# Tag Info

1

Let $B$ be a two-dimensional Banach algebra. Pick $x \in B \setminus \mathbb{C}\cdot e$. Since $\{e,x\}$ is a basis of $B$, we have $$x^2 = c\cdot e + d\cdot x$$ for some $c,d\in \mathbb{C}$. Now let $\lambda\in\mathbb{C}$ a zero of $z^2 - d\cdot z - c = 0$, then $$(x-\lambda\cdot e)^2 = x^2 - 2\lambda x + \lambda^2\cdot e = (d-2\lambda)x + (c+\lambda^2)e ... 1 Any finite dimensional Banach algebra of dimension not a square is not isomorphic to B(X) for any Banach space X. 1 Yes, the ideal must be proper. Otherwise \mathcal{B}/\mathcal{M} = \{0\}, and there is no element of norm 1 in the quotient at all. 1 By definition, F_\phi = \phi\circ F. The linear map \phi is defined on all of \mathcal{A}, so F_\phi = \phi\circ F is defined wherever F is defined. So where is F defined? By definition F(\zeta) = (1 - \zeta a)^{-1}, and this is only defined where 1 - \zeta a is invertible. If \zeta = 0, you have 1 - \zeta a = 1 is certainly invertible. ... 2 Denote by int(\sigma_B(x)) the interior of \sigma_B(x). Since \sigma_B(x) is a closed set, we have \sigma_B(x)=d(\sigma_B(x))\cup int(\sigma_B(x)); and since d(\sigma_B(x))\subset d(\sigma_A(x))\subset\sigma_A(x), it follows that \sigma_B(x)\setminus \sigma_A(x)=\sigma_B(x)\cap \sigma_A(x)^c is equal to int(\sigma_B(x))\cap \sigma_A(x)^c. Hence ... 2 No. If \sigma(ab)=\{ 0\}, then \sigma(ba)\subset\{ 0\} since, as you mentioned it, \sigma(ba)\setminus\{ 0\}=\sigma(ab)\setminus\{0\}=\emptyset. But \sigma(ba) is nonempty, so you must have \sigma(ba)=\{ 0\}. What is possible is that \sigma(ab) contains 0 and \sigma(ba) does not. For example, take A=\mathcal L(\ell^2), the algebra of all ... 0 Given any x\in E, there exists a rank-operator T with Tx=x (use Hahn-Banach to construct a bounded functional f with f(x)=1, and then define Ty=f(y)x). Then$$ S_\alpha x=S_\alpha Tx\to Tx=x.  This shows part 1. For part 2, let $X\subset E$ be compact. Fix $\varepsilon>0$. Then there exist $x_1,\ldots,x_n$ such that the balls of radius ...

3

Indeed, $J$ is an ideal of $A_0$. Any product of two elements of $A_0$ belongs to $J$, since $(f\cdot g)'(0) = f'(0)g(0) + f(0)g'(0) = f'(0)\cdot 0 + 0\cdot g'(0) = 0$, and hence $J$ is non-modular, as we have $au\in J$ for all $a,u\in A_0$, and hence $a-au = a-ua \in J \iff a\in J$. Also, $J$ is a $1$-codimensional (closed) linear subspace of $A_0$, hence ...

1

Disclaimer: this is an answer to the above series of comments by Student. It was too long for a comment. OK for "first" and "second". For "third", you shouldn't start by writing down $\frac1{1-a}$, since you precisely want to show that this exists. (By the way, the notation $\frac1{1-a}$ is usually "forbidden" in a vector-valued setting: it is generally ...

1

I thought I'd make the comments by Daniel Fischer into an answer. Here goes: All that remains to be done now is to show that $I(x)$ is closed. We show $I^c (x)$ is open. To this end, let $f \in I^c(x)$. Then $f(x) \neq 0$. Since the norm on $A$ dominates $\|\cdot\|_\infty$, we know that there exists $C \in \mathbb R$ such that $\|f\|_\infty \le C \|f\|$ ...

2

Solution: The problem appears to me to be false. Let $B$ be the Banach algebra of continuous functions on the annulus $\mathscr{A}=\{ z \in \mathbb{C} : 1/2 \le |z| \le 1\}$ with $\|b\|=\sup_{z \in \mathscr{A}}|b(z)|$. Let $A$ be the subalgebra of $\mathscr{A}$ generated as the closure in $B$ of polynomials in $z$. Let $b(z)=z$. Then $\sigma_{B}(b) = ... 0 No. A homomorphism should map$0$to$0.$But$p(T)=0$does not imply$p=0.$Take e.g.$T=T^*$with finite spectrum$\{\lambda_1,\dots,\lambda_n\}$and$p(x)=(x-\lambda_1)\dots(x-\lambda_n).$Note, that$p(x)\mapsto p(T)$is a homomorphism$\mathbb C[x]\to B(H).$3 Liouville's theorem (the one about bounded entire functions) is in complex analysis only proved for complex-valued functions. The generalisation to$\mathbb{C}^n$-valued functions is immediate, but for general Banach-space-valued (or Banach-algebra-valued) functions, it needs to be proved before it is used. The proof in that general case is by reducing it to ... 1 The algebra is commutative, and isomorphic to$C(X)$where$X$is the maximal ideal space. Under the Gelfand Transform, the operator itself goes to the function$f(z)=z\$. Have a look at continuos functional calculus in your book- they should prove this statement over there.

Top 50 recent answers are included