Proof We may assume that $\|x\|\leq 1.$ Let $p_1,q_1\in A_+^{\mathbf {1}}$ such that $\pi (p_1)=p, \pi (q_1)=q.$ Since $p,q$ are projections, we also have $\pi (f_{1/2}(p_1)) =f_{1/2}(\pi (p_1)) =p,$ and $\pi ({{f_{1/2}}}(q_1)) =f_{1/2}(\pi (q_1))=q.$ Note $x^*x\leq q.$ By [Reference Pedersen30, Proposition 1.5.10], there exists $y\in A^{\mathbf {1}}$ such that $\pi (y)=x$ and $y^*y\le f_{1/2}(q_1).$ Put $\tilde x=f_{1/2}(p_1)y.$ Then

$$ \begin{align*}\pi(\tilde x)=px=x, \tilde x \tilde x^*=f_{1/2}(p_1)yy^*f_{1/2}(p_1)\le f_{1/4}(p_1) \end{align*} $$

and $\tilde x^*\tilde x\le y^*y\le f_{1/2}(q_1).$ Set $\tilde p=f_{1/8}(p_1)$ and $\tilde q=f_{1/8}(q_1).$ Then $\pi (\tilde p)=p, \pi (\tilde q)=q.$ The facts that $f_{1/8}(p_1)f_{1/4}(p_1)=f_{1/4}(p_1)$ and $f_{1/8}(q_1)f_{1/4}(q_1)=f_{1/4}(q_1)$ imply that $\tilde p\tilde x=\tilde x=\tilde x\tilde q.$ Moreover, if $p=q,$ we can take $p_1=q_1,$ and hence $\tilde p=\tilde q.$

Proof Let $B=l^\infty (A)/I_T(A)$ and $\pi :l^\infty (A)\to B$ be the quotient map. Assume $p\in B$ is a projection, $u\in B$ satisfies $u^*u=p$ , and $uu^*\leq p.$ We need to show $uu^*=p.$

By Lemma 3.2, there are $a=\{a_1,a_2,...\}\in l^\infty (A)_+^{\mathbf {1}}$ and $v=\{v_1,v_2,...\}\in l^\infty (A)$ such that $\pi (a)=p, \pi (v)=u,$ and $av=v=va.$ Since $\pi (a) = p=\pi (v^*v),$ we have $\lim _{n\to \infty }\|a_n-v_n^*v_n\|_{_{2,T}}=0.$ By Lemma 3.1, $\|a_n-v_nv_n^*\|_{_{2,T}}=\|a_n-v_n^*v_n\|_{_{2,T}}\to 0 (n\to \infty ).$ Hence,

$$ \begin{align*}{{p-uu^*=\pi(\{a_1-v_1v_1^*, a_2-v_2v_2^*,...\})=0,}} \end{align*} $$

which shows p is a finite projection.

Proof It follows from the first paragraph of Definition 2.14 that there exists $c\in \mathrm {Her}(a)_+$ with $\|c\|=1$ such that $d_\tau (c)<\varepsilon ^2$ for all $\tau \in T.$ Define $b=a^{1/2}(1-f_{1/4}(c))a^{1/2}.$ Then $0\le b\le a.$ It follows from (e2.17) that

$$ \begin{align*}\|a-b\|_{_{2,T}}=\|a^{1/2}f_{1/4}(c)a^{1/2}\|_{_{2,T}} \leq \|f_{1/4}(c)\|_{_{2,T}}\leq (d_\tau(c))^{1/2}\leq \varepsilon. \end{align*} $$

For all $\tau \in T,$

$$ \begin{align*}\kern-6pt\begin{array}{lll} d_\tau(b)\!\!\!\! &\qquad\ =&\!\! d_{\tau}(a^{1/2}(1-f_{1/4}(c))a^{1/2}) = d_{\tau}((1-f_{1/4}(c){\kern-1pt})^{1/2} a (1-f_{1/4}(c){\kern-1pt})^{1/2}) \\\!\!\!\! &\qquad\ <&\!\! d_{\tau}((1-f_{1/4}(c))^{1/2} a (1-f_{1/4}(c))^{1/2})+ d_\tau(f_{1/2}(c)) \\\!\!\!\! &\overset{\mathrm{(orthogonality)}}{=}&\!\! d_{\tau}((1-f_{1/4}(c))^{1/2} a (1-f_{1/4}(c))^{1/2} +f_{1/2}(c)) \le d_\tau(a). \end{array}\end{align*} $$

Proof Set $B:=l^\infty (A)/I_{_{\overline {QT(A)}^w}}$ and let $\Pi :l^\infty (A)\to B$ denote the quotient map.

