I’m getting the impression that the bra-ket notation is more useful for linear ON THE LEFT!

An approximate unit for a C*-algebra is an increasing net $\{u_\lambda\}_{\lambda\in\Lambda}$ of positive elements in the closed unit ball of $A$ such that $a= \lim_{\lambda }au_\lambda=\lim_\lambda u_\lambda a$ for all $a\in A$ .

Example

Let $H$ be a Hilbert space with infinite orthonormal basis $\{e_n\}$ . The C*-algebra $K(H)$ is now non-unital. If $P_n$ is the projection onto $\langle e_1,\dots,e_n\rangle$ , then the increasing sequence $\{P_n\}\subset K(H)$ is an approximate unit for $K(H)$ . It will suffice to show that $T=\lim_np_nT$ if $T\in F(H)$ , since $F(H)$ is dense in $K(H)$ . Now if $T\in F(H)$ , there exist $x_1,\dots,x_m$ , $y_1,\dots,y_m\in H$ such that:

$T=\sum_{k=1}^m|x_k\rangle\langle y_k|$ .

Hence,

$P_nT=\sum_{k=1}^m|P_nx_k\rangle\langle y_k|$ .

Since $\lim_n P_nx=x$ for all $x\in H$ , therefore for each $k$:

$\lim_{n\rightarrow \infty}\||P_nx_k\rangle\langle y_k-|x_k\rangle\langle y_k|\|=\lim_{n\rightarrow \infty} \|P_nx_k-x_k\|\|y_k\|=0$ .

Hence, $\lim_{n}P_nT=T$ .

Let $A$ be a C*-algebra and denote by $\Lambda=A^+\cap B_1(\mathbf{0})$ . This set is a poset under the partial order of $A_{\text{SA}}$ . In fact, $\Lambda$ is also upwards directed; that is, if $a,b\in\Lambda$ , then there exists $c\in\Lambda$ such that $a,b\leq c$ . Suppose $a\in \Lambda$ (Murphy actually takes $a\in A^+$ but we can do without this — in particular I want to take $\|a\|<1$ to give me the required invertibility: but $1+a\in G(\tilde{A})$ when $a\in A^+$ .), then $1+a\in G(\tilde{A})$ ( I’m going to assume it is and take $a\in A^+$ after all because then $1+a\in G(\tilde{A})$ .). We hence have that $a(1+a)^{-1}=1-(1+a)^{-1}$ . We claim

$a,b\in A^+$ and $a\leq b\Rightarrow a(1+a)^{-1}\leq b(1+b)^{-1}$ (*)

Indeed if $0\leq a\leq b$ , then $1+a\leq 1+b$ which implies that $(1+a)^{-1}\geq (1+b)^{-1}$ , by Theorem 2.2.5, and therefore $1-(1+a)^{-1}\leq 1-(1+b)^{-1}$ ; thus proving the claim.

Note that if $a\in A^+$ then $a(1+a)^{-1}\in \Lambda$ . Suppose then that $a,b\in\Lambda$ . Let $a'=a(1-a)^{-1}$ , $b'=b(1-b)^{-1}$ and $c=(a'+b')(1+a'+b')^{-1}$ . Then $c\in \Lambda$ , and since $a'\leq a'+b'$ , we have that $a=a'(1+a')^{-1}\leq c$ by (*). Similarly $b\leq c$ and $\Lambda$ is upwards-directed as claimed.

Theorem 3.1.1

Every C*-algebra admits an approximate unit. Indeed, if $\Lambda$ is the upwards-directed set $A^+\cap B_1(\mathbf{0})$ and $u_\lambda=\lambda$ for all $\lambda\in \Lambda$ , then $\{u_\lambda\}_{\lambda\in\Lambda}$ is an approximate unit for $A$ (called the canonical approximate unit).

Proof

We need to show that $a=\lim_\lambda u_\lambda a$ for each $a\in A$ . Since $\Lambda$ linearly spans $A$ (The positive elements span the hermitian elements and the hermitian elements span the C*-algebra.), we can reduce to the case when $a\in\Lambda$ .

