# Does monotonicity of diamond distance hold for intermediate channels?

It is well known that $$\|\mathcal{E} \circ \mathcal{F} - \mathcal{E}\|_\lozenge \leq \|\mathcal{F} - \mathcal{I}\|_\lozenge$$.

What if I have $$\|\mathcal{A} \circ \mathcal{E} \circ \mathcal{F} - \mathcal{A} \circ \mathcal{E}\|_\lozenge$$? Does this obey

$$\|\mathcal{A} \circ \mathcal{E} \circ \mathcal{F} - \mathcal{A} \circ \mathcal{E}\|_\lozenge \leq \|\mathcal{A} \circ \mathcal{F} - \mathcal{A} \circ \mathcal{I}\|_\lozenge$$

Here I am concerned only with CPTP maps.

TL;DR: No. The suggested bound fails to hold for any norm. Briefly, if we choose $$\mathcal{A}=\mathcal{F}$$ to be an idempotent channel then the right-hand side vanishes. However, if we choose $$\mathcal{E}$$ to be the Hadamard and $$\mathcal{A}=\mathcal{F}$$ to be the completely dephasing channel (which is idempotent) then the left-hand side remains positive.
Let's represent a quantum state in Bloch coordinates $$(x, y, z)$$, so that \begin{align} \rho=\frac12(I+xX+yY+zZ).\tag1 \end{align} Then a quantum channel is completely described by its action on a general real $$3$$-vector $$(x, y, z)$$.
Let's choose $$\mathcal{A}$$ and $$\mathcal{F}$$ to be the completely dephasing channel$$^1$$ \begin{align} \mathcal{A}(x, y, z)=\mathcal{F}(x, y, z)=(0, 0, z).\tag2 \end{align} We note that $$\mathcal{A}\circ\mathcal{F}=\mathcal{A}=\mathcal{A}\circ\mathcal{I}$$, so \begin{align} \|\mathcal{A}\circ\mathcal{F}-\mathcal{A}\circ\mathcal{I}\|_\diamond=0.\tag3 \end{align} Further, let's choose $$\mathcal{E}$$ to be the unitary Hadamard channel$$^2$$ \begin{align} \mathcal{E}(x, y, z)=(z, -y, x).\tag4 \end{align} Then \begin{align} (\mathcal{A}\circ\mathcal{E}\circ\mathcal{F})(x, y, z)&=\mathcal{A}(\mathcal{E}(\mathcal{F}(x, y, z)))\tag5\\ &=\mathcal{A}(\mathcal{E}(0, 0, z))\tag6\\ &=\mathcal{A}(z, 0, 0)\tag7\\ &=(0, 0, 0)\tag8 \end{align} so $$\mathcal{A}\circ\mathcal{E}\circ\mathcal{F}$$ is the completely depolarizing channel$$^3$$. Similarly, \begin{align} (\mathcal{A}\circ\mathcal{E})(x, y, z)&=\mathcal{A}(z, -y, x)\tag9\\ &=(0, 0, x)\tag{10} \end{align} so $$\mathcal{A}\circ\mathcal{E}\circ\mathcal{F}\neq\mathcal{A}\circ\mathcal{E}$$. But then the positive definiteness of the diamond norm implies that \begin{align} \|\mathcal{A}\circ\mathcal{E}\circ\mathcal{F}-\mathcal{A}\circ\mathcal{E}\|_\diamond > 0.\tag{11} \end{align} Finally, combining $$(3)$$ and $$(11)$$, we obtain \begin{align} \|\mathcal{A}\circ\mathcal{E}\circ\mathcal{F}-\mathcal{A}\circ\mathcal{E}\|_\diamond > \|\mathcal{A}\circ\mathcal{F}-\mathcal{A}\circ\mathcal{I}\|_\diamond\tag{12}. \end{align} Note that the only property of the norm used in the proof is that $$\|a-b\|=0\iff a=b$$. This is satisfied by every norm, so the counterexample proves that the bound suggested in the question fails to hold for any norm, not just the diamond norm.
$$^1$$ Kraus representation $$\mathcal{A}(\rho)=\mathcal{F}(\rho)=\frac12\rho+\frac12 Z\rho Z$$.
$$^2$$ Kraus representation $$\mathcal{E}(\rho)=H\rho H$$.
$$^3$$ Kraus representation $$(\mathcal{A}\circ\mathcal{E}\circ\mathcal{F})(\rho)=\frac14\rho+\frac14 X\rho X+\frac14 Y\rho Y+\frac14 Z\rho Z$$.