Proof For each $\phi \in \mathcal {A}^{*}$ with $\phi (1)=1$ and $\|\phi \|=1$ we have

$$ \begin{align*} \mathrm{dist}(\phi(S),V(T))\leq|\phi(S)-\phi(T)|\leq\|S-T\| \end{align*} $$

and, likewise,

$$ \begin{align*} \mathrm{dist}(V(S),\phi(T))\leq|\phi(S)-\phi(T)|\leq\|S-T\|. \end{align*} $$

So by definition of the Hausdorff distance, we have

$$ \begin{align*} d_{H}(V(S),V(T))=\max\Big\{\sup_{z\in V(S)}\mathrm{dist}(z,V(T)),\sup_{z\in V(T)}\mathrm{dist}(V(S),z)\Big\}\leq\|S-T\| \end{align*} $$

as desired.

Proof Taking the polynomial $p(z)=z$ shows (i).

(ii) Assume the contrary, i.e., that $\Psi (\cdot )$ is not lower semicontinuous at some point T in $\mathcal {A}$ . Then there exists $C\in (0,\infty )$ , an $\varepsilon>0$ and a convergent sequence $T_{k}\to T$ such that $\Psi (T)>C+\varepsilon $ and $\Psi (T_{k})\leq C$ for all $k\geq 1$ . Note that C is finite, regardless of whether $\Psi (T)$ is finite or not. Further, there exists a polynomial $p\in \mathbb {C}[z]$ such that $\sup _{V(T)}|p|=1$ and $\|p(T)\|\geq C+\varepsilon $ . Now fix a number

$$ \begin{align*} 0<\delta<\sqrt{\frac{C+\varepsilon}{C}}-1. \end{align*} $$

Since $V(T_{k})\to V(T)$ with respect to the Hausdorff metric, there exists an integer $N_{1}\geq 1$ such that $\sup _{V(T_{k})}|p|\leq 1+\delta $ for all $k\geq N_{1}$ . Since $p(T_{k})\to p(T)$ , there also is an integer $N_{2}\geq 1$ such that $\|p(T_{k})\|\geq (1+\delta )^{-1}\|p(T)\|$ for all $k\geq N_{2}$ . Thus, for $k\geq N:=\max \{N_{1},N_{2}\}$ we have

$$ \begin{align*} \Psi(T_{k})(1+\delta)\geq\Psi(T_{k})\sup_{V(T_{k})}|p|\geq\|p(T_{k})\|\geq\frac{\|p(T)\|}{1+\delta}\geq\frac{C+\varepsilon}{1+\delta}. \end{align*} $$

However, this implies that $\Psi (T_{k})>C$ for all $k\geq N$ , which is a contradiction.

(iii) follows directly from (ii). To see (iv) define the mapping $\Lambda :{\mathbb {C}}[z]\ni p (z)\mapsto p(\alpha z+\beta )\in {\mathbb {C}}[z]$ for fixed $\alpha \neq 0$ and $\beta \in {\mathbb {C}}$ and note that it is bijective. Hence, by (2.7),

$$ \begin{align*}\Psi(\alpha T+\beta I)= \inf\{C>0: \| \Lambda( p) (T) \| \leq C \sup_{V(T)} |\Lambda(p)|,\ p\in{\mathbb{C}}[\lambda]\setminus\{ 0\} \}=\Psi(T). \end{align*} $$

(v) Let us fix $C>1$ . Then the set $\mathcal S:=\{ T\in \mathcal A: \Psi (T)> C \}$ is open as the function $\Psi (\cdot )$ is lower semi-continuous. Fix $\lambda _0\in {\mathbb {C}}$ and take any T with $\Psi (T)=C_0\in [C,\infty ]$ . Then $\Psi (\alpha T+ \lambda _0 I)=C_0$ for any $\alpha \neq 0$ , hence $\lambda _0 I$ is in the closure of $\mathcal S$ . As $\Psi (I)=1$ we have that $\lambda _0I$ is on the boundary of $\mathcal S$ . Statement (vi) is also a consequence of this reasoning.

Proof (i) Since for any $\theta \in \mathbb {R}$ , $\alpha>0$ we have

$$ \begin{align*} \|I_{X'}+\alpha {\mathrm{e}}^{i\theta}T'\|_{\mathcal{B}(X')}=\|I_{X}+\alpha {\mathrm{e}}^{i\theta}T\|_{\mathcal{B}(X)}, \end{align*} $$

the result follows from (2.4).

(ii) For $\zeta \in \mathbb {C}$ we trivially have

$$ \begin{align*} \Psi(\zeta I_{X'},\mathcal{B}(X'))=\Psi(\zeta I_{X},\mathcal{B}(X))=1. \end{align*} $$

Assume that T is not a scalar multiple of $I_{X}$ . By (i) we have

$$ \begin{align*} \Psi(T',\mathcal{B}(X'))&=\sup_{\substack{f\in\mathbb{C}[z] \\ f\neq0}}\frac{\|f(T')\|_{\mathcal{B}(X')}}{ \|f\|_{\infty,V(T',\mathcal{B}(X'))}}\\ &=\sup_{\substack{f\in\mathbb{C}[z] \\ f\neq0}}\frac{\|f(T)'\|_{\mathcal{B}(X')}}{\|f\|_{\infty,V(T',\mathcal{B}(X'))}}\\ &=\sup_{\substack{f\in\mathbb{C}[z] \\ f\neq0}}\frac{\|f(T)\|_{\mathcal{B}(X)}}{\|f\|_{\infty,V(T,\mathcal{B}(X))}}\\ &=\Psi(T,\mathcal{B}(X)) \end{align*} $$

as desired.

The inequality in statement (iii) is now obvious. Further, if X is reflexive, then $T\mapsto T'$ is surjective and the equality follows.

