Let $A$ be a C*-algebra, for a linear map $\phi: A\rightarrow M_{n}(\mathbb{C})$, we define a linear functional $\bar{\phi}$ on $M_{n}(A)$ by $$\bar{\phi}([a_{i,j}])=\Sigma_{i,j=1}^{n}\phi(a_{i,j})_{i,j}.$$ The notation $(a_{i,j})_{i,j}$ means the $(i, j)^{th}$ entry of the matrix $\phi(a_{i,j}).$
Meanwhile, let a net $\bar{\phi_{\lambda}}$ of positive normal linear functionals which converges pointwise to $\bar{\phi}$. Then the corresponding maps $\phi_{\lambda}$ are normal and converge to $\phi$ in the point-norm topology (i.e., $||\phi_{\lambda}(x)-\phi(x)||\rightarrow 0$ for all $x\in A$)
My question is
How to verify $\phi_{\lambda}$ are normal ?
How to verify $\phi_{\lambda}$ converge to $\phi$ in the point-norm topology?
1) Consider a norm bounded net $\{a_t\}\subset A$ of selfadjoint elements with $a_t\nearrow a$. Then, in $M_n(A)$, $a_t\otimes E_{kk}\nearrow a\otimes E_{kk}$ (we have $(a-a_t)\otimes E_{kk}\geq0$, and it is easy to see that $a\otimes E_{kk}$ is the supremum of the net).
Note that for the assertion we are trying to prove, the indices $\lambda $ play no role (the assertion is about each $\phi_\lambda$). So we just work with one linear map $\phi$ with $\bar\phi$ a positive normal functional . Since a positive functional is completely positive, we have, for any $x\in M_n(A)$, $$ |\bar\phi(x)|^2=\bar\phi(x)^*\bar\phi(x)\leq\bar\phi(x^*x)\leq\bar\phi(|x|^{1/2}\|x\|\,|x|^{1/2})=\|x\|\,\bar\phi(|x|). $$ So, as $a-a_t\geq0$, and the net $\{a_t\}$ is bounded, $$ |\bar\phi((a-a_t)\otimes E_{jk})|^2\leq\|a-a_t\|\,\bar\phi(|(a-a_t)\otimes E_{jk}|)\leq K\,\,\bar\phi(|(a-a_t)\otimes E_{jk}|)\\ =K\,\bar\phi(((a-a_t)^2\otimes E_{kk})^{1/2})=K\,\bar\phi((a-a_t)\otimes E_{kk})\to0. $$ So $\bar\phi(a_t\otimes E_{kj})\to\bar\phi(a\otimes E_{kj})$ for all $k,j$. Then $$ \phi(a_t)_{kj}=\bar\phi(a_t\otimes E_{kj})\to\bar\phi(a\otimes E_{kj})=\phi(a)_{kj}. $$ So each entry of $\phi(a_t)$ converges to the corresponding entry of $\phi(a)$. As there are finitely many matrix entries, we get that $\phi(a_t)\to\phi(a)$ in norm. As $\phi$ is positive (actually, completely positive, and this follows from the complete positivity of $\bar\phi$), we get that $\phi(a_t)\leq\phi(a)$, and so $\phi(a_t)\nearrow\phi(a)$.
2) For each $a\in A$, we have $\bar\phi_\lambda(a\otimes E_{kj})\to\bar\phi(a\otimes E_{kj})$. This translates into $\phi_\lambda(a)_{kj}\to\phi(a)_{kj}$, for al $k,j$. Once again we can use that entrywise convergence of matrices is equivalent to norm convergence, so $$ \|\phi_\lambda(a)-\phi(a)\|\to0. $$