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## New answers tagged banach-algebras

2

The first thing to show is that the decomposition is unique. That is, if $f$ is continuous on $\mathbb{R}$ has such a representation, then $d$ and $k$ are unique ($k$ is unique as an element of $L^1[0,\infty)$.) Equivalently, if $f=d+\int_{0}^{\infty}e^{ixt}k(t)dt$ is the $0$ function on $\mathbb{R}$, then $d=0$ and $k=0$ as an element of $L^1[0,\infty)$. ...

2

A typo slipped in; a $k$ became $j$ for no reason. Fixing that, you're almost there, re showing it's a Banach algebra: \begin{align}\dots=\max\limits_{0 \leq t \leq 1} \sum_{k=0}^{n}\sum_{j=0}^{k}{\dfrac{|f^{(k-j)}(t)}{(k-j)!}\dfrac{g^{(j)}(t)|}{j!}} &=\max\limits_{0 \leq t \leq 1} \sum_{j=0}^{n}\dfrac{|g^{(j)}(t)|}{j!}\sum_{k=j}^{n}\dfrac{|f^{(k-... 1 Given a (nonzero) multiplicative linear functional \chi on A and x,y\in A, we have \begin{align*} \hat{xy}(\chi)&=\chi(xy)=\chi(x)\chi(y)=\hat{x}(\chi)\hat{y}(\chi) \\ \hat{(x+y)}(\chi)&=\chi(x+y)=\chi(x)+\chi(y)=\hat{x}(\chi)+\hat{y}(\chi) \end{align*} Furthermore, for any scalar \alpha, we have \hat{(\alpha x)}(\chi)=\chi(\alpha x)=\...

3

Linear-multiplicative functionals (aka characters) on complex Banach algebras are automatically continuous, so their kernels are closed. (You will find a slick proof of this fact on p. 181 of Allan's and Dales' Introduction to Banach Spaces and Algebras.) However, in the non-unital case it may well happen that a maximal ideal is dense. The right notion to ...

4

The kernels of nonzero homomorphisms to $\mathbb C$ are modular ideals, terminology that might help you find more references. Without any further restriction on the algebras, using the zero product is a way to provide trivial counterexamples. E.g., take $\mathbb C$ with the $0$ product, which has maximal ideal $\{0\}$ and no nonzero homomorphisms to $\... 9 Suppose$a$is an element of the algebra, and$\chi_1(a) = \lambda$. Then$a - \lambda \mathbf{1} \in \ker \chi_1 = \ker \chi_2$so $$0 = \chi_2(a - \lambda \mathbf{1}) = \chi_2(a) - \lambda$$ 1 Say$K=\{0,1\}$, with the discrete topology. Let$A=C(K)$, but with the non-standard norm$||f||=\max(|f(0)|,2|f(1)|)$. 3 Not in general: Take$A = C^1[0,1]$of continuously differentiable functions on$[0,1]$with the norm$\|f\| := \|f\|_{\infty} + \|f'\|_{\infty}$. For each$t\in [0,1]$, the evaluation map$\tau_t : A \to \mathbb{C}$given by$f\mapsto f(t)$induces a continuous map $$[0,1] \to M_A \text{ given by } t\mapsto \tau_t$$ One can then show that this map is a ... 0 I can't think of any interesting consequences offhand. Now, assuming that the spectral radius of every element of$A$is equal to the norm does have an interesting consequence - in fact it's easy to see that that condition is equivalent to saying$A$is isometrically isomorphic to a closed subalgebra of$C(K)$for some compact Hausdorff space$K$. 4 Ben, the answer is no. Note that if$X$is reflexive, then$B(X)$is isometric to$B(X^*)$via$T\mapsto T^*$. Note that this map is an anti-isomorphism of Banach algebras. As for less trivial examples,$B(\ell_p)$is Banach-space isomorphic to$B(L_p)$as well to$B(X)$for any other separable, infinite-dimensional$\mathscr{L}_p$-space and$\ell_p$is ... 3 There's no simple description of the spectrum of the algebra of bounded holomorphic functions in the disk (known as$H^\infty$). It's an Axiom-of-Choice-ish thing. If$|z|<1$then$f\mapsto f(z)$is a complex homomorphism, so the open disk is contained in the spectrum in a natural way. The Corona Theorem says that the disk is dense. This is one of the ... 2 Edit: This is an answer to the previous version of this question. This algebra is complete; it is a simple application of Morera's theorem which you may use to show that the uniform limit of such functions is actually holomorphic. This algebra is traditionally denoted by$H^\infty$and is highly non-separable. For this reason, the maximal ideal space of$H^...

0

If a complex finite dimensional algebra is a C^* algebra, then it is isomorphic to a direct product of matrix algebras. So any algebra which is not of this form gives you an example.

2

Let $A$ be the algebra of all $2\times2$ matrices over $\mathbb{R}$ (or $\mathbb{C}$) of the form $$\begin{bmatrix} a & b \\ 0 & c \end{bmatrix}$$ Then $A$ is a Banach algebra, which is noncommutative, and is not a $C^*$-algebra.

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http://arxiv.org/abs/math/9909027 is Planar Algebras I by Jones.

3

I don't know about "geometrical view", but the condition essentially follows from assuming that multiplication is continuous. Note the "essentially"; we'll see below what I mean by that. Say we have an algebra with a norm, and multiplication is (jointly) continuous. Continuity at $(0,0)$ shows that there exists $\delta>0$ so that ||xy||\le1\quad(||x||,|...

0

I could solve the problems: We show that $\pi$ is open onto its image $\pi(S)$. Let $\{s\} \subseteq S$, then we have $\{\delta_s\} = \pi(s) \cap (\operatorname{ev}_{1_{\{s\}}})^{-1}(\mathbb C \setminus \{0\})$, where $\operatorname{ev}_{1_{\{s\}}}(\varphi) = \varphi(1_{\{s\}})$ for all $\varphi \in G_S$. Hence $\{\delta_s\}$ is open in $\pi(S)$ and $\pi(S)... 2 Any$*$-homomorphism between C$^*$-algebras is contractive. This is standard (i.e., it appears in every book on the subject) and is due to three things: The C$^*$-identity$\|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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