Proof Since $CAC=A^{*}$ and $CA^{*}C=A$ , one can see that $C(\mathcal {R}(A))=\mathcal {R}(A^{*})$ . Likewise, we have $C(\mathcal {R}(B))=\mathcal {R}(B^{*})$ . It follows immediately that $\mathcal {R}(A^{*})\subset \mathcal {R}(B^{*})$ . Thus $\mathcal {R}(B^{*})^\bot \subset \mathcal {R}(A^{*})^\bot $ , that is, $\ker {B}\subset \ker {A}$ .

On the other hand, $\mathcal {R}(A)\subset \mathcal {R}(B)$ implies $\mathcal {R}(B)^{\bot }\subset \mathcal {R}(A)^{\bot }$ and, equivalently, $\ker {B^{*}}\subset \ker {A^{*}}.$

Proof of Theorem 1.2

“ $\Longrightarrow $ ”. Assume that $C_{\mathcal {R}(T)}=T\mathcal {S}_C T$ . We shall show that $\mathcal {R}(T)$ is closed.

Assume that $\{x_n\}_{n=1}^\infty \subset {\mathcal {H}}$ and $Tx_n\rightarrow y_0\in {\mathcal {H}}$ . It suffices to show that $y_0\in \mathcal {R}(T)$ . Note that $T\in C_{\mathcal {R}(T)}$ . Thus there exists $A\in \mathcal {S}_C$ such that $T=TAT$ . Hence, $TATx_n\rightarrow y_0$ and $TATx_n\rightarrow TAy_0$ , which implies $y_0=TAy_0\in \mathcal {R}(T).$

“ $\Longleftarrow $ ”. Now assume that $\mathcal {R}(T)$ is closed. Since $T\mathcal {S}_C T \subseteq C_{\mathcal {R}(T)}$ is clear, it remains to show $ C_{\mathcal {R}(T)} \subseteq T\mathcal {S}_C T$ .

Denote $\mathcal {N}=(\ker {T})^{\bot }$ and $\mathcal {M}=\mathcal {R}(T)$ . For $x\in \mathcal {N}$ , define $Fx=Tx$ . Since $\mathcal {M}$ is norm-closed, F is an invertible operator from $\mathcal {N}$ into $\mathcal {M}$ . Define $S=P_{\mathcal {N}}F^{-1}P_{\mathcal {M}}$ . One can verify that $TS=P_{\mathcal {M}}$ and $ST=P_{\mathcal {N}}$ .

Since $T\in \mathcal {S}_C$ , one can see that $C(\ker T)=\ker T^{*}$ , that is, $C(\mathcal {N}^\bot )=\mathcal {M}^\bot $ . Thus $C(\mathcal {N})=\mathcal {M}$ and $C(\mathcal {M})=\mathcal {N}$ . Denote $C_1=C|_{\mathcal {M}}$ . Then $C_1:\mathcal {M}\longrightarrow \mathcal {N} $ is conjugate-linear, invertible, and isometric. One can check that $Cx=C^{-1}x= C_1^{-1}x$ for all $x\in \mathcal {N}.$

In fact, for any $x\in \mathcal { M} $ , note that

$$ \begin{align*}TCx=TC_1x=FC_1x, \ \ \ CT^{*}x=CF^{*}x=C_1^{-1}F^{*}x. \end{align*} $$

Thus $FC_1=C_1^{-1}F^{*}$ , ${(F^{-1})}^{*}C_1=C_1^{-1} F^{-1}$ and $C_1{(F^{-1})}^{*}= F^{-1}C_1^{-1}$ .

For any $x\in {\mathcal {H}}$ , there is the unique decomposition $x=x_1+x_2$ , where $x_1\in \mathcal {N}$ and $x_2\in \mathcal {N}^{\bot }$ . So $Cx_1\in \mathcal {M}$ and $Cx_2\in \mathcal {M}^{\bot }$ . By Claim 1, we have

$$ \begin{align*} CSCx&=C(P_{\mathcal{N}}F^{-1}P_{\mathcal{M}})(Cx_1+Cx_2)\\ &=C(P_{\mathcal{N}}F^{-1}P_{\mathcal{M}})Cx_1\\ &=C(P_{\mathcal{N}}F^{-1}P_{\mathcal{M}})C_1^{-1}x_1\\ &=C(P_{\mathcal{N}}F^{-1} C_1^{-1}x_1)\\ &=C_1^{-1} (F^{-1} C_1^{-1}x_1) \\ &=(C_1^{-1}F^{-1} C_1^{-1})x_1= {(F^{-1})}^{*} x_1\\ &= P_{\mathcal{M}}{(F^{-1})}^{*} P_{\mathcal{N}}x =S^{*}x. \end{align*} $$

This proves $S\in \mathcal {S}_C.$

Set $X=SAS$ . Thus one can easily check that $X\in \mathcal {S}_C$ . We shall prove that $TXT=A$ . Note that $\mathcal {R}(A)\subset \mathcal {R} (T)= \mathcal {M}$ . It follows that

$$ \begin{align*}TXT=TSAST=P_{\mathcal{M}}AP_{\mathcal{N}}=AP_{\mathcal{N}}.\end{align*} $$

It suffices to prove that $AP_{\mathcal {N}}=A$ , that is, $A(I-P_{\mathcal {N}})=0$ .

Since $\mathcal {R}(A)\subset \mathcal {R}(T)$ , it follows from Lemma 2.1 that $\ker {T}\subset \ker {A}$ . That is, $\mathcal {N}^{\bot }\subset \ker A$ . Therefore we conclude that $A(I-P_{\mathcal {N}})=0$ .

Proof It suffices to prove the necessity.

“ $\Longrightarrow $ ”. By Theorem 1.2, we need only deal with the case that $\mathcal {R}(A)$ is closed.

Using Theorem 1.2 again, we can find $Z\in \mathcal {S}_C$ such that $A=AZA$ . Since $\mathcal {R} (A)\subset \mathcal {R} (T)$ , by Douglas’ range inclusion theorem, $A=TY$ for some $Y\in \mathcal {B(H)}$ . Noting that $A,T\in \mathcal {S}_C$ , we have

$$ \begin{align*} A&=CA^{*}C=C(TY)^{*}C=CY^{*}T^{*}C\\ &=(CY^{*}C)(CT^{*}C)=(CY^{*}C)T. \end{align*} $$

Then

$$ \begin{align*}A=AZA=(TY)Z(CY^{*}C)T=T[YZ(CY^{*}C)]T.\end{align*} $$

One can verify that $X:=YZ(CY^{*}C)\in \mathcal {S}_C$ , which completes the proof.

