I am trying to prove something that seemed simple to me at first sight but apparently it is giving me a hard time. Here is [the same question on MSE](https://math.stackexchange.com/questions/4108555/bounded-sequence-in-spatial-tensor-product-and-boundedness-of-simple-tensor-summ).


Let $E\subset A$ be a finite dimensional operator system with Hamel basis $\{x\_1,\dots,x\_n\}$ and let $B$ be an arbitrary $C^\*$-algebra. I am trying to show that, if $\|\cdot\|$ is the minimal norm on $E\odot B$ (i.e. the spatial norm) and $E\otimes B:=\overline{E\odot B}^{\|\cdot\|}$, then every element of $E\otimes B$ is written uniquely as $\sum\_{i=1}^nx\_i\otimes b\_i$. This is of course equivalent to showing that $E\odot B$ is complete with the minimal norm.


It is trivial to verify the claim for the elements of the algebraic tensor product $E\odot B$, by the following lemma:



> 
> If $X,Y$ are vector spaces and $\{a\_1,\dots,a\_n\}\subset X$ are linearly independent, then
> $$\sum\_{i=1}^na\_i\otimes b\_i=0\implies b\_1=\dots=b\_n=0.$$
> 
> 
> 


This lemma also takes care of uniqueness for us. But what about elements of $E\otimes B$ in general? Of course, the ideal thing would be to establish an inequality of the form $\|x\_j\otimes b\_j\|\leq C\|\sum\_{i=1}^nx\_i\otimes b\_i\|$, but how?


I tried to consider $B\oplus\dots\oplus B\to (E\odot B,\|\cdot\|)$, $(b\_1,\dots,b\_n)\mapsto\sum\_{i=1}^nx\_i\otimes b\_i$ and observed that this is bounded and bijective. The annoying thing: if I knew that $E\odot B$ is complete (which is what I want to show), then I could apply the open mapping theorem to deduce that the inverse is bounded, hence obtain the desired estimate.


I also tried an induction argument on $\dim(E)$ but it didn't work out.



Elements of the minimal tensor product of a finite dimensional operator system and a $C^*$-algebra
One way to see it is to use the dual basis of $\{x\_1,\dots,x\_n\}$. Let $\{\varphi\_1,\dots,\varphi\_n\} \subset E^{\ast}$ be functionals such that $\varphi\_i(x\_j)=\delta\_{ij}$. Because functionals are automatically completely bounded, we get bounded maps $\Phi\_i:= \varphi\_i \otimes Id : E\otimes B \to B$. On the subspace $E\odot B$ we have $\Phi\_i(\sum\_{j=1}^{n} x\_{j} \otimes b\_{j}) = b\_{i}$, so $\|b\_{i}\| \leqslant \|\Phi\_i\|\cdot \|\sum\_{j=1}^{n} x\_{j} \otimes b\_{j}\|$. As a consequence, if a sequence $y(m):= \sum\_{i=1}^{n} x\_{i} \otimes b\_i(m) \in E\odot B$ converges in $E \otimes B$ then for each $i\in \{1,\dots, n\}$ the sequence $b\_i(m)$ converges in $B$. Let $B\_i = \lim\_{m\to\infty} b\_i(m)$. Then $\lim\_{m\to \infty} \sum\_{i=1}^{n} x\_i \otimes b\_i(m) = \sum\_{i=1}^{n} x\_i \otimes B\_i \in E \odot B$, hence $E\odot B$ is equal to $E\otimes B$.