Let $a\in \Lambda$ and $\varepsilon>0$ . Let $\varphi:C^*(a)\rightarrow C_0(\Phi(A))$ be the Gelfand transformation. If $f=\varphi(a)$ , then $K=\{\omega\in \Phi(A):|f(\omega)|\geq \varepsilon\}$ is compact, and therefore by Urysohn’s Lemma there is a continuous function $g:\Phi(A)\rightarrow [0,1]$ of compact support such that $g(\omega)=1$ for all $\omega\in K$ . Choose $\delta>0$ such that $\delta<1$ and $1-\delta<\varepsilon$ . Then $\|f-\delta gf\|\leq \varepsilon$ (I don’t see this: suppose, roughly, that the supremum is found ‘at’ $\omega\in K$ . Then

$\underbrace{\|f(\omega)\|}_{\geq \varepsilon}\underbrace{\|1-\delta g(\omega)\|}_{=1-\delta}$

that is I can’t make any conclusion. Now if the supremum is found at $\omega$ ‘outside’ $K$ then:

$\underbrace{\|f(\omega)\|}_{<\varepsilon}\underbrace{\|1-\delta g(\omega)\|}_{\in[1-\delta,1]}$

This is unresolved by my supervisor and I but we can prove this result using different techniques so no panic — approximate units still exist!). If $\lambda_0=\varphi^{-1}(\delta g)$ , then $\lambda_0\in\Lambda$ and $\|a-u_{\lambda_0} a\|\leq \varepsilon$ . Now suppose that $\lambda\in\Lambda$ and $\lambda\geq \lambda_0$ . Then $1-u_\lambda\leq 1-u_{\lambda_0}$ , so $a(1-u_\lambda)a\leq a(1-u_{\lambda_0})a$ (Th.2.2.5 (2): we can pre- and post-multiply inequalities by an element and it’s adjoint.). Hence,

$\|a-u_\lambda a\|^2=\|(1-u_\lambda)^{1/2}(1-u_\lambda)^{1/2}a\|^2\leq \|(1-u_\lambda)^{1/2}a\|^2$

(the first inequality here uses submultiplicativity and the C*-equation along with the the result $\|1-u_\lambda\|\leq1$ . To see that this is true, not that $\|u_\lambda\|<1$ so that $\sigma(u_\lambda)\subset [0,1]$ . Therefore $\sigma(1-u_\lambda)\subset (0,1]\Rightarrow \|1-u_\lambda\|\leq1$ . The next equality is just the C*-equation:)

$\|(1-u_\lambda)^{1/2}a\|^2=\|a(1-u_\lambda)a\|\leq \|a(1-u_{\lambda_0})a\|\leq \|(1-u_{\lambda_0})a\|\leq \varepsilon$ .

(The first and second inequalities here use Th. 2.2.5 (3) ( $a\leq b\Rightarrow \|a\|\leq\|b\|$ ) and submultiplicativity.) This shows that $a=\lim_\lambda u_\lambda a$ $\bullet$

Remark

If a C*-algebra $A$ is separable, then it admits an approximate unit which is a sequence. For in this case, there exist finite sets

$F_1\subset F_2\subset \cdots \subset F_n\subset\cdots$

such that $F=\cup_n F_n$ is dense in $A$ . Let $\{u_\lambda\}_{\lambda\in\Lambda}$ be any approximate unit for $A$ . If $\varepsilon>0$ , and $F_n=\{a_1,\dots,a_m\}$ say, then there exist $\lambda_1\dots\lambda_m\in\Lambda$ such that $\|a_j-a_j u_\lambda\|<\varepsilon$ if $\lambda\geq \lambda_j$ . Choose $\lambda_\varepsilon\in\Lambda$ such that $\lambda_\varepsilon\geq\lambda_1,\dots,\lambda_m$ . Then $\|a-au_\lambda\|<\varepsilon$ for all $a\in F_n$ and all $\lambda\geq\lambda_\varepsilon$ . Hence, if $n\in\mathbb{N}$ and $\varepsilon=1/n$ , then there exists $\lambda_n=\lambda_\varepsilon\in\Lambda$ such that $\|a-au_{\lambda_n}\|<1/n$ for all $a\in F_n$ . Also, we can choose the $\lambda_n$ such that $\lambda_n\leq\lambda_{n+1}$ for all $n$ . Consequently, $\lim_n\|a-au_{\lambda_n}\|=0$ , for all $a\in F$ , and since $F$ is dense in $A$ , this also holds for all $a\in A$ . Therefore $\{u_{\lambda_n}\}_{n=1}^\infty$ is an approximate unit for $A$ .

Theorem 3.1.2

If $L$ is a closed left ideal in a C*-algebra $A$ , then there is an increasing net $\{u_\lambda\}_{\lambda\in\Lambda}$ of positive elements in the closed unit ball of $L$ such that $a=\lim_\lambda au_\lambda$ for all $a\in L$ .

