extension of a trace on the positive cone of a $*$-algebra Let $A$ be a $*$-algebra. Denote $A^+$ by the positive self-adjoint elements in $A$. If $\tau$ is a tracial state of $A^+$, can we construct a tracial state $\tilde{\tau}$ on $A$ such that $\tilde{\tau}|_{A}=\tau$?
 A: $\newcommand{\Asa}{A_{\text{sa}}}\newcommand{\tt}{\tilde \tau }$There are few strong results on positivity that apply to  general $^*$-algebras, so let me restrict to C*-algebras
here.  Among the crucial tools at our disposal is  the fact that every self-adjoint element decomposes  as a
difference os positive elements.
Definition.  Given a C$^*$-algebra $A$, a   state  on $A_+$ is a function $\tau :A_+\to {\mathbb R}_+$ that is additive ($\tau (a+b)=\tau (a)+\tau (b)$)
and   positively homogeneous $(\tau (\lambda a)=\lambda \tau (a)$ for $\lambda \geq 0$).  If moreover $\tau (a^*a)=\tau (aa^*)$ for all $a$ in $A$, then $\tau $ is called a tracial
state.
Theorem.  Any state $\tau $ on $A_+$ admits a unique linear extension $\tt$ to $A$.  If $\tau $ is moreover a tracial state, then $\tt$ is a trace.
Proof. Denote by $\Asa$ the set of all self-adjoint elements in $A$.   For each $a$ in $\Asa$, write $a=b-c$, with
$b,c\in  A_+$, and set
$$
  \varphi (a) = \tau (b)-\tau (c).
  $$
If $b'$ and $c'$ satisfy the same properties as above then
$$
  b-c=b'-c' \quad\Rightarrow\quad
  b+c'=b'+c  \quad\Rightarrow\quad
  \tau (b)-\tau (c) =    \tau (b')-\tau (c'),
  $$
so we see tht $\varphi $ is well defined and it is easy to see that it becomes a real-linear functional
$$
  \varphi :\Asa\to \mathbb R.
  $$
Now,  given any  $a$ in $A$,  observe that   both
$$
  \text{Re}(a):= {a+a^*\over 2}, \quad \text {and} \quad   \text{Im}(a):= {a-a^*\over 2i}
  $$
lie in $\Asa$, so we may define
$$
  \tt (a) = \varphi (\text{Re}(a)) + i \varphi (\text{Im}(a)),
  $$
and one may then prove that   $\tt $ is a complex linear functional on $A$ extending $\tau $.  The uniqueness of
$\tt $ is clear.
Assuming that $\tau $ is a tracial state, we need to show that $\tt (ab)= \tt (ba)$, for every $a$ and $b$ in $A$.
For this, observe that if $u$ is a unitary element in the multiplier of $A$, and $a\in A_+$, then
$$
  \tt(u^*au) =   \tt(u^*a^{1/2}a^{1/2}u) =  \tt(a^{1/2}uu^*a^{1/2})=  \tt(a).
  $$
By linearity we therefore also have that
$$
  \tt(u^*au) =    \tt(a),
  $$
for every element $a$ in $A$. It $u$ is again assumed to be a unitary  multiplier, then
$$
  \tt(ua) =    \tt(u^*(ua)u) =   \tt(au).
  $$
Any element $b$ in $A$ may be written as the linear combination of four unitary multipliers,  so it follows that
$$
  \tt(ba) = \tt(ab).
  $$
QED.
