Q:

Given a channel $\Phi(X)=\sum_k c_k(X)\sigma_k$, are there always $F_k\ge0$ such that $\Phi(X)=\sum_k \operatorname{tr}(F_k X)\sigma_k$?

Fix a finite number of states $\sigma_k$, and consider a channel of the form
$$\Phi(X)=\sum_k c_{k}(X)\sigma_k.$$
For $\Phi$ to be linear and trace-preserving we must have:
$$c_k(X+X') = c_k(X) + c_k(X'), \qquad \sum_k c_k(X)=1.$$
In other words, the coefficients must be linear functionals $c_k\in\mathrm{Lin}(\mathcal X)^*$ for all $k$.
Does this imply that there must be some positive operators $F_k\ge0$ such that $c_k(X)=\operatorname{Tr}(F_k X)$ for all $k$ (which in turn would imply $\sum_k F_k=I$ and thus that $\{F_k\}_k$ is a POVM)? What's a good way to show this?

A:

The answer is no.
To this end, pick a linearly independent set $\{\sigma_k\}$ which spans the full matrix space (over $\mathbb C)$, that is, a basis.  (This is always possible, as the positive operators span the hermitian ones over $\mathbb R$.)
Then pick a dual basis $\sigma'_\ell$ such that
$$
\mathrm{tr}[\sigma'_\ell \sigma_k]=\delta_{k\ell}\ .
$$
Then,
$$
\Phi(X) = \sum_k \mathrm{tr}[\sigma'_k X]\,\sigma_k
$$
is the identity channel, which cannot be written as a POVM $F_k\ge0$ followed by a preparation of $\sigma_k$ (as that channel would be entanglement breaking).
(Note that this shows that the dual basis $\sigma'_\ell$ has non-positive elements. This is not surprising, since otherwise the scalar product $\mathrm{tr}[\sigma'_\ell\sigma_k]\ge0$ for all $k,\ell$.)