Proof Since $V(T)_{\varepsilon d}$ is convex, its compact boundary $\partial V(T)_{\varepsilon d}$ is locally the graph of a Lipschitz function and hence it is rectifiable. The first statement follows now directly from the Cauchy integral formula (or, more precisely, from the Riesz–Dunford calculus) and from the estimate (2.9). The fact that $V(T)_1$ is a $2$ -spectral set follows from taking $\varepsilon =1/2$ , so that $\varepsilon d\leq \varepsilon \cdot 2\left \| T\right \|\leq 1$ .

To see the second statement let $T\in \mathcal {B}(X)$ , $\varepsilon>0$ and $p(z)=\sum _{k=0}^{n}a_{k}z^{k}$ . Further, let $c=(1+\varepsilon )\|T\|$ and consider

$$\begin{align*}\ \left\|\sum_{k=0}^{n}a_{k}T^{k}\right\|=\left\|\sum_{k=0}^{n}a_{k}c^{k}(\tfrac{1}{c}T)^{k}\right\|\leq \left(\sum_{k=0}^{n}|a_{k}c^{k}|^{2}\right)^{\frac{1}{2}}\left(\sum_{k=0}^{\infty}(1+\varepsilon)^{-2k}\right)^{\frac{1}{2}}=\|f\|_{H^{2}(\mathbb{D})}C_{\varepsilon}, \end{align*}$$

where $f(z)=\sum _{k=0}^{n}a_{k}c^{k}z^{k}$ for $z\in \mathbb {D}$ , the norm $\left \|{\cdot }\right \|_{H^{2}(\mathbb {D})}$ refers to the norm of the Hardy space $H^{2}(\mathbb {D})$ and $C_{\varepsilon }$ to the constant above. Finally, we use that $H^{\infty }(\mathbb {D})$ , the space of bounded analytic functions on $\mathbb {D}$ , embeds continuously into $H^{2}(\mathbb {D})$ . More precisely,

$$ \begin{align*}\|f\|_{H^{2}(\mathbb D)}\leq \sup_{|z|<1}|f(z)|=\sup_{|z|<c}|p(z)|, \end{align*} $$

which finishes the proof.

Proof Suppose that $\lambda \in \partial V(T)$ is not semisimple. Then the Jordan normal form of T contains a Jordan block J corresponding to $\lambda $ of size s, where $1 < s \leq n$ . Since $\lambda \in \partial V(T)$ , we can find a sequence $(\lambda _{i})_{i\in \mathbb {N}}$ in $\mathbb {C}\setminus V(T)$ such that $\lambda _{i}\to \lambda $ and $d_{i}:=d(\lambda _{i},V(T))=|\lambda _{i}-\lambda |$ . Let $e_{s}$ be the sth standard basis vector of $\mathbb {C}^{s}$ and let $\left \|{\cdot }\right \|_1$ denote the operator norm induced by the $\ell _1$ -norm on ${\mathbb {C}}^s$ . Note that

$$ \begin{align*} \|(\lambda_{i}-J)^{-1}e_{s}\|_{1}= \sum_{k=1}^{s}\frac{1}{d_{i}^{k}}. \end{align*} $$

On the other hand,

$$ \begin{align*} \|(\lambda_{i}-J)^{-1}e_{s}\|_{1}\leq\|(\lambda_{i}-J)^{-1}\|_{1}\leq C\|(\lambda_{i}-T)^{-1}\|\leq\frac{C}{d_{i}}, \end{align*} $$

where $C>0$ depends only on the similarity transformation for the Jordan decomposition of T and the equivalence between the norms $\left \|{\cdot }\right \|_{1}$ and $\left \|{\cdot }\right \|$ . The last inequality follows directly from (2.9). We deduce that

$$ \begin{align*} 1+\frac{1}{d_{i}}+\cdots+\frac{1}{d_{i}^{s-1}}\leq C, \end{align*} $$

which contradicts $d_{i}\to 0$ .

Proof (of Theorem 4.1) Since T generates a finite-dimensional subalgebra, due to (3.3) it is enough to consider only finite-dimensional unital Banach algebras $\mathcal {A}$ . Furthermore, any such algebra $\mathcal {A}$ can be isometrically embedded in $\mathcal {B(A)}$ . Therefore, once again by (3.3), it is enough to consider only unital Banach algebras of operators on finite-dimensional spaces. In other words, it is enough to show that for fixed $n\in \mathbb N$ , fixed $T\in {\mathbb {C}}^{n,n}$ and fixed unital Banach algebra norm $\left \|{\cdot }\right \|$ on ${\mathbb {C}}^{n,n}$ one has $\Psi (T, \mathcal B({\mathbb {C}}^n, \left \|\cdot \right \|)) < \infty $ . Note that for some constant C (depending on the norm and hence implicitly on the dimension n) we have that $\left \|{p(T)}\right \|\leq C \left \|{p(T)}\right \|_2$ for any polynomial $p(z)$ . Hence, we subsequently reduce the proof to showing that

$$ \begin{align*}\left\|{p(T)}\right\|_2\leq C_2 \sup_{V(T, \mathcal B({\mathbb{C}}^n, \left\|\cdot\right\|))}|p|,\quad p\in{\mathbb{C}}[z], \end{align*} $$

with some constant $C_2$ possibly dependent on T. Let $T_J$ denote the Jordan form of T and let S denote the corresponding similarity transformation. We write

$$ \begin{align*}T_J=\text{diag}(\lambda_1,\dots,\lambda_r)\oplus R,\quad \sigma(T)=\{\lambda_1,\dots,\lambda_r\}\cup\sigma(R), \end{align*} $$

