Proof Let $\{ e_n(z) = z^n/\sqrt {n!} \}_{n \ge 0} $ be the standard orthonormal basis of the Fock space $\mathcal {F}^2.$ Consider a function

$$ \begin{align*} f=\sum_{n=0}^{\infty}c_ne_n\in\mathcal F^2. \end{align*} $$

Using the action of $V_g$ on the sequence $(e_n)$ ,

$$ \begin{align*} V_ge_n = \frac{b}{\sqrt{n+1}}e_{n+1} + 2a\sqrt{\frac{n+1}{n+2}}e_{n+2}, \end{align*} $$

and collecting the coefficients corresponding to the same basis vectors, we obtain

$$ \begin{align*} V_gf = bc_0e_1 + \sum_{n=0}^{\infty} \bigg( 2a\sqrt{\frac{n+1}{n+2}}c_n + \frac{b}{\sqrt{n+2}}c_{n+1} \bigg)e_{n+2}. \end{align*} $$

By orthonormality of the sequence $(e_n)$ , it follows that

$$ \begin{align*} \|V_gf\|^2 = |b|^2|c_0|^2 + \sum_{n=0}^{\infty} \left| 2a\sqrt{\frac{n+1}{n+2}}c_n + \frac{b}{\sqrt{n+2}}c_{n+1} \right|^2. \end{align*} $$

We now establish the identity

(2.1) $$ \begin{align} \left(4|a|^2+|b|^2\right)\|f\|^2-\|V_gf\|^2 =\sum_{n=0}^{\infty} \Big| \frac{2a}{\sqrt{n+2}}c_n - b\sqrt{\frac{n+1}{n+2}}c_{n+1} \Big|^2. \end{align} $$

Indeed, expanding the right-hand side of (2.1) gives

$$ \begin{align*} \sum_{n=0}^{\infty} \frac{4|a|^2}{n+2}|c_n|^2 + \sum_{n=0}^{\infty} |b|^2\frac{n+1}{n+2}|c_{n+1}|^2 - 4\mathrm{Re} \sum_{n=0}^{\infty} a\overline b \frac{\sqrt{n+1}}{n+2} c_n\overline{c_{n+1}}. \end{align*} $$

After shifting the index in the second sum, this becomes

$$ \begin{align*} \sum_{n=0}^{\infty} \left( \frac{4|a|^2}{n+2} + |b|^2\frac{n}{n+1} \right)|c_n|^2 - 4\mathrm{Re} \sum_{n=0}^{\infty} a\overline b \frac{\sqrt{n+1}}{n+2} c_n\overline{c_{n+1}}, \end{align*} $$

where the term containing $|b|^2$ is understood to be zero when $n=0$ . This is precisely the expression obtained by expanding the left-hand side of (2.1).

Since the right-hand side of (2.1) is nonnegative, we have

$$ \begin{align*} \|V_gf\|^2 \leq \left(4|a|^2+|b|^2\right)\|f\|^2 \end{align*} $$

from which the upper estimate for the norm follows.

To show that this estimate is sharp, suppose first that $b\neq 0$ . For each $n\geq 0$ , define the sequence

$$ \begin{align*} c_n = \exp\Big(-\frac{2|a|^2}{|b|^2}\Big) \frac{(2a/b)^n}{\sqrt{n!}}. \end{align*} $$

Then,

$$ \begin{align*} \sum_{n=0}^{\infty}|c_n|^2 = \exp\left(-\frac{4|a|^2}{|b|^2}\right) \sum_{n=0}^{\infty} \frac{(4|a|^2/|b|^2)^n}{n!} =1, \end{align*} $$

and hence $ f=\sum _{n=0}^{\infty }c_ne_n $ is a unit-norm function in $\mathcal F^2$ . Furthermore, $ b\sqrt {n+1}\,c_{n+1} = 2ac_n $ for every $n\geq 0,$ which implies that every term on the right-hand side of (2.1) vanishes. Hence, $ \|V_gf\|^2=4|a|^2+|b|^2, $ which yields the lower estimate for the norm.

