An exercise for positive map on operator system Definition. Let $A$ be a unital C*-algebra. If $A$ has a unit $1$ and $S$ is a self-adjoint ($S^*=S$) subsapce of $A$ containing $1$, then we call $S$ an $operator$ $system$.
Question: Let $S$ be an operator system, $B$ be a C*-algebra, and $\phi: S\rightarrow B$ be a positive map. How to prove that $\phi$ extends to a positive map on the norm closure of $S$?
This is an exercise from Page 21 of the book Completely Bounded Maps and Operator Theory by V. Paulsen.

The question is basically resolved in the comments; bounded maps extend in a norm-preserving way to the closure and to see positivity of the extension, take a positive element of the closure, find a sequence of positive elements from the space that converge there and apply $\psi$ and then use the fact that the positive elements in the target $C^*$-algebra is a closed set. The question becomes how to find a sequence of positive elements from $S$ that converges to $x$.
If $S$ was a $*$-subalgebra, then this would be easy: since $x\ge0\in\overline{S}$ we also have $x^{1/2}\in \overline{S}$ (since in this case $\overline{S}$ is a $C^*$-algebra) so we find a sequence $(x_n)\subset S$ with $x_n\to x^{1/2}$ and then $S\ni x_n^*x_n\to (x^{1/2})^*(x^{1/2})=x$ are a sequence of such positive elements.
This technique fails when we move to operator systems. But, the standard technique (which you will see used in Paulsen's book in many "flavors") is this: let $S$ be an operator system and $x\ge0$ an element of $\overline{S}$. Find a sequence $(x_n)\subset S$ with $x_n\to x$. Then $x_n^*\to x$, so $\frac{x_n+x_n^*}{2}\to x$, and the elements $\frac{x_n+x_n^*}{2}$ are self-adjoint elements of $S$. So we can assume without loss of generality that $(x_n)$ are self-adjoint. Set $\varepsilon_n:=\|x_n-x\|$. Then since for any self-adjoint we have $-\|h\|1\le h\le\|h\|1$, we have
$$-\varepsilon_n1\le x_n-x\le \varepsilon_n1$$
and thus $0\le x_n+\varepsilon_n1-x$. But $x\ge0$, so $(x_n+\varepsilon_n1-x)+x\ge0$, i.e. $x_n+\varepsilon_n1\ge0$. But $x_n+\varepsilon_n1$ are elements of $S$ and since $x_n\to x$ and $\varepsilon_n\to0$, we have $x_n+\varepsilon_n1\to x$.
Another way that this $+\varepsilon1$ technique is used (in Paulsen's book) is the following: assume you have a sequence of positive elements $(x_n)$ that converge to some $x$ but you need invertible elements; just take $(\varepsilon_n)\subset(0,1)$ with $\varepsilon_n\to0$ and take $x_n'=x_n+\varepsilon_n1$. Then $x_n'\to x$ and $x_n'\ge\varepsilon_n1$, giving you invertibility.