Proof

Set $B=L\cap L^*$ . Since $B$ is a C*-algebra, it admits an approximate unit, $\{u_\lambda\}_{\lambda\in\Lambda}$ say, by the previous theorem. If $a\in L$ , then $a^*a\in B$ , so

$\lim_\lambda a^*a(1-u_\lambda)=0.$

Hence,

and therefore $\|a-au_\lambda\|=0$ $\bullet$

(Where the last inequality uses submultiplicatively — and again requires that $\|1-u_\lambda\|<1$ . Murphy refers to this afterwards — he states that in this proof we worked in the unitisation $\tilde{A}$ of $A$ .)

Theorem 3.1.3

If $I$ is a closed ideal in a C*-algebra $A$ , then $I$ is self-adjoint and therefore a C^*-subalgebra of $A$ . If $\{u_\lambda\}_{\lambda\in\Lambda}$ is an approximate unit for $I$ , then for each $a\in A$ :

$\|a+I\|=\lim_\lambda\|a-u_\lambda a\|=\lim_\lambda\|a-au_\lambda\|.$

Proof

By the previous theorem there is an increasing net $\{u_\lambda\}_{\lambda\in\Lambda}$ of positive elements in the closed unit ball of $I$ such that $a=\lim_\lambda au_\lambda$ for all $a\in I$ . Hence $a^*=\lim_\lambda u_\lambda a^*$ , so $a^*\in I$ , because all of the elements $u_\lambda$ belong to $I$ . Therefore $I$ is self-adjoint.

Suppose that $\{u_\lambda\}_{\lambda\in\Lambda}$ is an arbitrary approximate unit of $I$ , that $a\in A$ and that $\varepsilon>0$ . There is an element $b\in I$ such that:

$\|a+b\|<\|a+I\|+\frac{\varepsilon}{2}$ .

Since $b=\lim_\lambda u_\lambda b$ , there exists $\lambda_o\in\Lambda$ such that $\|b-u_\lambda b\|<\varepsilon/2$ for all $\lambda\geq\lambda_0$ , and therefore (triangle inequality after a trick and the dreaded $\|1-u_\lambda\|<1$ are the first two inequalities):

$\|a-au_\lambda\|\leq \|(1-u_\lambda)(a+b)\|+\|b-u_\lambda b\|$

$\leq \|a+b\|+\|b-u_\lambda b\|$

$<\|a+I\|+\varepsilon/2+\varepsilon/2$ .

It follows that $\|a+I\|=\lim_\lambda\|a-a_\lambda a\|$ , and therefore also

$\|a+I\|=\|a^*+I\|=\lim_\lambda\|a^*-u_\lambda a^*\|=\lim_\lambda \|a-au_\lambda\|$ $\bullet$

Remark

Let $I$ be a closed ideal in a C*-algebra $A$ , and $J$ a closed ideal in $I$ . Then $J$ is also an ideal in $A$ . To show this we need only show that $ab\in J$ and $ba\in J$ whenever $a\in A$ and $b\in J^+$ (since $J$ is a C*-algebra, $J^+$ linearly spans $J$ ). If $\{u_\lambda\}$ is an approximate unit for $I$ , then $b^{1/2}=\lim_\lambda u_\lambda b^{1/2}$ because $b^{1/2}\in I$ . Hence, $ab=\lim_\lambda au_\lambda b^{1/2}b^{1/2}$ , so $ab\in J$ because $b^{1/2}\in J$ , $au_\lambda b^{1/2}\in I$ , and $J$ is an ideal in $I$ . Therefore $a^*b\in J$ also, so $ba\in J$ , since $J$ is self-adjoint.

Theorem 3.1.4

If $I$ is a closed ideal of a C*-algebra $A$ , then the quotient $A/I$ is a C*-algebra under its usual operations and the quotient norm.

Proof

Let $\{u_\lambda\}$ be an approximate unit for $I$ . If $a\in A$ and $b\in I$ , then

$\|a+I\|^2=\lim_\lambda \|a-au_\lambda\|^2$

$=\lim_\lambda \|(1-u_\lambda)a^*a(1-u_\lambda)\|$

$\leq \sup_\lambda\|(1-u_\lambda)(a^*a+b)(1-u_\lambda)\|+\lim_\lambda \|(1-u_\lambda)b(1-u_\lambda)\|$

$\leq \|a^*a+b\|+\lim_\lambda \|b-u_\lambda b\|=\|a^*a+b\|$ .