Let $p,q,r\in B$ be projections with $p,q\leq r,$ and assume that there is a partial isometry $v{\kern-1pt}\in{\kern-1pt} B$ such that $v^*v{\kern-1pt}={\kern-1pt}p, vv^*{\kern-1pt}={\kern-1pt}q.$ By Lemma 3.2, there are $e{\kern-1pt}={\kern-1pt}\{e_1,e_2,...\}{\kern-1pt}\in{\kern-1pt} l^\infty (A)_+^{\mathbf {1}}$ and $w=\{w_1,w_2,...\}\in l^\infty (A)$ such that $\pi (e)=r, \pi (w)=v,$ and $ew=w=we.$ Then, by Lemma 3.1, we have $ d_\tau (f_{1/4}(e_n-w^*_nw_n))= d_\tau (f_{1/4}(e_n-w_nw_n^*)) $ for all $\tau \in \overline {QT(A)}^w$ and $n\in \mathbb {N}.$ By Proposition 3.4, for each $n\in \mathbb {N},$ there is $b_n\in A_+^{\mathbf {1}}$ such that

(e3.3) $$ \begin{align} &\|f_{1/4}(e_n-w_n^*w_n)-b_n\|_{_{2,\overline{QT(A)}^w}}<1/n, \text{ and } \end{align} $$

(e3.4) $$ \begin{align} & d_\tau(b_n)<d_\tau(f_{1/4}(e_n-w_n^*w_n))= d_\tau(f_{1/4}(e_n-w_nw_n^*)) \text{ for all } \tau\in \overline{QT(A)}^w. \end{align} $$

Since A has strict comparison, we have $b_n\lesssim f_{1/4}(e_n-w_nw_n^*).$ By [Reference Rørdam40, Proposition 2.4(iv)], for each $n\in \mathbb {N},$ there is $x_n'\in A$ such that

(e3.5) $$ \begin{align} (x_n')^*(x_n')=f_{1/n}(b_n)\,\,\,\mathrm{and}\,\,\, x_n'(x_n')^*\in \mathrm{Her}(f_{1/4}(e_n-w_nw_n^*)). \end{align} $$

Note that $f_{1/n}(b)(b-1/n)_+=(b-1/n)_+.$ Choose $x_n=x_n'(b-1/n)_+^{1/2}.$ Then

(e3.6) $$ \begin{align} x_n^*x_n=(b_n-1/n)_+ \quad \text{ and } \quad x_nx_n^*\in \mathrm{Her}(f_{1/4}(e_n-w_nw_n^*)). \end{align} $$

Note $\|x_n\|^2=\|x_n^*x_n\|=\|(b_n-1/n)_+\|\leq 1.$ The second part of (e3.6) implies

(e3.7) $$ \begin{align} x_nx_n^*\leq f_{1/8}(e_n-w_nw_n^*). \end{align} $$

Let $c_n=e_n-w^*_nw_n$ and $d_n=e_n-w_nw_n^*$ ( $n\in \mathbb {N}$ ). Let $x=\{x_1,x_2,...\}, b= \{b_1,b_2,...\}, c=\{c_1,c_2,...\},$ and $d=\{d_1,d_2,...\}\in l^\infty (A).$ Then

(e3.8) $$ \begin{align} \Pi(x)^*\Pi(x)\overset{({3.3})}{=}\Pi(b) \overset{({3.3})}{=} \Pi(f_{1/4}(c)) =f_{1/4}(\Pi(c)) =f_{1/4}(r-p) =r-p, \end{align} $$

and

(e3.9) $$ \begin{align} \Pi(x)\Pi(x)^*\overset{({3.7})}{\leq} \Pi(f_{1/8}(d)) =f_{1/8}(\Pi(d)) =f_{1/8}(r-q) =r-q. \end{align} $$

Let $y=v+\Pi (x).$ Note that $v\Pi (x)^*=vp(r-p)\Pi (x)^*=0$ and $\Pi (x)v^*=(v\Pi (x)^*)^*=0.$ Also, note that $v^*\Pi (x)=v^*q(r-q)\Pi (x)=0$ and $\Pi (x)^*v=(v^*\Pi (x))^*=0.$ Then we compute (using also (e3.8) and (e3.9)) that

$$ \begin{align*}{{r=y^*y\sim yy^*= \Pi(x)\Pi(x)^*+q \leq r.}}\end{align*} $$

By Proposition 3.3, r is a finite projection. Hence, $\Pi (x)\Pi (x)^*+q=r.$ Consequently, $\Pi (x)\Pi (x)^*=r-q.$ Together with (e3.8), we obtain $r-p\sim r-q.$ The theorem then follows.

Proof Let $\tau \in S.$ Let $O(\tau )$ be any open neighborhood of $\tau .$ For any $\varepsilon>0,$ there is $\delta>0$ such that

(e4.10) $$ \begin{align} \sup\{d_t(a)-t(f_{\delta}(a)):t\in O(\tau)\cap S\}<\omega_{O(\tau)}(a){{|_S}} +\varepsilon. \end{align} $$

Since $a\sim b,$ there exists a sequence $x_n\in A\otimes {\cal K}$ such that $x_nx_n^*\to a$ and $x_n^*x_n\in \mathrm {Her}(b)=\overline {b(A\otimes {\cal K})b}.$ Since $a^{1/m}aa^{1/m}\to a$ as $m\to \infty ,$ replacing $x_n$ by $a^{1/m(n)}x_n$ for some subsequence $\{m(n)\},$ we may assume that $x_nx_n^*\in \mathrm {Her}(a).$ Note that, for any $\delta>0,$

$$ \begin{align*}\lim_{n\to\infty}\|f_\delta(x_nx_n^*)-f_\delta(a)\|=0. \end{align*} $$

