bmatrix}\|
&= \left\|\begin{bmatrix}\pi(x_1)&\cdots&\pi(x_n)\\0&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&0 \end{bmatrix} \right\|\\[0.3cm]
&=\|\pi^{(n)}(\tilde x)\|\\[0.3cm]
&\leq\|\tilde x\|=\|x\|.
\end{align}

My first answer was a more "operator systems" approach, that I left below.
By Wittstock's Theorem we may assume that $\pi$ is defined on all of $B(H)$, completely contractive, and completely positive.
\begin{align}
\|x^*\|^2
&=\sup\big\{\|(x_1^*\xi,\ldots,x_n^*\xi)\|^2:\ \|\xi\|=1\big\}\\[0.3cm]
&=\sup\Big\{\Big\langle\Big(\sum_jx_jx_j^*\Big)\xi,\xi\Big\rangle:\ \|\xi\|=1\Big\}\\[0.3cm]
&=\Big\|\sum_jx_jx_j^*\Big\|.
\end{align}
It is an easy exercise that $\|x\|=\|x^*\|$. We also have Kadison's Inequality
$$
\pi(a)\pi(a)^*\leq\pi(aa^*),
$$
and that if $0\leq a\leq b$, then $\|a\|\leq\|b\|$.
Now, writing $\tilde\pi(x)=\big(\pi(x_1),\ldots,\pi(x_n)\big)$,
\begin{align}
\|\tilde\pi(x)\|^2
&=\|\tilde\pi(x)^*\|^2=\Big\|\sum_j\pi(x_j)\pi(x_j)^*\Big\|\\[0.3cm]
&\leq\Big\|\sum_j\pi(x_jx_j)^*\Big\|=\Big\|\pi\Big(\sum_jx_jx_j^*\Big)\Big\|\\[0.3cm]
&\leq\Big\|\sum_jx_jx_j^*\Big\|=\|x^*\|^2=\|x\|^2.\\[0.3cm]
\end{align}