The first equality is by the previous theorem (the second is the C*-equation. The first inequality is a trick and a triangle inequality. The second inequality again uses $\|1-u_\lambda\|\leq 1$ with a bit of submultiplicativity). Therefore, $\|a+I\|^2\leq \|a^*a+I\|$ ; so by Lemma 2.1.3 (if a Banach algebra with an involution satisfies $\|a\|^2\leq \|a^*a\|$ then it is a C*-algebra), $A/I$ is a C*-algebra $\bullet$

Theorem 3.1.5

If $\varphi:A\rightarrow B$ is an injective *-homomorphism between C*-algebras $A$ and $B$ , then $\varphi$ is necessarily isometric.

Proof

It suffices to show that $\|\varphi(a)\|^2=\|a\|^2$ , that is $\|\varphi(a^*a)\|=\|a^*a\|$ . Thus we may suppose that $A$ is abelian (restrict to $C(a^*a)$ if necessary), and that $B$ is abelian (replace $B$ by $\overline{\varphi(A)}$ if necessary). Moreover, by extending $\varphi:A\rightarrow B$ to $\tilde{\varphi}:\tilde{A}\rightarrow\tilde{B}$ if necessary, we may further assume the $A$ , $B$ and $\varphi$ are unital.

If $\tau$ is a character on $B$ , then $\tau\circ\varphi$ is one on $A$ . Clearly the map

$\varphi':\Phi(B)\rightarrow \Phi(A)$ , $\tau\mapsto \tau\circ\varphi$

is continuous (as the composition of continuous maps — homomorphisms are always continuous). Hence, $\varphi'(\Phi(B))$ is compact, because $\Phi(B)$ is compact (Th. 1.3.5: If $A$ is unital and abelian, $\Phi(A)$ is compact and Hausdorff), and therefore $\varphi'(\Phi(B))$ is closed in $\Phi(A)$ . If $\varphi'(\Phi(B))\neq \Phi(A)$ , then by Urshohn’s Lemma there is a non-zero continuous function $f:\Phi(A)\rightarrow \mathbb{C}$ that vanishes on $\varphi'(\Phi(B))$ . By the Gelfand representation, $f=\hat{a}$ for some $a\in A$ . Hence, for each $\tau\in\Phi(B)$ , $\tau(\varphi(a))=\hat{a}(\tau\circ\varphi)=0$ . Therefore $\varphi(a)=0$ , so $a=0$ (O.K. — with injectivity of $\varphi$ ). But this implies that $f$ is zero, a contradiction. The only way to avoid this is to have $\varphi'(\Phi(B))=\Phi(A)$ . Hence, for each $a\in A$ ,

$\|a\|=\|\hat{a}\|_\infty=\sup_{\tau\in \Phi(A)}|\tau(a)|=\sup_{\tau\in\Phi(B)}|\tau(\varphi(a))|=\|\varphi(a)\|$ .

Thus $\varphi$ is isometric $\bullet$

Theorem 3.1.6

If $\varphi:A\rightarrow B$ is a *-homomorphism between C*-algebras, then $\varphi(A)$ is a C*-subalgebra of $B$ .

Proof

The map

$A/\text{ker }\varphi\rightarrow B$ , $a+\text{ker }\varphi\mapsto \varphi(a)$ ,

is an injective *-homomorphism between C*-algebras and is therefore isometric. It’s image is $\Phi(A)$ , so this space is necessarily complete and therefore closed in $B$ $\bullet$

Theorem 3.1.7

Let $B$ and $I$ respectively be a C*-subalgebra and a closed ideal of in a C*-algebra $A$ . Then $B+I$ is a C*-subalgebra of $A$ .

Proof

We show only that $B+I$ is complete, because the rest is trivial. Since $I$ is complete we need only prove that the quotient $(B+I)/I$ is complete ( $A/I$ , $I$ complete implies $A$ complete — a quick calculation). The intersection $B\cap I$ is a closed ideal in $B$ and the map $\varphi$ from $B/(B\cap I)$ to $A/I$ defined by setting $\varphi(b+B\cap I)=b+I$ is a *-homomorphism with range $(B+I)/I$ . By the previous theorem, $(B+I)/I$ is a C*-algebra, and hence complete $\bullet$

Remark

The map

$\varphi: B/(B\cap I)\rightarrow (B+I)/I$ , $b+B\cap I\mapsto b+I$ ,

in the preceding proof is in fact clearly a *-isomorphism.

