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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 $\... 4 No. For any such sequence$\{p_n\}$with each$p_n$of finite rank, take$A=B(\mathcal H)$and $$B=\overline{ \{b\in B(\mathcal H):\ \exists n,\ b=p_nbp_n\}}.$$ Then$p_nAp_n=p_nBp_n$for all$n$, but$B$is separable while$A$is non-separable. 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 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 If$\delta\leq1$, then$x^*x$and$xx^*$are invertible. In particular, we can do the polar decomposition$x=u(x^*x)^{1/2}$and we will have$u\in A$. Also,$u$is a unitary because $$u^*u=(x(x^*x)^{-1/2})^*(x(x^*x)^{1/2} =(x^*x)^{-1/2}x^*x(x^*x)^{-1/2}=1,$$ $$uu^*=x(x^*x)^{-1/2}(x^*x)^{-1/2}x^*=x(x^*x)^{-1}x^*=1.$$ This last equality is not obvious, but ... 2 Yes. What you do is show that if$A$is not simple, then there is a non-faithful irrep. If$J\subset A$is a non-trivial ideal, then consider an irreducible representation of$A/J$into$B(H_J)$; then$A\to A/J\to B(H_J)$is an irreducible representation of$A$with kernel that at least contains$J$, so it is not faithful. 2 You can't find a positive$\gamma$such that$\gamma \leq \frac{\langle Lu\mid u \rangle}{\langle u\mid u \rangle}$if$\langle Lu\mid u \rangle=0$. So to find a counterexample, you might look for positive operators such that$Lu$is sometimes$0$. The simplest example is$L=0$, the operator that sends everything to$0$. 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^...

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Yes, this follows from the monotone convergence theorem for nets (see Theorem IV.15 of Reed and Simon's Functional Analysis, for instance) Let $\mu$ be a regular Borel measure on a compact Hausdorff space $X$ and let $(f_{\alpha})$ be an increasing net of continuous functions converging pointwise to $f$. If $\sup \|f_{\alpha}\|_1 < \infty$, then $\... 2 Your questions reveal a typo that Valette was not aware of. The third line of the proof of Proposition 3.3.7 should be:$$\pi_n(\text{GL}_\infty(A)) = \pi_{n - 1}(\Omega \text{GL}_\infty(A)) = \pi_{n - 1}(\text{GL}_\infty(SA)),$$where$\Omega X$is the loop space of$X$. The first equality is a basic property of loop spaces, see e.g. here. The second follows ... 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$L^1[0,\infty)$. ... 1 If all you assume about your involution is that it's an involution. that is,$(x+y)^*=x^*+y^*$,$(xy)^*=x^*y^*$,$x^{**}=x$and$(cx)^*=\overline cx^*$, then most of what you expect doesn't follow. In particular you assume above that$\phi(x^*)=\overline{\phi(x)}$, and that doesn't follow: Consider$C([-1,1])$. Define $$f^*(t)=\overline{f(-t).}$$ That's an ... 1 These results follow from some quite general results about surjective homomorphisms between$C^{\ast}$-algebras which I will state without proof. In the following,$A$and$B$are unital$C^{\ast}$-algebras and$\varphi :A\to B$is a unital surjective$\ast$-homomorphism. [Proposition 4.3.14 in Higson & Roe's Analytic K-homology ] If$f: [0,1] \to ...

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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 Somehow, after I post here a question the solution comes to my mind... So, I'll use the following argument:almost unitaries are close to a unitary element Now, it's enough to show that: If A is a unital C^* algebra and A=\overline{\bigcup_{k\in \mathbb{N}} A_k} ,where each A_k is a unital (same unit of A) C^* subalgebra, then for any unitary ... 1 Ignoring your original question: Note that A = \lim_k A_k so K_1(A) = \lim_k K_1(A_k). But K_1(A_k) = 0 since K_1(M_n(\mathbb C)) = 0 for each natural number n. This is a general argument that AF algebras have trivial K_1-group. 1 There is a typo in the statement: it should say e\precsim f. The rest looks ok to me. As the proof says, one uses Theorem 1.8: there exists central z with ze\precsim zf and (1-z)e\succsim (1-z)f. The proof shows that (1-z)e\sim (1-z)f. Then e=ze+(1-z)e\precsim zf+(1-z)f=f. $$1 Another similar approach (inspired by Martin Argerami) is the following: If x \in A and x^*x, xx^* are invertible, then x is invertible. To see this, check that (x^*x)^{-1}x^* = x^*(xx^*)^{-1}. This is then the inverse of x. Now x = u \sqrt{x^*x} for u unitary since x is invertible. Then the last argument of Martin applies, i.e. that \... 1 Already \mathscr{B}(H)^* or even L_\infty^* are intractable. However, \mathscr{M}^{**} is a von Neumann algebra too, so \mathscr{M}^* is indeed a predual of the second dual. This is the best we can say in that generality. Maybe you should rethink your question. 1 Note that the two equalities A=AA^*A and A^*=A^*AA^* are the same, since you can obtain one from the other by taking adjoints. Assume first that A=AA^*A. By multiplying by A^* on the left, we get$$ A^*A=(A^*A)^2.$$It follows that the eigenvalues of A^*A all satisfy the equation \lambda=\lambda^2, so only 0 and 1 are possible. Conversely,... 1 Some notation is missing, like for example you need to write your C^*-algebra as C(X) for a locally compact Hausdorff space. For your converse, if \phi_\omega is the state given by evaluation at \omega and \phi_\omega=\alpha\varphi+(1-\alpha)\psi, by Riesz-Markov you get an equality of measures$$\mu_\omega=\alpha\mu_\varphi+(1-\alpha)\mu_\psi.$$... 1 No. The von Neumann algebra B(H) is separable (in the sot topology, say); it follows that any von Neumann algebra R\subset B(H) is separable. This is not totally straightforward, but it is simple: A countable dense subset of B(H) is given by the complex-rational linear combinations of matrix units (because its commutant is trivial); Separability can ... 1 (You don't say how you got the second equality; since it is not trivial, I'm not sure how you did it and so it is done below) Since A^*A is positive and compact, it is orthogonally diagonalizable (spectral theorem): A^*A=U^*D^2U for some unitary U and D diagonal with diagonal s_1(A),s_2(A),\ldots Assume s_1(A)\geq s_2(A)\geq \cdots Since U ... 1 1\iff4 : Assume 1. Given x\in N_+, there exists nonzero y\in N_\tau^+ with y\leq x. Now use Zorn to find a maximal ordered family \{y_j\}\subset N_\tau^+ with y_j\leq x for all j. As the net is bounded, it has a sup, say y=\lim_{sot}y_j. Then y=x, because otherwise a nonzero element of N+\tau^+ below y-x contradicts the maximality. ... 1 To answer the new request from the OP (expressed in a commentary), let us see that if R is a ring and e_i\in R are nonzero idempotent and commutative such that e_1+\ldots+e_n=1 then they are in fact orthogonal, provided that R is free of torsion up to n-1: I have thought of a proof by backwards induction. First observe that$$e_1\ldots e_n = e_1\...

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A unitary operator is a diagonalizable operator whose eigenvalues all have unit norm. If we switch into the eigenvector basis of U, we get a matrix like: \begin{bmatrix}e^{ia}&0&0\\0&e^{ib}&0\\0&0&e^{ic}\\\end{bmatrix} which is obviously the exponential of a diagonal hermitian matrix.

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