where $\lambda _1,\dots ,\lambda _r$ are the semisimple eigenvalues written with their multiplicities and R consists of all nontrivial Jordan blocks (possibly one of these two parts constituting $T_J$ might be void). Recall that the algebraic numerical range $V(T, \mathcal B({\mathbb {C}}^n, \left \|\cdot \right \|))$ has the property that all eigenvalues of T on its boundary are semisimple by Lemma 4.2. Hence, the eigenvalues of R (if there are any) lie inside the interior of $V(T, \mathcal B({\mathbb {C}}^n, \left \|\cdot \right \|))$ , by Lemma (4.2). The boundary $\partial V(T,\mathcal {B}({\mathbb {C}}^n,\left \|{\cdot }\right \|))$ is rectifiable. Estimating in a routine way the Cauchy integral formula we receive $\left \|{p(R)}\right \|\leq C_1 \sup _{V(T, \mathcal B({\mathbb {C}}^n, \left \|\cdot \right \|))}|p|$ for any polynomial p and some constant $C_1$ , depending on the maximum of the norm of the resolvent of R on $\partial V(T, \mathcal B({\mathbb {C}}^n, \left \|\cdot \right \|))$ . Therefore,

$$ \begin{align*}\left\|{p(T)}\right\|_2\leq \left\|{S}\right\|_2\left\|{S^{-1}}\right\|_2 \left\|{p(T_J)}\right\|_2 \end{align*} $$

and

$$ \begin{align*} \left\|{p(T_J)}\right\|_2&=\max ( |p(\lambda_1)|,\dots, |p(\lambda_r)|,\left\|{ p(R)}\right\|) \\ &\leq \max(1,C_1) \sup_{V(T, \mathcal B({\mathbb{C}}^n, \left\|\cdot\right\|))}|p|, \end{align*} $$

for all $p\in {\mathbb {C}}[z]$ , from which we obtain the constant $C_2$ .

Proof The first inequality follows from the fact that $\left \|{J_n}\right \|\leq 1$ . To see the second one observe that for any polynomial f of degree $n-1$ with coefficients $\alpha _{0},\alpha _{1},\alpha _{2},\ldots ,\alpha _{n-1}$ the polynomial functional calculus of $J_{n}$ is given by

$$ \begin{align*} f(J_{n})= \begin{bmatrix} \alpha_{0}&\alpha_{1}&\alpha_{2}&\cdots&\cdots&\alpha_{n-1}\\ &\alpha_{0}&\alpha_{1}&&&\vdots\\ &&\ddots&\ddots&&\vdots\\ &&&\ddots&\alpha_{1}&\alpha_{2}\\ &&&&\alpha_{0}&\alpha_{1}\\ &&&&&\alpha_{0} \end{bmatrix}. \end{align*} $$

From this it quickly follows that

$$ \begin{align*} \|f(J_{n})\|_{p}\geq\|f(J_{n})e_{n}\|_{p}=\begin{cases} (\sum_{j=0}^{n-1}|\alpha_{j}|^{p})^{\frac{1}{p}}&p<\infty\\ \max_{j=0}^{n-1}|\alpha_{j}|&p=\infty \end{cases} \end{align*} $$

and, under the Banach space isomorphism $(\mathbb {C}^{n},\left \|{\cdot }\right \|_{p})'=(\mathbb {C}^{n},\left \|{\cdot }\right \|_{q})$ ,

$$ \begin{align*} \|f(J_{n})\|_{p}=\|f(J_{n})'\|_{q}\geq\|f(J_{n})'e_{1}\|_{q}=\begin{cases} (\sum_{j=0}^{n-1}|\alpha_{j}|^{q})^{\frac{1}{q}}&q<\infty\\ \max_{j=0}^{n-1}|\alpha_{j}|&q=\infty \end{cases}. \end{align*} $$

Applying this to the polynomials from (4.1), we conclude that

$$ \begin{align*} \sup_{\substack{f\in\mathbb{C}[z] \\ f\neq0}}\frac{\|f(J_{n})\|_{p}}{\|f\|_{\infty,\overline{\mathbb{D}}}}\geq\frac{\|f_{n}(J_{n})\|_{p}}{\|f_{n}\|_{\infty,\overline{\mathbb{D}}}}\geq\frac{\max\{n^{\frac{1}{p}},n^{\frac{1}{q}}\}}{\sqrt{6}\sqrt{n}}=\frac{1}{\sqrt{6}}n^{\max\{\frac{1}{p},\frac{1}{q}\}-\frac{1}{2}} \end{align*} $$

as desired.

Proof Assume first that $\Psi _{\mathcal B}$ is finite. Let $f\in C(X,\mathcal B)$ . Note that each pair $(x,\phi )$ , where $x\in X$ and $\phi \in \mathcal B'$ with $\left \|{\phi }\right \|=1=\phi (I)$ , constitutes a functional $f\mapsto \phi (f(x))$ in the dual space $C(X,\mathcal B)'$ . Therefore,

$$ \begin{align*} \left\|{p(f)}\right\|_{C(X,\mathcal{ B})} &= \sup_{x\in X} \|p(f) (x)\| \\ &= \sup_{x\in X} \|p(f (x))\| \\ &\leq \Psi_{\mathcal B}\sup_{x\in X} \sup_{\substack{\phi\in \mathcal{ B}'\\ \left\|\phi\right\|=1=\phi(I)}} |p(\phi(f(x))| \end{align*} $$

which shows the inequality $\Psi _{C(X,\mathcal B)}\leq \Psi _{\mathcal B}$ . The reverse inequality, and the case ${\Psi _{\mathcal B}=\infty} $ follow by identifying $\mathcal B$ with constant functions in $C(X,\mathcal B)$ and applying (3.3).

The second statement follows from identifying $\mathcal A$ with the algebra $ C(X,{\mathbb {C}})$ and Proposition 3.1(i).

