Suppose that we are given a Hilbert space $H$ with an orthogonal basis $(e_i)_{i\in I}$ and let $P_i$ denote the projection of $H$ onto $\mathbb{C}e_i$. Then we can consider the map $$\varphi(T)=\sum_{i\in I}P_iTP_i,\quad T\in B(H).$$ It is easy to see (by means of Tomiyama's theorem) that this map is a conditional expectation of $B(H)$ onto the commutant of $\{P_i\,|\,i\in I\}$. My question is, is $\varphi$ normal?
It would indeed seem so, but I am not sure whether my argument holds. One way of seeing that $\varphi$ is normal is by showing that $\omega\circ\varphi$ is an ultraweakly continuous functional if $\omega\in B(H)_*$ is an ultraweakly continuous functional itself, but since $B(H)_*$ is norm-closed, isn't it then enough just to show that $\omega\circ\varphi\in B(H)_*$ for all SOT-continuous linear functionals $\omega$ on $B(H)$, since these are norm-dense in $B(H)_*$?
As you say, one needs to test that $\omega\circ\varphi$ is normal for all normal states $\omega$. I see two immediate ways to achieve the proof from here:
1) Use that $\omega=\text{Tr}(H\cdot)$ for some positive trace-class $H$. Then, if $T_j\nearrow T$, $$ \omega\circ\varphi(T_j)=\text{Tr}(H\sum_iP_iT_jP_i)=\text{Tr}(H^{1/2}\sum_iP_iT_jP_iH^{1/2})=\text{Tr}(\sum_iH^{1/2}P_iT_jP_iH^{1/2}). $$ Here one would use a basis for the range of $H^{1/2}$, write the trace in terms of that basis, and use that $H$ is trace-class (allowing you to consider finite sums) to deduce normality.
2) Assume that $T_j\nearrow T$. The point to use here is that $\text{Tr}\circ\varphi=\text{Tr}$. Given any normal state $\omega$, we write $\omega=\text{Tr}(H\cdot)$ and then, as $\text{Tr}(\varphi(H)\cdot)$ is also a normal state, $$ \omega(\varphi(T_j))=\text{Tr}(H\varphi(T_j))=\text{Tr}(\varphi(H\,\varphi(T_j)))=\text{Tr}(\varphi(H)\varphi(T_j))=\text{Tr}(\varphi(H)T_j)\to\text{Tr}(\varphi(H)T)=\text{Tr}(H\varphi(T))=\omega(\varphi(T)). $$