If $b=0$ , then

$$ \begin{align*} V_ge_n = 2a\sqrt{\frac{n+1}{n+2}}e_{n+2}, \end{align*} $$

and hence

$$ \begin{align*} \|V_g\| \geq \mathop{\mathrm{sup}}_{n\geq0}\|V_ge_n\| = 2|a|\mathop{\mathrm{sup}}_{n\geq0}\sqrt{\frac{n+1}{n+2}} = 2|a|. \end{align*} $$

The reverse inequality to this follows from (2.1).

To compute the essential norm, we begin by decomposing the operator as $ V_g = V_{g_1} + V_{g_2}, $ where $g_1(w)= aw^2$ and $g_2(w)= bw+c$ . First, since $V_{g_2}$ is compact, we have

$$ \begin{align*} \|V_g\|_e = \|V_{g_1}+ V_{g_2}\|_e =\|V_{g_1}\|_e. \end{align*} $$

Next, we compute the essential norm of $V_{g_1}$ . Again, using the basis $\{e_n\}$ ,

$$ \begin{align*} \|V_{g_1} e_n\| = \|2a\,\sqrt{\frac{n+1}{n+2}}\,e_{n+2}\|=2|a|\sqrt{\frac{n+1}{n+2}}. \end{align*} $$

Since $\{e_n\}$ is an orthonormal sequence, it converges weakly to zero. Thus, $\|Ke_n\|\to 0$ for every compact operator K on $\mathcal F^2$ . Therefore,

$$ \begin{align*} \|V_{g_1}-K\| \geq \limsup_{n\to\infty}\|(V_{g_1}-K)e_n\| = \limsup_{n\to\infty}\|V_{g_1}e_n\| = 2|a|. \end{align*} $$

Taking the infimum over all compact operators K, we obtain $ \|V_{g_1}\|_e\geq 2|a|. $ On the other hand, from the boundedness estimate established in the first part, we have $ \|V_{g_1}\|_e \le \|V_{g_1}\|= 2|a| $ from which the claim follows.

Proof (i) We may first compute $V_g K_w$ as

$$ \begin{align*} V_g K_w(z) &= \int_0^z (2 a \zeta + b) K_w(\zeta) d\zeta = 2a \int_0^z \zeta e^{\bar{w}\zeta} d\zeta + b \int_0^z e^{\bar{w} \zeta} d\zeta\\& = 2a\bigg( \frac{z e^{\bar{w} z}}{\bar{w}} - \frac{e^{\bar{w} z} - 1}{\bar{w}^2}\bigg) +b \bigg(\frac{e^{\bar{w} z} - 1}{\bar{w}}\bigg) \end{align*} $$

for each $w\in \mathbb {C}$ . Then, the Berezin transform of $V_g$ at each nonzero $w \in \mathbb {C}$ becomes

(2.3) $$ \begin{align} \widetilde{V_g}(w) &= \frac{\big\langle V_g K_w, K_w \big\rangle}{\|K_w\|^2}= \frac{V_g K_w(w)}{e^{|w|^2}} = 2a \frac{w}{\bar{w}} - \frac{2a}{\bar{w}^2} \Big(1 - e^{-|w|^2}\Big) + \frac{b}{\bar{w}} \Big(1 - e^{-|w|^2}\Big)\nonumber\\ &=2a \frac{w}{\bar{w}} + \frac{1 - e^{-|w|^2}}{\bar{w}^2}(b\bar{w} - 2a). \end{align} $$

For $w=0$ , it follows that $ V_g K_0(z) = \int _0^z (2 a \zeta + b) d\zeta = az^2 + bz$ , and hence $ \widetilde {V_g}(0) = 0$ . From this and (2.3), the claim follows.