Proof Let us recall that the quotient of a C*-algebra by a closed *-ideal is again a C*-algebra. Thus the Calkin algebra is a C*-algebra. Then the inequality $ \Psi _{ \mathcal {C}( H ) }\leq \Psi _{\mathrm {Cro}}$ follows from (5.2) and (5.1) via the fact that $\mathcal {C}(H)$ can be embedded in a C*-subalgebra of $\mathcal B(\tilde H)$ for some Hilbert space $\tilde H$ . The reverse inequality follows again from (5.2) and (5.1) and the fact that the algebra $\mathcal B( {\mathbb {C}}^d, \left \|{\cdot }\right \|_2)$ can be embedded in a C*-subalgebra of the Calkin algebra $\Psi _{ \mathcal {C}( H ) }$ . Below we present a simple proof of the latter fact, referring also to [Reference Farah, Katsimpas and Vaccaro25] for a rich theory of embedings of C*-algebras into the Calkin algebra.

Note that it is enough to embed $\mathcal B( {\mathbb {C}}^d, \left \|{\cdot }\right \|_2)$ in $\mathcal C(H_0)$ for some separable Hilbert space $H_0$ . We define $H_{0}:=\ell ^{2}\otimes \mathbb {C}^{d}$ and let $\pi (T)=[I_{\ell ^{2}}\otimes T]\in \mathcal {C}(H_0)$ . Let $P_k$ be an orthogonal projection on the first k basis vectors of $\ell ^2$ and let $Q_k=(I_{\ell ^2} - P_k)\otimes I_{{\mathbb {C}}^d}$ . By [Reference Müller37, Proposition 6], we have that

$$ \begin{align*}\left\|{ \pi(T) }\right\|_{\mathcal{C}(H_0)}= \lim_{k\to\infty} \left\|{ Q_k(I_{\ell^2}\otimes T) Q_k }\right\|_{\mathcal{B}(H_0)}=\left\|{T}\right\|_{\mathcal B(H_0)}. \end{align*} $$

Hence, the mapping $\pi $ is an isometry, it is also clearly linear, multiplicative, and preserves the adjoint and identity.

Let us show now (5.4). By [Reference Chui, Smith, Smith and Ward18] (see also [Reference Müller37]) there exists $K\in \mathcal {K}(H)$ such that $V(T+\mathcal {K}(H))=V(T+K)$ . It follows that

$$ \begin{align*} \|p([ T] )\|_{\mathcal C(H)}&=\|[ p(T)]\|_{\mathcal C(H)}\\ &\leq\|p(T+K)\|_{\mathcal{B}(H)}\\ &\leq\Psi(T+K)\cdot {\textstyle \sup_{V(T+K)}|p|}\\ &=\Psi(T+K)\cdot {\textstyle\sup_{V([T] )}|p|} \end{align*} $$

for any polynomial p.

Proof Suppose that $p\in [1,\infty )$ with Hölder conjugate $q\in (1,\infty ]$ , then $(\ell ^{p})'=\ell ^{q}$ (i.e., $\ell ^{q}$ is the dual space of $\ell ^{p}$ ) with $R'=L$ and using Lemma 3.2 we deduce that

$$ \begin{align*} V(R,\mathcal{B}(\ell^{p}))&=V(R',\mathcal{B}((\ell^{p})'))=V(L,\mathcal{B}(\ell^{q})),\\ \Psi(R,\mathcal{B}(\ell^{p}))&=\Psi(R',\mathcal{B}((\ell^{p})'))=\Psi(L,\mathcal{B}(\ell^{q})). \end{align*} $$

For $p=\infty $ we have $\ell ^{\infty }=(\ell ^{1})'$ with $R=L'$ and therefore

$$ \begin{align*} V(R,\mathcal{B}(\ell^{\infty}))&=V(L',\mathcal{B}((\ell^{1})'))=V(L,\mathcal{B}(\ell^{1})),\\ \Psi(R,\mathcal{B}(\ell^{\infty}))&=\Psi(L',\mathcal{B}((\ell^{1})'))=\Psi(L,\mathcal{B}(\ell^{1})). \end{align*} $$

Hence it suffices to prove both (i) and (ii) only for the left-shift L.

(i) Assume $p\in [1,\infty )$ . Since $\nu (L,\mathcal {B}(\ell ^{p}))\leq \|L\|_{p}=1$ , we have $V(L,\mathcal {B}(\ell ^{p}))\subseteq \overline {\mathbb {D}}$ . To see that the reverse inclusion holds, define for each $\theta \in \mathbb {R}$ and $n\in \mathbb {N}$ the vector

$$ \begin{align*} x_{\theta,n}:=n^{-\tfrac{1}{p}}(1,{\mathrm{e}}^{-i\theta},{\mathrm{e}}^{-2i\theta},\ldots,{\mathrm{e}}^{-(n-1)i\theta},0,\ldots)\in\ell^{p} \end{align*} $$

and observe that

$$ \begin{align*} \lim_{\alpha\downarrow0}\frac{\|I_{\ell^{p}}+\alpha {\mathrm{e}}^{i\theta} L\|_{p}-1}{\alpha}&\geq\lim_{\alpha\downarrow0}\frac{\|(I_{\ell^{p}}+\alpha {\mathrm{e}}^{i\theta} L)x_{\theta,n}\|_{p}-1}{\alpha}\\ &=\lim_{\alpha\downarrow0}\frac{n^{-\frac{1}{p}}(1+(n-1)(1+\alpha)^{p})^{\frac{1}{p}}-1}{\alpha}=\frac{n-1}{n}, \end{align*} $$

which after letting $n\to \infty $ yields $\sup \operatorname{Re}{\mathrm{e}}^{i\theta } V(L,\mathcal {B}(\ell ^{p}))\geq 1$ by the equality (2.6). The case $p=\infty $ follows from a similar but more direct argument using the vector $x_{\theta }:=(1,{\mathrm{e}}^{-i\theta },0,\ldots )$ for $\theta \in \mathbb {R}$ .

