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If we didn't have the $1/n$ term, what's the standard example here? Let $T$ be the left shift on $\ell^2$, so $T^n\rightarrow 0$ strongly, but $T^*$ is the right shift, an isometry. To deal with the …

Let $T\in NA(X,Y)$ so there is $x\in X, \|x\|=1, \|T(x)\| = \|T\|$. By Hahn-Banach there is $f\in Y^*, \|f\|=1, f(T(x)) = \|T(x)\| = \|T\|$. But $f(T(x)) = T^*(f)(x)$ so $$ \|T^*\| = \|T\| = |T^*(f) …

So I think this works. Fix $1<r<\infty$ and let $r'$ be the conjugate index to $r$ so $1/r + 1/r' = 1$. Let $$ \tau = \sum_n S_n \otimes T_n \in \mathcal{A}(\ell^q,\ell^p) \widehat\otimes \mathcal{A …

The following seems like overkill to me; I'd like to see a solution with less machinery. So I just give a sketch. As $\newcommand{\im}{\operatorname{Im}} \im(M)^\perp = \ker(M)$ we may reduce to th …

For the 1st part, the answer is "yes". Let $T,S$ be bounded operators on $H$ and $K$ respectively. As $(T\otimes S)^*(T\otimes S) = |T|^2\otimes|S|^2$ it follows that $|T\otimes S| = |T|\otimes |S|$ …

The limits are always the same. As $\mathcal K = S_2(H)$ is a Hilbert space, and as only the Schur product is involved anywhere, we can infact identify $\mathcal K$ with $\ell^2 = \ell^2(\mathbb N)$ …

One can say something positive, though it may not be useful. You say you are willing to "be flexible about what map means". Well, you could define the maps you are interested in to be those which ex …

Since Ryan's book is mentioned in the original question, here's some pointers for an answer based on this source: See Section 3.4 for what the dual space of $X\otimes_\epsilon Y$ is. If you combine …

(This is really a very long comment...) I think maybe the actual question comes about because some of the terminology in this area is hazy. Let $T:X\supseteq D(T)\rightarrow X$ be a linear operator …

The following is an abstract (Banach) algebraic take on Werner's construction. Let $A=\ell^1(\mathbb Z)$ with convolution (but in general $A$ is any Banach algebra). We turn the dual space $A^*$ int …

It is quite hard to answer this question, as I do not know exactly how $\phi$ is defined, nor what we "know" about the spectrum of a self-adjoint operator. I think standard presentations of this circ …

I think you need to be very careful about hypotheses. Looking in Lance's book, we have Lemma 9.8: Suppose $t:E\rightarrow E$ is densely-defined and self-adjoint. Then $t$ is regular if and only if …

For $\xi,\eta\in H$ let $\theta_{\xi,\eta}$ be the rank-one operator $\theta_{\xi,\eta}(\gamma) = (\gamma|\eta) \xi$ for $\gamma\in H$. Let $(e_i)$ be an orthonormal sequence in $H$, set $S_i = \thet …

Here's an example showing that $T$ can be trace-class but $T|_{\ell^1}$ is not compact. Let $(x_n)$ be a sequence of vectors in $\ell^2$ with disjoint supports, $\sum_n \|x_n\|_2 \leq 1$ and $\|x_n\| …

This is far from an area I am an expert in, but I would keep in mind the following class of examples. Let $A$ be any Banach $*$-algebra. So $A$ is a Banach algebra, and $A$ is a $*$-algebra, and the …