Since S is compact, by (2) of Proposition 2.10, $\sup \{\|t|_{\mathrm {Her}(a)}\|:t\in S\}<\infty .$ It follows that there is $m\in \mathbb {N}$ such that

(e4.11) $$ \begin{align} \sup\{|t(f_\delta(a))-t(f_\delta(x_mx_m^*))|:t\in S\}<\varepsilon. \end{align} $$

Note that $d_t(a)=d_t(b)$ for all $t\in S$ because of $a\sim b.$ Also, note that $t(f_\delta (x_mx_m^*))=t(f_\delta (x_m^*x_m))$ for all $t\in S.$ Then

(e4.12) $$ \begin{align} \omega_{O(\tau)}(b)|_S &\leq \sup\{d_t(b)-t(f_\delta(x_m^*x_m)):t\in O(\tau)\cap S\} \end{align} $$

(e4.13) $$ \begin{align} &= \sup\{d_t(a)-t(f_\delta(x_mx_m^*)):t\in O(\tau)\cap S\} \end{align} $$

(e4.14) $$ \begin{align} &\leq \sup\{d_t(a)-t(f_\delta(a)):t\in O(\tau)\cap S\} \end{align} $$

(e4.15) $$ \begin{align} &\quad + \sup\{|t(f_\delta(a))-t(f_\delta(x_mx_m^*))|: t\in O(\tau)\cap S\} \end{align} $$

(e4.16) $$ \begin{align} &\leq \omega_{O(\tau)}(a){{|_S}} +\varepsilon +\varepsilon. \end{align} $$

Since $\varepsilon $ is arbitrary, we have $\omega _{O(\tau )}(b){{|_S}}\leq \omega _{O(\tau )}(a){{|_S}}.$ Exactly the same argument shows that $\omega _{O(\tau )}(a)|_S\leq \omega _{O(\tau )}(b){{|_S}}.$ Hence, $\omega _{O(\tau )}(a)|_S=\omega _{O(\tau )}(b){{|_S}}.$ Since $O(\tau )$ is an arbitrary open neighborhood of $\tau ,$ we have

$$ \begin{align*}\omega(b)(\tau)|_S =\inf \{\omega_{O(\tau)}(b)|_S:\tau\in O(\tau){{\cap S}}\} = \inf \{\omega_{O(\tau)}(a){{|_S}}:\tau\in O(\tau){{\cap S}}\} =\omega(a)(\tau)|_S. \end{align*} $$

For the last identity in the proposition, we note that, by (e4.9),

$$ \begin{align*}\omega(a)|_S =\sup\{\omega(a)(\tau)|_S: \tau\in S\} =\sup\{\omega(b)(\tau)|_S: \tau\in S\}=\omega(b)|_S.\\[-34pt] \end{align*} $$

Proof (1) For the inequality on the left, let $\varepsilon>0.$ Fix $\tau \in S.$ Choose an open neighborhood $O(\tau )$ of $\tau $ in S such that

(e4.20) $$ \begin{align} &\omega_{O(\tau)}(a+b)|_S \le \omega(a+b)(\tau)|_S+\varepsilon, \mbox{ and } \end{align} $$

(e4.21) $$ \begin{align} &\sup\{d_t(b):t\in O(\tau)\} \le \overline{d_\tau}(b)|_S+\varepsilon. \end{align} $$

Note that there is $\delta>0$ such that

(e4.22) $$ \begin{align} \sup\{d_t(a+b) -t(f_\delta(a+b)):t\in O(\tau)\} \le \omega_{O(\tau)}(a+b){{|_S}}+\varepsilon. \end{align} $$

Note that $a+b\approx _{\delta /2} (a-\delta /2)_++b.$ By [Reference Rørdam40, Proposition 2.2], we have $f_{\delta }(a+b)\lesssim (a-\delta /2)_++b.$ Then, for any $t\in {{O(\tau )}},$ we have

(e4.23) $$ \begin{align} t(f_{\delta}(a+b))&\leq d_t(f_{\delta}(a+b)) \leq d_t((a-\delta/2)_++b) \leq d_t((a-\delta/2)_+)+d_t(b) \end{align} $$

(e4.24) $$ \begin{align} &\overset{({4.21})}{\leq} t(f_{\delta/2}(a))+\overline{d_\tau}(b)|_S+\varepsilon. \end{align} $$

Then, for $t\in O(\tau ),$ $d_t(a)+t(f_{\delta }(a+b)) \leq d_t(a+b)+t(f_{\delta /2}(a))+\overline {d_\tau }(b)|_S+\varepsilon .$ It follows that

(e4.25) $$ \begin{align} d_t(a)-t(f_{\delta/2}(a)) &\leq d_t(a+b)-t(f_{\delta}(a+b)) +\overline{d_\tau}(b)|_S+\varepsilon \end{align} $$

(e4.26) $$ \begin{align} &\overset{({4.22})}{\leq} \omega_{O(\tau)}(a+b)(\tau)|_S +\overline{d_\tau}(b)|_S+2\varepsilon \end{align} $$

(e4.27) $$ \begin{align}&\overset{({4.20})}{\leq} \omega(a+b)(\tau)|_S +\overline{d_\tau}(b)|_S+3\varepsilon\,\,\,{{\,\,\,\mathrm{for\,\,\,all}\,\,\, t\in O(\tau).}} \end{align} $$

Hence,

(e4.28) $$ \begin{align} \omega(a)(\tau)|_S &\leq \omega_{O(\tau)}(a)(\tau) \leq \sup\{d_t(a)-t(f_{\delta/2}(a)):t\in O(\tau)\} \end{align} $$

(e4.29) $$ \begin{align} &\leq \omega(a+b)(\tau)|_S +\overline{d_\tau}(b)|_S+3\varepsilon. \end{align} $$

Let $\varepsilon \to 0,$ then we have the desired inequality.

