The following essentially implies the equivalence of Anantharaman-Delaroche's compact approximation property (page 337 of [Link](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-171/issue-2/Amenable-correspondences-and-approximation-properties-for-von-Neumann-algebras/pjm/1102368918.full)) and the Haagerup approximation property.


Let $M$ be a type ${II}\_{1}$ factor with trace $\tau$. Let $\Omega$ denote the standard unit cyclic trace vector in $L^{2}(M)$ (associated to the element $1\in M$). If $\phi:M \rightarrow M$ is a normal completely positive map, we naturally associate an operator $T\_{\phi}\in B(L^{2}(M))$ extending $T\_{\phi}(x\Omega)=\phi(x)\Omega$.


If the map $x\mapsto \phi(x) \Omega$ is a compact linear map from $M$ with the operator norm into $L^{2}(M)$, is the operator $T\_{\phi}$ compact?



Are the compact and Haagerup approximation properties equivalent?
*(I removed my first answer as it contained an egregious mistake, pointed out by Yemon; here's a second attempt)*


I think that $T\_\phi$ may fail to be compact.


Fix a sequence of projections $\{p\_k\}$ in $M$, pairwise orthogonal, with $\tau(p\_k)=2^{-k}$ ($\tau$ the trace in $M$) and sum 1. Now define 
$$
\phi:M\to M,\ \ \ \mbox{ given by } \phi(x)=\sum\_k 2^k\tau(xp\_k)p\_k.
$$
(the series converges strongly because all of its terms are positive and any partial sum is bounded by $\|x\|$). This map is ucp (it is an infinite sum of cp), and since it commutes with $\tau$, it is normal. Let us also define the maps
$$
\phi\_n=\sum\_{k=1}^n 2^k\tau(xp\_k)p\_k.
$$
The maps $x\mapsto \phi\_n(x)\Omega$ are all finite-rank.


If $\|x\|\leq1$, we have, using that the set $\{p\_k\}$ is orthogonal in $L^2(M)$,
$$
\left\|\phi(x)-\phi\_n(x)\right\|\_2^2=\left\|\sum\_{k>n}2^k\tau(xp\_k)p\_k\right\|\_2^2
=\sum\_{k>n}|\tau(xp\_k)|^2\leq\sum\_{k>n}\tau(p\_k)^2\leq\frac1{3\times 4^n}
$$
This shows that $\|\phi-\phi\_n\|<4^{-n}$ in $B(L^2(M))$: so the map $x\mapsto \phi(x)\Omega$ is compact. 


Now consider the orthonormal set $\{2^kp\_k\}$ in the unit ball of $L^2(M)$. Since $\phi(p\_k)=p\_k$, we get that $T\_\phi(2^kp\_k)=2^kp\_k$; so the range of $T\_\phi$ contains an orthonormal set: $T\_\phi$ is not compact.