(ii) The case $p=2$ follows directly from von Neumann’s inequality as $V(L,\mathcal {B}(\ell ^{2}))=\overline {\mathbb {D}}$ and $\|L\|_{2}=1$ . Assume $p\in [1,\infty ]\setminus \{2\}$ and let $n\in \mathbb {N}$ be arbitrary. Let $P_{n}\colon \ell ^{p}\to (\mathbb {C}^{n},\left \|{\cdot }\right \|_{p})$ and $Q_{n}\colon (\mathbb {C}^{n},\left \|{\cdot }\right \|_{p})\to \ell ^{p}$ be defined by

$$ \begin{align*} P_{n}(x_{1},\ldots,x_{n},x_{n+1},\ldots):=(x_{1},\ldots,x_{n}),\qquad Q_{n}(x_{1},\ldots,x_{n}):=(x_{1},\ldots,x_{n},0,\ldots). \end{align*} $$

Clearly, $P_{n}Q_{n}=I_{\mathbb {C}^{n}}$ and $P_{n}LQ_{n}=J_{n}$ and $\|P_{n}\|=\|Q_{n}\|=1$ . For every $f\in \mathbb {C}[z]$ , we infer

$$ \begin{align*} P_{n}f(L)Q_{n}=f(P_{n}LQ_{n})=f(J_{n}) \end{align*} $$

and therefore $\|f(L)\|_{p}\geq \|f(J_{n})\|_{p}$ . It follows that

$$ \begin{align*} \Psi(L,\mathcal{B}(\ell^{p}))\geq\sup_{\substack{f\in\mathbb{C}[z] \\ f\neq0}}\frac{\|f(J_{n})\|_{p}}{\|f\|_{\infty,\overline{\mathbb{D}}}}\geq\frac{1}{\sqrt{6}} n^{\max\{\frac{1}{p},\frac{1}{q}\}-\frac{1}{2}}, \end{align*} $$

by Theorem 4.3. Thus $p\neq 2$ implies $\Psi (L,\mathcal {B}(\ell ^{p}))=\infty $ as desired.

Proof Let us fix a number $k \in \mathbb {N}$ . By property (3) of a combinatorial Banach space, there exists $n\geq k$ such that there exists a subset $(k_1,\dots , k_n) \in \mathcal {F}$ . Thanks to the spreading property we have that

(6.3) $$ \begin{align} (k_n, k_n +1, \dots, k_n +n-1)\in\mathcal F. \end{align} $$

Let $P_n$ denote the projection onto the coordinates $k_n+1,\dots , k_n+n$ and R be the right shift, both defined on $c_{00}$ . Define the following linear operator

$$ \begin{align*}S_n = P_n \circ R,\quad S_n( (x_{j})_{j=1}^\infty)=(\underbrace{0,\dots,0}_{k_n}, x_{k_n},\dots,x_{k_n+n-1},0,0,\dots ). \end{align*} $$

Observe that $\left \|{S_n}\right \|\leq 1$ , hence, it extends to a bounded operator of norm not greater than 1 on the whole space $\mathcal S$ . Indeed, we have

$$ \begin{align*}\left\|{S_nx}\right\|_{\mathcal{S}} \leq \left\|{S_nx}\right\|_{\ell_1}=\sum_{i=0}^{n-1}|x_{k_n+i}|\leq \left\|{x}\right\|_{\mathcal S}, \end{align*} $$

where the last inequality follows due to (6.3).

Further, observe that $\left \|{S_n+\lambda I}\right \|_{\mathcal {S}} \leq |\lambda | +1$ for $\lambda \in {\mathbb {C}}$ . In fact, we have equality. To see this, take $x= e_{k_n}=(0, \dots , 0,1, 0,\dots )$ (unit on the $k_n$ th position) and note that it is a unit vector, due to (6.3). Hence,

$$ \begin{align*}\left\|{S_n+\lambda I}\right\|_{\mathcal S} =|\lambda| +1,\quad \lambda\in{\mathbb{C}}, \end{align*} $$

again thanks to (6.3). From this, together with (2.3), we obtain that $ V(S_n)$ is the closed unit disk.

Let $f_n$ be the polynomials as in (4.1). From the form of $S_n$ we see that

$$ \begin{align*}f_{n+1}(S_n)e_{k_n} = (\underbrace{0,\dots, 0}_{k_n-1}, \alpha_0, \alpha_1, \dots, \alpha_{n-1}, 0, 0, \dots), \end{align*} $$

where $f_{n+1}(z)=\alpha _{0}+\alpha _{1}z+\dots +\alpha _{n}z^{n}$ and $\alpha _j\in \{-1,1\}$ ( $j=0,\dots ,n$ ). Hence, using (6.3) for the final time,

$$ \begin{align*}\left\|{f_{n+1}(S_n)}\right\|_{\mathcal{S}} \geq n, \end{align*} $$

while $\sup _{z\in \overline {\mathbb {D}}}|f_{n+1}(z)| \leq \sqrt {6} \sqrt {n+1}$ , which shows that $\Psi _{\mathcal {B}(\mathcal S)}=\infty $ .