Proof Denote $X=x\otimes (Cx)$ and $Y=x\otimes (Cy)+y\otimes (Cx)$ . It is easy to verify that $X\in \mathcal {S}_C$ and $\overline {\mathcal {R}(X)}=\mathcal {R}(X)\subset \mathcal {R}(T)$ . By Corollary 2.2, we have $X\in T\mathcal {S}_C T$ . Likewise we have $Y\in T\mathcal {S}_C T$ .

Proof of Theorem 1.3

Since it is clear that $T\mathcal {S}_C T\subset C_{\mathcal {R}(T)}$ , it suffices to prove that $C_{\mathcal {R}(T)}\subset \overline {T\mathcal {S}_C T}$ . Fix an operator A in $C_{\mathcal {R}(T)}$ . We shall prove that $A\in \overline {T \mathcal {S}_C T}$ . This follows readily from the following claim.

In fact, if the preceding claim holds, then, by Corollary 2.2, $B\in T\mathcal {S}_C T$ and $\text {dist}(A, T\mathcal {S}_C T)<\varepsilon $ . Since $\varepsilon>0$ was arbitrary, we deduce that $A\in \overline {T\mathcal {S}_C T}$ .

By Lemma 2.7, $A=CJP$ , where $P=|A|$ and J is a partial conjugation supported on $\overline {\mathcal {R}(P)}$ with $JP=PJ$ .

Now we fix an $\varepsilon>0$ and define three functions $f, g, h $ on $[0,\|A\|]$ as

$$ \begin{align*}f(t)=\begin{cases} 0,& t\in[0,\varepsilon/2), \\ t,& t\in[\varepsilon/2, \|A\|],\end{cases} \ \ h(t)=\begin{cases} 0,& t\in[0,\varepsilon/2), \\ 1/t,& t\in[\varepsilon/2, \|A\|], \end{cases} \ \ \ g=fh.\end{align*} $$

It is easy to see that:

1. (i) $f(P), g(P),$ and $h(P)$ are positive operators commuting with J,

2. (ii) $g(P)$ is an orthogonal projection with $\mathcal {R}(f(P))=\mathcal {R}(g(P))=\mathcal {R}(h(P))$ , and

3. (iii) $\mathcal {R}(g(P))$ reduces J.

Define $\tilde {J}=Jg(P)$ . Then $\tilde {J}$ is a partial anti-conjugation supported on $\mathcal {R}(h(P))$ commuting with $h(P)$ . Set $B=CJf(P)$ . Note that $Jf(P)=\tilde {J}P$ . It follows that ${B=C\tilde {J}f(P)}$ . By Lemma 2.7, we have $B\in \mathcal {S}_C$ . Note that

$$ \begin{align*}\overline{\mathcal{R}(f(P))}=\mathcal{R}(f(P))\subset \mathcal{R}(P).\end{align*} $$

Since J is supported on $\overline {\mathcal {R}(P)}$ , it follows that

$$ \begin{align*}\overline{\mathcal{R}(B)}= \mathcal{R}(B)\subset\mathcal{R}(A).\end{align*} $$

Note that

$$ \begin{align*}\|A-B\|=\|CJP-CJf(P)\|\leq \|P-f(P)\|<\varepsilon.\end{align*} $$

This completes the proof.

Proof Let $T=U|T|$ be the polar decomposition of T, where U is a partial isometry. So $\mathcal {R}(T)=\mathcal {R} (U|T|U^{*})$ .

Denote $P=CU|T|U^{*}C$ . Clearly, P is positive and hence a complex symmetric operator. Then we can find a conjugation J on ${\mathcal {H}}$ so that $JP=PJ$ . Set $A=CJP$ . Then, by [Reference Gilbreath and Wogen17, Theorem 3.1], $A\in \mathcal {S}_C$ and

$$ \begin{align*}\mathcal{R}(A)=\mathcal{R}(CJP)=C(\mathcal{R} (JP))=C(\mathcal{R} (P))=\mathcal{R} (U|T|U^{*}C)=\mathcal{R}(T).\end{align*} $$

This ends the proof.

Proof (i) $\Longrightarrow $ (ii). This is obvious.

(ii) $\Longrightarrow $ (i). For any $x\in \mathcal {R}(A)$ , by Corollary 2.4, we have

$$ \begin{align*}x\otimes (Cx)\in C_{\mathcal{R}(A)}\subset C_{\mathcal{R}(B)},\end{align*} $$

which implies $x\in \mathcal {R}(B)$ . Hence $\mathcal {R}(A)\subset \mathcal {R}(B)$ .

(i) $\Longrightarrow $ (iii). By Douglas’ range inclusion theorem, $\mathcal {R}(A)\subset \mathcal {R}(B)$ implies $A=BZ$ for some $Z\in \mathcal {B(H)}$ . Note that

$$ \begin{align*}A^t=CA^{*}C=C(BZ)^{*}C=CZ^{*}B^{*}C=(CZ^{*}C)(CB^{*}C)=Z^tB^t.\end{align*} $$

Thus, for any $X\in \mathcal {S}_C$ , we have

$$\begin{align*}AXA^t=(BZ)X(Z^tB^t) =B(ZXZ^t)B^t.\end{align*}$$

One can verify that $ZXZ^t\in \mathcal {S}_C$ . Thus $AXA^t\in B\mathcal {S}_C B^t$ .

(iii) $\Longrightarrow $ (i). For $x\in {\mathcal {H}}$ , denote $y=Ax$ . We shall prove that $y\in \mathcal {R}(B)$ . Note that

$$ \begin{align*} y\otimes (Cy)&=(Ax)\otimes(CAx)=(Ax)\otimes[(CAC)(Cx)]\\ &=A[x\otimes (Cx)](CAC)^{*} =A[x\otimes (Cx)]A^t\in A\mathcal{S}_C A^t. \end{align*} $$

So $y\otimes (Cy)\in B\mathcal {S}_C B^t$ . Then $y\otimes (Cy)=B X B^t$ for some $X\in \mathcal {S}_C$ . It follows that $y\in \mathcal {R}(B)$ .

