# Tag Info

5

In a unital Banach algebra the set of invertible elements is open and nonvoid, hence the set of noninvertible elements is closed ( and also nonvoid, since it contains zero). Any proper ideal is contained in the set of noninvertible elements. Hence, no dense proper ideals in a unital Banach algebra.

4

It is true for $B(H)$ also. Another way to see it for $M_n$ when $n>1$ that may be extended in spirit to $B(H)$ is to note that $M_n$ is simple, and the kernel of a multiplicative linear functional would be an ideal of codimension $1$. If $H$ is infinite dimensional then $B(H)$ is not simple, but it does have a unique maximal ideal, and the codimension ...

3

Here's a direct argument. Since $\Phi$ is onto and multiplicative, $\phi(I)=1$. Write $I=P+Q$ for two orthogonal equivalent projections. Then, as $PQ=0$, we have $\Phi(P)\Phi(Q)=0$, so at least one of them is zero. But as $P=V^*V$, $Q=VV^*$, we get $\Phi(P)=\Phi(Q)$, so both are zero. Then $1=\Phi(P+Q)=\Phi(P)+\Phi(Q)=0$, a contradiction.

3

Any finite rank operator is a compact operator, and it's a known result that the only points in the spectrum of a compact operator are the eigenvalues. See theorem 35.17 here for the general statement.

3

I think you have already done the hard part. Assume $h \neq 0$ and fix some $a \in \mathcal{A}$ with $h(a) \neq 0$. For each $i$, we have $h(e_i) h(a) = h(e_i a)$, so $$h(e_i) = \frac{h(e_ia)}{h(a)} \to 1.$$ Thus there is a net of elements of norm $\leq 1$ whose images tend to $1$. It follows that $\|h\| \geq 1$.

3

I am assuming that you are working with the operator norm on $B(H)$ and with the unique $C^*$-norm on $M_2(B(H))$ given by identifying this space with $B(H \oplus H)$. The short answer is that in general there is no equation in the norms of $A$, $B$, $C$, and $D$ that results in $\Vert T \Vert$. To see this, we can consider the case where $H= \mathbb C$ ...

