Proof of Theorem 1.2

The proof is divided into two steps.

Step 1. $w(CT)=\|A\|$ .

We denote $B=\frac {T-CT^*C}{2}.$ One can check that $T=A+B$ , $CAC=A^*$ and ${CBC=-B^*}$ . For each $x\in {\mathcal {H}}$ , we have

$$ \begin{align*}\langle CBx,x\rangle=\langle Cx,Bx\rangle=\langle B^*Cx,x\rangle=-\langle CBx,x\rangle,\end{align*} $$

which implies that $\langle CBx,x\rangle =0.$ Hence

$$\begin{align*}\langle CTx,x\rangle=\langle CAx,x\rangle+\langle CBx,x\rangle=\langle CAx,x\rangle,\end{align*}$$

which implies that $W(CT)=W(CA)$ and $w(CT)=w(CA)$ . Denote $\lambda =\|A\|$ . By Lemma 2.1, we can find a sequence of unit vectors $f_n$ such that $\|Af_n-\lambda Cf_n\|\rightarrow 0$ . Thus

$$ \begin{align*} &|\langle CAf_n,f_n\rangle| = |\langle Cf_n,Af_n\rangle|\\ =&|\langle Cf_n,\lambda Cf_n\rangle+\langle Cf_n,(A-\lambda C)f_n\rangle|\rightarrow\lambda=\|A\|, \end{align*} $$

which implies that $w(CA)\geq \|A\|$ . Since the converse is obvious, we conclude that $w(CT)=w(CA)=\|A\|$ .

Note that $W(CT)$ is circularly symmetric. It follows from $w(CT)=\|A\|$ that either $W(CT)=\left \{z\in \mathbb {C}: |z|<\|A\|\right \}$ or $W(CT)=\left \{z\in \mathbb {C}: |z|\leq \|A\|\right \}$ .

Step 2. $W(CT)=\left \{z\in \mathbb {C}: |z|\leq \|A\|\right \}$ if and only if A is norm-attaining.

“ $\Longrightarrow $ .” By Step 1, we have $W(CT)=W(CA)$ . It follows that $\|A\|\in W(CA)$ . Then there exists a unit vector $x\in {\mathcal {H}}$ such that

$$ \begin{align*}\langle Cx,Ax\rangle=\langle CAx,x\rangle=\|A\|.\end{align*} $$

Thus $\|A\|=\langle Cx,Ax\rangle \leq \|Ax\|$ , which implies $\|A\|=\|Ax\|$ .

“ $\Longleftarrow $ .” We assume that $\|Ax\|=\|A\|$ for some unit vector $x\in {\mathcal {H}}$ . This implies that $\||A|x\|=\|A\|=\||A|\|$ . Thus $\|A\|$ is an eigenvalue of $|A|$ . By Lemma 2.1, there exists a unit vector $f\in {\mathcal {H}}$ such that $Af=\|A\| Cf$ . Thus

$$ \begin{align*}\langle CAf,f\rangle = \langle Cf,Af\rangle = \langle Cf,\|A\| Cf\rangle = \|A\|,\end{align*} $$

which implies that $\|A\|\in W(CA)=W(CT)$ . Since $W(CT)$ is circularly symmetric, we conclude that $W(CT)=\left \{z\in \mathbb {C}: |z|\leq \|A\|\right \}$ . This completes the proof.

Proof of Theorem 1.1

Choose a conjugation C on ${\mathcal {H}}$ and set $T=CR$ . Then $T\in \mathcal {B(H)}$ and $R=CT$ . By Theorem 1.1, we have either

$$ \begin{align*}W(R)= \left\{z\in\mathbb{C}: |z|< \frac{\|T+CT^*C\|}{2}\right\}\end{align*} $$

or

$$ \begin{align*}W(R)=\left\{z\in\mathbb{C}: |z|\leq\frac{\|T+CT^*C\|}{2}\right\},\end{align*} $$

and the latter holds if and only if $T+CT^*C$ is norm-attaining. Note that $R=CT$ and $R^\#=T^*C$ . Thus

$$ \begin{align*}\frac{T+CT^*C}{2}=\frac{C(CT+T^*C)}{2}=\frac{C(R+R^\#)}{2}.\end{align*} $$

This implies that $T+CT^*C$ is norm-attaining if and only if so is $R+R^\#$ . Thus the desired result follows readily.

Proof (i) Assume that $\varphi (z)=\sum _{k=-\infty }^\infty a_k z^k$ and denote $\theta =\frac {\varphi +\psi }{2}$ . For $i,j\geq 0$ , compute to see that $\langle T_\psi z^i,z^j\rangle = \left \langle \sum _{k} a_kz^{-k+i},z^j\right \rangle =a_{i-j}$ and

$$ \begin{align*} \langle CT_\varphi^* C z^i,z^j\rangle=\langle CT_\varphi^* z^i,z^j\rangle=\langle C z^j,T_\varphi^* z^i\rangle=\langle z^j,T_\varphi^* z^i\rangle=\langle T_\varphi z^j, z^i\rangle=a_{i-j}. \end{align*} $$

It follows that $CT_\varphi ^* C= T_\psi $ . Then

$$ \begin{align*}\frac{T_\varphi+CT_\varphi^*C}{2}=\frac{T_\varphi+T_\psi}{2}= T_\theta\end{align*} $$

and, by Theorem 1.2, we have $w(CT_\varphi ) =\frac {\|T_\varphi +CT_\varphi ^*C\|}{2}=\|T_\theta \| = \|\theta \|_\infty .$ Moreover, $W(CT_\varphi )$ is closed if and only if $T_\theta $ is norm-attaining.

