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I assume that $A$ is a $C^\star$-algebra and that $A_0$ is a subset of $A$? Then the generated $C^\star$-subalgebra is the norm-closure of the linear span of the multplicative closure of the $\star$-closure of $A_0$. That is, it is the norm-closure of the generated $*$-subalgebra $\{\sum_i \lambda_i a^i_1 \dotsc a^i_n : \lambda_i \in \mathbb{C}, a^i_k \in ...


1

No, even if the group is compact, Abelian and separable. For instance, the Cantor group $\{0,1\}^{\omega_1}$, where $\omega_1$ is the first uncountable cardinal, is a good counter-example. This group is obviously compact and Abelian. It is also separable by the Hewitt–Marczewski–Pondiczery theorem. However $L_1(\{0,1\}^{\omega_1})$ is non-separable. ...


0

No, because any discrete group is locally compact and admits a Haar measure that assigns the measure 1 to any element. $L^1$ on a non-countable set with such measure isn’t separable. BTW I doubt that we necessarily have separability even for compact groups. Compactness is not a very strong condition to infer “nice” properties of function spaces. There are ...


0

Sure, you can approximate continuous functions by analytic ones (in the $\sup$ norm) on a compact. There is no difference whether are they ℝ-valued, vector-valued, or matrix-valued, as long as “analytic” denotes analyticity over ℝ not anything stronger. What means “unitary function”, simply $S^1 \to {\mathrm U}(n)$? You can approximate it as well, because ...


2

No. For example, $S_3$ contains a characteristic, abelian subgroup not contained in the centre of the group. Both of the automorphisms of this subgroup extend to (inner) automorphisms $G$.


1

Hint. $$ \|a_nb_n - ab\| \le \|a_nb_n - a_nb\| + \|a_nb - ab\|\le \|a_n\|\|b_n - b\| + \|a_n - a\|\|b\| $$


1

In general the answer is no. If $G$ is compact and abelian then $I$ is a left closed ideal with bounded approximate identity iff $I=L_1(G)*\mu$ for some idempotent $\mu\in M(G)$ iff $I$ is a kernel of coset ring of closed subsets in dual group of $G$. Taking a kernel of the set that doesn't belong to that coset ring you get (not much explicit) example of ...


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For every $d\in A$ such that $\|d\|=1$ one has $$\zeta(a)\leq\|bd\|+\|a-b\|.$$ So $$\zeta(a)\leq\zeta(b)+\|a-b\|.$$ Using a symmetric reasoning we get $$\zeta(b)\leq\zeta(a)+\|a-b\|,$$ and the inequality is proved.


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Let $P$ be an $L$-projection. From $$\lVert x\rVert = \lVert Px\rVert + \lVert x-Px\rVert$$ for all $x\in X$ it follows that $\lVert P\rVert \leqslant 1$ and $\lVert I-P\rVert \leqslant 1$, and hence $$\lVert P^\ast f\rVert = \lVert f\circ P\rVert \leqslant \lVert f\rVert\cdot \lVert P\rVert \leqslant \lVert f\rVert,$$ and analogously $\lVert (I-P)^\ast ...



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