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Let $A$ be the algebra of all bounded functions from $[0,\;1]$ to $\mathbb{C}$.

For $f\in A,\;$ $\omega_{f}$ is the standard oscillation function.. Each of the following two (equivalent) norms on $A$, defines a Banach algebra structure on $A$.

$$\parallel f \parallel=\parallel f\parallel_{\infty}+ \parallel \omega_{f}\parallel_{\infty}$$ or $$\parallel f \parallel=\parallel f\parallel_{\infty}+ \int_{[0,\;1]} \omega_{f}(x)dx$$ The later is well defined, since the oscillation function is a bounded measurable function.

Questions:

  1. Let $X$ be the Gelfand spectrum of $A$. What is the topological structure of this compact nonmetrizable disconnected Haussdorf space, precisely ? Is it homeomorphic to a known space?

  2. Assume that $B$ and $C$ are two $C^{*}$-algebras which are embedded isometrically into $A$. Must their $C^{*}$-tensor product be embedded in $A$, too?

  3. Is there a $C^{*}$-norm on $A$ which is equivalent to the above norms? In particular is $A$, semi-simple?

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    $\begingroup$ Since $\Vert\omega_f\Vert_\infty \leq 2\Vert f\Vert_\infty$, both of the norms $\Vert\cdot\Vert$ that you define above are equivalent to the usual sup norm, and so $A$ is isomorphic as a Banach algebra to $\ell^\infty([0,1])$. Thus for Q1 and Q3 you do not get anything new. $\endgroup$
    – Yemon Choi
    Commented Dec 7, 2014 at 17:22
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    $\begingroup$ AliTaghavi: if X is any indexing set then $\ell^\infty(X)=C(\beta X_d)$ where $\beta X_d$ is the Stone-Cech compactification of $X$ in its discrete topology. $\endgroup$
    – Yemon Choi
    Commented Dec 7, 2014 at 18:13
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    $\begingroup$ Ali, I suggest rewriting the question using topological terms only. No sophisticated language of operator algebras is necessary here. $\endgroup$
    – Tomasz Kania
    Commented Dec 7, 2014 at 19:36
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    $\begingroup$ @AliTaghavi Consider $\ell^1({\bf Z}/3{\bf Z})$ and ${\bf C}^3$ with $\ell^\infty$-norm. These Banach algebras are isomorphic, but not isometric as Banach spaces. $\endgroup$
    – Yemon Choi
    Commented Dec 8, 2014 at 16:37
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    $\begingroup$ @AliTaghavi Yes (this was just to get unital examples) $\endgroup$
    – Yemon Choi
    Commented Dec 8, 2014 at 16:47

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This is an answer to Q2, because as pointed out by Yemon, Q1 and Q3 follow easily from the fact that this algebra is isomorphic to $C(\beta [0,1]_d)$. It is enough to prove then that if two compact spaces $X$ and $Y$ are continuous images of $\beta \kappa$ for some cardinal $\kappa$, then so is $X\times Y$.

Let $X$ be a compact Hausdorff space. Let $D$ be a dense subset of $X$ endowed with the discrete topology. The identity map

$$\iota\colon D \to X$$

is continuous. By the universal property of the Stone–Čech functor $\beta$, $\iota$ extends (uniquely) to a continuous map

$$\beta \iota \colon \beta D \to X.$$ Since continuous images of compact spaces are compact, $\beta \iota$ is surjective. Hence we arrive at the following conclusion.

Conclusion. Let $X$ be a compact Hausdorff space with density character $\kappa$. Then $X$ is a continuous image of $\beta \kappa$. Consequently, since the product of two compact spaces with character $\kappa$ is compact and has density character $\kappa$, if $X$ are $Y$ are continuous, Hausdorff images of $\beta \kappa$, then so is $X\times Y$.

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  • $\begingroup$ Hi Tomek, I think in your first line you mean Q1 and Q3. $\endgroup$
    – Yemon Choi
    Commented Dec 8, 2014 at 13:15

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