(ii) Still we denote $\theta =\frac {\varphi +\psi }{2}$ .

“ $\Longleftarrow $ .” If $\theta (z)=\beta \zeta (z)\zeta (\overline {z})$ for some inner function $\zeta $ and some complex number $\beta $ , then it is easy to check that $\|\theta \|_\infty =|\beta |$ , $\overline {\zeta (\overline {z})}\in H^2$ with norm one and

$$ \begin{align*}\|T_\theta (\overline{\zeta(\overline{z})})\|=\|\beta\zeta\|=|\beta|=\|\theta\|_\infty=\|T_\theta\|.\end{align*} $$

That is, $T_\theta $ is norm-attaining, which implies $W(CT_\varphi )$ is closed.

“ $\Longrightarrow $ .” Assume that $W(CT_\varphi )$ is closed. By Theorem 1.2, we deduce that $T_\theta $ is norm-attaining. Without loss of generality, we may assume that $\|T_\theta \|=1$ . By Lemma 2.2, there are inner functions $\theta _1, {\theta _2}$ such that $\theta =\theta _1\overline {\theta _2}$ and

$$ \begin{align*}\{x\in H^2: \|T_\theta x\|=\|x\|\}=\theta_2 H^2.\end{align*} $$

Denote $\widetilde {\theta }_1(z)=\overline {\theta _1(\overline {z})}$ and $\widetilde {\theta }_2(z)=\overline {\theta _2(\overline {z})}$ . Then

$$ \begin{align*}\theta(\overline{z})=\theta_1(\overline{z})\overline{\theta_2(\overline{z})}=\overline{\widetilde{\theta}_1({z})} {\widetilde{\theta}_2({z})}.\end{align*} $$

Denote $\widetilde {\theta }(z)=\theta (\overline {z})$ . Then, using Lemma 2.2 again, we obtain

$$ \begin{align*}\{x\in H^2: \|T_{\widetilde{\theta}} x\|=\|x\|\}=\widetilde{\theta}_1 H^2.\end{align*} $$

On the other hand, since $\theta (z)=\frac {\varphi (z)+\psi (z)}{2}=\frac {\varphi (z)+\varphi (\overline {z})}{2}=\theta (\overline {z})=\widetilde {\theta }(z)$ , we deduce that $\theta _2 H^2=\widetilde {\theta }_1 H^2$ . Thus we can find a unimodular number $\beta $ such that

$$ \begin{align*}\theta_2(z)=\beta\widetilde{\theta}_1(z)=\beta\overline{\theta_1(\overline{z})}.\end{align*} $$

It follows that $\frac {\varphi (z)+\varphi (\overline {z})}{2}=\theta (z)=\overline {\beta }\theta _1({z}){\theta _1(\overline {z})}$ .

Proof (i) Let $T=UP$ be the polar decomposition of T, where $P=|T|$ and U is a partial isometry. Denote $r=\|T\|$ .

Case 1. $r\in \sigma _p(P)$ .

Choose a unit vector $x\in \ker (P-r)$ and set $y=Ux$ . Then $\|y\|=1$ . By Lemma 3.1, we can construct a conjugation C on ${\mathcal {H}}$ such that $Cx=y$ . Hence

$$ \begin{align*}\langle CTx,x\rangle=\langle Cx,Tx\rangle=\langle Cx,UPx\rangle=\langle y,UPx\rangle=r\langle y,Ux\rangle=r\langle y,y\rangle=r= \|T\|,\end{align*} $$

which means that $w(CT)=\|T\|$ .

Case 2. $r\notin \sigma _p(P)$ .

Since $r=\|P\|$ and $r\in \sigma (P)$ , we deduce that r is an accumulation point of $\sigma (P)$ . Denote by $E_P(\cdot )$ the projection-valued spectral measure associated with P. Then we can find pairwise disjoint nonempty subsets $\Delta _1, \Delta _2, \Delta _3, \ldots $ of $\sigma (P)\setminus \{r\}$ such that ${\mathcal {H}}_i:=\text {ran~} E_P(\Delta _i)\ne \{0\}$ and $\sup \Delta _{i} \leq \inf \Delta _{i+1}\rightarrow r$ for all $i\geq 1$ . Clearly, ${\mathcal {H}}=\oplus _{i=1}^\infty {\mathcal {H}}_i$ . For each i, choose a unit vector $e_i\in {\mathcal {H}}_i$ and denote by $P_i$ the projection of ${\mathcal {H}}$ onto $\oplus _{j=1}^i {\mathcal {H}}_j$ . Then one can check that $\|Pe_i\|\rightarrow r$ and $P_i{\longrightarrow } I$ in the strong operator topology.

Set $n_1=1$ . Denote by $Q_1$ the projection of ${\mathcal {H}}$ onto $\vee \{e_1,Ue_1\}$ . Here $\vee $ denotes the closed linear span.

Since both $\{e_i\}_{i=2}^\infty $ and $\{Ue_i\}_{i=2}^\infty $ are orthonormal, they converge to $0$ in the weak topology. Note that $Q_1$ is of finite rank. Then we deduce that $\lim _iQ_1e_i=0$ and $\lim _iQ_1Ue_i=0$ . We can choose $n_2>n_1$ such that $\|Q_1e_{n_2}\|+\|Q_1Ue_{n_2}\|<\frac {1}{2}$ .