Proof By Lemma 2.9, there exists $A\in \mathcal {S}_C$ with $\mathcal {R}(A)=\mathcal {R}(T)$ . Then $C_{\mathcal {R}(A)}=C_{\mathcal {R}(T)}$ and, by Proposition 2.10, we have

$$ \begin{align*}T\mathcal{S}_C T^t=A\mathcal{S}_C A^t=A\mathcal{S}_C A.\end{align*} $$

Then $C_{\mathcal {R}(T)}=T\mathcal {S}_C T^t$ if and only if $C_{\mathcal {R}(A)}=A\mathcal {S}_C A$ . The desired result follows readily from Theorem 1.2.

Proof of Theorem 1.4

Denote ${\mathcal {M}}=\mathcal {R}(A)$ and $\mathcal {N}=\mathcal {R}(B)$ . By Theorem 1.3, $\overline {A\mathcal {S}_C A}=\overline {C_{{\mathcal {M}}}}$ and $\overline {B\mathcal {S}_C B}=\overline {C_{\mathcal {N}}}$ .

“ $\Longrightarrow $ ”. Choose a closed subspace $\mathcal {M}_1$ of ${\mathcal {M}}$ and an orthonormal basis $\{e_i\}_{i\in \Lambda }$ of $\mathcal {M}_1$ . Define $U=\sum _{i\in \Lambda } e_i\otimes (ce:i)$ . Then one can check that $U\in \mathcal {S}_C$ is a partial isometry with $\mathcal {R}(U)=\mathcal {M}_1$ , which implies $U\in C_{\mathcal {M}}\subset \overline {C_{\mathcal {N}}}$ . By the hypothesis, there exist $X_n\in C_{\mathcal {N}}$ , $n\geq 1$ , such that $X_n\rightarrow U$ .

By Corollary 2.8, we can find $\{Y_n\}\subset \mathcal {S}_C$ with $\mathcal {R}(Y_n)=\overline {\mathcal {R}(Y_n)}\subset \mathcal {N}$ such that $\|Y_n-X_n\|<\frac {1}{n}, n\geq 1$ . Then $Y_n\in C_{\mathcal {N}}$ and $Y_n\rightarrow U$ . Furthermore, we have $Y_nY_n^{*}\rightarrow UU^{*}$ .

Note that $UU^{*}=P_{\mathcal {M}_1}$ and $\mathcal {R}(Y_nY_n^{*})=\mathcal {R}(Y_n)$ is closed, $n\geq 1$ . There exists $\delta>0$ such that $\sigma (Y_nY_n^{*})\subset \{0\}\cup [\delta , \infty )$ . Define a continuous function f on $\{0\}\cup (\delta /2, \infty )$ as

$$ \begin{align*}f(t)=\begin{cases} 0,& t=0, \\ 1,& t\in(\delta/2, \infty).\end{cases}\end{align*} $$

Then $f(Y_nY_n^{*})\rightarrow f(UU^{*})=P_{\mathcal {M}_1}$ . Note that $f(Y_nY_n^{*})$ is an orthogonal projection with

$$ \begin{align*}\mathcal{R} (f(Y_nY_n^{*}))=\mathcal{R} (Y_nY_n^{*})=\mathcal{R}(Y_n)\subset\mathcal{N}.\end{align*} $$

This proves the necessity.

“ $\Longleftarrow $ ”. Clearly, it suffices to prove $C_{{\mathcal {M}}}\subset \overline {C_{\mathcal {N}}}$ . By Corollary 2.8, it suffices to prove that any $X\in \mathcal {S}_C$ with $\mathcal {R}(X)=\overline {\mathcal {R}(X)}\subset {\mathcal {M}}$ satisfies $X\in \overline {C_{\mathcal {N}}}$ .

Denote $\mathcal {M}_1=\mathcal {R}(X)$ and, for convenience, we write P for $P_{\mathcal {M}_1}$ . By Corollary 2.11, we have $X\in C_{\mathcal {M}_1}=P\mathcal {S}_C P^t$ . So $X=PYP^t$ for some $Y\in \mathcal {S}_C$ . On the other hand, by the hypothesis, we can find orthogonal projections $\{P_n\}$ with $\mathcal {R}(P_n)\subset \mathcal {N}$ such that $P_n\rightarrow P$ . So $P_nYP_n^t\rightarrow PYP^t=X$ . Noting that $P_nYP_n^t \in \mathcal {S}_C$ and $\mathcal {R}(P_nYP_n^t)\subset \mathcal {R}(P_n)\subset \mathcal {N}$ , we conclude that $P_nYP_n^t \in C_{\mathcal {N}}$ and $X\in \overline {C_{\mathcal {N}}}$ .

Proof One can easily check that $C_{\overline {\mathcal {M}}}$ is closed in the weak operator topology. It is clear that

$$ \begin{align*}[C_{\mathcal{M}}\cap\mathcal{K(H)}] \subset[\overline{C_{\mathcal{M}}}\cap\mathcal{K(H)}]\subset [C_{\overline{\mathcal{M}}}\cap\mathcal{K(H)}].\end{align*} $$

So it suffices to prove $[C_{\overline {\mathcal {M}}}\cap \mathcal {K(H)}] \subset \overline { C_{\mathcal {M}}\cap \mathcal {K(H)} }$ .

Now assume that $A\in C_{\overline {\mathcal {M}}}\cap \mathcal {K(H)}$ . We shall prove $A\in \overline {C_{\mathcal {M}}\cap \mathcal {K(H)}}$ . By [Reference Garcia and Putinar15, Theorem 3], A can be written as

$$ \begin{align*}A=\sum_{n=1}^\infty \lambda_n(Ce_n)\otimes e_n,\end{align*} $$

where $e_n$ are certain orthonormal eigenvectors of $|A| =|AA^{*}|^{1/2}$ and the $\lambda _n$ are the nonzero eigenvalues of $|A|$ , repeated according to multiplicity. Clearly, $\lambda _n\rightarrow 0$ and $Ce_n\in \mathcal {R}(A)\subset \overline {\mathcal {M}}$ . Then

$$ \begin{align*}\sum_{n=1}^m \lambda_n(Ce_n)\otimes e_n\longrightarrow A\ \ \ \ \ \ (m\rightarrow\infty).\end{align*} $$

So it remains to prove that $(Ce_n)\otimes e_n\in \overline {C_{\mathcal {M}}\cap \mathcal {K(H)}}$ for each $n\geq 1$ .