(ii) Suppose first that $b=0$ and $a\neq 0$ . In this case,

$$ \begin{align*} \widetilde{V_g}(re^{i\theta}) = 2a\left( 1-\frac{1-e^{-r^2}}{r^2} \right)e^{2i\theta}. \end{align*} $$

The function

$$ \begin{align*} r\longmapsto 1-\frac{1-e^{-r^2}}{r^2} \end{align*} $$

is continuous and strictly increasing from $0$ to $1$ on $(0,\infty )$ . As $\theta $ varies, $e^{2i\theta }$ traces the unit circle. Therefore, $ \mathrm{Ber}(V_g) = \{z\in \mathbb C:|z|<2|a|\} $ .

Suppose next that $a=0$ . Then,

$$ \begin{align*} \widetilde{V_g}(re^{i\theta}) = b\frac{1-e^{-r^2}}{r}e^{i\theta}. \end{align*} $$

Set

$$ \begin{align*} \rho_0 = \max_{r>0}\frac{1-e^{-r^2}}{r}. \end{align*} $$

The function appearing in this expression is continuous, tends to zero both as $r\to~0$ and as $r\to \infty $ , and attains its maximum at some point in $(0,\infty )$ . Its range is consequently $[0,\rho _0]$ . Since $e^{i\theta }$ traces the unit circle, it follows that $ \mathrm{Ber}(V_g) = \{z\in \mathbb C:|z|\leq |b|\rho _0\} $ .

(iii) By Theorem 2.1, $ \|V_g\|_{\mathrm{ber}} = \mathrm {sup}_{w\in \mathbb C}\|V_gk_w\| \leq \|V_g\| = \sqrt {4|a|^2+|b|^2}. $ To show that this estimate is sharp, suppose first that $b\neq 0$ and set $ w_0=\frac {2\overline a}{\overline b}. $ The normalized reproducing kernel $k_{w_0}$ has the expansion $ k_{w_0}=\sum _{n=0}^{\infty }c_ne_n, $ where

$$ \begin{align*} c_n = \exp\left(\!\!-\frac{\!2|a|^2}{|b|^2}\right) \frac{(2a/b)^n}{\sqrt{n!}}. \end{align*} $$

A direct computation gives $ b\sqrt {n+1}\,c_{n+1}=2ac_n, \ n\geq 0. $ It follows that

$$ \begin{align*} \frac{2a}{\sqrt{n+2}}c_n - b\sqrt{\frac{n+1}{n+2}}c_{n+1} = \frac{1}{\sqrt{n+2}} \left(2ac_n-b\sqrt{n+1}\,c_{n+1}\right) = 0 \end{align*} $$

for every $n\geq 0$ . Therefore, applying identity (2.1) to $f=k_{w_0}$ yields

$$ \begin{align*} \left(4|a|^2+|b|^2\right)\|k_{w_0}\|^2 - \|V_gk_{w_0}\|^2 = 0\ \text{and}\ \|V_gk_{w_0}\|^2=4|a|^2+|b|^2. \end{align*} $$

Hence,

$$ \begin{align*} \|V_g\|_{\mathrm{ber}} \geq \|V_gk_{w_0}\| = \sqrt{4|a|^2+|b|^2}, \end{align*} $$

which proves the required lower estimate.

Suppose now that $b=0$ . In this case, we have

$$ \begin{align*} \|V_gk_w\|^2 = 4|a|^2e^{-|w|^2} \sum_{n=0}^{\infty} \frac{n+1}{n+2}\frac{|w|^{2n}}{n!}. \end{align*} $$

Since

$$ \begin{align*} \frac{n+1}{n+2} = 1-\frac{1}{n+2}, \end{align*} $$

we obtain

$$ \begin{align*} \|V_gk_w\|^2 = 4|a|^2 \left( 1- e^{-|w|^2} \sum_{n=0}^{\infty} \frac{|w|^{2n}}{(n+2)n!} \right). \end{align*} $$

Furthermore,

$$ \begin{align*} e^{-|w|^2} \sum_{n=0}^{\infty} \frac{|w|^{2n}}{(n+2)n!} = \int_0^1te^{-|w|^2(1-t)}\,dt \longrightarrow0 \end{align*} $$

as $|w|\to \infty $ . Therefore, $ \mathrm {sup}_{w\in \mathbb C}\|V_gk_w\| = 2|a|. $ Since $b=0$ , this gives $ \|V_g\|_{\mathrm{ber}} = 2|a| = \sqrt {4|a|^2+|b|^2} = \|V_g\|$ .

