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Let $g \in G$. Then the translation $x \to gx$ defines an action from $G$ onto itself. This action extends to a continuous action from $\beta G$ to $\beta G$.


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False. The closed span of $e_k$ for $k \ge n$ is invariant.


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The following formula for $\|T^{-1}\|$ is relevant for the question posted. Let $(\mathcal E, \|\cdot\|_{\mathcal E})$ and $(\mathcal F, \|\cdot\|_{\mathcal F})$ be Banach spaces and let $\mathcal L(\mathcal E,\mathcal F)$ be the space of all bounded operators from $\mathcal E$ into $\mathcal F$. Let $T \in \mathcal L(\mathcal E,\mathcal F)$. The following ...


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You have, since $0\leq q\leq I$ and $0\leq p\leq I$, $$ -I\leq -q\leq p-q\leq I-q\leq I. $$ So, as you mentioned, it follows that $\sigma(p-q)\subset[-1,1]$. Note also that the argument does not use that $p,q$ are projections, only that they are positive elements of the unit ball.


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(the argument below is extracted from Lemmas 7.2.13 and 7.2.14 of Kadison-Ringrose; the relevant more general theorems are Theorem 7.2.15 and Corollary 7.2.16) If $x,y\in A'$ are selfadjoint, then we can find $\{a_n\}, \{b_n\}\subset A$, selfadjoint, with $a_n\xi\to x\xi$ and $b_n\xi\to y\xi$. Indeed, since $\xi$ is cyclic we can get $c_n$ in $A$ with $c_n\...


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The only solution I can see uses Tomita-Takesaki theory: Let $M=A''$ be the von Neumann algebra generated by $A$, so that $M$ is also abelian, i.e., $M\subseteq M'$. Note that $M'=A'''=A'$, so we are done if we show that $M'=M$ (in fact, this implies that $M$ is maximal abelian). We already have one inclusion $M\subseteq M'$. Since $\xi$ is cyclic for $A$, ...


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I just want to add a couple of observations to Jose Brox's answer. Firstly your question had the additional assumption that $A$ has an anti-involution $*$ fixing each of the $x_i$, but this doesn't make a difference. Indeed consider unital $\mathbb C$-algebra with presentation $$ F=\mathbb C\langle x_1,\ldots,x_n \mid x_i^2=x_i,\;x_1+\ldots+x_n=1\rangle. $...


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Your first inequality is usually known as the Kadison-Schwarz inequality. It only requires $2$-positivity. Claim. $A=\begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\geq0$ if and only if $a^*a\leq b$. Proof. If $A\geq0$, then for any $\xi\in H$, $$ \langle (b-a^*a)\xi,\xi\rangle=\left\langle \begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\,\begin{...


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I don't know enough to give you a very authoritative answer. But, as far as I can tell, there is no "general theory" of non-selfadjoint subalgebras of $B(H)$ the way that there is a theory for c$^*$-algebras or von Neumann algebras. There is a rather complete classification of nest algebras, and some generalizations. The original source for non-selfadjoint ...


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This is exactly the uniqueness in the polar decomposition. You have, since $v $ is a partial isometry, $$\tag {2}{\text {ran}\,v^*v}= {\text {ran}\,v^*}=(\ker v)^\perp=(\ker y)^\perp=\overline {\text {ran}\,y}. $$ Suppose that $w,z $ gives another such decomposition of $x $. Let $p=v^*v=w^*w $. Then, since $py=y$, we have $$ y^2=y^*y=y^*py=y^*v^*vy=x^*x=|x|...


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Here $$g=\text{diag}\,(1,-1)=\begin{bmatrix}1&0\\0&-1\end{bmatrix}.$$ You can easily check that $$ M_2(A)^{(0)}=\{a:\ gag=a\},\ \ \ \ M_2(A)^{(1)}=\{a:\ gag=-a\}. $$ The standard even grading on $A\otimes\mathbb K$ is obtained by doing the above on $M_2(A\otimes\mathbb K)$ (diagonal matrices and matrices with diagonal zero). How it looks like ...


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Not true as stated. Let $y$ be the operator on $\ell_2$ that maps each sequence $(t_j)_{j\in\mathbb{N}}$ to $(2^{-j}t_j)_{j\in\mathbb{N}}$. This is a positive operator with zero kernel. Let $c$ be the orthogonal projection onto the orthogonal complement of the vector $z = (1,1/2,1/3,\dots)$. Since $z\notin \operatorname{ran} y$, it follows that $cy$ also has ...


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Note that in your definition of $I_n$ you have some confusing use of $n$, which is fixed but also appears in $\cup_n\{x_n\}$, which is probably better written as $\overline{\{x_k:\ k\in\mathbb N\}}$. I'm not particularly comfortable with the way you argue that $I\ne A$. It is enough to show that $I$ cannot contain any function that is nonzero on $\overline{...


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Your proof is fine. What you are missing to work at points other than zero is the following lemma: Lemma. Let $f:X\to\mathbb C$ with $X$ a compact subset of $\mathbb R$, $\varepsilon>0$ and $R>0$. Then there exists $\delta=\delta(\varepsilon,R)>0$ such that if $a,b\in A^+$, with $\|a\|+\|b\|<R$, with $\sigma(a)\cup\sigma(b)\subset X$, and such ...


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Invertible in $qAq$ is not the same as invertible in $A$. For instance, $q$ is invertible in $qAq$ (it is the unit there). That $a$ is invertible in $qAq$ means that there exists $y\in qAq$ such that $ay=ay=q$. So, you have $x=va$; then $$ v=vq=vay=xy\in A. $$ About the spectrum, in general you cannot say anything: if $A=M_2(\mathbb C)$ and $$ p=\begin{...



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