Since $Ce_n\in \overline {{\mathcal {M}}}$ , there exists a sequence $\{x_k\}_{k=1}^\infty \subset {\mathcal {M}}$ converging to $Ce_n$ . Clearly, $x_k\otimes (Cx_k)\in C_{{\mathcal {M}}}\cap \mathcal {K(H)}$ for all $k\geq 1$ . Note that $\{x_k\otimes (Cx_k)\}$ converges to ${(Ce_n)\otimes e_n}$ in the norm topology. We deduce that $(Ce_n)\otimes e_n \in \overline {C_{\mathcal {M}}\cap \mathcal {K(H)}}$ . Thus we complete the proof.

Proof of Theorem 1.5

We denote $\mathcal {M}={\mathcal {R}(T)}$ . It is trivial to see that $C_{\overline {{\mathcal {M}}}}$ is closed in both the weak operator topology and the weak* topology; moreover,

On the other hand, since $C_{{\mathcal {M}}}$ is a subspace of $\mathcal {B(H)}$ , it is clear that . Then it suffices to show $C_{\overline {{\mathcal {M}}}}\subset \overline {C_{{\mathcal {M}}}}^{w*}$ .

We need only consider the case that $\mathcal {M}$ is not closed. In this case, we can choose an orthonormal basis $\{e_i\}_{i=1}^\infty $ of $\overline {\mathcal {M}}$ . Define $U\in \mathcal {B(H)}$ as

$$\begin{align*}U=\sum_{i=1}^\infty e_{i}\otimes (ce_{i}).\end{align*}$$

Thus U is a partial isometry lying in $\mathcal {S}_C$ , $\mathcal {R}(U)=\overline {\mathcal {M}}$ and $(\ker U)^\bot =C(\overline {\mathcal {M}})$ . So $U\in C_{\overline {{\mathcal {M}}}}$ and it follows from Theorem 1.2 that $C_{\overline {{\mathcal {M}}}}=U\mathcal {S}_C U$ .

Fix an operator $A\in \mathcal {S}_C$ . Then it suffices to prove $UAU\in \overline {C_{{\mathcal {M}}}}^{w*}$ .

For the conjugation C, there is an orthonormal basis $\{f_i\}$ of ${\mathcal {H}}$ such that $Cf_i=f_i$ for all i (see [Reference Garcia, Prodan and Putinar13, Lemma 2.11]). For each $n\geq 1$ , set $P_n=\sum _{i=1}^nf_i\otimes f_i$ . It is easy to see $P_n\in \mathcal {S}_C$ . Noting that $P_nAP_n\in \mathcal {S}_C$ and, as $n\rightarrow \infty $ , $P_nAP_n$ converges to A in the . So $UP_nAP_nU$ converges to $UAU$ in the . Now it remains to prove the following claim.

In fact, if the preceding claim holds, then, by [Reference Conway4, Proposition 20.3], $UAU\in \overline {C_{{\mathcal {M}}}}^{w*}$ .

Set $a_{j,i}=\langle Af_i, f_j\rangle $ for all $i,j\in \{1,2,\dots \}$ . Since $A\in \mathcal {S}_C$ , we have $a_{i,j}=a_{j,i}.$ Then

$$ \begin{align*}P_nAP_n=\sum_{i=1}^n a_{i,i} f_i\otimes f_i +\sum_{1\leq i<j\leq n} a_{i,j}(f_i\otimes f_j+f_j\otimes f_i) \end{align*} $$

and

(1) $$ \begin{align} &UP_nAP_nU \end{align} $$

(2) $$ \begin{align} &= \sum_{i=1}^n a_{i,i} (Uf_i)\otimes (U^{*}f_i) +\sum_{1\leq i<j\leq n} a_{i,j}[(Uf_i)\otimes (U^{*}f_j)+(Uf_j)\otimes (U^{*}f_i)] \end{align} $$

(3) $$ \begin{align} =& \sum_{i=1}^n a_{i,i} (Uf_i)\otimes (U^{*}Cf_i) +\sum_{1\leq i<j\leq n} a_{i,j}[(Uf_i)\otimes (U^{*}Cf_j)+(Uf_j)\otimes (U^{*}Cf_i)] \end{align} $$

(4) $$ \begin{align} =& \sum_{i=1}^n a_{i,i} (Uf_i)\otimes (CUf_i) +\sum_{1\leq i<j\leq n} a_{i,j}[(Uf_i)\otimes (CUf_j)+(Uf_j)\otimes (CUf_i)]. \end{align} $$

Note that $Uf_i, Uf_j\in \mathcal {R}(U)=\overline {{\mathcal {M}}}$ .

For any $x,y\in \overline {{\mathcal {M}}}$ , it follows from Lemma 2.14 that $x\otimes (Cy)+y\otimes (Cx)\in \overline {C_{{\mathcal {M}}}}.$ In view of (1), we obtain $UP_nAP_nU\in \overline {C_{{\mathcal {M}}}}\subset \overline {C_{{\mathcal {M}}}}^{w*}$ . This proves Claim and completes the proof.

Proof “(i) $\Longrightarrow $ (iii), (iv) and (viii)”. Since $\mathcal {R}(T)$ is norm-closed, one can easily verify that $C_{\mathcal {R}(T)}$ is closed in the wot and, by Theorem 1.2, we have $C_{\mathcal {R}(T)}=T\mathcal {S}_C T$ . Hence $T\mathcal {S}_C T$ is closed in the wot.

“(viii) $\Longrightarrow $ (vii)” and “(iii) $\Longrightarrow $ (ii)”. Both are obvious.

“(vii) $\Longrightarrow $ (i)”. Since $T\mathcal {S}_C T$ is norm closed, by Theorem 1.3, $ T\mathcal {S}_C T= \overline {C_{\mathcal {R}(T)}} \supseteq C_{\mathcal {R}(T)} $ . Hence, $T\mathcal {S}_C T=C_{\mathcal {R}(T)}$ and, by Theorem 1.2, $\mathcal {R}(T)$ is closed.

“(ii) $\Longrightarrow $ (i)”. Choose an $x\in \overline {\mathcal {R}(T)}$ . Then there exists a sequence $\{x_n\}\in \mathcal {R}(T)$ converging to x. Clearly, $x_n\otimes (Cx_n)\in C_{\mathcal {R}(T)}$ for all n. Note that $\{x_n\otimes (Cx_n)\}$ converges to $x\otimes (Cx)$ in norm. Since $C_{\mathcal {R}(T)}$ is norm-closed, we deduce that $x\otimes (Cx)\in C_{\mathcal {R}(T)}$ . Thus $x\in \mathcal {R}(T)$ . This shows that ${\mathcal {R}(T)}=\overline {\mathcal {R}(T)}.$

“(i) $\Longrightarrow $ (vi)”. By Theorem 1.2, we have $T\mathcal {S}_C T=C_{\mathcal {R}(T)}$ .

