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3

We do not need to consider $\Omega(A)\cup \{0\}$, but it makes things easier, since that is a compact (Hausdorff, if that isn't part of your definition of compact) space. Every open subspace of a compact (Hausdorff) space is locally compact, and since the weak* topology is Hausdorff, singletons are closed, whence $$\Omega(A) = (\Omega(A)\cup \{0\}) ...


3

There are $2^{2^{\aleph_0}}$ automorphisms of $\mathbb C$, a mind-boggling cardinal, and of those only the identity and conjugation can be explicitly described. The others depend on set-theoretical wizardry (axiom of choice). What is much more interesting is to take a field extension $k\subset K$ and to study the group $Aut(K/k)$ of automorphisms ...


2

Apparently you forgot about the negative powers. If for example we consider $a \in \ell^1(\mathbb{Z})$ given by $a_n = 2^{-\lvert n\rvert}$, the corresponding Laurent series $$f(z) = \sum_{n=-\infty}^{+\infty} a_n z^n = \sum_{n=1}^\infty a_{-n}z^{-n} + \sum_{n=0}^\infty a_nz^n = \sum_{n=1}^\infty \frac{1}{(2z)^n} + \sum_{n=0}^\infty ...


2

Example 3 is correct, for $\lambda \notin S^1$, $z \mapsto \frac{1}{z-\lambda}$ is a continuous function on the unit circle, and hence $p(z) - \lambda$ is invertible. On the other hand, for $\lambda\in S^1$, $p(z)-\lambda$ has a zero on $S^1$ and is therefore not invertible, thus $\sigma_C(p) = S^1$. Example 4, however, is not correct. The spectrum of an ...


2

Of course not. For example the Gelfand spectrum of Banach algebra is locally compact, and it is compact iff $A$ is unital. There a lot of examples in Banach homology where existence of unit plays significant role. Another example which is more related to Banach space geometry than to Banach algebras: the Banach algebra $c_0$ have no extreme points while its ...


2

Is there a mean value theorem for differentiable functions $[a,b]\to \mathbb R^2$? Yes, there are several theorems that are called mean value theorem for that situation (and the more general of maps to $\mathbb{R}^n$). One mean value theorem is Let $f \colon [a,b] \to \mathbb{R}^n$ a differentiable function. If $\lVert f'(t)\rVert \leqslant M$ for ...


2

The claim is that for any $a \in A$, $\sigma(a) = \{\chi(a) \mid \chi \in \Omega(A)\} \cup \{0\}$. Hence, to prove this, you take any element $a \in A$, and show that for that fixed element $a$, $\sigma(a) \subseteq \{\chi(a) \mid \chi \in \Omega(A)\} \cup \{0\}$ and $\sigma(a) \supseteq \{\chi(a) \mid \chi \in \Omega(A)\} \cup \{0\}$. So, you must fix a ...


2

I don't understand how $C_0(U)$ can be a subset of $C(X)$ Unless $U = X$, it isn't strictly a subset. However, we have a canonical isometric embedding $\eta\colon C_0(U) \hookrightarrow C(X)$ given by $$\eta(f) = x \mapsto \begin{cases} f(x) &, x \in U\\ 0 &, x \notin U. \end{cases}$$ Since $f(x) \to 0$ as $X$ approaches $\partial U$ for every ...


2

The first thing in the proof concerning $A$ is $b - \lambda \notin \operatorname{Inv}(A)$, which just means $\lambda \in \sigma_A(b)$. Next, it is observed that the terms of the sequence $(\lambda_n)$ do not belong to $\sigma_A(b)$ - namely, it is stated that $b - \lambda_n \in \operatorname{Inv}(A)$ - which says that $\lambda$ is not an interior point of ...


2

Ultimately, it is just a matter of convention and how a particular author wants to write things. Authors usually choose the path that makes their writing more succinct and easier to read. The example you gave is a common one: Make a product of objects... shouldn't the original objects be subobjects of the new object? Depending on your decision, the ...


2

First, an answer to your question. No, one typically does not assume that Banach subalgebras are unital. One of the standard examples of a Banach algebra is $C(X)$ where $X$ is compact, and one of the standard examples of a Banach subalgebra of $C(X)$ (an ideal, in fact) is is $C_0(U)$ where $U \subseteq X$ is an open set. But $C_0(U)$ is not unital. ...


2

A character is a linear functional with some additional properties ($\varphi(a\cdot b) = \varphi(a)\varphi(b)$ and $\varphi(1) = 1$). Then the fact that a linear functional $\lambda \colon E \to \mathbb{K}$ on a topological vector space $E$ over $\mathbb{K}\in \{\mathbb{C},\mathbb{R}\}$ is continuous if and only if its kernel is closed yields the continuity ...


1

Note that the expression $a - \tau(a)$ means, in this context, $a - \tau(a)1_A$, where $1_A \in A$ is the unit element of $A$. Clearly $I + \Bbb C 1_A \subset A$. For any $a \in A$, $a - \tau(a)1_A \in I$, that is, for some $b \in I$, $a - \tau(a)1_A = b$. Then $a = b + \tau(a)1_A \in I + \Bbb C 1_A$ since $\tau(a) \in \Bbb C$, so $A \subset I + \Bbb C ...


1

No. Take $A=c_0$. Then $c_0$ is complemented in $\ell_\infty(c_0)$ (just project down on the very first coordinate). Note that $c_0$ is never complemented in a dual Banach space as this is equivalent to being complemented in $c_0^{**}\cong \ell_\infty$, which is apparently not the case. See t.b.'s answer for Complementability of von Neumann algebras ...



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