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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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