“(vi) $\Longrightarrow $ (v)”. Assume that $T\mathcal {S}_C T=C_{\mathcal {M}}$ for some subspace $\mathcal {M}$ of ${\mathcal {H}}$ . It is clear that $T\mathcal {S}_C T\subset C_{\mathcal {R}(T)}$ . So $C_{\mathcal {M}}\subset C_{\mathcal {R}(T)}$ . Then, for any $x\in \mathcal {M},$ we have $x\otimes (Cx)\in C_{\mathcal {M}}\subset C_{\mathcal {R}(T)}$ and $x\in \mathcal {R}(T)$ . This shows $\mathcal {M}\subset \mathcal {R}(T)$ .

On the other hand, if $x\in \mathcal {R}(T)$ , then, by Corollary 2.4, $x\otimes (Cx)\in T\mathcal {S}_C T=C_{\mathcal {M}}$ , which implies $x\in \mathcal {M}$ . Hence $\mathcal {R}(T)\subset \mathcal {M}$ . We conclude that $\mathcal {R}(T)=\mathcal {M}$ and $T\mathcal {S}_C T=C_{\mathcal {R}(T)}$ .

“(v) $\Longrightarrow $ (i)”. Assume that $C_{\mathcal {R}(T)}=A\mathcal {S}_C A$ . Since $A\mathcal {S}_C A\subset C_{\mathcal {R}(A)}$ , it follows that $C_{\mathcal {R}(T)} \subset C_{\mathcal {R}(A)}$ . By Proposition 2.10, we have $ {\mathcal {R}(T)} \subset {\mathcal {R}(A)}$ .

On the other hand, if $x\in \mathcal {R}(A)$ , then, by Corollary 2.4, $x\otimes (Cx)\in A\mathcal {S}_C A=C_{\mathcal {R}(T)}$ , which implies $x\in \mathcal {R}(T)$ . Hence $ {\mathcal {R}(A)} \subset {\mathcal {R}(T)}$ and ${\mathcal {R}(A)} ={\mathcal {R}(T)}$ . By Proposition 2.10, we have $A\mathcal {S}_C A=T\mathcal {S}_C T$ . So $C_{\mathcal {R}(T)}=T\mathcal {S}_C T$ . By Theorem 1.2, we deduce that $\mathcal {R}(T)$ is closed.

“(iv) $\Longrightarrow $ (i)”. Denote ${\mathcal {M}}=\mathcal {R}(T)$ . We choose an orthonormal basis $\{e_i\}_{i=1}^\infty $ of $\overline {\mathcal {M}}$ . Define $U\in \mathcal {B(H)}$ as

$$\begin{align*}U=\sum_{i=1}^\infty e_{i}\otimes (ce_{i}).\end{align*}$$

Thus $U\in \mathcal {B(H)}$ is a partial isometry, $\mathcal {R}(U)=\overline {\mathcal {M}}$ and $ \mathcal {R}(U^{*})=(\ker U)^\bot =C(\overline {\mathcal {M}})$ . It is easy to check that $U\in \mathcal {S}_C$ . So $U\in C_{\overline {\mathcal {M}}}$ .

By the hypothesis, we can find $\{T_n\}\subset C_{\mathcal {M}}$ such that $T_n\longrightarrow U$ .

Since $\mathcal {R}(T_n)\subset \mathcal {M} \subset \overline {\mathcal {M}}=\mathcal {R}(U)$ , by Lemma 2.1, we have

$$ \begin{align*}(\ker T_n)^\bot=\overline{{\mathcal{R}(T_n^{*})}}\subset \mathcal{R}(U^{*})=(\ker U)^\bot=C(\overline{\mathcal{M}}).\end{align*} $$

Denote $\widetilde {U}=U|_{C(\overline {\mathcal {M}})}$ and $\widetilde {T}_n=(T_n)|_{C(\overline {\mathcal {M}})}$ . Thus $\widetilde {U},\widetilde {T}_n: C(\overline {\mathcal {M}})\rightarrow \overline {\mathcal {M}}$ are bounded linear operators and $\widetilde {U}$ is invertible.

Since $T_n\rightarrow U$ , it follows that $\widetilde {T}_n\rightarrow \widetilde {U}$ . By the stability of invertibility, we deduce that $\widetilde {T}_n$ is invertible for n large enough. This implies that $\mathcal {R}(T_n)=\mathcal {R}(\widetilde {T}_n)=\overline {\mathcal {M}}$ for n large enough. Since $\mathcal {R}(T_n)\subset \mathcal {M}$ , we obtain $\overline {\mathcal {M}}=\mathcal {M}$ .

Proof The implications (iv) $\Longrightarrow $ (iii) $\Longrightarrow $ (ii) is obvious.

“(ii) $\Longrightarrow $ (iv)”. Using a similar argument as in the proof for (ii) $\Longrightarrow $ (i) of Proposition 2.15, one can see that $\mathcal {M}$ is closed.

“(i) $\Longrightarrow $ (iv)”. Assume that $C_{\mathcal {M}}=T\mathcal {S}_C T$ for some $T\in \mathcal {S}_C$ . From the proof for (vi) $\Longrightarrow $ (v) in Proposition 2.15, one can see that $T\mathcal {S}_C T=C_{\mathcal {R}(T)}$ and $\mathcal {M}=\mathcal {R}(T)$ . By Theorem 1.2, we deduce that ${\mathcal {M}}=\mathcal {R}(T)$ is closed.

“(iv) $\Longrightarrow $ (i)”. By Lemma 2.9, we can choose an operator $T\in \mathcal {S}_C$ with $\mathcal {R}(T)={\mathcal {M}}$ . Then, by Theorem 1.2, $T\mathcal {S}_C T=C_{\mathcal {R}(T)}=C_{\mathcal {M}}$ .

Proof Let E be the projection-valued spectral measure for T and assume that $\sigma (T)\subset [a,b]$ . We fix a positive integer n. Then there exist pairwise disjoint Borel subsets $\Delta _1,\Delta _2,\dots ,\Delta _n$ and $a_1,a_2,\dots ,a_n\in [a,b]$ such that

$$ \begin{align*}[a,b]=\cup_{i=1}^n \Delta_i \ \ \ \text{and}\ \ \ \Delta_i\subset[a_i-\frac{b-a}{2n}, a_i+\frac{b-a}{2n}].\end{align*} $$

For $i=1,2,\dots ,n$ , denote ${\mathcal {H}}_i=\mathcal {R} (E(\Delta _i))$ . Then each ${\mathcal {H}}_i$ reduces T and ${\mathcal {H}}=\oplus _{i=1}^n{\mathcal {H}}_i$ . Put $T_i=T|_{{\mathcal {H}}_i}$ . Then $T=\oplus _{i=1}^nT_i$ . For each i, choose $\lambda _i\in \Delta _i$ . Since $\sigma (T_i)\subset \Delta _i^-$ , it follows that $\|T_i-\lambda _i\|\leq (b-a)/n$ .

