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3

Yes, this is standard. Let $T\in A$ selfadjoint, i.e. with $T=T^*$. Note first that $\phi(T)$ is real: since $T+\|T\|\,\text{id}$ is positive, we have that $$ \phi(T)+\|T||\in\mathbb R, $$ so $\phi(T)\in \mathbb R$. Now, as $-T+\|T\|\,\text{id}\geq0$, we get $-\phi(T)+\|T\|\geq0$, so $$\phi(T)\leq\|T\|.$$ Since $-T$ is also selfadjoint, we can also get ...


1

The spectrum of $a$ is a compact set that does not contain $0$. So there is a disk $D$ around $0$ with $D\cap\sigma(a)=\emptyset$. Thus, on $\sigma(a)$, $f:t\longmapsto 1/t$ is continuous, so $f\in C(\sigma(a))$. Then $f(a)\in C^*(a)$ via the Gelfand transform. Or, even easier, you could check that $f$ is analytic on $\sigma(a)$, and so $a^{-1}$ belongs ...


1

Let $\{\phi\}$ be a net of pure states on $A$ and assume that it is $w^*$-convergent to $\phi$. Clearly $\phi$ is a positive linear functional on $A$. To prove $\phi$ is a pure state, it is enough to show that support of $\phi$ is a minimal projection in $A^{**}$. Assume $q$ is a non-trivial projection majorized by the support of $\phi$. We have that ...


2

That $\Gamma(y)$ should be closed doesn't follow from the fact that $\Gamma$ is a closed subalgebra. Example. Let $\mathbb N^+ = \mathbb N \setminus \{0\}$ be the set of positive integers. Consider $X = \ell^2(\mathbb N^+)$ and let $R : X \to X$ be the following weighted right shift: $$ (Rx)(n) = \begin{cases} \quad 0, & \quad\text{if $n = 1$}; ...


3

A maximal ideal is by definition a proper ideal (which happens to not be strictly contained in any other proper ideal). If $I$ is an ideal and an element $x\in I$ is invertible, then $x^{-1}x=e\in I$ so $a=a\cdot e\in I$ for all $a\in A$, so $I=A$. So no proper ideal (and in particular, no maximal ideal) can contain an invertible element.


2

Let $A$ be a finite-dimensional Banach algebra. The group $K_0(A)$ only depends on the underlying ring structure of $A$. In what follows, $A$ can be any finite-dimensional algebra over an arbitrary field $k$, well $A\neq 0$. The map $\{\text{finitely generated projective $A$-modules}\}\mapsto \mathbb Z$ given by $P\mapsto\dim_kP$ is well defined (being $A$ ...


1

$\newcommand{\ip}[2]{\left\langle #1,#2\right\rangle}$ $\newcommand{\norm}[1]{\left|\left| #1\right|\right|}$ I think your confusion is due to notation used in Pythagorean theorem: we don't actually require any norm-property of $||\cdot||$ induced by the inner product in proving it. It's only after we recognise $||\cdot||$ as a norm that it has its usual ...


2

As Gro-Tsen commented, the fact that $\Gamma$ is an isometry implies its image is closed because $\mathfrak{U}$ is a complete metric space (since it is closed in $C(X)$). Any subset of a metric space that is complete as a metric space is automatically closed, since a sequence that converged to a point outside the set would be a Cauchy sequence with no limit ...


0

It can be easily shown that $B(X)$ has always nontrivial zero divisor. Thus every Banach algebra which does not have zero divisor is another example. For example the disc algebra, the algebra of all holomorphic functions on the unit disk with continuous extension to the boundary.


0

Let $\{e_n\}$ be an orthonormal sequence in $H$ and denote $p_n$ by the rank one projection onto $\mathbb{C}e_n$. Then the positive element $x=\sum \frac{1}{n}p_n$ works. To see it, assume $x$ is a convex combination of projections $q_1,\cdots,q_k$, namely $x=t_1q_1+\cdots t_kq_k$. Then there is $j$ such that $q_jH$ contains an infinite subset $E$ of the ...


0

Your problem is solved by the following important fact which has been proved by Arens & Kadison 1n 1968 (see their paper: Pure state and approximate identities) Theorem (Arense & Kadison). Let $a$ be an strictly positive in a C*-algebra $A$. The $\{a^{\frac{1}{n}}\}$ forms an approximate identity for $A$. In your problem, you assumed $\sigma(a)$ ...


0

You have $$ (1-p)a(1-p)\leq(1-p)p(1-p)=0. $$ Thus $$0=(1-p)a(1-p)=(a^{1/2}(1-p))^*(a^{1/2}(1-p)),$$ and $a^{1/2}(1-p)=0$, from where $a(1-p)$. Taking adjoints, $(1-p)a=0$. If follows that $a=pap$.


3

This is far from being true, indeed, every symmetric operator has only real eigenvalues: If $\psi\in\ker(T-\lambda),\,\psi\neq 0$, then $$ \lambda\|\psi\|^2=\langle T\psi,\psi\rangle=\langle \psi,T\psi\rangle=\bar\lambda\|\psi\|^2, $$ hence $\lambda=\bar\lambda$. Now every symmetric operator that is not self-adjoint yields a counterexample to your ...


0

Let $f\in A=C_0(0,1)$. Then $f(t)>0$ for all $t$ if and only if $f$ is strictly positive. Assume first that $f(t)>0$ for all $t$. Fix, initially, $g\in C_0(0,1)$ such that $g(t)=0$ for all $t\in (0,b)\cup(1-b,1)$ for some $b>0$. As $[b,1-b]$ is compact and $f>0$ on $[b,1-b]$, there exists $\delta>0$ with $f(t)\geq\delta$ for all $t\in ...


0

Let $a(t)>0$ for all $t\in (0,1)$ (eg $1-2|t-1/2|$). Then for each $n \in \mathbb{N}$, there exists a function $c_n \in C_0(0,1)$ so that $c_n(t)a(t)=1$ whenever $t \in I_n :=[\frac{1}{n},1-\frac{1}{n}]$. This consideration shows that for any $f \in C_0(0,1)$ $(a c_n f c_n a)(t)=f(t)$ whenever $t \in I_n$. The construction implies that $f \in ...



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