Denote by $Q_2$ the projection of ${\mathcal {H}}$ onto $\vee \{e_i,Ue_i: i=n_1, n_2\}$ . Clearly, $Q_2$ is of finite rank. Then we can choose $n_3>n_2$ such that $\|Q_2e_{n_3}\|+\|Q_2Ue_{n_3}\|<\frac {1}{2^2}$ .

Recursively, we can choose a subsequence $\{n_i\}_{i=1}^\infty $ of positive integers such that $\|Q_ie_{n_{i+1}}\|+\|Q_iUe_{n_{i+1}}\|<\frac {1}{2^i}$ for $i=1,2,\ldots $ , where $Q_i$ the projection of ${\mathcal {H}}$ onto $\vee \{e_j,Ue_j: j=n_1, n_2,\ldots , n_i \}$ .

Denote ${\mathcal {K}}_1=\text {ran~} Q_1$ and ${\mathcal {K}}_i=\text {ran~} Q_i\ominus \text {ran~} Q_{i-1}$ for each $i\geq 2$ . Moreover, denote ${\mathcal {K}}_0={\mathcal {H}}\ominus \left (\oplus _{i\geq 1}{\mathcal {K}}_i\right ).$ Denote

$$ \begin{align*}x_1=e_{n_1}, \ \ y_1=Ue_{n_1}; \ \ x_i=\frac{(I-Q_{i-1})e_{n_i}}{\|(I-Q_{i-1})e_{n_i}\|}, \ \ y_i=\frac{(I-Q_{i-1})Ue_{n_i}}{\|(I-Q_{i-1})Ue_{n_i}\|}, \ \ i\geq 2.\end{align*} $$

For each $i\geq 1$ , define a conjugation $C_i$ on ${\mathcal {K}}_i$ satisfying $C_i x_i=y_i$ . Arbitrarily choose a conjugation $C_0$ on ${\mathcal {K}}_0$ and set $C=\oplus _{i=0}^\infty C_i$ . Then C is a conjugation on ${\mathcal {H}}$ . Now it remains to check that $w(CT)\geq \|T\|$ .

Compute to see that

$$ \begin{align*} \langle CTx_i, x_i\rangle= \langle Cx_i, Tx_i\rangle =\langle Cx_i, UPx_i\rangle= \langle y_i, UPx_i\rangle=\langle U^*y_i,Px_i\rangle. \end{align*} $$

Note that $\|x_i-e_{n_i}\|+\|y_i-Ue_{n_i}\|\rightarrow 0$ . Hence $\lim _i|\langle U^*y_i,Px_i\rangle - \langle U^*Ue_{n_i},Pe_{n_i}\rangle |=0.$ Since it is obvious that $\langle U^*Ue_{n_i},Pe_{n_i}\rangle =\langle e_{n_i},Pe_{n_i}\rangle \rightarrow \|P\|=r$ . We conclude that

$$ \begin{align*}\langle CTx_i, x_i\rangle=\langle U^*y_i,Px_i\rangle\rightarrow r.\end{align*} $$

This shows that $w(CT)\geq \|T\|$ , which completes the proof.

(ii) If $T=0$ , then the result is clear. In the sequel, we only consider the case that $T\ne 0$ .

“ $\Longrightarrow $ .” Assume that C is a conjugation on ${\mathcal {H}}$ such that $W(CT)=B(0,\|T\|)^-$ . Then there exists a unit vector $x\in {\mathcal {H}}$ such that $\langle CTx,x\rangle =\|T\|$ . Noting that $|\langle CTx,x\rangle |=|\langle Cx,Tx\rangle |\leq \|Tx\|$ , we obtain $\|T\|\leq \|Tx\|$ , which implies $\|T\|=\|Tx\|$ .

“ $\Longleftarrow $ .” We assume that $\|Tx\|=\|T\|$ for some unit vector $x\in {\mathcal {H}}$ . Set $y=\frac {Tx}{\|Tx\|}$ . By Lemma 3.1, we can find a conjugation C on ${\mathcal {H}}$ such that $Cx=y$ . Then

$$ \begin{align*}\langle CTx, x\rangle=\langle Cx, Tx\rangle=\langle \frac{Tx}{\|Tx\|}, Tx\rangle=\|Tx\|=\|T\|.\end{align*} $$

It follows that $\|T\|\in W(CT)$ and $w(CT)=\|T\|$ . Since $W(CT)$ is circularly symmetric, we conclude that $W(CT)=B(0,\|T\|)^-$ .

Proof of Theorem 1.4

Without loss of generality, we assume that $T\ne 0$ . Denote $\Gamma =\{w(CT): C\in {\mathcal {B}_{c}(\mathcal {H})}\}$ . The proof is reduced to proving several claims.

Claim 1. $\max \Gamma =\|T\|$ .

For each unit vector $x\in {\mathcal {H}}$ , we have $|\langle CTx,x\rangle |\leq \|CTx\|\leq \|T\|$ , which implies $\sup \Gamma \leq \|T\|$ . From Lemma 3.2 (i), one can see that $\|T\|\in \Gamma $ and $\sup \Gamma \geq \|T\|$ . Hence we deduce that $\max \Gamma =\|T\|$ .

Claim 2. $\inf \Gamma =\kappa (T)$ .

For each $C\in {\mathcal {B}_{c}(\mathcal {H})}$ , note that $C(\frac {T-CT^*C}{2})C=\frac {CTC-T^*}{2}=-(\frac {T-CT^*C}{2})^*$ , that is, $\frac {T-CT^*C}{2}\in {SSO}$ . Then, by Theorem 1.2, we have

$$ \begin{align*}w(CT)=\frac{\|T+CT^*C\|}{2}=\left\|T-\frac{T-CT^*C}{2}\right\|\geq\text{dist}(T,{SSO}).\end{align*} $$

Since C is arbitrary, it follows that $\inf \Gamma \geq \text {dist}(T, {SSO})=\kappa (T)$ .