Fix an i with $i=1,\dots ,n$ . It suffices to prove $CE(\Delta _i)C=E(\Delta _i)$ . Note that there exists a sequence $\{p_k\}_{k=1}^\infty $ of polynomials with real coefficients such that $p_k(T)\rightarrow E(\Delta _i)$ in the strong operator topology. Since $Cp_k(T)C=p_k(T)$ for all k, it follows immediately that $CE(\Delta _i)C=E(\Delta _i)$ . This proves the claim.

For each i, denote $C_i=C|_{{\mathcal {H}}_i}$ . Then $C_i$ is a conjugation on ${\mathcal {H}}_i$ and, from $CTC=T^{*}=T$ , we obtain $C_iT_iC_i=T_i$ .

Since ${\mathcal {H}}=\oplus _{i=1}^n{\mathcal {H}}_i$ , we may assume that $e=\sum _{i=1}^n e_i$ with $e_i\in {\mathcal {H}}_i$ . Then, by Claim, $Ce_i\in {\mathcal {H}}_i$ . For each i with $1\leq i\leq n$ , put ${\mathcal {M}}_i=\vee \{e_i, Ce_i\}$ . Then ${\mathcal {M}}_i$ is a subspace of ${\mathcal {H}}_i$ , reducing C, and $1\leq \dim {\mathcal {M}}_i\leq 2$ . Then, relative to the decomposition ${\mathcal {H}}_i={\mathcal {M}}_i\oplus ({\mathcal {H}}_i\ominus {\mathcal {M}}_i)$ , $T_i$ can be written as

$$ \begin{align*}T_i=\begin{bmatrix} A_i&F_i\\ G_i&B_i \end{bmatrix}\begin{matrix} {\mathcal{M}}_i\\ {\mathcal{H}}_i\ominus {\mathcal{M}}_i \end{matrix}.\end{align*} $$

Note that $G_i=F_i^{*}$ , since $T_i$ is self-adjoint.

Since $C_i({\mathcal {M}}_i)={\mathcal {M}}_i$ , relative to the decomposition ${\mathcal {H}}_i={\mathcal {M}}_i\oplus ({\mathcal {H}}_i\ominus {\mathcal {M}}_i)$ , $C_i$ can be written as

$$ \begin{align*}C_i=\begin{bmatrix} C_{i,1}&0\\ 0&C_{i,2} \end{bmatrix}\begin{matrix} {\mathcal{M}}_i\\ {\mathcal{H}}_i\ominus {\mathcal{M}}_i \end{matrix}.\end{align*} $$

Then both $C_{i,1}$ and $C_{i,2}$ are conjugations. It follows readily from $C_iT_iC_i=T_i$ that $C_iK_iC_i=K_i$ , where

$$ \begin{align*}K_i:=\begin{bmatrix} 0&F_i\\ G_i&0 \end{bmatrix}\begin{matrix} {\mathcal{M}}_i\\ {\mathcal{H}}_i\ominus {\mathcal{M}}_i \end{matrix}.\end{align*} $$

Clearly, $K_i$ is self-adjoint, $\text {rank} K_i\leq 4$ and

$$ \begin{align*} \|K_i\|&=\max\{ \|F_i\|, \|G_i\| \} \leq \left\|\begin{bmatrix} A_i-\lambda_i&F_i\\ G_i&B_i-\lambda_i \end{bmatrix}\right\| =\|T_i-\lambda_i\|\leq(b-a)/n. \end{align*} $$

Set $K=-\oplus _{i=1}^n K_i$ . Thus K is a self-adjoint, finite-rank operator, $CKC=K$ ,

$$ \begin{align*}T+K=\oplus_{i=1}^n (T_i-K_i)=\oplus_{i=1}^n (A_i\oplus B_i)\end{align*} $$

and

$$ \begin{align*}\|K\|_p\leq (\text{rank} K)^{1/p} \|K\| \leq (4n)^{1/p}\frac{b-a}{n}\leq \frac{4(b-a)}{n^{1/q}},\end{align*} $$

where $q=1/(1-1/p)$ . Thus, for any $\varepsilon>0$ , there exists n large enough such that $\|K\|_p<\varepsilon $ .

Denote by P the orthogonal projection of ${\mathcal {H}}$ onto $\oplus _{i=1}^n {\mathcal {M}}_i$ . Thus $e\in \mathcal {R}(P)$ , $\text {rank} P\leq 2n$ and $(T+K)P=P(T+K)$ ; indeed,

$$ \begin{align*}(T+K)|_{\mathcal{R}(P)}=\oplus_{i=1}^n A_i.\end{align*} $$

Noting that $CPC=P$ and $CKC=K$ , that are, $P,K\in \mathcal {S}_C$ , we complete the proof.

Proof of Theorem 1.6

Since C is a conjugation, we can find an orthonormal basis $\{e_n\}_{n=1}^\infty $ of ${\mathcal {H}}$ such that $Ce_n=e_n$ for all n.

Now fix an $\varepsilon>0$ and $p>1$ . By Lemma 3.1, we can find a finite-rank, self-adjoint operator $K_1\in \mathcal {S}_C$ with $\|K_1\|_p<\varepsilon /2$ and a finite-rank orthogonal projection $P_1\in \mathcal {S}_C$ such that $e_1\in \mathcal {R}(P_1)$ and $P_1(T+K_1)=(T+K_1)P_1$ . Denote ${\mathcal {H}}_1=\mathcal {R}(P_1)$ and $\widetilde {{\mathcal {H}}_1}={\mathcal {H}}\ominus {\mathcal {H}}_1$ . Then, with respect to the decomposition ${\mathcal {H}}={\mathcal {H}}_1\oplus \widetilde {{\mathcal {H}}_1}$ ,

(5) $$ \begin{align} T+K_1=\begin{bmatrix} T_1&0\\ 0& \widetilde{T_1} \end{bmatrix},\ \ \ C=\begin{bmatrix} C_1&0\\ 0& \widetilde{C_1} \end{bmatrix}, \end{align} $$

where $C_1T_1C_1=T_1$ and $\widetilde {C_1}\widetilde {T_1}\widetilde {C_1}=\widetilde {T_1}$ .