Now we turn to the inequality on the right. By definition, for any $\varepsilon>0,$ there are open neighborhood $O(\tau )$ of $\tau $ in $S,$ and $\delta>0$ such that

(e4.30) $$ \begin{align} & \sup\{d_t(b):t\in O(\tau)\} \le \overline{d_\tau}(b)+\varepsilon, \mbox{ and } \end{align} $$

(e4.31) $$ \begin{align} & \sup\{d_t(a)-t(f_\delta(a)):t\in O(\tau)\}{{\le}} \omega_{O(\tau)}(a){{|_S}}+\varepsilon/2 \le \omega(a)(\tau)|_S+\varepsilon. \end{align} $$

Note that $a\in \mathrm {Her}(a+b),$ then there is $\eta>0$ such that $f_\delta (a)\approx _{\varepsilon /(s+1)} f_\eta (a+b)f_\delta (a) f_\eta (a+b).$ Hence, for any $t\in O(\tau ),$ by [Reference Blackadar and Handelman2, Corollary II.2.5(iii)],

(e4.32) $$ \begin{align} d_t(a+b)-t(f_\eta(a+b)) &\leq d_t(b)+d_t(a)-t(f_\eta(a+b)f_\delta(a) f_\eta(a+b)) \nonumber\\ &\leq d_t(b)+d_t(a)-t(f_\delta(a))+\varepsilon \end{align} $$

(e4.33) $$ \begin{align}&\overset{({4.30}),({4.31})}{\leq} \overline{d_\tau}(b)+\omega(a)(\tau)|_S +3\varepsilon. \end{align} $$

Hence,

(e4.34) $$ \begin{align}\nonumber \omega(a+b)(\tau)|_S &\leq \sup\{d_t(a+b)-t(f_\eta(a+b)):t\in O(\tau)\}\\ &\leq \overline{d_\tau}(b)+\omega(a)(\tau)|_S+3\varepsilon. \end{align} $$

Let $\varepsilon \to 0,$ (1) then follows.

For (2), we have, for any $1/2>\varepsilon >0,$ since $a\perp b,$

(e4.35) $$ \begin{align} d_\tau(a+b)-\tau(f_\varepsilon(a+b))=(d_\tau(a)-\tau(f_\varepsilon(a))+(d_\tau(b)-\tau(f_\varepsilon(b)) \end{align} $$

for all $\tau \in S.$ Thus,

(e4.36) $$ \begin{align} &\omega(a+b)|_S\le \omega(a)|_S+\omega(b)|_S\,\,\,\mathrm{and}\,\,\, \end{align} $$

(e4.37) $$ \begin{align} &\omega(a+b)(\tau)|_S\le \omega(a)(\tau)|_S+\omega{{(b)}}(\tau)|_S\,\,\,\mathrm{for\,\,\,all}\,\,\, \tau\in S. \end{align} $$

Hence, the inequality on the right in (e4.18) holds.

Now we turn to the inequality on the left of (e4.18). Since $a\bot b,$ for all $\tau \in S$ and all $\eta>0,$

(e4.38) $$ \begin{align} d_\tau(a)-\tau(f_\eta(a))\leq ( d_\tau(a)-\tau(f_\eta(a)))+ ( d_\tau(b)-\tau(f_\eta(b))) = d_\tau(a+b)-\tau(f_\eta(a+b)). \end{align} $$

Thus, $ \omega (a)|_S \leq \sup \{d_\tau (a)-f_{\eta }(a): \tau \in S\} \leq \sup \{d_\tau (a+b)-f_{\eta }(a+b): \tau \in S\}. $ Since $\eta $ can be arbitrary small, we have

(e4.39) $$ \begin{align} \omega(a)|_S\le \inf_{\eta>0}\sup\{d_\tau(a+b)-f_{\eta}(a+b): \tau\in S\}=\omega(a+b)|_S. \end{align} $$

Similarly, $\omega (b)|_S\le \omega (a+b)|_S.$ Thus, the inequality on the left of (e4.18) holds. The estimates (e4.19) can be checked similarly.

For (3), recall that $\mathrm {Her}(a)$ has continuous scale if and only if $e_{m(n)}-e_n\stackrel {c.}{\to } 0$ for any $m(n)>n,$ where $e_n=f_{1/2^n}(a), n\in \mathbb {N}.$ Suppose that $\omega ^c(a)=0.$ Then, for each $n\in \mathbb {N}$ and any $m(n)\ge n,$

(e4.40) $$ \begin{align} e_{m(n)}-e_n\lesssim g_{1/n}(a)\stackrel{c.}{\to} 0. \end{align} $$

It follows that $\mathrm {Her}(a)$ has continuous scale.

