TITLE: Non-unital Russo-Dye Theorem
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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.