Now apply Lemma 3.1 again to the self-adjoint operator $\widetilde {T_1}$ and the vector $(I-P_1)e_2$ to get a finite-rank, self-adjoint operator $\widetilde {K_2}\in \mathcal {S}_{\widetilde {C_1}}$ and two subspaces ${\mathcal {H}}_2,\widetilde {{\mathcal {H}}_2}$ of $\widetilde {{\mathcal {H}}_1}$ such that

$$ \begin{align*}\|\widetilde{K_2}\|_p<\varepsilon/4, \ \ (I-P_1)e_2\in{\mathcal{H}}_2, \ \ \dim{\mathcal{H}}_2<\infty, \ \ \widetilde{{\mathcal{H}}_1}={\mathcal{H}}_2\oplus\widetilde{{\mathcal{H}}_2}\end{align*} $$

and with respect to which

$$\begin{align*}\widetilde{T_1}+\widetilde{K_2}=\begin{bmatrix} T_2&0\\ 0& \widetilde{T_2} \end{bmatrix},\ \ \ \widetilde{C_1}=\begin{bmatrix} C_2&0\\ 0& \widetilde{C_2} \end{bmatrix},\end{align*}$$

where $C_2T_2C_2=T_2$ and $\widetilde {C_2}\widetilde {T_2}\widetilde {C_2}=\widetilde {T_2}$ . In view of (5), we can find a finite-rank, self-adjoint operator $K_2\in \mathcal {S}_C$ with $\|K_2\|_p<\varepsilon /4$ such that

(6) $$ \begin{align} T+K_1+K_2=\begin{bmatrix} T_1&0&0\\ 0&T_2&0\\ 0& 0&\widetilde{T_2} \end{bmatrix}\begin{matrix} {\mathcal{H}}_1\\ {\mathcal{H}}_2\\ \widetilde{{\mathcal{H}}_2} \end{matrix},\ \ \ C=\begin{bmatrix} C_1&0&0\\ 0&C_2&0\\ 0&0 &\widetilde{C_2} \end{bmatrix}\begin{matrix} {\mathcal{H}}_1\\ {\mathcal{H}}_2\\ \widetilde{{\mathcal{H}}_2} \end{matrix}. \end{align} $$

Note that $e_1,e_2\in {\mathcal {H}}_1\oplus {\mathcal {H}}_2$ , $C_iT_iC_i=T_i$ $(i=1,2),$ and $\widetilde {C_2}\widetilde {T_2}\widetilde {C_2}=\widetilde {T_2}$ .

By induction, we can find a sequence of self-adjoint, finite-rank operators $\{K_i\}_{i=1}^\infty \subset \mathcal {S}_C$ and a sequence of pairwise orthogonal, finite-dimensional subspaces $\{{\mathcal {H}}_i\}$ such that for each $n\geq 1,$

1. (i) $e_1,\dots ,e_n\in \oplus _{i=1}^n{\mathcal {H}}_i$ ,

2. (ii) $\|K_n\|_p<\varepsilon /2^n$ , and

3. (iii) relative to the decomposition ${\mathcal {H}}=(\oplus _{i=1}^n {\mathcal {H}}_i)\oplus \widetilde {{\mathcal {H}}_n}$ ,

   (7) $$ \begin{align} T+\sum_{i=1}^nK_i=(\oplus_{i=1}^n T_i)\oplus \widetilde{T_n},\ \ \ \ C=(\oplus_{i=1}^n C_i)\oplus \widetilde{C_n}, \end{align} $$
   where $\widetilde {{\mathcal {H}}_n}={\mathcal {H}}\ominus (\oplus _{i=1}^n {\mathcal {H}}_i)$ and
   $$ \begin{align*}\widetilde{C_n}\widetilde{T_n}\widetilde{C_n}=\widetilde{T_n}, \ \ \ C_iT_iC_i=T_i, \ \ \ i=1,2,\dots, n. \end{align*} $$

Since $\{e_i\}$ is an orthonormal basis of ${\mathcal {H}}$ , it follows from statement (i) that ${\mathcal {H}}=\oplus _{i=1}^\infty {\mathcal {H}}_i$ and hence $C=\oplus _{i=1}^\infty C_i$ . Set $K=\sum _{i=1}^\infty K_i$ . Then $K\in \mathcal {B}_p\mathcal {(H)}$ is self-adjoint, $\|K\|_p<\varepsilon $ and $T+K=\oplus _{i=1}^\infty T_i$ . Note that each $T_i$ is self-adjoint and acting on a finite-dimensional space. Thus $T+K$ is diagonal. Since $C_iT_iC_i=T_i$ for all i, it follows that $C(T+K)C=T+K$ . Set $D=T+K$ . This completes the proof.

Proof Assume that $\sigma _p(T)=\{\lambda _i: i\in \Lambda \}$ . Set

$$ \begin{align*}{\mathcal{H}}_0=\vee_{i\in\Lambda}\ker(T-\lambda_i)\ \ \ \text{and}\ \ \ {\mathcal{H}}_1={\mathcal{H}}_0^\bot.\end{align*} $$

Since T is self-adjoint, we deduce that ${\mathcal {H}}_0$ reduces T. Then

$$ \begin{align*}T=\begin{bmatrix} T_0&0 \\ 0&T_1 \end{bmatrix}\begin{matrix} {\mathcal{H}}_0\\ {\mathcal{H}}_1 \end{matrix}.\end{align*} $$

Clearly, $T_0$ is diagonal and $\sigma _p(T_1)=\emptyset $ . Since $CTC=C$ , one can see $C({\mathcal {H}}_0)={\mathcal {H}}_0$ , which implies $C({\mathcal {H}}_1)={\mathcal {H}}_1$ and

$$ \begin{align*}C=\begin{bmatrix} C_0&0 \\ 0&C_1 \end{bmatrix}\begin{matrix} {\mathcal{H}}_0\\ {\mathcal{H}}_1 \end{matrix}.\end{align*} $$

It is easy to check that each $C_i$ is a conjugation and $T_i\in {\mathcal S}_{C_i}$ .