Conversely, suppose that $\mathrm {Her}(a)$ has continuous scale. For any $d\in A_+\setminus \{0\},$ choose $n_0\in \mathbb {N}$ such that, for any $m(n)>n\ge n_0,$

(e4.41) $$ \begin{align} e_{m(n)}-e_n\lesssim d. \end{align} $$

Suppose that $k_0>2^{n_0}.$ Fix $k\ge k_0.$ For any $\varepsilon \in (0,1/4),$ there is $m(n)>n_0$ such that

(e4.42) $$ \begin{align} f_\varepsilon(g_{1/k}(a))\lesssim e_{m(n)}-e_{n_0}. \end{align} $$

In other words, for any $\varepsilon \in (0, 1/4), f_\varepsilon (g_{1/k}(a))\lesssim d.$ It follows that $g_{1/k}(a)\lesssim d$ (for any $k\ge k_0$ ). This proves (3).

Proof We may assume that $\omega (a)|_S<\infty .$ There exists $\delta _0>0$ such that, for all $0<\eta \le 2\delta _0,$

(e4.44) $$ \begin{align} d_\tau(a)-\tau(f_{\eta}(a))<\omega(a)+\varepsilon/2\,\,\,\mathrm{for\,\,\,all}\,\,\, \tau\in S. \end{align} $$

Then, there exists $\sigma _0>0$ such that, if $0<\delta <\delta _0,$

(e4.45) $$ \begin{align} d_\tau(f_\delta(a))-\tau(f_{\sigma_0}(f_\delta(a)))\le d_\tau(a)-\tau(f_{2\delta_0}(a))<\omega(a)+\varepsilon/2 \end{align} $$

for all $\tau \in S.$ Note that $d_\tau (f_\delta (a))\le \tau (f_{\delta /2}(a))$ for all $\tau \in \widetilde {QT}(A).$ It follows that (see (2) of Proposition 2.10)

$$ \begin{align*}\nonumber \omega(f_\delta(a))|_S<\omega(a)|_S+\varepsilon.\\[-34pt] \end{align*} $$

Proof Fix $0<\delta <1/2.$ For any $\varepsilon>0,$ there exists $\tau _0\in S$ such that

(e4.48) $$ \begin{align} d_{\tau_0}(a)-\tau_0(f_{\delta/2}(a))>\omega(a)|_S-\varepsilon/4. \end{align} $$

For this $\tau _0,$ choose $0<\eta <\delta /2$ such that $ d_{\tau _0}(a)-\tau _0(f_\eta (a))<\varepsilon /4. $ Then

(e4.49) $$ \begin{align} \tau_0(f_\eta(a))-d_{\tau_0}(f_\delta(a)) &> d_{\tau_0}(a)-{\tau_0}(f_{\delta/2}(a))-( d_{\tau_0}(a)-\tau_0(f_\eta(a))) \end{align} $$

(e4.50) $$ \begin{align} &>\omega(a)|_S-\varepsilon/4-\varepsilon/4. \end{align} $$

Hence,

(e4.51) $$ \begin{align} \sup\{{{\tau(f_\eta(a))}}-d_\tau(f_\delta(a)):\tau\in S \} \ge \omega(a)|_S-\varepsilon/2. \end{align} $$

To see the second inequality, choose $n_0\in \mathbb {N}$ such that

(e4.52) $$ \begin{align} \|a^{1/n_0}f_\eta(a)-f_\eta(a)\|<\varepsilon/4. \end{align} $$

It follows that, for all $\tau \in S,$

(e4.53) $$ \begin{align} \tau(a^{1/n_0})-d_\tau(f_\delta(a))&\ge \tau(a^{1/n_0}f_\eta(a))-d_\tau(f_\delta(a)) \end{align} $$

(e4.54) $$ \begin{align} &\ge \tau(f_\eta(a))-d_\tau(f_\delta(a))-\varepsilon/4. \end{align} $$

Therefore, by (e4.51), $ \sup \{\tau (a^{1/n_0})-d_\tau (f_\delta (a)): \tau \in S \}\ge \omega (a)|_S-\varepsilon. $

Proof Recall that $B={{\mathrm {Her}(a)}}$ and $\Pi _S: l^\infty (B)\to l^\infty (B)/I_{_{S,B}}$ is the quotient map.

