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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 sp …

In this setting, this is just the maximum principle (or Hadamard's three-line lemma) in complex analysis. Namely you can define the operator $C_\lambda$ for every complex number $\lambda$ with real pa …

When $H,K$ are finite dimensional, this is well explained in Pisier's book Introduction to operator space theory in the chapter on Haagerup tensor product, and your space is just the completely bounde …

Yes. This follows from the uniform convexity of the Schatten $p$ classes. Indeed, for every unitary diagonal operator $u$, we have $\| (x + u x u^*)/2\|_p \geq \| E( x+uxu^*)/2\|_p = \|E(x)\|p \geq 1- …

In the vocabulary of Aleksandrov and Peller, you are asking for a proof that the function $t \in [0,a]\mapsto t^r$ is operator $r$-Hölder for any $0<r<1$. There might be an easy proof in this particul …

I guess that the answer is no in general. More precisely what I consider as the discrete version of your question has a negative answer. I guess that one should be able to find a couterexample to your …

The term "operator Lipschitz function" is definitely not reserved to the Hilbert-Schmidt norm. On the opposite, I would say that it is mostly used for the operator norm (but not only, see for example …

No, because if $f$ is invariant under $U_{T,2}$ then $|f|^2$ is invariant under $U_{T,1}$. More generally the Mazur map $f \in L^p(X,\mu) \mapsto sgn(f) |f|^{p/q} \in L^q(X,\mu)$ is a homeomorphism i …

$k$ being compactly generated, you can as well assume that $k$ is a smooth function defined on $\mathbb{T}^2$ and $Op(k)$ acts on $L^2(\mathbb T)$ (for $\mathbb{T} = \mathbb R/\mathbb Z$ the unit circ …

No, there does not necessarily exist such an $x$. For example, if $M$ is a $II_1$ factor with trace $\tau$, $\mathcal{H} = L^2(M,\tau)$ and $\xi = 1$ (the identity of $M$, seen in $L^2(M,\tau)$), then …

The answer is no: there is no bounded projection from $B(H)$ onto $V$. For a proof, see for example Theorem 5.12 in Peller's book Hankel operators and their applications. If you replace $B(H)$ by the …

If a simple $C^*$-algebra admits an infinite projection $p$ (ie a projection that is equivalent to a proper subprojection $q$), then it does not carry any tracial trace and in particular it provides a …

A good reference is Pisier's monograph Similarity problems and completely bounded maps. The first chapter is devoted to the von Neumann inequality and its generalizations to two and more contractions. …

The answer to (ii) is positive. Here is a construction, which relies on graphs with spectral gap. For simplicity I write things for trees, but one can probably do the same for more general graphs with …

By the uniform bound on $\|A^{(N)}\|$ and linearity, SOT convergence follows from the $\ell^2$-norm convergence, for every $i$, of the $i$-th column $A^{(N)} e_i$ to the $i$-th column $A e_i$. This co …