On the other hand, we choose an operator $A\in {SSO}$ . Thus $A=-CA^*C$ for some $C\in {\mathcal {B}_{c}(\mathcal {H})}$ . For $x\in {\mathcal {H}}$ , we have

$$ \begin{align*}\langle CAx,x\rangle=\langle Cx,Ax\rangle=\langle A^*Cx,x\rangle=-\langle CAx,x\rangle,\end{align*} $$

which implies that $\langle CAx,x\rangle =0$ and hence

$$ \begin{align*}\langle C(T-A)x,x\rangle=\langle CTx,x\rangle-\langle CAx,x\rangle=\langle CTx,x\rangle.\end{align*} $$

It follows that

$$ \begin{align*}\inf\Gamma\leq w(CT)=w(C(T-A))\leq\|T-A\|.\end{align*} $$

Since A was arbitrarily chosen in $ {SSO}$ , we conclude that $\inf \Gamma \leq \text {dist}(T, {SSO})=\kappa (T).$ This proves Claim 2.

Claim 3. ${\mathcal {B}_{c}(\mathcal {H})}$ is arcwise connected.

Choose a conjugation $C_0$ on ${\mathcal {H}}$ . Then, by [Reference Garcia and Putinar10, Lemma 1], we can find an orthonormal basis $\{e_n\}_{n=1}^\infty $ such that $C_0e_n=e_n$ for all n. If $C\in {\mathcal {B}_{c}(\mathcal {H})}$ and $\{f_n\}_{n=1}^\infty $ is an orthonormal basis of ${\mathcal {H}}$ such that $Cf_n=f_n$ for all n, then we can find a unitary operator $U\in \mathcal {B(H)}$ such that $Uf_n=e_n$ for all n. Then one can verify that $U^*C_0U=C$ . This shows that ${\mathcal {B}_{c}(\mathcal {H})}=\{U^*C_0U: U\in \mathcal {B(H)}$ is unitary $\}$ . Since all unitary operators on ${\mathcal {H}}$ constitute an arcwise connected subset of $\mathcal {B(H)}$ , it follows that ${\mathcal {B}_{c}(\mathcal {H})}$ is arcwise connected.

Claim 4. $\left (\kappa (T),\|T\|\right ]\subset \Gamma $ .

We define $\Phi : {\mathcal {B}_{c}(\mathcal {H})} \longrightarrow [0,\infty )$ as $\Phi (C)=w(CT)$ for $C\in {\mathcal {B}_{c}(\mathcal {H})}$ . By Theorem 1.2, we have $\Phi (C)=\frac {\|T+CT^*C\|}{2}$ for $C\in {\mathcal {B}_{c}(\mathcal {H})}$ . This shows that $\Phi $ is continuous. Since ${\mathcal {B}_{c}(\mathcal {H})}$ is connected, it follows that

$$ \begin{align*}\Phi({\mathcal{B}_{c}(\mathcal{H})})=\{w(CT): C\in{\mathcal{B}_{c}(\mathcal{H})}\}\subset[0,\infty)\end{align*} $$

is connected. In view of Claims 1 and 2, we conclude that Claim 4 holds. This completes the proof.

Proof Arbitrarily choose a conjugation C on ${\mathcal {H}}$ . Then we can find a unit vector x such that $Cx=x$ . Thus

$$ \begin{align*}|\langle CTx,x\rangle|=|\langle Cx,Tx\rangle|=|\langle x,Tx\rangle|\geq \inf\{|z|:z\in W(T)\}.\end{align*} $$

Since C is arbitrary, we deduce that $\kappa (T)\geq \inf \{|z|:z\in W(T)\}$ .

If T is positive, then it is well known that $\overline {W(T)}=[\inf \sigma (T),\|T\|]$ . Then the desired result follows readily.

Proof “ $\Longrightarrow $ .” Since $\kappa (N)=0$ , that is, $N\in \overline { {SSO}}$ , we can choose $\{T_n\}_{n=1}^\infty \subset \mathcal {B(H)}$ and $\{C_n\}_{n=1}^\infty \subset {\mathcal {B}_{c}(\mathcal {H})}$ such that $T_n+C_nT_n^*C_n=0$ for all n and $T_n\rightarrow N$ . It follows immediately that $N+C_nN^*C_n\rightarrow 0$ . For each $\lambda \in \mathbb {C}$ , we have

(3.1) $$ \begin{align} C_n(N+\lambda)^*C_n\rightarrow -(N-\lambda),\ \ \ \ \ -C_n(N-\lambda)C_n\rightarrow (N+\lambda)^*, \end{align} $$

which implies that $N-\lambda $ is invertible if and only if $N+\lambda $ is invertible. Hence $\sigma (N)=-\sigma (N)$ ; moreover, $\lambda \in \operatorname {iso}\sigma (N)$ if and only if $-\lambda \in \operatorname {iso}\sigma (N)$ .

For each function f on $\sigma (N)$ , we define $\widetilde {f}(z)=\overline {f(-z)}$ for $z\in \sigma (N)$ .

Claim. If f is a continuous function on $\sigma (N)$ , then $\lim _n C_nf(N)C_n=\widetilde {f}(N)$ .