Proof Before we proceed with the proof let us note that the operator norm of $T\in \mathcal B(\ell _1, \left \|\cdot \right \|_1)$ can be calculated similarly to $\ell ^1$ -matrix norm. Namely, let $e_1,e_2,\dots $ be the canonical basis of $\ell ^1$ and let $e_1^*,e_2^*,\dots $ denote their dual operators (i.e., the coefficient functionals corresponding to the Schauder basis $(e_{j})_{j=1}^{\infty }$ ). Define $t_{k,j} = e_k^* Te_j$ . Then the norm is given by

$$ \begin{align*}\left\|{T}\right\|_1 = \sup_{j\in\mathbb N} \ \sum_{k\in\mathbb N} |t_{k,j}|.\end{align*} $$

Now let us fix an angle $\theta \in [0,2\pi )$ . Our goal is to find the supporting hyperplane $H_{\theta }$ for the set $V(T)$ using formulas (2.4)–(2.6). To simplify calculations let us rotate the coordinate complex plane by $\theta $ , so that

$$ \begin{align*}T' := [t_{i,j}']_{i,j\in\mathbb N} := {\mathrm{e}}^{-i\theta}T, \qquad H' := H_0(T') = H_{\theta}(T), \qquad r':= r_0(T') = r_{\theta}(T).\end{align*} $$

Then the distance $r'$ can be expressed as

$$ \begin{align*} r' & = \lim_{\alpha\downarrow 0}\frac{\left\|{I + \alpha T'}\right\|_{1} - 1}{\alpha} \\ & = \lim_{\alpha\downarrow 0} \sup _{j\in \mathbb N} \left\{ \frac{|1+ \alpha t_{j,j}'| - 1 + \sum_{k=1, k\neq j}^{\infty} | \alpha t_{k,j}'|}{\alpha} \right\}.\end{align*} $$

For each $\alpha $ let us choose a sequence of indices $(j_m)_{m\in \mathbb N}$ approaching the supremum above and define a function

$$ \begin{align*}f(\alpha,m):= \frac{|1+ \alpha t_{j_m,j_m}'| - 1}{\alpha} + \sum_{k=1, k\neq j_m}^{\infty} |t_{k,j_m}'|, \qquad a\in(0,1),\ m\in\mathbb N.\end{align*} $$

Note that the function $g(\alpha ):=\frac {1}{\alpha }(\left \|{I+\alpha T}\right \| - 1)$ is decreasing, as for any $\alpha <\beta $ we have

$$ \begin{align*}g(\alpha) = \frac{\left\|{\frac{\beta}{\alpha} I + \beta T}\right\|_{1} - \frac{\beta}{\alpha} }{\beta} \leq \frac{\left\|{I + \beta T}\right\|_{1} + \left\|{\left( \frac{\beta}{\alpha} - 1\right)I}\right\| -\frac{\beta}{\alpha} }{\beta} = g(\beta). \end{align*} $$

Hence, we infer that

$$ \begin{align*}r' = \lim_{n\to\infty}\lim_{m\to\infty} f(1/n,m). \end{align*} $$

Next we show that $f(1/n,m)$ satisfies the conditions of the Moore–Osgood theorem to switch the order of the limits. Note that $\lim \limits _{m\to \infty } f(1/n,m) = \frac {\left \|{I + 1/n T'}\right \|_{1} - 1}{1/n} <\infty $ for each $n\in \mathbb N$ by the definition. To find the other limit observe that for an arbitrary complex number $z=a+bi$ , we have

$$ \begin{align*}\lim_{n\to\infty}(|z + n| - n) = \lim_{n\to\infty} \frac{a^2 + 2an + b^2}{\sqrt{(a+n)^2+b^2} + n} = a = \text{Re}(z). \end{align*} $$

And consequently,

(7.1) $$ \begin{align} \lim_{n\to\infty} f(1/n,m) = \sum_{k=1, k\neq j_m}^{\infty} |t_{k,j_m}'| + \text{Re}(t_{j_m,j_m}'). \end{align} $$

Let us show that $\lim \limits _{n\to \infty } f(1/n,m)$ is also uniform in m. Observe that $|t^{\prime }_{k,j}|\leq \|T\|_1$ for all $k,j$ . For $\varepsilon>0$ let us choose $n>\frac {2}{\varepsilon }M^2$ , where $M:= \max \{1,\left \|{T}\right \|_1\}$ . For simplicity let $t^{\prime }_{j_m,j_m} = a_m+ib_m$ be the decomposition into real and imaginary parts. Then

$$ \begin{align*} 0\leq f(1/n, m) - \lim_{n\to\infty} f(1/n,m) &= \sqrt{(a_m+n)^2+b_m^2} - n - a_m \\ &= \frac{b_m^2}{\sqrt{(a_m+n)^2+b_m^2} + n+a_m} \leq \frac{M^2}{n-M} \leq \varepsilon \end{align*} $$

for all $m\in \mathbb N$ . Hence, we can change the order of the limits in the definition of $r'$ , which together with the equality (7.1) provides

$$ \begin{align*}r' = \lim_{m\to\infty} \left( \sum_{k=1, k\neq j_m}^{\infty} |t_{k,j_m}'| + \text{Re}(t_{j_m,j_m}') \right) = \sup _{j\in \mathbb N} \left\{ \sum_{k=1, k\neq j}^{\infty} | t_{k,j}'| + \text{Re}( t_{j,j} ')\right\}. \end{align*} $$

Let us now fix j and consider the Gershgorin disk corresponding to the jth column of $T'$ , i.e., $D(t^{\prime }_{j,j}, \sum _{k=1, k\neq j}^{\infty } | t_{k,j}'|)$ . Let us look at its vertical tangent lines. If the Gershgorin disk is just a point there is only one such line passing through $t_{j,j}$ , let us call it $l_{j,\theta }$ . Otherwise there are two such lines. Let us denote their touch points as $p_1$ and $p_2$ and without loss of generality assume that $\text {Re}(p_1) < \text {Re}(p_2)$ . Then we denote the line corresponding to $p_2$ by $l_{j, \theta }$ (see Figure 1).

Figure 1: Illustration of the Gershgorin disk $D(t^{\prime }_{j,j}, \sum _{k=1, k\neq j}^{\infty } | t_{k,j}'|)$ together the tangent line $l_{j,\theta }$ . Its radius and the distance from 0 to its center is highlighted. The left picture shows an example when $\text {Re}(t_{j,j}')$ is positive and the right one corresponds to the negative case.

