Let $\mathcal{H}$ be a finite-dimensional Hilbert space and $\mathcal{A}\subseteq\mathcal{B(\mathcal{H})}$ be an operator system. Suppose $T_n$ is a collection of all $n$-by-$n$ matrices equipped with the trace class norm. Let $\varphi:\mathcal{A}\widehat{\otimes} T_n\rightarrow\mathbb{C}$ be a linear functional with $\Vert\varphi\Vert=1$. Then it was concluded that, by the Hahn-Banach theorem, there exists an extension $\hat{\varphi}:\mathcal{B{\mathcal{(H)}}}\widehat{\otimes} T_n\rightarrow\mathbb{C}$ of $\varphi$ such that $\Vert\hat{\varphi}\Vert=1$. My question is how does this conclusion follow?

This doubt comes while reading Theorem 4 of the paper Extension of positive maps (see page 3, line 9).

It is to be noted that the conclusion would follow easily, by the Hahn-Banach theorem, if the inclusion $\mathcal{A}\widehat{\otimes} T_n\rightarrow\mathcal{B}\mathcal{(H)}\widehat{\otimes} T_n$ is isometric, that is, if the projective norm on $\mathcal{B}\mathcal{(H)}\widehat{\otimes} T_n$ is an extension of the projective norm of $\mathcal{A}\widehat{\otimes} T_n$. But, in general, the inclusion $\mathcal{A}\widehat{\otimes} T_n\rightarrow \mathcal{B\mathcal{(H)}}\widehat{\otimes} T_n$ is norm decreasing. The book "A theory of cross space" by Schatten contains an example (Corollary 3.5, page 57) that the above inclusion is not isometric in general.

However, I think the author means the inclusion $\mathcal{A}\widehat{\otimes} T_n\rightarrow\mathcal{B}\mathcal{(H)}\widehat{\otimes} T_n$ is isometric in this special situation, that is, when $\mathcal{A}$ is an operator system and $\mathcal{H}$ is finite-dimensional. If so, it will lead to the conclusion written in the first paragraph immediately using the Hahn-Banach theorem.

I tried to prove this in various ways but could not able to prove it.

Any comment is highly appreciated. Thanks in advance.

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    $\begingroup$ I don't understand what your question is. The linked paper makes no such claim, and is very careful to point out that you need extra hypotheses to extend functionals. $\endgroup$ Jul 1, 2022 at 13:21
  • $\begingroup$ @MatthewDaws I have edited the question. Please read it now. $\endgroup$
    – Piku
    Jul 1, 2022 at 15:41
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    $\begingroup$ You have now edited away the reference to where "it was concluded that" there exists an extension. $\endgroup$ Jul 1, 2022 at 18:09
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    $\begingroup$ @Piku: The point is, it is unclear right now if you are saying "I read this specific fact asserted somewhere and I want to know why it's true" or if you are saying "I am wondering if the following might be true." $\endgroup$ Jul 1, 2022 at 19:01
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    $\begingroup$ @Piku: in that case, it would be really helpful for everyone trying to answer your question if you cited specifically where you read it. $\endgroup$ Jul 1, 2022 at 19:20

1 Answer 1


Your interpretation of what the paper you link to says is confused.

  • The paper considers only positive maps $\phi:A\rightarrow M_n$;
  • The paper does not consider $A \widehat\otimes M_n$ but rather $A \widehat\otimes T_n$ where $T_n$ is the $n\times n$ matrices with the trace class norm.

The paper (or really an "addenda") is hard to read, because you need access to Stormer's book, reference [6]. However, this all said, I agree with you that the argument seems wrong. The only thing [6, Lemma 4.2.2] shows is that the dual of $A \widehat\otimes T_n$ is the bounded linear maps $A\rightarrow M_n$. Of course, this says nothing about the inclusion

$$ A \widehat\otimes T_n \rightarrow B(H) \widehat\otimes T_n $$

and it does seem to simply be asserted without proof (in the paper you link) that this inclusion is an isometry, not merely norm-decreasing.

I'm afraid I don't have more to say right now; but I think your doubts are reasonable.


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