Note that if $g(z)=\alpha z^i\overline {z}^j$ , then

$$ \begin{align*}C_ng(N)C_n=C_n(\alpha N^i{N^*}^j)C_n\rightarrow \overline{\alpha}(-N^*)^i(-N)^j=\widetilde{g}(N).\end{align*} $$

Then one can see that $C_nf(N)C_n\rightarrow \widetilde {f}(N)$ for each continuous function f on $\sigma (N)$ .

Choose a point $\lambda \in \operatorname {iso}\sigma (N)$ . Define a continuous function g on $\sigma (N)$ as

$$ \begin{align*}h(z)=\begin{cases} 1,& z=\lambda,\\ 0,& z\in\sigma(N)\setminus\{\lambda\}. \end{cases}\end{align*} $$

Then

$$ \begin{align*}\widetilde{h}(z)=\begin{cases} 1,& z=-\lambda,\\ 0,& z\in\sigma(N)\setminus\{-\lambda\}. \end{cases}\end{align*} $$

Then, by Claim, we deduce that $\lim _n C_nh(N)C_n=\widetilde {h}(N)$ . Since ${h}(N)$ and $\widetilde {h}(N)$ are orthogonal projections, it follows from [Reference Murphy22, Lemma 6.2.1] that $\text {rank}~ \widetilde {h}(N)=\text {rank}~ {h}(N)$ . Noting that $\text {ran~} h(N)=\ker (N-\lambda )$ and $\text {ran~} \widetilde {h}(N)=\ker (N+\lambda )$ , we conclude that $\dim \ker (N-\lambda )=\dim \ker (N+\lambda )$ . This proves the necessity.

“ $\Longleftarrow $ .” We first consider the case that $\ker N=\{0\}$ .

By the hypothesis, we can find a sequence $\{\lambda _i\}_{i=1}^\infty \subset \sigma (N)$ satisfying

1. (a) $\{\lambda _i: i\geq 1\}$ is dense in $\sigma (N)$ ;

2. (b) if $i\geq 1$ and $\lambda _i\in \operatorname {iso}\sigma (N)$ , then $\text {card}\{j\geq 1: \lambda _j=\lambda _i\}=\dim \ker (N-\lambda _i)$ , where $\text {card}(\cdot )$ denotes cardinality;

3. (c) if $i\geq 1$ and $\lambda _i\notin \operatorname {iso}\sigma (N)$ , then $\text {card}\{j\geq 1: \lambda _j=\lambda _i\}=\infty $ ;

4. (d) $\text {card}\{j\geq 1: \lambda _j=\lambda _i\}=\text {card}\{j\geq 1: \lambda _j=-\lambda _i\}$ for every $i\geq 1$ .

Choose an orthonormal basis $\{e_i\}_{i=1}^\infty $ of ${\mathcal {H}}$ and define $T\in \mathcal {B(H)}$ as

$$ \begin{align*}Te_i=\lambda_ie_i,\ \ \ \ i=1,2,3,\ldots.\end{align*} $$

One can check that $\sigma (T)=\sigma (N)$ and $\dim \ker (T-z)=\dim \ker (N-z)$ for each $z\in \operatorname {iso}\sigma (N)$ . Then, by [Reference Conway8, Proposition 39.10], we deduce that T is approximately unitarily equivalent to N, that is, there are unitary operators $U_n: {\mathcal {H}}\rightarrow {\mathcal {H}}$ such that $U_nTU_n^*\rightarrow N$ .

Next we shall show that T is skew symmetric (which implies $\kappa (N)= \text {dist}(N,SSO) =0$ ). In fact, in view of (a)–(d), we can find a bijective map $\tau $ on the set $\mathbb {N}$ of positive integers such that $\tau ^{-1}=\tau $ and $\lambda _{\tau (i)}=-\lambda _i$ for every $i=1,2,3,\ldots .$ We define a conjugation C on ${\mathcal {H}}$ as $C(\sum _i a_ie_i)=\sum _i \overline {a_i}e_{\tau (i)}$ . Then one can check that

$$ \begin{align*}CTCe_i=CTe_{\tau(i)}=C(\lambda_{\tau(i)}e_{\tau(i)})=\overline{\lambda_{\tau(i)}}e_{i}=-\overline{\lambda_{i}}e_{i}=-T^*e_i, \ \ \ i=1,2,\ldots.\end{align*} $$

Hence T is skew symmetric relative to C.

Now we consider the case that $\ker N\ne \{0\}$ . Assume that

$$\begin{align*}N=\begin{bmatrix} 0&0 \\ 0&\widetilde{N} \end{bmatrix}\begin{matrix} \ker N\\ (\ker N)^\bot \end{matrix}.\end{align*}$$

Thus $\ker \widetilde {N}=\{0\}$ , $\sigma (\widetilde {N})=-\sigma (\widetilde {N})$ and

$$ \begin{align*}\dim\ker(\widetilde{N}-z)=\dim\ker(\widetilde{N}+z)\end{align*} $$

for every $z\in \operatorname {iso}\sigma (\widetilde {N})$ .

By the proof in the case that $\ker N=\{0\}$ , one can see that $\widetilde {N}$ is a norm limit of skew symmetric operators. Then so is N. Hence we conclude that $\kappa (N)= \text {dist}(N, {SSO}) =0$ . This completes the proof.