For (1), there is, for each $k\in \mathbb {N},$ a sequence $\{c_n^{(k)}\}\in l^\infty ({{\mathrm {Ped}(\mathrm {Her}(a))}})$ with $0\le c_n^{(k)}\le 1$ and $\lim _{n\to \infty }\omega (c_n^{(k)})|_S=0$ such that

(e4.60) $$ \begin{align} \|\Pi_S(\iota(a)-\{c_n^{(k)}\})\|<1/k. \end{align} $$

Therefore, for each $k\in \mathbb {N},$ there is $n(k)\in \mathbb {N}$ such that

(e4.61) $$ \begin{align} \|a-c_{n(k)}^{(k)}\|_{_{2,S}} <2/k\,\,\,\mathrm{and}\,\,\, \omega(c_{n(k)}^{(k)})<1/k. \end{align} $$

Define $b_k=c_{n(k)}^{(k)},\,\, k\in \mathbb {N}.$ Then $0\le b_k\le 1, b_k\in {{\mathrm {Ped}(\mathrm {Her}(a))}}$ and $\lim _{n\to \infty } \omega (b_n)|_S=0.$ Moreover,

(e4.62) $$ \begin{align} \|a-b_n\|_{_{2, S}}<2/n\,\,\,\mathrm{for\,\,\,all}\,\,\, n\in \mathbb{N}. \end{align} $$

It follows that ${{\|}}\Pi _S(\iota (a)-\{b_n\})\|=0,$ and (e4.58) holds. A similar proof above shows that, if $\Omega ^T_T(a)=0$ implies that there is $b_n\in \mathrm {Ped}(\mathrm {Her}(a))_+^{\mathbf {1}}$ such that (e4.58) holds.

For (2), we note that (e4.58) implies that $\iota (a)-\{b_n\}\in I_{_{S}}.$ So, if $\Omega ^T_T(a)|_S=0,$ then $\|\Pi _S(\iota (a)-\{b_n\})\|=0.$ Hence, $\Omega ^T(a)|_S=0.$ The converse also holds.

To show (3) holds, suppose that there are $c_n^{(k)}\in {{\mathrm {Ped}}}(\mathrm {Her}(a))$ such that $0\le c_n^{(k)}\le M, \lim _{n\to \infty }\omega (c_n^{(k)}){{|_S}}=0$ and

(e4.63) $$ \begin{align} \|\Pi_S(\iota(a)-\{c_n^{(k)}\}{{)}}\|<1/k\,\,\,\mathrm{for\,\,\,all}\,\,\, k\in \mathbb{N}. \end{align} $$

By the proof of (1), one obtains $b_n\in {{\mathrm {Ped}}}(\mathrm {Her}(a))_+$ with $0\le b_n\le M$ such that

(e4.64) $$ \begin{align} \lim_{n\to\infty}\omega(b_n)|_S=0\,\,\,\mathrm{and}\,\,\, \Pi_S(\iota(a))=\Pi_S(\{b_n\}). \end{align} $$

Define $g\in C_0((0, \|a\|+M])$ by $g(t)=\|a\|$ if $t\in [\|a\|, \|a\|+M]$ and $g(t)=t$ if $t\in [0,\|a\|].$ Since M is fixed and $g(a)=a,$ we have

(e4.65) $$ \begin{align} \Pi_S(a)=\Pi_S(\{g(b_n)\}). \end{align} $$

Put $c_n=g(b_n).$ Then $c_n\in {{\mathrm {Ped}}}(\mathrm {Her}(a))_+$ and $\|c_n\|\le \|a\|.$ Since $c_n=g(b_n)\sim b_n,$ then $\lim _{n\to \infty }\omega (c_n){|_{S}}=0.$ Therefore, by (e4.65), $\Omega ^T(a){{|_S}}=0.$

(4) Since $\{a_n\}\in (I_{_{S}})_+^{\mathbf {1}}, \lim _{n\to \infty }\|a_n\|_{_{2, S}}=0.$ Choose $\delta _n:=\sqrt {\|a_n\|_{2,S}+1/n}$ and $b_n=(a_n-\delta _n)_+, n\in \mathbb {N}.$ Then $\|b_n\|\le 1, n\in \mathbb {N}, \|b_n\|_{_{2, S}}\le \|a_n\|_{_{2, S}}\to 0$ and $\lim _{n\to \infty } \|a_n-b_n\|=0.$

Note that

(e4.66) $$ \begin{align} f_\eta((x-\delta_n)_+^2)\le \chi_{[\delta_n,+\infty)}(x)\le (1/\delta_n^2)x^2\,\,\,\mathrm{for\,\,\,all}\,\,\, x \in\mathbb{R}_+, \end{align} $$

where $\eta \in (0,1)$ and $\chi _{[\delta _n,+\infty )}$ is the characteristic function of $[\delta _n,+\infty ).$ Then, for all $\tau \in S,$

(e4.67) $$ \begin{align} d_\tau( b_n)=d_\tau(b_n^2) =\sup_{\eta>0}\tau(f_\eta({{(a_n-\delta_n)_+^2}})) \le (1/\delta_n^2)\tau(a_n^2) \le (1/\delta_n^2) \|a_n\|^2_{_{2,S}} =\|a_n\|_{_{2, S}}. \end{align} $$

It follows that $\lim _{n\to \infty }\sup \{d_\tau (b_n):\tau \in S\}=0.$

Proof It follows from (1) of Proposition 2.10 that there is $L\in \mathbb {R}_+$ such that

$$ \begin{align*}S_2\subset \{r s: s\in S_1 \,\,\,\mathrm{and}\,\,\, r\in [0, L]\}.\end{align*} $$

