Additivity of degradable and anti-degradable quantum capacities

Question

It is known that for degradable channels $\mathcal{N}$ and $\mathcal{M}$, the single-letter quantum capacity is aditive (Potential Capacities of Quantum Channels), i.e. \begin{equation} Q^{(1)}(\mathcal{N}\otimes\mathcal{M}) = Q^{(1)}(\mathcal{N}) + Q^{(1)}(\mathcal{M}). \end{equation}

I am unsure if this kind of results can be extrapolated to the setting of anti-degradable channels, that is, is it true that for $\mathcal{N}$ and $\mathcal{M}$ anti-degradable the quantum capacity equals zero (additive)? i.e. \begin{equation} Q^{(1)}(\mathcal{N}\otimes\mathcal{M}) = Q^{(1)}(\mathcal{N}) + Q^{(1)}(\mathcal{M})=0? \end{equation} Moreover, would the following be also true: let $\mathcal{N}$ be a degradable channel and $\mathcal{M}$ an anti-degradable channel, then, \begin{equation} Q^{(1)}(\mathcal{N}\otimes\mathcal{M}) = Q^{(1)}(\mathcal{N}) + Q^{(1)}(\mathcal{M})=Q^{(1)}(\mathcal{N})? \end{equation}

I have been looking in the literature for results of this type and have not found anything. Also, if this are not true in general, is it possible that they are true for particular channel, for example, if we have two amplitude damping channels, one with the damping parameter in the degradable region and the other with the damping parameter in the antidegradable region.

Answer 1

I have been able to find the answer to this question, so I will post it myself for anyone that would be interested.

The result is proven in Useful States and Entanglement Distillation by Leditzky, Datta and Smith. Specifically, proposition 13 of such article discusses the second question of this post. The proof is based on the fact that every antidegradable channel has a degradable extension with vanishing quantum capacity. It is easy to use the same argument to prove the first question of the post too.