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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 $c_n\... 1 The only solution I can see uses Tomita-Takesaki theory: Let$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$, ... 1 I just want to add a couple of observations to Jose Brox's answer. Firstly your question had the additional assumption that$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 $$F=\mathbb C\langle x_1,\ldots,x_n \mid x_i^2=x_i,\;x_1+\ldots+x_n=1\rangle. ... 1 Your first inequality is usually known as the Kadison-Schwarz inequality. It only requires 2-positivity. Claim. A=\begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\geq0 if and only if a^*a\leq b. Proof. If A\geq0, then for any \xi\in H,$$ \langle (b-a^*a)\xi,\xi\rangle=\left\langle \begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\,\begin{... 1 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 ... 1 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 $$y^2=y^*y=y^*py=y^*v^*vy=x^*x=|x|... 1 Here$$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\}. $$The standard even grading on A\otimes\mathbb K is obtained by doing the above on M_2(A\otimes\mathbb K) (diagonal matrices and matrices with diagonal zero). How it looks like ... 1 Not true as stated. Let y be the operator on \ell_2 that maps each sequence (t_j)_{j\in\mathbb{N}} to (2^{-j}t_j)_{j\in\mathbb{N}}. This is a positive operator with zero kernel. Let c be the orthogonal projection onto the orthogonal complement of the vector z = (1,1/2,1/3,\dots). Since z\notin \operatorname{ran} y, it follows that cy also has ... 1 Note that in your definition of I_n you have some confusing use of n, which is fixed but also appears in \cup_n\{x_n\}, which is probably better written as \overline{\{x_k:\ k\in\mathbb N\}}. I'm not particularly comfortable with the way you argue that I\ne A. It is enough to show that I cannot contain any function that is nonzero on \overline{... 1 Your proof is fine. What you are missing to work at points other than zero is the following lemma: Lemma. Let f:X\to\mathbb C with X a compact subset of \mathbb R, \varepsilon>0 and R>0. Then there exists \delta=\delta(\varepsilon,R)>0 such that if a,b\in A^+, with \|a\|+\|b\|<R, with \sigma(a)\cup\sigma(b)\subset X, and such ... 1 Invertible in qAq is not the same as invertible in A. For instance, q is invertible in qAq (it is the unit there). That a is invertible in qAq means that there exists y\in qAq such that ay=ay=q. So, you have x=va; then$$ 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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