If $\Omega ^T(a)|_{S_1}=0,$ then, by Proposition 4.8, there exists a sequence $b_n\in {{\mathrm {Ped}(\mathrm {Her}(a))}}$ with $\|b_n\|\le \|a\|$ such that $\lim _{n\to \infty }\omega (b_n)|_{S_1}=0$ and $\lim _{n\to \infty }\|a-b_n\|_{_{2, S_1}}=0.$ It follows that

$$ \begin{align*}\lim_{n\to\infty}\omega(b_n)|_{S_2}\le \lim_{n\to\infty}L\cdot \omega(b_n)|_{S_1}=0\,\,\,\mathrm{and}\,\,\, \lim_{n\to\infty}\|a-b_n\|_{_{2, S_2}}\le \lim_{n\to\infty}L\|a-b_n\|_{_{2, S_1}}=0.\end{align*} $$

Then $\Omega ^T(a)|_{S_2}=0.$ Exactly the same argument shows that if $\Omega _C^T(a)|_{S_1}=0$ (or $\Omega ^N(a)|_{S_1}=0$ ), then $\Omega _C^T(a)|_{S_2}=0$ (or $\Omega ^N(a)|_{S_2}=0$ ).

To see the last statement, we note that $(S_1)|_{\mathrm {Her}(a)}$ is bounded (see (2) of Proposition 2.10). In other words, there is $L>0$ such that $(S_1)|_{\mathrm {Her}(a)}\subset \{r\cdot \tau : \tau \in S_a, r\in [0, L]\}.$

Proof Since $e\in \mathrm {Her}(a)_+^{\mathbf {1}},$ we compute that

(e4.70) $$ \begin{align} \lim_{n\to\infty}\|(a+1/n)^{-1/2} a^{1/2} e a^{1/2}(a+1/n)^{-1/2}-e\|=0. \end{align} $$

It follows that $e\sim a^{1/2}ea^{1/2}\sim e^{1/2}ae^{1/2}\le e.$ Therefore, by Proposition 4.2, (e4.68) and (e4.69) hold.

Proof Fix $a\in \mathrm {Ped}(A\otimes {\cal K})_+\setminus \{0\}.$ Let $\varepsilon>0.$ If there is a subsequence $\{n_k\}\subset \mathbb {N}$ such that $f_{1/4n_k}(a)-f_{{{1/}}n_k}(a)=0,$ then $f_{1/4n_k}(a)$ is a projection. Consequently, $\omega (f_{1/4n_k}(a))=0$ for all $k\in \mathbb {N}.$ Note that $af_{1/4n_k}(a)\sim f_{1/4n_k}(a).$ It follows that $\omega (af_{1/4n_k}(a))=0, k\in \mathbb {N}.$ Since $\lim _{k\to \infty }\|a-af_{1/4n_k}(a)\|=0, \Omega ^T(a)=0.$ Hence, (e4.71) holds.

Next, we assume that for some $n_0\in \mathbb {N}, f_{1/4n}(a)-f_{{1/n}}(a)\not =0$ for all $n\ge n_0.$ Moreover,

(e4.72) $$ \begin{align} \sup\{d_\tau(a)-\tau(f_{1/n}(a)): \tau\in \overline{QT(A)}^w\}<\omega(a)+\varepsilon/2{{\,\,\,\mathrm{for\,\,\,all}\,\,\, n\ge n_0.}} \end{align} $$

For the rest of this proof, we will assume $n\geq n_0.$ Since the map $\Gamma $ is surjective, there is $c_n\in (A\otimes {\cal K})_+^{\mathbf {1}}$ such that $d_\tau (c_n)=\tau (f_{1/n}(a))$ for all $\tau \in \overline {QT(A)}^w.$ Since $\widehat {f_{1/n}(a)}$ is continuous on $\overline {QT(A)}^w, \omega (c_n)=0.$ Choose $0<\delta _n<1$ such that (by also Lemma 4.5),

(e4.73) $$ \begin{align} d_\tau(c_n)-\tau(f_{\delta_n}(c_n))<1/2^n\,\,\,\mathrm{for\,\,\,all}\,\,\, {{\tau\in}}\overline{QT(A)}^w{{\,\,\,\mathrm{and}\,\,\, \omega(f_{\delta_n}(c_n))<1/2^n.}} \end{align} $$

Note that we have

(e4.74) $$ \begin{align} {{d_\tau(c_n)=}}\tau(f_{1/n}(a){{)}} <d_\tau(f_{1/4n}(a{{))}}\,\,\,\mathrm{for\,\,\,all}\,\,\, \tau\in \overline{QT(A)}^w. \end{align} $$

Since A has strict comparison, $c_n\lesssim f_{1/4n}(a).$ By Proposition 2.4 of [Reference Rørdam40], there is $x_n\in A\otimes {\cal K}$ such that

(e4.75) $$ \begin{align} x_n^*x_n=f_{{{\delta_n}} }(c_n)\,\,\,\mathrm{and}\,\,\, x_nx_n^*\in \mathrm{Her}(f_{1/4n}(a)). \end{align} $$

Put $b_n=a^{1/2}x_nx_n^*a^{1/2}.$ Then $\|b_n\| \le \|a\|$ and $b_n\in \mathrm {Her}(f_{1/4n}(a)).$ By Lemma 4.10 and (e4.73), $\omega (b_n)=\omega (x_nx_n^*)=\omega (x_n^*x_n)=\omega (f_{\delta _n}(c_n))\leq 1/2^n\to 0.$ Note that