Proof Assume that $\{U_n\}_{n=1}^\infty \subset \mathcal {B(H)}$ are unitary operators with $U_nAU_n^*\rightarrow B$ . Note that ${SSO}$ is unitarily invariant. For any $X\in { {SSO}}$ , we have

$$\begin{align*}\|B-X\|=\lim_n\|U_nAU_n^*-X\|=\lim_n\|A-U_n^*X U_n\|\geq\text{dist}(A,{SSO}).\end{align*}$$

It follows that $\text {dist}(B, {SSO})\geq \text {dist}(A, {SSO})$ . By the symmetry, one can see that $\text {dist}(B, {SSO})\leq \text {dist}(A, {SSO})$ , which implies $\kappa (A)=\kappa (B)$ .

Proof “ $\Longleftarrow $ .” Without loss of generality, we assume that $\sigma _e(N)=\{1\}$ . Since N is normal, then, by the BDF Theorem, we can find a compact operator K such that $N=I+K$ .

Fix a conjugation C on ${\mathcal {H}}$ . We shall show that $w(CN)=1$ . By [Reference Garcia and Putinar10, Lemma 1], we can find an orthonormal basis $\{e_i\}_{i=1}^\infty $ such that $Ce_i=e_i$ for all i. Then $\{e_i\}_{i=1}^\infty $ converges in the weak topology to $0$ , that is, $\langle e_i,x\rangle \rightarrow 0$ for every $x\in {\mathcal {H}}$ . Since K is compact, we deduce that $\lim _i\|Ke_i\|=0$ . Hence

$$ \begin{align*}\langle CNe_i, e_i\rangle=\langle Ce_i, Ne_i\rangle=\langle e_i, Ne_i\rangle=\langle e_i, (I+K)e_i\rangle=1+\langle e_i, Ke_i\rangle\rightarrow 1.\end{align*} $$

It follows that $w(CN)\geq 1$ . Noting that $\|CN\|=1$ , we obtain $w(CN)=1$ .

“ $\Longrightarrow $ .” Since $w(CN)=1$ for every conjugation C, we have $\inf _Cw(CN)=1$ . By Lemma 3.6, we may directly assume that N is a diagonal operator, say $N=\text {diag}\{\lambda _1, \lambda _2, \lambda _3,\ldots \}$ relative to an orthonormal basis $\{e_i\}$ of ${\mathcal {H}}$ . For convenience, we denote $\mathbb {D}=\{z\in \mathbb {C}: |z|<1\}$ .

Claim 1. $\sigma _e(N)\nsubseteq \mathbb {D}$ .

Otherwise, we have $r<1$ , where $r:=\sup \{|z|: z\in \sigma _e(N)\}$ . So $\text {card}\{i\geq 1: |\lambda _i|>\frac {1+r}{2}\}<\infty $ . Without loss of generality, we assume that $|\lambda _i|>\frac {1+r}{2}$ for $i=1,2,\ldots ,s$ and $|\lambda _i|\leq \frac {1+r}{2}$ for $i>s$ .

We define a conjugation C on ${\mathcal {H}}$ satisfying

$$\begin{align*}Ce_i=\begin{cases} e_{i+s},& 1\leq i \leq s,\\ e_{i-s},& s+1\leq i\leq 2s,\\ e_i, & i\geq 2s+1. \end{cases}\end{align*}$$

Then for any unit vector $x\in {\mathcal {H}}$ with $x=\sum _ia_i e_i$ , we have

$$ \begin{align*}Cx=\sum_{i=1}^s \overline{a_i} e_{s+i}+ \sum_{i=s+1}^{2s} \overline{a_i} e_{i-s} +\sum_{i=2s+1}^{\infty} \overline{a_i} e_{i}\end{align*} $$

and

$$ \begin{align*} |\langle CNx,x\rangle|&=|\langle Cx,Nx\rangle|=|\langle Cx,\sum_ia_i \lambda_i e_i\rangle|\\ &=\left|\sum_{i=1}^s \overline{a_ia_{i+s}\lambda_{i+s}} + \sum_{i=s+1}^{2s} \overline{a_ia_{i-s}\lambda_{i-s}} +\sum_{i=2s+1}^{\infty} \overline{a_i}^2\overline{\lambda_i } \right|\\ &\leq\sum_{i=1}^s \left|a_i a_{i+s}{\lambda_{i+s}} \right| + \sum_{j=1}^{s} \left|a_{j+s}{a_{j}}{\lambda_{j}}\right| +\sum_{i=2s+1}^{\infty} |a_i|^2|{\lambda_i }|\\ &\leq \frac{1+r}{2}\sum_{i=1}^s \left|a_i{a_{i+s}}\right| + \sum_{i=1}^{s} \left|a_{i+s}{a_{i}} \right| +\frac{1+r}{2}\sum_{i=2s+1}^{\infty} |a_i|^2 \\ &\leq(1+\frac{1+r}{2})\sum_{i=1}^s\left|a_i{a_{i+s}}\right| +\frac{1+r}{2}\sum_{i=2s+1}^{\infty} |a_i|^2\\ &\leq \frac{1}{2}(1+\frac{1+r}{2})\sum_{i=1}^s(|a_i|^2+|{a_{i+s}}|^2) +\frac{1+r}{2}\sum_{i=2s+1}^{\infty} |a_i|^2\\ & = \frac{3+r}{4} \sum_{i=1}^{2s} |a_i|^2 +\frac{1+r}{2}\sum_{i=2s+1}^{\infty} |a_i|^2\leq \frac{3+r}{4}<1. \end{align*} $$

Since x is a unit vector arbitrarily chosen in ${\mathcal {H}}$ , this shows that $w(CN)<1$ , a contradiction. This proves Claim 1.