We return to the topic of multiplier algebras, because we can now say a little more about them using the results of this section.

Suppose that $I$ is a closed ideal in a C*-algebra $A$ . If $a\in A$ , define $L_a$ and $R_a\in B(I)$ by setting $L_a(b)=ab$ and $R_a(b)=ba$ . $(L_a,R_a)$ is a double centraliser on $I$ and the map:

$\varphi:A\rightarrow M(A)$ , $a\mapsto (L_a,R_a)$ ,

is a *-homomorphism. Recall that we identified $I$ as a closed ideal in $M(I)$ by identifying $a$ with $(L_a,R_a)$ if $a\in I$ (We have that $\varphi$ is a homomorphism as

$\varphi(a)\varphi(b)=(L_a,R_a)(L_bR_b)$

$=(L_aL_b,R_bR_a)=(L_{ab},R_{ab})=\varphi(ab)$ .

Also

). Hence, $\varphi$ is an extension of the inclusion map $I\mapsto M(I)$ .

If $I_1,I_2,\dots,I_n$ are sets in $A$ , we define $I_1I_2\cdots I_n$ to be the closed linear span of all products $a_1a_2\cdots a_n$ , where $a_j\in I$ . If $I$ , $J$ are closed ideals in $A$ , then $I\cap J=IJ$ . The inclusion $IJ\subset I\cap J$ is obvious. To show the reverse inclusion we need only show that if $a$ is a positive element of $I\cap J$ , then $a\in IJ$ . Suppose then that $a\in (I\cap J)^+$ . Hence $a^{1/2}\in I\cap J$ . If $\{u_\lambda\}_{\lambda\in\Lambda}$ is an approximate unit for $I$ , then $a=\lim_\lambda (u_\lambda a^{1/2})a^{1/2}$ , and since $u_\lambda a^{1/2}\in I$ for all $\lambda\in\Lambda$ , we get $a\in IJ$ , as required.

Let $I$ be a closed ideal in $A$ . We say $I$ is essential if $aI=0\Rightarrow a=0$ . From the previous observations, it is easy to check that $I$ is essential in $A$ iff $I\cap J\neq \{\mathbf{0}\}$ for all non-zero closed ideals $J$ in $A$ (is every element contained in some ideal — how about $AaA$ — but we need a unit for this (and it’s probably not always proper).).

Every C*-algebra $I$ is an essential ideal in it’s multiplier algebra $M(I)$ (a quick calculation).

Theorem 3.1.8

Let $I$ be a closed ideal in a C*-algebra $A$ . Then there is a unique *-homomorphism $\varphi:A\rightarrow M(I)$ extending the inclusion $I\rightarrow M(I)$ . Moreover, $\varphi$ is injective if $I$ is essential in $A$ .

Proof

We have already seen above that the inclusion map $I\rightarrow M(I)$ admits a *-homomorphic extension $\varphi: A\rightarrow M(I)$ . Suppose that $\psi:A\rightarrow M(I)$ is another such extension. If $a\in A$ , $b\in I$ , then $\varphi(a)b=\varphi(ab)=ab=\psi(ab)=\psi(a)b$ . Hence $(\varphi(a)-\psi(a))I=0$ , so $\varphi(a)=\psi(a)$ , since $I$ is essential in $M(I)$ . Thus $\psi=\varphi$ .

Suppose now that $I$ is essential in $A$ and let $a\in\text{ker }\varphi$ . Then $aI=L_a(I)=0$ , so $a=0$ . Thus, $\varphi$ is injective $\bullet$

Theorem 3.1.8 tells us that the multiplier algebra $M(I)$ of $I$ is the largest unital C*-algebra containing $I$ as an essential closed ideal.

Example

If $H$ is a Hilbert space, then $K(H)$ is an essential ideal in $B(H)$ . For if $T\in B(H)$ such that $T \,K(H)=0$ , then for all $x\in H$ we have that $|Tx\rangle\langle x|=T(|x\rangle\langle x|)=0$ , so $Tx=0$ . By Theorem 3.1.8, the inclusion map $K(H)\rightarrow M(K(H))$ extends uniquely to an injective *-homomorphism $\varphi:B(H)\rightarrow M(K(H))$ . We show that $\varphi$ is surjective, that is, a *-isomorphism. Suppose that $(L,R)\in M(K(H))$ , and fix a unit vector $e\in H^1$ . The linear map:

$T:H\rightarrow H$ , $x\mapsto (L(|e\rangle\langle e|))(e)$ ,

is bounded, since

$\|Tx\|\leq \|L(|x\rangle\langle e|)\|\leq \|L\|\||x\rangle\langle e|\|=\|L\|\|x\|$ .