By Theorem 1.6, we can find $K_1\in {\mathcal S}_{C_1}$ with $\|K_1\|<\frac {\varepsilon }{2}$ such that $T_1+K_1$ is diagonal. Moreover, by the upper semi-continuity of spectrum, it can be required that $\sigma (T_1+K_1)\subset \sigma (T_1)+B(0,\frac {\varepsilon }{2})$ .

Since $T_1+K_1$ is diagonal, without loss of generality, we may assume that $\sigma _p(T_1+K_1)=\{z_i: i\geq 1\}$ . Denote $\mathcal {K}_i=\ker (T_1+K_1-z_i)$ for $i\geq 1$ . Then

$$ \begin{align*}T_1+K_1=\oplus_{i=1}^\infty z_i I_i\end{align*} $$

relative to the decomposition ${\mathcal {H}}_1=\oplus _{i=1}^\infty \mathcal {K}_i$ , where $I_i$ is the identity operator on $\mathcal {K}_i$ . Since $T_1+K_1\in \mathcal {S}_C$ is diagonal, it follows that

$$ \begin{align*}C_1(\ker(T_1+K_1-z_i))=\ker(T_1+K_1-z_i),\ \ \ \ i=1,2,3,\dots.\end{align*} $$

Then each $\mathcal {K}_i$ reduces $C_1$ . We may assume that $C_1=\oplus _{i=1}^\infty C_i'$ , where $C_i'=C_1|_{\mathcal {K}_i}$ is a conjugation on $\mathcal {K}_i$ .

Fix an $i\geq 1$ . If $z_i\in \sigma (T_1)$ , then we set $\widetilde {z_i}=z_i$ . If $z_i\notin \sigma (T_1)$ , then $z_i\notin \sigma _e(T_1)=\sigma _e(T_1+K_1)$ and hence $\dim \mathcal {K}_i<\infty $ ; noting that $z_i\in \sigma (T_1)+B(0,\frac {\varepsilon }{2})$ , we can find $\widetilde {z_i}\in \sigma (T_1)$ such that $|z_i-\widetilde {z_i}|=\text {dist}(z_i,\sigma (T_1))<\varepsilon /2$ .

Set $K_2=\oplus _{i=1}^\infty (\widetilde {z_i}-z_i) I_i $ . Then $K_2\in {\mathcal S}_{C_1}$ , $\|K_2\|\leq \varepsilon /2$ and

$$ \begin{align*}T_1+K_1+K_2=\oplus_{i=1}^\infty \widetilde{z_i} I_i.\end{align*} $$

Clearly, $\sigma (T_1+K_1+K_2)=\{\widetilde {z_i}: i\geq 1 \}^-\subseteq \sigma (T_1)$ .

It suffices to show $\lim _i|z_i-\widetilde {z_i}|=0.$ Otherwise, we can choose a subsequence $\{i_k\}_{k=1}^\infty $ of $\mathbb {N}$ such that $z_{i_k}\notin \sigma (T_1)$ , $z_{i_k}\rightarrow z_0$ and

$$ \begin{align*}\inf_k \text{dist}(z_{i_k},\sigma(T_1))>0.\end{align*} $$

Hence, $\text {dist}(z_{0},\sigma (T_1))>0$ and

$$ \begin{align*}z_0\in\sigma_e(T_1+K_1)=\sigma_e(T_1)\subset \sigma(T_1),\end{align*} $$

a contradiction.

Set $K=0\oplus (K_1+K_2)$ . Then $K\in \mathcal {S}_C$ is compact with $\|K\|<\varepsilon $ and

$$ \begin{align*}T+K=T_0\oplus (T_1+K_1+K_2).\end{align*} $$

By the preceding discussion, we have

$$ \begin{align*}\sigma(T+K)=\sigma(T_0)\cup \sigma(T_1+K_1+K_2)\subset \sigma(T_0)\cup \sigma(T_1)= \sigma(T).\end{align*} $$

It remains to show $\sigma (T)\subset \sigma (T+K)$ .

Note that K is compact. Hence

$$ \begin{align*}\sigma_e(T)=\sigma_e(T+K)\subset \sigma(T+K).\end{align*} $$

If $z\in \sigma (T)\setminus \sigma _e(T)$ , then $z\in \sigma _p(T)$ and $z=\lambda _i\in \sigma (T_0)\subset \sigma (T+K)$ for some $i\in \Lambda $ . Thus we have proved that $\sigma (T)\subset \sigma (T+K)$ , which implies $\sigma (T)= \sigma (T+K)$ .

Proof Denote by ${\mathcal {A}}$ the von Neumann algebra generated by $N_1$ , $\dots $ , $N_n$ . It is easy to see $\mathcal {A}\subset \mathcal {S}_C$ . By [Reference Davidson5, Lemma II.2.8], we can find a self-adjoint operator $A\in \mathcal {B(H)}$ generating ${\mathcal {A}}$ and $N_1$ , $\dots $ , $N_n\in C^{*}(A)$ . Clearly, $A\in \mathcal {S}_C$ . We can find continuous functions $f_1,\dots ,f_n$ on $\mathbb {R}$ such that $N_i=f_i(A)$ , $i=1,\dots ,n.$

Now fix an $\varepsilon>0$ . In view of [Reference Conway4, Lemma 39.5], there exists $\delta>0$ such that $\sup _{1\leq i\leq n}\|f_i(X)-f_i(Y)\|<\varepsilon $ for all self-adjoint operators $X,Y$ satisfying $\|X\|\leq \|A\|, \|Y\|\leq \|A\|$ and $\|X-Y\|<\delta $ .

By Corollary 3.2, we can find self-adjoint, diagonal $D\in \mathcal {S}_C$ with $\sigma (D)=\sigma (A)$ such that $D-A\in \mathcal {K(H)}$ and $\|D-A\|< \delta $ . Then $\|D\|=\|A\|$ and

$$ \begin{align*}\sup_{1\leq i\leq n}\|N_i-f_i(D)\|=\sup_{1\leq i \leq n}\|f_i(A)-f_i(D)\|<\varepsilon.\end{align*} $$

Note that $f_1(D)$ , $f_2(D)$ , $\dots $ , $f_n(D)$ are commuting diagonal operators. Also, since $D\in \mathcal {S}_C$ , it follows that $f_1(D),f_2(D)$ , $\dots $ , $f_n(D)$ $\in $ $\mathcal {S}_C$ .

Note that $A-D\in \mathcal {K(H)}$ . Using the functional calculus for self-adjoint operators, one can easily prove that $f_i(A)-f_i(D)\in \mathcal {K(H)}$ , $i=1,2,\dots ,n$ . This completes the proof.