Proof (i) For $w=re^{i\theta }$ , formula (2.3) gives

(2.5) $$ \begin{align} \widetilde{V_g}(re^{i\theta}) = 2a\Big( 1-\frac{1-e^{-r^2}}{r^2} \Big)e^{2i\theta} + b\frac{1-e^{-r^2}}{r}e^{i\theta}. \end{align} $$

For fixed $r>0$ , the triangle inequality yields

$$ \begin{align*} \left| \widetilde{V_g}(re^{i\theta}) \right| \leq 2|a|\Big( 1-\frac{1-e^{-r^2}}{r^2} \Big) + |b|\frac{1-e^{-r^2}}{r} = M_{(a,b)}(r). \end{align*} $$

This estimate is sharp. Indeed, if $ab\neq 0$ , choose $ \theta =\arg b-\arg a \ \mathrm {mod}2\pi. $ Then, $ \arg a+2\theta = 2\arg b-\arg a $ and $ \arg b+\theta = 2\arg b-\arg a, $ so the two terms in (2.5) have the same argument. The cases in which $a=0$ or $b=0$ are immediate. Therefore,

$$ \begin{align*} \max_{\theta\in[0,2\pi)} \left| \widetilde{V_g}(re^{i\theta}) \right| = M_{(a,b)}(r), \end{align*} $$

and hence $ \mathrm{ber}(V_g)=\mathrm {sup}_{r>0}M_{(a,b)}(r). $

(ii) If $b=0$ , then

$$ \begin{align*} M_{(a,0)}(r) = 2|a|\Big( 1-\frac{1-e^{-r^2}}{r^2} \Big) \end{align*} $$

and setting

$$ \begin{align*} F(r) = 1-\frac{1-e^{-r^2}}{r^2}, \end{align*} $$

a direct computation gives

$$ \begin{align*} F'(r) = \frac{2}{r^3} \left( 1-(1+r^2)e^{-r^2} \right). \end{align*} $$

Since $e^{r^2}>1+r^2$ for every $r>0$ , we have $ 1-(1+r^2)e^{-r^2}>0. $ Thus, F is strictly increasing on $(0,\infty )$ . Moreover,

$$ \begin{align*} \lim_{r\to0}F(r)=0 \ \text{ while }\ \lim_{r\to\infty}F(r)=1. \end{align*} $$

It follows that $ \mathrm{ber}(V_g)=2|a|. $ Since $F(r)<1$ for every finite r, this supremum is not attained when $a\neq 0$ .

(iii) Suppose next that $a=0$ and $b\neq 0$ . In this case,

$$ \begin{align*} M_{(0,b)}(r) = |b|\frac{1-e^{-r^2}}{r}, \end{align*} $$

and differentiation yields

$$ \begin{align*} M_{(0,b)}'(r) = \frac{|b|}{r^2} \left( (2r^2+1)e^{-r^2}-1 \right). \end{align*} $$

Thus, the critical points are determined by the equation $ e^{r^2}=2r^2+1. $ Setting again $ q(x)=e^x-2x-1, \ x\geq 0, $ we have $ q(0)=0, \ q'(x)=e^x-2, $ and $ q"(x)=e^x>0. $ Hence, q decreases initially, has a unique minimum at $x=\log 2$ , and then increases to $\infty $ . It follows that, besides the solution $x=0$ , the equation $q(x)=0$ has exactly one positive solution $x_0$ . Therefore, $ r_0=\sqrt {x_0} $ is the unique positive critical point of $M_{(0,b)}$ . Moreover, $ M_{(0,b)}'(r)>0 $ for $0<r<r_0$ , whereas $ M_{(0,b)}'(r)<0 $ for $r>r_0$ . Thus, $r_0$ is the unique global maximizer and

$$ \begin{align*} \mathrm{ber}(V_g) = |b|\frac{1-e^{-r_0^2}}{r_0}. \end{align*} $$

