Is a functional positive on a dense set bounded?

Question

Assume we have a functional $f$ defined on a $C^*$-algebra $A$. I know that there is a proper dense subset $D\subset A$ such that for every positive element $a\in D$ we have $f(a)\ge 0$.

Can we conclude that $f$ is bounded on $A$?

Answer 1

No. Let $A=C[0,1]\oplus C[0,1]$, and let $$ D=\{(a,b):\ a,b\in C[0,1],\ \text{Im}\, a\ne0,\ \text{Im}\,b\ne0\}\cup\{(1,0)\}. $$ Let $\psi$ be an unbounded linear functional on $C[0,1]$. Define $$ \phi(a,b)=\psi(b). $$ Then $\phi$ is positive on $D$ (because $\phi(1,0)=0$, and $(1,0)$ is the only positive element of $D$), and it is not bounded.