2

Every element of $B(A_0)$ is the limit of polynomials of the form $$p(A_0)=\sum_{k=0}^na_kA_0^k,\quad n\in\mathbb N,\,\,a_k\in\mathbb C.$$ Hence the first and second bullets hold. Note that $$\|B\|=\sup_{\|x\|=\|y\|=1}(x,By),$$ hence, if $A$ is self-adjoint, then $$\|A^2\|=\sup_{\|x\|=\|y\|=1}(x,A^2y)=\sup_{\|x\|=\|y\|=1}(Ax,Ay)\ge ... 2 Yes, your argument is correct. 2 I find it easy to prove that it is a normed algebra, so proving that it is complete is all that remains to do; You need to see that if (f_n) is a Cauchy sequence in B_T, then for every t\in T, the sequence \bigl(f_n(t)\bigr) is a Cauchy sequence in \mathbb{C}, which follows directly from \lvert f_n(t) - f_m(t)\rvert \leqslant \lVert f_n - ... 2 The right naming here would not be "Heine-Borel" but "Banach-Alaoglu". 2 The simplest example would be a nonzero algebra with any involution and zero multiplication. 2 Any normal operator T gives rise to some spectral measure E:Bor(\sigma(T))\to\mathcal{P}(H) which maps Borel subsets of the spectrum of T into orthogonal projections in H. If you take Borel subset A\subset\sigma(T), then E(A) is called a spectral projection. Search spectral theorem on this site. 2 If a and b are self-adjoint then so are a^2 + b^2 and i ( ab - ba). Let's check the last one:$$[i(ab - ba)]^* = (-i) ( (ab)^* - (ba)^*) = -i ( b^* a^* - a^* b^*) = -i( b a - a b) = i(ab - ba)$$2 What involution do you consider? If just complex conjugation, then x\mapsto \overline{x} is not even differentiable. If you consider f\mapsto f^* where f^*(z) = \overline{f(\overline{z})} then it does not satisfy the C*-identity. 1 I'm assuming a unit 1. The resolvent (x-\lambda 1)^{-1} is uniformly bounded near \infty because$$ (x-\lambda 1)^{-1} = -\sum_{n=0}^{\infty}\frac{1}{\lambda^{n}}x^{n},\;\;\; |\lambda| > r_{\sigma}(x). $$If U is any open set containing \sigma(x), then M=\sup_{\lambda\in\mathbb{C}\setminus U}\|(x-\lambda 1)^{-1}\| < \infty because the ... 1 Well, x^* x is clearly self-adjoint since (x^* x)^* = x^* x^{**} = x^* x. (Here we just use x^{**}=x and (xy)^* = y^* x^*, which are rules in the definition of a *-algebra.) Notice that your expression x^* x = a^2 + b^2 + i(ab-ba) doesn't have the property that c:=ab-ba is self-adjoint, in fact, c^*=-c. This implies that ic is self-adjoint ... 1 Without loss of generality, you can assume that H is an infinite-dimensional, separable Hilbert space. Let (e_n)_{n=1}^\infty be an orthonormal basis for H. Let$$P_n = e_1\otimes e_1 + \ldots e_n\otimes e_n.$$Then (P_n)_{n=1}^\infty is a sequence of finite-rank projections converging pointwise to the identity. Consequently, (I-P_n)_{n=1}^\infty ... 1 You have (x-\lambda e)q(x)=(x^{n}-\lambda^{n}e) for a polynomial q. Suppose (x^{n}-\lambda^{n}e) is invertible. Then x-\lambda e is invertible because$$ (x-\lambda e)[q(x)(x^{n}-\lambda^{n}e)^{-1}]=e=[q(x)(x^{n}-\lambda^{n}e)^{-1}](x-\lambda e) $$Therefore, if (x-\lambda e) is not invertible, then (x^{n}-\lambda^{n}e) cannot be invertible ... 1 in Lemma 2 (p 521) only necessity is demonstrated, and the demonstration does no more than point to the general structure theorem for arbitrary commutative rings - that for a ring A and an ideal I the ideals over I in A are in 1-1 correspondence with the ideals of A/I. the answer to your question is, therefore, yes. 1 Extended discussion... (Find a draft of the solution below!) Query T.A.E. nicely showed Hadamard's criterion saying that the series:$$\sum_{k=0}^\infty A_k$$converges for \limsup_{k\to\infty}\|A_k\|^\frac{1}{k}<1 and certainly diverges for \limsup_{k\to\infty}\|A_k\|^\frac{1}{k}>1. Now, why does one need bounded linear functionals anyway? ... 1 Takesaki (and most authors) do this because it is technically simpler. It reduces directly to standard complex analysis. Otherwise, it would be necessary to develop a little of the theory of complex analysis with values in a Banach space and a little integration theory so as to express Cauchy's formula, or use a special method. The extension of complex ... 1 If you start with any topological space \Omega  then  C_b (\Omega) , the set of bounded continuous functions on \Omega , is a C ^*-algebra. But the Gelfand transform allows you to show that there exists a locally compact \Omega' with  C_b (\Omega)\simeq C_b (\Omega') . So when talking in abstract, you gain nothing by considering ... 1 As I already stated in the comment, the cyclic vectors are exactly those f \in L^2 for which f(x) \neq 0 holds for almost every x \in X. It is easy to see that if M := \{x \mid f(x) = 0\} has positive measure, then by \sigma-finiteness, there is some set M' \subset M of finite positive measure. But then$$ \overline{\{\phi \cdot f \mid \phi \in ...

1

Yes. Any function in $A(G)$ vanishes at infinity, whereas $B(G)$ contains the constant functions. So whenever $G$ is non-compact, $A(G)\subsetneq B(G)$. A related question is when $A(G) = B(G)\cap C_0(G)$. See e.g. http://arxiv.org/abs/1311.5400.

1

Take $\ell_1(\mathbb{Z})$ with the $*$-operatorion given by $\delta_n^* = \delta_{-n}$. Consider the element $\tfrac{1}{2}\delta_1 + \tfrac{1}{2}\delta_2$. It has norm one but its spectral radius is 1/2. Instead of $\mathbb{Z}$ you can take your favourite finite, non-trivial group to have a finite-dimensional example.

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