Are the compact and Haagerup approximation properties equivalent?

The following essentially implies the equivalence of Anantharaman-Delaroche's compact approximation property (page 337 of Link) 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?

(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.

• I don't think that the unit ball of L^\infty[0,1] is pre-compact as a subset of L^2[0,1] -- consider the sequence $e_n(t) =\exp(2\pi int)$. (This was pointed out to me, after a talk I gave, when I was momentarily hung up on an example of a non-compact map from $L^\infty[0,1]$ to $L^2[0,1]$. The map is completely continuous and weakly compact, though.) Commented Oct 9, 2010 at 5:00
• Yeah, I could have thought of that example. Thanks, Yemon! Commented Oct 9, 2010 at 5:20
• Will find time to check, Martin. Thanks for updating! Thanks to Yemon for coming back to this ancient thing and finding the bug! (I had a weird premonition to come and check MO today for some reason...good thing I did!) Commented Jan 23, 2020 at 17:24
• It's been a while :D Commented Jan 23, 2020 at 17:26
• No worries. It's been a while, though, this is my second answer ever on the SE network! Commented Mar 1, 2021 at 17:31