Claim 2. $\sigma _e(N)\cap \mathbb {D}=\emptyset $ .

In fact, if not, then there exists $r\in (0,1)$ such that $\dim \text {ran~} E_N(B(0,r))=\infty $ , where $E_N(\cdot )$ is the projection-valued spectral measure associated with N and $B(0,r)=\{z\in \mathbb {C}: |z|<r\}$ . By Claim 1, we also have $\dim {\mathcal {H}}\ominus \text {ran~} E_N(B(0,r))=\infty $ . Denote $M=\text {ran~} E_N(B(0,r))$ . Then, relative to the decomposition ${\mathcal {H}}=M\oplus M^\bot $ , N can be written as $N=N_1\oplus N_2$ , where $N_1$ and $N_2$ are normal. Clearly, $\sigma (N_1)\subset \overline {B(0,r)}$ and $\sigma (N_2)\subset \{z\in \mathbb {C}: r\leq |z|\leq 1\}$ .

Choose an antiunitary operator $D:M\rightarrow M^\bot $ and define a conjugation C on ${\mathcal {H}}$ as

$$\begin{align*}C=\begin{bmatrix} 0&D^{-1}\\ D&0 \end{bmatrix}\begin{matrix} M\\ M^\bot \end{matrix}.\end{align*}$$

For any $x\in M$ and $y\in M^\bot $ with $\|x\|^2+\|y\|^2=1$ , we have

$$ \begin{align*} \left|\langle CN(x+y), x+y\rangle\right|& =\left|\langle C(x+y), N(x+y)\rangle\right| \\ &=\left|\langle Dx+D^{-1}y, N_1x+N_2y\rangle\right|\\ &= \left|\langle Dx, N_2y\rangle+\langle D^{-1}y, N_1x\rangle\right| \\ &\leq \|N_2\|\|x\|\|y\|+\|N_1\|\|x\|\|y\|\\ &\leq \|x\|\|y\|+r\|x\|\|y\| \leq (1+r)\|x\|\|y\|\\ &\leq \frac{1+r}{2}(\|x\|^2+\|y\|^2)=\frac{1+r}{2}<1. \end{align*} $$

This shows that $w(CN)<1$ , a contradiction. This proves Claim 2.

Since $\|N\|=1$ , by Claims 1 and 2, we have $\sigma _e(N)\subset \{z\in \mathbb {C}: |z|=1\}$ .

Claim 3. $\text {card}~\sigma _e(N)<\infty $ .

Otherwise, we can find $\theta _1,\theta _2,\theta _3,\theta _4\in [0,2\pi )$ with $\theta _1<\theta _2<\theta _3<\theta _4$ such that $\sigma _e(N)\cap \Delta _i\ne \emptyset , i=1,2,3,4,$ where $\Delta _1=\{e^{\text {i}\theta }: \theta _4 -2\pi <\theta <\theta _1 \}$ and

$$ \begin{align*}\Delta_i= \{e^{\text{i}\theta}: \theta_{i-1} <\theta<\theta_i \}, \ \ \ i=2,3,4.\end{align*} $$

Set $\widetilde {\Delta }_1=\{re^{\text {i}\theta }: \theta _4-2\pi <\theta \leq \theta _1, r\in [0,1] \}$ and

$$ \begin{align*}\widetilde{\Delta}_i= \{re^{\text{i}\theta}: \theta_{i-1} <\theta\leq\theta_i, r\in(0,1] \}, \ \ \ i=2,3,4.\end{align*} $$

Then $\{\widetilde {\Delta }_i\}_{i=1}^4$ is a partition of the closed unit disc $\overline {\mathbb {D}}$ ; moreover, one can see that $\dim \text {ran~} E_N(\Delta _i)=\infty $ , $i=1,2,3,4$ . Hence

$$ \begin{align*}\Gamma_i:=\{j\geq 1: \lambda_j\in \widetilde{\Delta}_i\}\end{align*} $$

is infinite for each $1\leq i\leq 4$ . We choose a bijective map $\tau $ on the set $\mathbb {N}$ of positive integers satisfying $\tau ^{-1}=\tau $ , $\tau (\Gamma _1)=\Gamma _3$ and $\tau (\Gamma _2)=\Gamma _4$ . It is not difficult to verify that

$$ \begin{align*}\delta:=\sup\left\{\frac{|\lambda_i+\lambda_{\tau(i)}|}{2}: i\geq 1 \right\}<1.\end{align*} $$

For each $x=\sum _ia_ie_i\in {\mathcal {H}}$ , we define $Cx=\sum _i\overline {a_i}e_{\tau (i)}$ . Then one can verify that C is a conjugation on ${\mathcal {H}}$ . Then for any unit vector $x\in {\mathcal {H}}$ with $x=\sum _ia_i e_i$ , we have