If $x,y,z\in H$ , then

$(L_T(|x\rangle\langle y|))(z)=(|Tx\rangle\langle y|)(z)=\langle z,y\rangle (L(|x\rangle\langle e|))(e)$

$=(L(|x\rangle\langle e|))(\langle z,y\rangle e)=(L(| x\rangle\langle e|))(|e\rangle\langle y|)(z)$ .

Thus we may regard $B(H)$ as the multiplier algebra of $K(H)$ .

Example

If $X$ is a locally compact Hausdorff space, then it is easy to check that $C_0(X)$ is an essential ideal in the C*-algebra $C_b(X)$ . Therefore, by Theorem 3.1.8 there is a unique injective *-homomorphism $\varphi:C_b(X)\rightarrow M(C_0(X))$ extending the inclusion $C_0(X)\rightarrow C_0(X)$ . We show that $\varphi$ is surjective, that is, a *-isomorphism. To see this, it suffices to show that if $g\in M(C_0(X))$ is positive, then it is in the range of $\varphi$ . If $\{u_\lambda\}_{\lambda\in \Lambda}$ is an approximate unit for $C_0(X)$ , then for each $x\in X$ the net of real numbers $\{g(x)u_\lambda(x)\}$ is increasing and bounded above by $\|g\|$ , and therefore it converges to a number, $h(x)$ say. The function:

$f:X\rightarrow \mathbb{C}$ , $x\mapsto h(x)$ .

is bounded. Moreover, if $f\in C_0(X)$ , then $hf=gf$ , since $f=\lim_\lambda fu_\lambda$ . To see that $f$ is continuous, let $\{x_\mu\}_{\mu\in M}$ be a net in $X$ converging to a point $x$ . Let $K$ be a compact neighbourhood of $x$ in $X$ . To show that $h(x)=\lim_\mu h(x_\mu)$ , we may suppose $x_\mu\in K$ for all indices $\mu$ ,(there exists $\mu_0$ such that $x_\mu\in K$ for all indices $\mu\geq \mu_0$ , so, if necessary, replace the net $\{x_\mu\}_{\mu \in M}$ by the net $\{x_\mu\}$ ). Use Urysohn’s Lemma to choose a function $f\in C_0(X)$ such that $f=1$ on $K$ . Since $fh\in C_0(X)$ ,

$h(x)=fh(x)=\lim_\mu f(x_\mu)h(x_\mu)=\lim_\mu h(x_\mu)$ .

Therefore, $h$ is continuous, so $h\in C_b(X)$ . For $f$ an arbitrary function in $C_0(X)$ we have

$\varphi(h)f=\varphi(hf)=hf=gf$ ,

so $(\varphi(h)-g)C_0(X)=0$ . Consequently $g=\varphi(h)$ .

2 comments

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December 4, 2014 at 2:00 am

momumomu

if $(U_{n})$ is an approximate unit for a C^{∗}-algebra A. Is (U_{n})^2$ is an approximateunit for a C^{*}−algebra A? Thank in advance.

December 10, 2014 at 5:15 pm

J.P. McCarthy

http://math.stackexchange.com/questions/1050761/existence-of-a-approximate-unit-u-n2-for-a-c-algebra/1061102#1061102

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Ideals and Positive Functionals: 3.1 Ideals in C*-Algebras

Example

Theorem 3.1.1

Proof

Remark

Theorem 3.1.2

Proof

Theorem 3.1.3

Proof

Remark

Theorem 3.1.4

Proof

Theorem 3.1.5

Proof

Theorem 3.1.6

Proof

Theorem 3.1.7

Proof

Remark

Theorem 3.1.8

Proof

Example

Example

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Ideals and Positive Functionals: 3.1 Ideals in C*-Algebras

Example

Theorem 3.1.1

Proof

Remark

Theorem 3.1.2

Proof

Theorem 3.1.3

Proof

Remark

Theorem 3.1.4

Proof

Theorem 3.1.5

Proof

Theorem 3.1.6

Proof

Theorem 3.1.7

Proof

Remark

Theorem 3.1.8

Proof

Example

Example

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