Now it can be easily seen that the expression $\sum _{k=1, k\neq j}^{n} | t_{k,j}'| + \text {Re}(t_{j,j}')$ is equal to the distance from $0$ to the tangent line $l_{j,\theta }$ , which, in turn, is equal to $\text {Re}(p_2)$ . So, $r'$ is equal to the supremum of such distances from $0$ to $l_{j,\theta }$ over all $j\in \mathbb N$ . Hence, half-plane $H'$ contains all Gershgorin disks and, moreover, it is tangent to the closure of their convex hull. That means that $H' = H_{\theta }(T)$ is simultaneously a supporting half-plane for the algebraic numerical range $V(T)$ and for the convex hull of the disks. Since both these sets are convex, they must be equal.

Proof Observe that T can be written in the following from

$$ \begin{align*} T = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x& 1 \\ 0& x \end{bmatrix} \begin{bmatrix} d& -b \\ -c& a \end{bmatrix} , \quad a,b,c,d,x\in\mathbb{C}, \quad\text{with } ad-bc =1. \end{align*} $$

Since $\Psi (\alpha T + \beta I) = \Psi (T)$ (see Proposition 3.1), we can assume $x=0$ . Then T takes form

$$ \begin{align*} T = \begin{bmatrix} -ac & a^2 \\ -c^2 & ac \end{bmatrix} \end{align*} $$

and its algebraic numerical range $V(T)$ is given by the convex hull of two disks $D(-ac,|c|^2)$ and $D(ac,|a|^2)$ . Let us take a polynomial $p \in \mathbb {C} [z]$ such that $\sup \limits _{z\in V(T)} |p(z)|=1$ . By a straightforward computation, we have

$$ \begin{align*} p(T) = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} p(0) & p'(0) \\ 0& p(0) \end{bmatrix} \begin{bmatrix} d& -b \\ -c& a \end{bmatrix} = \begin{bmatrix} p(0) - ac p'(0)& a^2 p'(0) \\ -c^2 p'(0) & p(0) + ac p'(0) \end{bmatrix}. \end{align*} $$

and so $ ||p(T)||_1 = \max \{ |p(0) - ac p'(0)| + |c^2 p'(0)|,\ |p(0) + ac p'(0)| + |a^2 p'(0)| \} $ .

Notice that the disk $D\left (0,\frac {|a|^2 +|c|^2}{2}\right )$ is contained in the algebraic numerical range $V(T)$ . This can be seen by calculating the midline of the trapezoid formed by a common tangent to two disks $D(-ac,|c|^2)$ and $D(ac,|a|^2)$ and radii drawn to this tangent (see Figure 2). So, p is analytic in $D\left (0,\frac {|a|^2 +|c|^2}{2}\right )$ and bounded by 1. Hence, by Cauchy’s inequality for the Taylor series coefficients of a complex analytic function, we get $|p'(0)| \leq \frac {2}{|a|^2+ |c|^2}$ .

Figure 2: Illustration of the two disks $D(-ac,|c|^2)$ and $D(ac,|a|^2)$ with highlighted trapezoid formed by their common tangent and radii drawn to this tangent. The circle with dotted line illustrates the disk $D\left (0,\frac {|a|^2 +|c|^2}{2}\right )$ which is contained in the closure of the convex hull of $D(-ac,|c|^2)$ and $D(ac,|a|^2)$ .

Therefore,

$$ \begin{align*} \|p(T)\|_1 \leq |p(0)| + (|ac| + \max(|a|^2, |c|^2)) |p'(0)| \leq 1 + 2\frac{|ac|+ \max(|a|^2, |c|^2)}{|a|^2+|c|^2}. \end{align*} $$

Using the fact that the function $f(\alpha ,\beta )=\frac {\alpha \beta + \beta ^2}{\alpha ^2+\beta ^2}$ is bounded by $(1+\sqrt {2})/2$ for $0<\alpha \leq \beta $ we obtain

$$ \begin{align*}\|p(T)\|_1 \leq2+\sqrt{2} \end{align*} $$

as desired.

Proof The matrix T can be written in the form

$$ \begin{align*} T = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x& 0 \\ 0& y \end{bmatrix} \begin{bmatrix} d& -b \\ -c& a \end{bmatrix} , \quad a,b,c,d,x,y\in\mathbb{C}, \quad ad-bc =1. \end{align*} $$

As $\Psi (\lambda I) = 1$ , let us assume $x\neq 0$ , $x\neq y$ . Also, since $\Psi (T)=\Psi (\frac {1}{x}(T-yI))$ , we can assume $x=1$ and $y=0$ . Then

$$ \begin{align*} T = [a,c][d,-b]^T= \begin{bmatrix} ad & -ab \\ cd & -bc \end{bmatrix}. \end{align*} $$

Let us fix a polynomial $p \in \mathbb {C} [z]$ such that $\sup \limits _{z\in V(T)} |p(z)|=1$ . Notice that

$$ \begin{align*} p(T) = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} p(1) & 0\\ 0& p(0) \end{bmatrix} \begin{bmatrix} d& -b \\ -c& a \end{bmatrix} =(p(1)-p(0))T+p(0)I, \end{align*} $$

and hence by the triangle inequality

(7.2) $$ \begin{align} \left\|{p(T)}\right\|_1 \leq |p(1)-p(0)| \left\|{T}\right\|_1 +|p(0)|. \end{align} $$

Since $\sup _{V(T)}|p|\leq 1$ , we clearly have $|p(0)-p(1)|\leq 2$ and $|p(0)|\leq 1$ . Thus, if $\|T\|_{1}\leq 6$ , then

$$ \begin{align*} \left\|{p(T)}\right\|_1 \leq 2\cdot 6 +1 = 13. \end{align*} $$

Now assume that $\left \|{T}\right \|_1> 6$ . Observe that the algebraic numerical range $V(T)$ is given by the convex hull of Gershgorin disks $D_1:=D(ad, |cd|)$ and $D_2:=D(-bc,|ab|)=D(1-ad,|ab|)$ . Each of the eigenvalues $0$ and $1$ of T belongs to at least one of the disks $D_{1}$ and $D_{2}$ by the Gershgorin circle theorem.

