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4

Let $J $ be an index for the cardinality of an orthonormal basis of $H $. Then $H $ is isometrically isomorphic to $\ell^2 (J) $, so it is enough to discuss the problem on this latter space. Define the product $fg $ pointwise, i.e. $fg (j):=f (j)g (j) $. The question is whether this product stays in $\ell^2$, and whether the norm is submultiplicative. We ...


3

Another Banach-algebra structure on the Hilbert space $\mathsf{hs}(H)$ of Hilbert-Schmidt operators on a Hilbert space $H$ is just operator multiplication (composition). There is a natural involution on this algebra but it does not make it a C*-algebra.


3

You are almost done: \begin{align} ((\lambda m+\mu n)(a))^*b&=\overline{\lambda}m(a)^*b+\overline{\mu}n(a)^*b=\overline{\lambda}a^*m^*(b)+ \overline{\mu}a^*n^*(b)=a^*(\bar\lambda m^*(b)+\bar\mu n^*(b))\\ &=a^*(\bar\lambda m^* +\bar\mu n^*)(b). \end{align} But now you know that $((\lambda m+\mu n)(a))^*b=a^*(\lambda m+\mu n)^*(b)$. If you look at the ...


2

The notation $\prod_{i\in I}S_i$ denotes a set of functions. By definition, $f\in\prod_{i\in I}S_i$ if (i) $f$ is a function with domain $I$ and (ii) $f(i)\in S_i$ for every $i\in I$. So $\phi\in\prod_{a\in A}sp(a)$. Because $\phi$ is a function with domain $A$ and $\phi(a)\in sp(a)$ for every $a\in A$. Come to think of it, that raises an obvious ...


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As noted in my comment, $E$ has to be a Hilbert space. To see this, fix a linear functional $\varphi \in E'$ with $\Vert \varphi \Vert = 1$. For $z \in E$, define $$ A_z : E \to E, x\mapsto \varphi(x) \cdot z. $$ It is not hard to see that $E \to B(E), z \mapsto A_z$ is linear and isometric. Thus, if $B(E) = H$ is a Hilbert space, we see that $E$ is ...


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I know a very special case, and I guess it is extensible. Suppose that $H$ is Hilbert space such that $H\cong (H_1\hat{\otimes}(H_2)^*)^*$, where $\hat{\otimes}$ is projective tensor product, and $H_1,H_2$ is some Hilbert spaces. For any Hilbert space $\mathcal{H}$, always $\mathcal{H}^{**}=\mathcal{H}$. Also for two Banach space $E,F$ always ...


2

Suppose that $A_1=\{a\in A: \|a\|<1\}$ and $B=\{ b\in A: \|be\|<1\}$. Obviously $Be\subset A_1$, and $S_e(Be)=T_e(Be)$. Hence $S_e(Be)\subset T_e(A_1)$, and consequently $\overline{S_e(Be)}^w\subset \overline{T_e(A_1)}^w$. But $\overline{T_e(A_1)}^w$ is weakly compact from assumption, and so $\overline{S_e(Be)}^w$ is weakly compact. But ...


2

First, you can forget $S,T,U$, and $V$. Say $A$ is that $2\times 2$ matrix with operator entries. Suppose you could prove $$||\alpha F||\le||\alpha||\,||F||\quad(i)$$for all $F\in L^p\oplus L^p$. Then it would follow that $$||(\alpha A)F||=||\alpha(AF)||\le ||\alpha||\,||AF||\le||\alpha||\,||A||\,||F||,$$which is exactly what you want. So you only need to ...


2

The fact that $A$ is weak$^*$-dense in $A^{**}$ is basic functional analysis. I will be surprised if there is a functional analysis book that doesn't contain this result. For your second question, it is not true as you stated it: $\pi$ cannot be any faithful representation but it is rather a very special one, the universal representation. The way it ...


2

This is example of a Banach algebra ( also a ring) that has not minimal ideal. But I don't have any idea that it has left or right minimal ideal . I found it from this link Let $\Delta=\{z\in \mathcal{C}| |z|\leq 1\}$. Suppose that $A(\Delta)$ be the set of all elements $C(\Delta)$ which are analytic on the interior of $\Delta$. $A(\Delta)$ is closed ...


1

The full quote is: If $\partial\mathbb{D}=\{z\in\mathbb{C}:|z|=1\}$, let $B=$ the uniform closure of the polynomials in $C(\partial\mathbb{D})$. This means: consider the set of all continuous functions on $\partial\mathbb{D}$, equipped with the uniform norm $\|f\|=\sup_{\partial \mathbb{D}}|f|$. This space is denoted by $C(\partial\mathbb{D})$. ...


1

This is immediate from the definitions, plus the ordinary scalar-valued Cauchy's Integral Formula. Suppose $\Lambda\in X^*$, and let $g=\Lambda\circ f$. So $g$ is analytic (by definition or not, depending on which definition of "analytic" you took). CIF shows that $$\Lambda(n(\gamma;\lambda)f(\lambda))=n(\gamma;\lambda)g(\lambda)=\frac1{2\pi ...


1

It's not so much that the closed unit ball of $B(H)$ is never compact in the strong operator topology, but it is not compact in general. More precisely, it is compact if and only if $H$ is finite dimensional. For convenience, let $S$ denote the closed unit ball of $B(H)$. If $H$ is finite dimensional, then $B(H)$ is a finite dimensional normed space, so ...


1

For your second question, the answer is always yes. More generally, given any Banach algebra $A$, let $\tilde{A}=A\oplus \mathbb{C}$ be its unitization (with norm $\|(a,z)\|=\|a\|+|z|$). Given $a\in A$, let $L_a\in B(\tilde{A})$ be left multiplication by $a$. Then $a\mapsto L_a$ is an isometric isomorphism from $A$ to a subalgebra of $B(\tilde{A})$. ...


1

Perhaps an easier approach is as follows: Let $R$ be any unital commutative ring, then there is a one-to-one correspondence between ideals in $R$ and ideals in $M_2(R)$. In fact, if $J \subset M_2(R)$ is an ideal, then $$ I = \{a \in R : \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \in J\} $$ is the corresponding ideal in $R$ such that $J = M_2(I)$. ...


1

The product $\prod_{i\in I}A_i$ of an indexed family of sets is, by definition, the set of all functions $f$ whose domain is the index set $I$ and which satisfy, for each index $i\in I$, the requirement that $f(i)\in A_i$. So the product in your question is the set of functions that assign, to each $a$ in your algebra, an element of its spectrum. The ...



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