(e4.76) $$ \begin{align} {{f_{1/8n}}}(a)(x_nx_n^*)=x_nx_n^*=x_nx_n^* {{f_{1/8n}}}(a). \end{align} $$

Put $a_n=a^{1/2}{{f_{1/8n}(a)}} a^{1/2}$ and $d_n= {{f_{1/8n}(a)}}-x_nx_n^*.$ Then $0\le d_n\le 1$ and $0\le a_n-b_n, n\in \mathbb {N}.$ For all $\tau \in \overline {QT(A)}^w,$ we compute that

(e4.77) $$ \begin{align} {{\tau((a_n-b_n)^2)}} &={{\tau(a^{1/2}d_nad_na^{1/2})\le \|a\|\tau(a^{1/2}d_n^2a^{1/2})}}\end{align} $$

(e4.78) $$ \begin{align} &\le \|a\|\tau(a^{1/2}d_na^{1/2})\le \|a\|^2 \tau( {{d_n}} ) \end{align} $$

(e4.79) $$ \begin{align} &{\stackrel{(4.76)}{=}} \|a\|^2(\tau( {{f_{1/8n}(a)}} )-\tau(x_nx_n^*))\le \|a\|^2(d_\tau(a)-\tau(x_n^*x_n)) \end{align} $$

(e4.80) $$ \begin{align} &=\|a\|^2{{(}}d_\tau(a)-\tau(f_{1/n}(a))+\tau(f_{1/n}(a))-\tau(f_{\delta_n}(c_n)){{)}} \end{align} $$

(e4.81) $$ \begin{align} &=\|a\|^2{{(}}d_\tau(a)-\tau(f_{1/n}(a))+d_\tau(c_n)-\tau(f_{\delta_n}(c_n)){{)}} \end{align} $$

(e4.82) $$ \begin{align} &\overset{{{(4.72)}, (4.73)}}{\le} \|a\|^2( (\omega(a)+\varepsilon/2)+1/2^n). \end{align} $$

Note that $\|a-a_n\|<1/n.$ It follows that, by (e2.17), (e2.19), and (e4.82),

(e4.83) $$ \begin{align} \|a-b_n\|_{_{2, \overline{QT(A)}^w}}^{2/3} &\le \|a-a_n\|^{2/3}_{_{2,\overline{QT(A)}^w}}+\|a_n-b_n\|^{2/3}_{_{2, \overline{QT(A)}^w}} \end{align} $$

(e4.84) $$ \begin{align} &<(1/n)^{2/3}\sup\{d_\tau(a)^{1/3}:\tau\in \overline{QT(A)}^w\}+(\omega(a)+\varepsilon/2+1/2^n)^{1/3}. \end{align} $$

Hence (by also (2) of Proposition 2.10),

(e4.85) $$ \begin{align} \limsup_{n\to\infty}\|a-b_n\|_{_{2, \overline{QT(A)}^w}}\le \|a\|\sqrt{\omega(a)+\varepsilon/2}. \end{align} $$

Let $\varepsilon \to 0.$ We then obtain (e4.71) (recall that $\lim _{n\to \infty }\omega (b_n)=0$ ).

Proof This follows from the definition immediately.

Proof For (1), let us assume that $\{e_n\}$ is a (not necessarily increasing) approximate identity for $\mathrm {Her}(a)$ such that $\lim _{n\to \infty }\omega (e_n)=0.$ Then

(e5.4) $$ \begin{align} \lim_{n\to\infty}\|a- e_n^{1/2}ae_n^{1/2}\|=0. \end{align} $$

By Lemma 4.10, $ \omega (e_n^{1/2}ae_n^{1/2})|_S\le \omega (e_n)|_S\to 0. $ Thus, $\Omega ^N(a)|_S=0.$

Conversely, suppose that $a\in (A\otimes {\cal K})_+^{\mathbf {1}}$ and $\{b_n\}\in l^\infty (\mathrm {Her}(a))_+$ such that

(e5.5) $$ \begin{align} \lim_{n\to\infty}\|a-b_n\|=0\,\,\,\mathrm{and}\,\,\, \lim_{n\to\infty}\omega(b_n)|_S=0. \end{align} $$

Note that $\lim _{n\to \infty }\|b_n\|=\|a\|\le 1.$ Let $g\in C([0,\infty ))_+^{\mathbf {1}}$ such that $g(t)=t$ if $t\in [0, 1]$ and $g(t)=1$ if $t>1.$ Then, for any $n\in \mathbb {N}, g(b_n)\sim b_n.$ We also have $\lim _{n\to \infty }\|g(a)-g(b_n)\|=0.$ But $g(a)=a.$ Then, replacing $b_n$ by $g(b_n),$ we may assume that $\|b_n\|\le 1.$ For each $k\in \mathbb {N},$ choose $ {{n_k}}$ such that

(e5.6) $$ \begin{align} \|b_{n_k}^{1/k}-a^{1/k}\|<1/k,\,\,k\in \mathbb{N}. \end{align} $$