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3

If your example $\mathcal{B}$ were a real Banach algebra instead of a complex Banach algebra, then you would be right that there are four connected components, since $\mathcal{B}$ can be identified with $\mathbb{R}^2$ and the invertible elements split into four quadrants. But over $\mathbb{C}$, you have the complement in $\mathbb{C}^2$ of two (complex) one-...


2

No. For example, say $$Ax=\sum\lambda_n\langle x,e_n\rangle e_n,$$where $(e_n)$ are orthomormal and $\lambda_n$ are scalars. Then $A$ is compact if $\lambda_n\to0$, while $A$ is trace class requires $\sum|\lambda_n|<\infty$.


2

The fact that $\overline{J}$ is an ideal follows from the continuity of the algebraic operations. For example, if $x \in \overline{J}$ and $y \in A$ then we can choose $x_n \in J$ such that $x_n \rightarrow x$ and then $yx_n \rightarrow yx$ and since $J$ is an ideal, $yx_n \in J$ and so $yx \in \overline{J}$. Similarly for right multiplication and addition. ...


2

Any $*$-homomorphism between C$^*$-algebras is contractive. This is standard (i.e., it appears in every book on the subject) and is due to three things: The C $^*$-identity $\|a\|^2=\|a^*a\|$, which reduces the problem to norms of positives; The equality $\|a\|=\text {spr}\, (a) $ for $a $ positive; The fact that a $*$-homomorphism reduces the spectral ...


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Seems to me the only example is $\Bbb C$. Say $\hat a$ is the Gelfand transform. Since $||\hat a||_\infty\le||a||$, similarly for $a^{-1}$, and $||\hat a||_\infty||1/\hat a||_\infty\ge1$ it follows that $|\hat a|$ is constant whenever $a$ is invertible. Hence, whether $a$ is invertible or not, $|\hat a-\lambda|$ is constant for every $\lambda$ not in the ...


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I recently wrote up a solution to this very problem, which I've copied below. Note that in this context, $\mathfrak A$ is the space of bounded operators on $B$. I hope you find this helpful. $ \newcommand{\f}{\mathfrak} \DeclareMathOperator{\rad}{r} \newcommand{\eps}{\varepsilon} \DeclareMathOperator{\spec}{spec} \newcommand{\lp}{\left(} \newcommand{\rp}{\...


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Well, $$||x||^{m^2}=(||x||^m)^m\le(K||x^m||)^m=K^m||x^m||^m\le K^mK||(x^m)^m||=K^{m+1}||x^{m^2}||.$$Now your argument for $n\le m$ also works for $n\le m^2$...



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