# Approximating the norm of an operator-valued linear function with operator inputs via a matrix-valued linear function

Let $\mathcal{H}$ and $\mathcal{K}$ be infinite-dimensional Hilbert spaces. Let $B_1, \ldots, B_k \in B(\mathcal{H}).$ Define $L: B(\mathcal{K})^k \rightarrow B(\mathcal{H}\otimes \mathcal{K})$ via the formula $$L(X_1,\ldots, X_k) = \sum B_i \otimes X_i.$$

For each $\varepsilon > 0,$ does there exist a finite rank projection $P \in B(\mathcal{H})$ such that $$\|L(X_1,\ldots, X_k )\| \leq (1+\varepsilon)\|(P \otimes I)L(X_1,\ldots, X_k )(P \otimes I)\|$$ for all $(X_1,\ldots, X_k ) \in B(\mathcal{K})^k?$

• What happens if you take $P$ to be projection onto the span of $\{x_1, \dots, x_k\}$, where $B_i$ almost attains its norm on $x_i$? – Nate Eldredge Apr 27 '15 at 6:07

No, there does not always exist such a finite rank projection.

Indeed, this implies that the linear space spanned by $B_1,\dots,B_k$ is exact as an operator space, and there are non-exact operator spaces. Actually one can check that the converse holds: if $B_1,\dots,B_n$ span an exact operator space, then such a $P$ exists (Edit: see the end of this answer for a proof). For a reference, I suggest Pisier's book operators spaces, or the original paper http://arxiv.org/abs/math/9308204.

To make this answer a bit more self-contained, recall that

• An operator space is (by definition) a subspace of $B(H)$ for a Hilbert space $H$.
• The completely bounded (cb) norm of a linear map $u \colon E \to F$ between two operator spaces is $\|u\|_{cb}$, the supremum over all $n$, all $x_1,\dots,x_k \in E$, all Hilbert spaces $K$ and all $a_1,\dots,a_n \in B(K)$ of $\frac{ \|\sum_i a_i \otimes u(x_i)\|}{ \|\sum_i a_i \otimes x_i\|}$.
• The cb-distance between $E$ and $F$ is $d_{cb}(E,F)$, the infimum over all linear bijections $u \colon E \to F$ of $\|u\|_{cb} \|u^{-1}\|_{cb}$.
• The exactness constant of a (finite dimensional) operator space $E$, denoted $ex(E)$, is the infimum, over all subspaces $F$ of a matrix algebra, of $d_{cb}(E,F)$.

With this terminology, it is immediate that a family $B_1,\dots,B_k$ for which your question has a positive answer satisfies that $ex(span(B_1,...,B_k))=1$. Indeed, your inequality is exactly that the map $u \colon B_i \mapsto P B_i P$ satisfies $\|u^{-1}\|_{cb} \leq 1+\varepsilon$, whereas the inequality $\|u\|_{cb} \leq 1$ is always true.

To completely answer your question, it remains to give examples of finite dimensional operator spaces with $ex(E)>1$.

The simplest one is probably the one that is denoted by $\max(\ell^1_k)$. It is the dual of $\ell^\infty_k$ in the category of operator spaces. Explicitely, it is linearly spanned by $B_1,\dots,B_k \in B(H)$ and satisfies that $$\| \sum X_k \otimes B_k\| = \sup_{U_1,\dots,U_k \in B(\ell^2), \|U_i\| \leq 1} \|\sum X_k \otimes U_k\|.$$ However, the existence of a family $B_1,\dots,B_k$ satisfying this equality is not obvious. There are only nonexplicit constructions based on the Hahn-Banach theorem. Look at Ruan's theorem in Pisier's book.

Another example is obtained using groups. Consider $G$ a group generated by a finite set $g_1,\dots,g_k$, and take $B_i=\lambda(g_i) \in B(\ell^2(G))$, the operator by left-translation by $g_i$. Assume furthermore that the $g_1$ is equal to the identity of $G$. Then $ex(span(B_1,\dots,B_k))=1$ if and only if the group $G$ is exact. And there are groups constructed by Gromov that are not exact, as proved by Ozawa in http://arxiv.org/abs/math/0002185.

As requested in the comments, let me show that such a $P$ exists if $B_1,\dots,B_n$ span an operator space $E$ with exactness constant $ex(E)=1$.

The tool is a lemma by Roger Smith (see Proposition 1.12 in Pisier's book) asserting that for a map $u$ from an operator space to $M_n(\mathbf C)$ the cb-norm of $u$ is equal to the supremum (denoted $\|u\|_n$ in general) of $\frac{ \|\sum_i a_i \otimes u(x_i)\|}{ \|\sum_i a_i \otimes x_i\|}$ over all $k$, all $x_1,\dots,x_k \in E$ and all $a_1,\dots,a_k \in M_n(\mathbf C)$.

First, it is easy to see that for every finite subset $F$ of $B(K)^k$ and each $\varepsilon>0$, there exists a finite dimensional projection $P$ such that $$\| L(X_1,\dots,X_k)\| \leq (1+\varepsilon) \| (P \otimes 1)L(X_1,\dots,X_k) (P \otimes 1)\| .$$

By the compactness of the unit ball of $M_n(\mathbf C)$, for every integer $n$ there exists a projection $P$ such that the preceding holds for all $X_1,\dots,X_n \in M_n(\mathbf C)$.

The problem is to have such a $P$ which does not depend on $n$, and this is where exactness is used: since $ex(E)=1$, there exists $n$, a subspace $F \subset M_n(\mathbf C)$ and a map $u \colon E \to F$ such that $\|u\|_{cb} \|u^{-1}\|_{cb} \leq 1+\varepsilon$. Let $P$ be projection as above for this value of $n$. If $u_P$ is the map sending $B_i$ to $P B_i P$, then I claim that $\|u_P^{-1}\|_{cb} \leq (1+\varepsilon)^2$, QED. Indeed: $$\|u_P^{-1}\|_{cb} \leq \|u^{-1}\|_{cb} \|u \circ u_P^{-1}\|_{cb} = \|u^{-1}\|_{cb} \|u \circ u_P^{-1}\|_{n} \leq \|u^{-1}\|_{cb} \|u\|_n \|u_P^{-1}\|_{n} \leq (1+\varepsilon)^2.$$ The equality is Smith's lemma, and the last inequality comes from the choice of $P$.

• In the exact case, how do you get the projections? – J. E. Pascoe Apr 27 '15 at 18:06
• I added the proof at the bottom of my answer. – Mikael de la Salle Apr 27 '15 at 19:30