Bounded sequence in spatial tensor product and boundedness of simple tensor summands Let $A$ and $\{B_n\}_{n=1}^\infty$ be $C^*$-algebras. Assume that $a_1,\dots, a_n$ are linearly independent vectors in $A$ and let $\{b_{1,m},\dots,b_{n,m}\}\subset B_m$ be such that
$$\sup_{m\geq1}\bigg{\|}\sum_{i=1}^na_i\otimes b_{i,m}\bigg{\|}<\infty,\;\;\;\;(\star)$$
where the norm is computed in the minimal (spatial) tensor product of $A,B$, namely $A\otimes B$.
I am trying to show that, for any $i=1,\dots,n$ we have that
$$\sup_{m\geq1}\|a_i\otimes b_{i,m}\|<\infty$$
but I am stuck. I know that $C^*$-norms on $A\odot B$ are cross norms, hence $\|a_i\otimes b_{i,m}\|=\|a_i\|\|b_{i,m}\|$, so the problem is equivalent to showing that $\sup_{m\geq1}\|b_{i,m}\|<\infty$, but I cannot see why this must follow from $(\star)$. Any ideas on what I am missing out? I also tried to get some inequality of the form $\|a_i\otimes b_{i,m}\|\leq C\cdot\|\sum_{j=1}^na_j\otimes b_{j,m}\|$ but I quickly gave up on this approach.
Edit: At the same time I am trying to show that, if $E$ is a finite dimensional operator system and $B$ is any $C^*$-algebra, then $E\odot B$ is closed with respect to the minimal norm, i.e. that any element of $E\otimes B$ can be written uniquely as $\sum_{i=1}^nx_i\otimes b_i$ where $\{x_i\}$ is a basis of $E$. While trying to prove this, I am encountering the same problem. An inequality of the form $\|x_i\otimes b_i\|\leq C\|\sum_jx_j\otimes b_j\|$ would be very useful, but I cannot see how to prove this.
 A: I am aware that this was already answered in MO based on CB maps and
Wittstock's Theorem.  So let me attempt to give a more elementary answer
which is perhaps more suited for MSE.
First let us represent  $A$ on a Hilbert space $H$, and let us also
represent all of the
$B_n$ on another Hilbert
space, say  $K$.  Then all we must prove is that, for every $j$, there is a constant $C$,
such that
$$
  \|b_j\|\leq C\Big \|\sum_{i=1}^na_i\otimes b_i\Big \|,
  \tag 1
  $$
for every $n$-tuple $(b_1,b_2,\ldots ,b_n)$ of operators on $K$.
To prove this, let $V$ be the linear span of the $a_i$, and let $\varphi $ be
the linear functional on $V$ determined by $\varphi (a_i) = \delta _{i, j}$.
Use Hahn-Banach to extend $\varphi $ to a WOT continuous linear functional on $B(H)$,
also denoted $\varphi $ by abuse of language,
and recall that such functionals are necessarily  of the form
$$
  \varphi (a) =  \text{tr}(aF),\quad (a\in  B(H)),
  $$
where $F$ is a finite rank operator.
Given $b_1,b_2,\ldots ,b_n\in B(H)$,  we then have that
$$
  (\phi\otimes \text{id})\Big (\sum_{i=1}^na_i\otimes b_i\Big ) = b_j,
  $$
so it only remains to prove that $\phi\otimes \text{id}$ is a bounded linear
functional on the algebraic tensor product $B(H)\otimes B(K)$,
relative to the operator norm  of $B(H\otimes K)$.
For this,  let $\{e_i\}_{i\in I}$ be an orthonormal  basis of $H$, such that a finite
subset $\{e_i\}_{i\in I_0}$ is a basis for $\text{Ran}(F) +   \text{Ker}(F)^\perp$.
The matrix of $F$ on that basis is then a finite matrix,  having nonzero
coefficients $F_{k, l}$ only when $(k, l)\in I_0\times I_0$.
Given any bounded operator $T$ on $H\otimes K$, we may write it as a block
matrix
$$
  T=(T_{k, l})_{(k, l)\in I\times I}
  $$
relative to the decomposition
$$
  H\otimes K = \bigoplus_{i\in I} e_i\otimes K.
  $$
The "block-trace"
$$
  \text{TR}(T) = \sum_{k\in I} T_{(k,k)} \in B(K),
  $$
is of course not always convergent, but it clearly involves only a finite
sum when computed on $T(F\otimes I)$, because the matrix of this operator  turns out to have zero entries $(k, l)$
whenever  $l$ is not in   $I_0$.  We may then define a "restricted"
block-trace map by
$$
  \text{TR}_0(T) = \sum_{k\in I_0} T_{(k,k)},
  $$
which clearly defines a bounded linear map from $B(H\otimes K)$ to $B(K)$.
To conclude,
it is not hard to prove that
$$
  \text{TR}_0(T(F\otimes I)) =   (\phi\otimes \text{id})(T),
  $$
whenever $T$ has the form
$$
  T= \sum_{i=1}^na_i\otimes b_i,
  $$
and this immediately implies (1).
