# Are quantum channels bounded linear maps?

I've been reading about quantum channels from a couple of sources and have some doubts regarding some mathematical perspectives and properties of quantum channels. I've listed them below:

1. It is known that quantum channels map density operators on Hilbert space (say $$\mathcal{H}_A$$ to another space (say, $$\mathcal{H}_B$$). Is it the standard convention to express any channel as a liner map $$\mathcal{N}:D(\mathcal{H}_A) \rightarrow D(\mathcal{H}_B)$$ or is $$\mathcal{N}:B(\mathcal{H}_A) \rightarrow B(\mathcal{H}_B)$$ more appropriate? Here $$D(\mathcal{H})$$ is the set of density operators on Hilbert space $$\mathcal{H}$$ while $$B(\mathcal{H}$$ is the space of bounded linear operators.

2. In some texts, some effort is made to differentiate channels as linear maps instead of linear operators. But aren't linear maps the same as linear operators acting on a higher dimension? Or are there some properties unique to linear maps to require such an emphasis?

3. Finally, are quantum channels bounded maps? Based on my naive reasoning, I think they should be bounded, given that they map density operators to density operators (which are known to be bounded). I wanted to know if there is a situation or an example for which this would fail.

It would be great if you could drop the link to those questions that have been answered already. Thanks!

To complement Danylo's great answer, let me go into a bit more details about the infinite-dimensional case and point out that the correct extension of a channel $$\mathcal{N}:D(\mathcal{H}_A) \rightarrow D(\mathcal{H}_B)$$ defined only on states is $$\mathcal{N}:B^1(\mathcal{H}_A) \rightarrow B^1(\mathcal{H}_B)$$ where $$B^1$$ denotes the trace class. Roughly speaking, the trace class is the largest subset of $$B(\mathcal H)$$ for which the expression $$\sum_j\langle g_j|A|g_j\rangle$$ converges absolutely for every orthonormal basis $$\{g_j\}_j$$ of $$\mathcal H$$, see this math.SE post or Proposition 16.18 in "Introduction to Functional Analysis" (1997) by Meise & Vogt; given that the trace is essential to the concept of density operators this seems quite reasonable. The claimed unique linear extension of $$\mathcal N$$ then comes from the fact that every matrix (trace-class operator) can be written as the linear combination of at most four states.
Moreover, every quantum channel (acting on the trace class) is bounded with operator norm $$\|\mathcal N\|_{1\to 1}=\sup_{A\in\mathcal B^1(\mathcal H_A),\|A\|_1=1}\|\mathcal N(A)\|_1=1$$ as well as diamond norm $$\|\mathcal N\|_\diamond:=\sup_{n\in\mathbb N}\|\mathcal N\otimes{\rm id}_n\|_{1\to 1}=1$$, where $$\|A\|_1:={\rm tr}(\sqrt{A^\dagger A})$$ is the usual trace norm (which turns $$\mathcal B^1$$ into a Banach space). The norms being equal to $$1$$ then is a consequence of the Russo-Dye theorem together with the usual Schrödinger-Heisenberg duality of channels. For further reading on channels and general quantum information in infinite dimensions refer, e.g., to Davies' book "Quantum Theory of Open Systems" (1976) or Heinosaari & Ziman, "The Mathematical Language of Quantum Theory" (2012).