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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 $$0 = \chi_2(a - \lambda \mathbf{1}) = \chi_2(a) - \lambda$$

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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 ...

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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 $\... 3 Linear-multiplicative functionals (aka characters) on complex Banach algebras are automatically continuous, so their kernels are closed. (You will find a slick proof of this fact on p. 181 of Allan's and Dales' Introduction to Banach Spaces and Algebras.) However, in the non-unital case it may well happen that a maximal ideal is dense. The right notion to ... 3 Not in general: Take$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 ... 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>0so that $$||xy||\le1\quad(||x||,|... 3 There's no simple description of the spectrum of the algebra of bounded holomorphic functions in the disk (known as H^\infty). It's an Axiom-of-Choice-ish thing. If |z|<1 then f\mapsto f(z) is a complex homomorphism, so the open disk is contained in the spectrum in a natural way. The Corona Theorem says that the disk is dense. This is one of the ... 2 Edit: This is an answer to the previous version of this question. This algebra is complete; it is a simple application of Morera's theorem which you may use to show that the uniform limit of such functions is actually holomorphic. This algebra is traditionally denoted by H^\infty and is highly non-separable. For this reason, the maximal ideal space of H^... 2 Let A be the algebra of all 2\times2 matrices over \mathbb{R} (or \mathbb{C}) of the form$$\begin{bmatrix} a & b \\ 0 & c \end{bmatrix} $$Then A is a Banach algebra, which is noncommutative, and is not a C^*-algebra. 2 Any *-homomorphism between C^*-algebras is contractive. This is standard (i.e., it appears in every book on the subject) and is due to three things: The C ^*-identity \|a\|^2=\|a^*a\|, which reduces the problem to norms of positives; The equality \|a\|=\text {spr}\, (a) for a positive; The fact that a *-homomorphism reduces the spectral ... 2 The first thing to show is that the decomposition is unique. That is, if f is continuous on \mathbb{R} has such a representation, then d and k are unique (k is unique as an element of L^1[0,\infty).) Equivalently, if f=d+\int_{0}^{\infty}e^{ixt}k(t)dt is the 0 function on \mathbb{R}, then d=0 and k=0 as an element of L^1[0,\infty). ... 2 A typo slipped in; a k became j for no reason. Fixing that, you're almost there, re showing it's a Banach algebra:$$\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)=\... 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)|)\$.

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