(iv) Suppose now that $ab\neq 0$ . Differentiating $M_{(a,b)}$ gives the equation

(2.6) $$ \begin{align} M_{(a,b)}'(r) = \frac{4|a|}{r^3} \left( 1-(1+r^2)e^{-r^2} \right) + \frac{|b|}{r^2} \left( (2r^2+1)e^{-r^2}-1 \right). \end{align} $$

The first term in (2.6) is strictly positive for every $r>0$ . By the definition of $r_0$ , the second term is nonnegative on $(0,r_0]$ . Consequently, $ M_{(a,b)}'(r)>0, $ for $0<r\leq r_0. $ Every critical point of $M_{(a,b)}$ must therefore belong to $(r_0,\infty )$ .

For $r>r_0$ , equation $M_{(a,b)}'(r)=0$ is equivalent to $ \frac {|b|}{2|a|} = Q(r), $ where

$$ \begin{align*} Q(r) = \frac{ 2\left(1-(1+r^2)e^{-r^2}\right) }{ r\left(1-(2r^2+1)e^{-r^2}\right) }. \end{align*} $$

The denominator is positive for $r>r_0$ . Differentiating Q and simplifying shows that $ Q'(r)<0, \ r>r_0. $ Indeed, after multiplying $Q'(r)$ by the positive quantity

$$ \begin{align*} \frac{r^2e^{2r^2}}{ 2\left(1-(2r^2+1)e^{-r^2}\right)^2 }, \end{align*} $$

its sign is the sign of $ -e^{2r^2} + \left ( 2+5r^2-2r^4 \right )e^{r^2} - \left ( 1+5r^2+2r^4 \right ), $ which is strictly negative for $r>r_0$ . To prove this, we argue as follows. Set $x=r^2$ and $x_0=r_0^2$ . By the definition of $r_0$ , $x_0$ is the unique positive solution of $ e^{x_0}=1+2x_0. $

Consider the function $ q(x)=e^x-1-2x. $ Since $ q'(x)=e^x-2, $ q is strictly increasing on $(\log 2,\infty )$ . Moreover, $ q(\log 2) = 1-2\log 2<0, $ so its positive zero satisfies $ x_0>\log 2>\frac 12. $ Therefore, for every $x>x_0$ , we have $ e^x>1+2x $ and $x>1/2$ .

For fixed $x>x_0$ , define the quadratic polynomial

$$ \begin{align*} P_x(y) = y^2 - \left( 2+5x-2x^2 \right)y + 1+5x+2x^2. \end{align*} $$

The expression under consideration can be written as

$$ \begin{align*} -e^{2x} + \left( 2+5x-2x^2 \right)e^x - \left( 1+5x+2x^2 \right) = -P_x(e^x). \end{align*} $$

It is therefore enough to prove that $P_x(e^x)>0$ .

Differentiating $P_x$ with respect to y gives $ P_x'(y) = 2y-\left (2+5x-2x^2\right ). $

For $y\geq 1+2x$ , we also have

$$ \begin{align*} P_x'(y) \geq 2(1+2x) - \left( 2+5x-2x^2 \right) = x(2x-1)>0. \end{align*} $$

Thus, $P_x$ is strictly increasing on $[1+2x,\infty ),$ and hence $ P_x(e^x)>P_x(1+2x) $ as $e^x>1+2x$ .

A direct calculation yields

$$ \begin{align*} P_x(1+2x) = (1+2x)^2 - \left( 2+5x-2x^2 \right)(1+2x) + 1+5x+2x^2 = 2x^2(2x-1)>0. \end{align*} $$

Consequently, $ P_x(e^x)>0 $ and $ -e^{2x} + \left ( 2+5x-2x^2 \right )e^x - \left ( 1+5x+2x^2 \right ) <0. $ Substituting $x=r^2$ gives

$$ \begin{align*} -e^{2r^2} + \left( 2+5r^2-2r^4 \right)e^{r^2} - \left( 1+5r^2+2r^4 \right) <0 \end{align*} $$

for every $r>r_0$ . Therefore, $ Q'(r)<0, \ r>r_0 $ as required.