Define a polynomial $q\in \mathbb C[z]$ as $q(z):=p(z)-p(0)$ . Notice that $q(0)=0$ and also $|q(z)| \leq |p(z)|+ |p(0)| \leq 2$ for $z\in V(T)$ . Let $r_0$ denote the maximal radius such that $D(0,r_0) \subseteq V(T)$ . If $r_0 \geq 1$ , then, by Schwarz’s lemma,

$$ \begin{align*} |p(1)- p(0)| = |q(1)| \leq \frac{2}{r_0}. \end{align*} $$

Otherwise, if $r_0<1$ , then the same inequality holds trivially, as then $|q(1)|\leq 2 \leq \frac {2}{r_0}$ .

Next we estimate $r_0$ . Draw the line l tangent to disks $D_1$ and $D_2$ which is closer to 0 (see Figure 3). Let $t_0$ and $t_{1/2}$ denote the orthogonal projections of $0$ and $1/2$ onto l. Consider the trapezoid formed by the centers of $D_{1}$ and $D_{2}$ , and the points where l meets the disks. Let us recall that $ad - bc = 1$ , so the midpoint of $ad$ and $-bc$ is $(ad - bc)/2 = 1/2$ . Then it is easy to see that the line segment with endpoints $1/2$ and $t_{1/2}$ is a midline of this trapezoid and hence $|t_{1/2} - 1/2 | = \frac {1}{2}(|ab|+|cd|)$ .

Figure 3: Illustration of the disks $D_1:=D(ad, |cd|)$ and $D_2:=D(-bc,|ab|)$ with highlighted trapezoid formed by their common tangent l closest to 0 and radii drawn to it.

Let $m_0$ denote the projection of $0$ onto the midline. Then we have $|t_0|=|t_{1/2}-m_0| = |t_{1/2} - 1/2| - |m_0-1/2|$ . It is easy to see that $|m_0-1/2|\leq 1/2$ as it is a leg in a right triangle with hypotenuse of length $1/2$ . Altogether we get

(7.3) $$ \begin{align} r_0 = |t_0|\geq \frac{|ab|+|cd|-1}{2} \end{align} $$

Depending on the values of $|a|$ , $|b|$ , $|c|$ and $|d|$ we now distinguish between several cases.

Case 1: $|a| \leq |c|$ and $|b| \leq |d|$ , or equivalently $0,1\in D_1$ . In this case, we have $\|T\|_{1}=|ad|+|cd|\leq 2|cd|$ . Inequality (7.2) together with (7.3) gives

$$ \begin{align*} \left\|{p(T)}\right\|_1 \leq 4 \frac{|ad|+|cd|}{|ab|+|cd|-1} + 1 \leq 4 \frac{2|cd|}{|cd|-1} + 1. \end{align*} $$

Since we also assumed $\|T\|_{1}>6$ , we have $|cd|>3$ and hence $\frac {2|cd|}{|cd|-1}<3$ . Altogether we get

$$ \begin{align*} \left\|{p(T)}\right\|_1 \leq 4\cdot 3+1=13. \end{align*} $$

Case 2: $|a|> |c|$ and $|b|> |d|$ , or equivalently $0,1\in D_2$ . This case is symmetrical to the previous Case 1. Indeed, we have $6<\left \|{T}\right \|=|ab|+|cb|\leq 2|ab|$ and

$$ \begin{align*} \left\|{p(T)}\right\|_1 \leq 4 \frac{|ab|+|bc|}{|ab|+|cd|-1} + 1 \leq 4 \frac{2|ab|}{|ab|-1} + 1 \leq 13. \end{align*} $$

Case 3: $|a| \leq |c|$ and $|b|> |d|$ , or equivalently $0\in D_1$ and $1\in D_2$ . In this case inequality (7.2) together with (7.3) gives

$$ \begin{align*} \left\|{p(T)}\right\|_1 \leq 4 \frac{|ab|+|bc|}{|ab|+|cd|-1} + 1 = 4 \frac{|ab|+|bc|}{(|ab|+|bc|)+(|cd|-|bc|)-1} + 1. \end{align*} $$

Let us estimate difference $|bc|-|cd|$ . Firstly $0<|bc|-|cd|$ due to our assumption $|b|>|d|$ . On the other hand, since $|a|\leq |c|$ and $ad-bc=1$ , by the triangle inequality we have $|bc|-|cd|\leq |bc|-|ad|\leq |bc-ad|=1$ . So,

$$ \begin{align*} \left\|{p(T)}\right\|_1 \leq 4 \frac{|ab|+|bc|}{|ab|+|bc|-2} + 1 \leq 7, \end{align*} $$

where in the last line we use the assumption $\left \|{T}\right \|=|ab|+|bc|>6$ which implies $\frac {\left \|{T}\right \|}{\left \|{T}\right \|-2}\leq \frac {3}{2}$ .

Case 4: $|a|> |c|$ and $|b| \leq |d|$ , or equivalently $0\in D_2$ and $1\in D_1$ . This case is symmetrical to Case 3. Similar reasoning gives $0<|ad|-|ab|\leq |ad|-|bc|\leq |ad-bc|=1$ and

$$ \begin{align*} \left\|{p(T)}\right\|_1 \leq 4 \frac{|ad|+|cd|}{(|ad|+|cd|)+(|ab|-|ad|)-1} + 1 \leq 4 \frac{|ad|+|cd|}{|ad|+|cd|-2} + 1 \leq 7. \end{align*} $$