$$ \begin{align*} |\langle CNx,x\rangle|&=|\langle Cx,Nx\rangle|=\left|\left\langle \sum_i\overline{a_i}e_{\tau(i)},\sum_ia_i \lambda_i e_i\right\rangle\right|\\ &=\left|\left\langle \sum_i\overline{a_{\tau(i)}}e_i,\sum_ia_i \lambda_i e_i\right\rangle\right|= \left| \sum_i\overline{\lambda_ia_i a_{\tau(i)}} \right|\\ &\leq \left| \sum_{{i\in\Gamma_1\cup\Gamma_3}} \lambda_ia_i a_{\tau(i)}\right|+\left| \sum_{{i\in\Gamma_2\cup\Gamma_4}} \lambda_ia_i a_{\tau(i)}\right|\\ & = \left| \sum_{i\in\Gamma_1} (\lambda_i+\lambda_{\tau(i)})a_i a_{\tau(i)}\right|+ \left| \sum_{i\in\Gamma_2} (\lambda_i+\lambda_{\tau(i)})a_i a_{\tau(i)}\right|\\ &\leq \sum_{i\in\Gamma_1} |\lambda_i+\lambda_{\tau(i)}||a_i a_{\tau(i)}|+ \sum_{i\in\Gamma_2} |\lambda_i+\lambda_{\tau(i)}||a_i a_{\tau(i)}|\\ &\leq \sum_{i\in\Gamma_1} \frac{|\lambda_i+\lambda_{\tau(i)}|}{2} (|a_i|^2+ |a_{\tau(i)}|^2)+ \sum_{i\in\Gamma_2} \frac{|\lambda_i+\lambda_{\tau(i)}|}{2} (|a_i|^2+ |a_{\tau(i)}|^2)\\ &\leq\sum_{i\in\Gamma_1} \delta (|a_i|^2+ |a_{\tau(i)}|^2)+\sum_{i\in\Gamma_2} \delta (|a_i|^2+ |a_{\tau(i)}|^2) =\delta<1. \end{align*} $$

This shows that $w(CN)<1$ , a contradiction. This proves Claim 3.

Now we shall conclude the proof for the necessity by proving $\text {card}~\sigma _e(N)=1$ . For a proof by contradiction, we assume that $\text {card}~\sigma _e(N)>1$ . Then we can choose $\theta _1,\theta _2\in [0,2\pi )$ with $\theta _1<\theta _2$ such that $e^{\text {i}\theta _1},e^{\text {i}\theta _2}\in \sigma _e(N)$ . Without loss of generality, we may assume that $\theta _1=0$ .

By Claim 3, $\text {card}~\sigma _e(N)<\infty $ . Thus we can find real numbers $\theta _3,\theta _4,\theta _5,\theta _6$ with $ \theta _1<\theta _3<\theta _4 <\theta _2< \theta _5<\theta _6<2\pi $ such that

$$ \begin{align*}\sigma(N)\cap \{re^{\text{i}\theta}: \theta\in(\theta_3,\theta_4)\cup(\theta_5,\theta_6), r\in(0,1] \} =\emptyset.\end{align*} $$

Denote

$$ \begin{align*}\Omega_1=\{re^{\text{i}\theta}: \theta\in[\theta_6-2\pi,\theta_3], r\in[0,1] \}\end{align*} $$

and

$$ \begin{align*}\Omega_2=\{re^{\text{i}\theta}: \theta\in[\theta_4,\theta_5], r\in(0,1] \}.\end{align*} $$

Then $\Omega _1 \cap \Omega _2=\emptyset $ , $\sigma (N) \subset [\Omega _1 \cup \Omega _2]$ and one can see that $\dim \text {ran~} E_N(\Omega _i)=\infty $ , since ${e^{\text {i}\theta _i}}\in \sigma _e(N)\cap \Omega _i$ for $i=1,2.$ Hence

$$ \begin{align*}\Lambda_i:=\{j\geq 1: \lambda_j\in \Omega_i\}\end{align*} $$

is infinite for each $i=1,2$ . We choose a bijective map $\tau $ on the set $\mathbb {N}$ of positive integers satisfying $\tau ^{-1}=\tau $ , $\tau (\Lambda _1)=\Lambda _2$ . It is not difficult to verify that

$$ \begin{align*}\gamma:=\sup\left\{\frac{|\lambda_i+\lambda_{\tau(i)}|}{2}: i\geq 1 \right\}<1.\end{align*} $$

For each $x=\sum _ia_ie_i\in {\mathcal {H}}$ , we define $Cx=\sum _i\overline {a_i}e_{\tau (i)}$ . Then one can verify that C is a conjugation on ${\mathcal {H}}$ . Then for any unit vector $x\in {\mathcal {H}}$ with $x=\sum _ia_i e_i$ , we have

$$ \begin{align*} |\langle CNx,x\rangle|&=|\langle Cx,Nx\rangle|=\left|\left\langle \sum_i\overline{a_i}e_{\tau(i)},\sum_ia_i \lambda_i e_i\right\rangle\right|\\ &=\left|\left\langle \sum_i\overline{a_{\tau(i)}}e_i,\sum_ia_i \lambda_i e_i\right\rangle\right|= \left| \sum_i\overline{\lambda_ia_i a_{\tau(i)}} \right|\\ &= \left| \sum_{{i\in\Lambda_1 }} \lambda_ia_i a_{\tau(i)} + \sum_{{i\in\Lambda_2}} \lambda_ia_i a_{\tau(i)}\right|\\ &= \left| \sum_{{i\in\Lambda_1 }} \lambda_ia_i a_{\tau(i)} + \sum_{{i\in\Lambda_1}} \lambda_{\tau(i)}a_{\tau(i)} a_{i}\right|\\ & = \left| \sum_{i\in\Lambda_1} (\lambda_i+\lambda_{\tau(i)})a_i a_{\tau(i)}\right| \\ &\leq \sum_{i\in\Lambda_1} \frac{|\lambda_i+\lambda_{\tau(i)}|}{2} (|a_i|^2+ |a_{\tau(i)}|^2) \\ &\leq\sum_{i\in\Lambda_1} \gamma (|a_i|^2+ |a_{\tau(i)}|^2) =\gamma<1. \end{align*} $$

This shows that $w(CN)<1$ , a contradiction. This completes the proof.