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Let $A$ be a C$^*$-algebra and let $\phi$ be a positive linear map from $A$ to $B(H)$ (bounded linear operators on Hilbert's space). If $A$ is unital, then the Russo-Dye Theorem implies that $\|\phi\|=\|\phi(1)\|$, from where it immediately follows that $$ \|\phi\| = \sup\big \{\|\phi(a)\|: a\geq 0,\ \|a\|\leq 1\big \}. \tag 1 $$

Question. Is (1) still valid in case $A$ is non-unital?

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  • $\begingroup$ Hint: Cauchy-Schwarz inequality. $\endgroup$
    – Mikael de la Salle
    Apr 13, 2021 at 7:37
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    $\begingroup$ The consequence of the Russo-Dye Theorem I am referring to may itself be proved by means of Cauchy-Schwartz-like inequalities (see Theorems 1.3.1(ii) and 1.3.3 in Erling Størmer's book "Positive Linear Maps of Operator Algebras"), but since these maps are not scalar valued, Cauchy-Schwartz is not as straightforward and often requires the elements involved to be normal. In fact there are open problems regarding whether or not these hold for hyponormal elements (see M.-D. Choi, Some assorted inequalities for positive linear maps on C$^*$-algebras. J. Oper. Theory 4(2), 271–285 (1980)). $\endgroup$
    – Black
    Apr 13, 2021 at 14:27
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    $\begingroup$ @YemonChoi, Well, I definitely do not think Russo-Dye is an overkill, and I think Størmer agrees, given his proof of Theorem 1.3.3. I think I now have a proof based on extending $\phi$ to the unitization of $A$, but I was hoping to find a one-liner, possibly based on Cauchy-Schwartz, as mentioned by Mikael. Yes, I know that $\|\phi(b)\|$ is the norm of an operator and that that $\langle Tx, y\rangle = \langle x, T^*y\rangle $, but still I cannot see how to get the conclusion from Cauchy-Schwartz. Would you be so kind as to give further details? $\endgroup$
    – Black
    Apr 13, 2021 at 15:23
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    $\begingroup$ @Black I am sorry, I misread your question and I thought that the assumption was that $\varphi$ is completely positive. And (as you observe in the comments), then the validity of (1) is indeed a consequence of Cauchy-Schwarz. $\endgroup$
    – Mikael de la Salle
    Apr 13, 2021 at 16:22
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    $\begingroup$ @Black My bad: I made the same misreading as Mikael. Thanks for your patience. $\endgroup$
    – Yemon Choi
    Apr 13, 2021 at 18:36

1 Answer 1

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As I mentioned in the comments, I'm hoping to find a one-liner based on Cauchy-Schwartz, as suggested by user Mikael de la Salle. However, while this is not available, let me present a proof I just found based on extending $\phi$ to the unitization of $A$. Let $$ S = \sup\big \{\|\phi(a)\|: a\geq 0,\ \|a\|\leq 1\big \}, $$ and let $\Phi $ be the extension of $\phi$ to the unitization $\tilde A$ defined by setting $\Phi (1)=SI_H$.

I claim that $\Phi $ is positive. To see this we must check that $$ \Phi \big ((a-\lambda 1)^*(a-\lambda 1)\big )\geq 0, \tag{$\star$} $$ for every $a$ in $A$, and every $\lambda \in {\mathbb C}$. The case $\lambda =0$ is clearly true, so we may assume that $\lambda \neq 0$. In the latter case, we may change variables by replacing $a$ with $\lambda a$, and then divide everithing by $|\lambda |^2$, leading to the following equivalent form of $(\star)$: $$ \Phi \big ((a-1)^*(a-1)\big )\geq 0, \tag{$\star\star$} $$ Observing that $$ 0\leq (a-1)^*(a-1) = a^*a-a^*-a+1, $$ and fixing an approximate identity $\{u_i\}_i$ for $A$, we have for all $i$ that $$ u_i(a^*+a -a^*a)u_i \leq u_i^2, $$ so $$ \phi\big (u_i(a^*+a -a^*a)u_i\big ) \leq \phi(u_i^2) \leq \|\phi(u_i^2)\|I_H \leq SI_H = \Phi (1). $$ Taking the limit as $i\to \infty $, the above yields $$ \phi (a^*+a -a^*a) \leq \Phi (1), $$ which is equivalent to $(\star\star)$, proving the claim.

By [1, Theorem 1.3.3], (which Størmer proves using Russo-Dye and Cauchy-Schwartz), we then deduce that $$ S = \|\Phi (1)\| = \|\Phi \| \geq \|\phi\| \geq S, $$ concluding the proof.

[1] Størmer, Erling, Positive linear maps of operator algebras, Springer 2013.

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