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## Hot answers tagged banach-algebras

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 ...

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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.

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 $C(\partial\mathbb{D})$. ...

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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 $J = M_2(I)$. ...

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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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