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The following essentially implies the equivalence of Anantharaman-Delaroche's compact approximation property (page 337 of http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1102368918) 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?

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up vote 4 down vote accepted

(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.

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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.) –  Yemon Choi Oct 9 '10 at 5:00
    
Yeah, I could have thought of that example. Thanks, Yemon! –  Martin Argerami Oct 9 '10 at 5:20
    
Excellent! Thank you! –  Jon Bannon Oct 10 '10 at 0:32
    
You are welcome! –  Martin Argerami Oct 10 '10 at 4:45
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