Now,

$$ \begin{align*} \lim_{r\to r_0}Q(r)=+\infty \ \text{and}\ \lim_{r\to\infty}Q(r)=0. \end{align*} $$

Therefore, the equation $ Q(r)=|b|/(2|a|)$ has a unique solution $r_{(a,b)}\in (r_0,\infty )$ . Equivalently, $r_{(a,b)}$ is the unique solution of (2.4).

Finally,

$$ \begin{align*} \lim_{r\to0}M_{(a,b)}(r)=0 \ \text{and}\ M_{(a,b)}(r) = 2|a| + \frac{|b|}{r} - \frac{2|a|}{r^2} + o\left(\frac{1}{r^2}\right) , \ r\to\infty. \end{align*} $$

In particular, $ M_{(a,b)}(r)>2|a| $ for all sufficiently large r. Since $M_{(a,b)}$ has only one critical point, this point is its unique global maximizer. Hence, $ \mathrm{ber}(V_g)=M_{(a,b)}(r_{(a,b)}).$

(v) It remains to prove the condition for equality of the Berezin number and the Berezin norm. If $b=0$ , then $ \mathrm{ber}(V_g) = 2|a| = \sqrt {4|a|^2+|b|^2} = \|V_g\|_{\mathrm{ber}}. $

Conversely, suppose that $b\neq 0$ . For each $w\in \mathbb C$ , the Cauchy–Schwarz inequality gives $ \left | \widetilde {V_g}(w) \right | = \left | \langle V_gk_w,k_w\rangle \right | \leq \|V_gk_w\|. $ Equality can hold only if $V_gk_w$ is a scalar multiple of $k_w$ . Suppose that $ V_gk_w=\lambda k_w. $ Since $ V_gk_w(0)=0 \ \text {and}\ k_w(0)=e^{-|w|^2/2}\neq 0, $ we obtain $\lambda =0$ . This would imply $V_gk_w=0$ , which is impossible because $ (V_gk_w)'(z) = (2az+b)k_w(z) $ is not identically zero. Therefore, $ \big | \widetilde {V_g}(w) \big | < \|V_gk_w\| $ for every $w\in \mathbb C$ and

$$ \begin{align*} \limsup_{|w|\to\infty} \big| \widetilde{V_g}(w) \big| = 2|a| < \sqrt{4|a|^2+|b|^2}. \end{align*} $$

Thus, outside a sufficiently large disk, the modulus of the Berezin transform is uniformly smaller than $\|V_g\|_{\mathrm{ber}}$ . On the remaining compact disk, it attains its maximum, and the pointwise strict inequality above shows that this maximum is also strictly smaller than $\|V_g\|_{\mathrm{ber}}$ , and hence $ \mathrm{ber}(V_g) < \sqrt {4|a|^2+|b|^2} = \|V_g\|_{\mathrm{ber}}. $

Proof Define the map $ \Phi :[0,\infty )\times [0,2\pi ]\longrightarrow \mathbb C $ by

$$ \begin{align*} \Phi(r,\theta) = A(r)e^{2i\theta}+B(r)e^{i\theta} \end{align*} $$

for $r>0$ , and set $ \Phi (0,\theta )=0 $ for every $\theta \in [0,2\pi ]$ . Since A and B are continuous on $(0,\infty )$ and satisfy

$$ \begin{align*} \lim_{r\to0}A(r)=0 \ \text{and}\ \lim_{r\to0}B(r)=0, \end{align*} $$

the map $\Phi $ is continuous on $[0,\infty )\times [0,2\pi ]$ . Moreover,

$$ \begin{align*} \mathrm{Ber}(V_g) = \Phi\big([0,\infty)\times[0,2\pi]\big). \end{align*} $$

We first show that every point in $\mathrm{Ber}(V_g)$ can be connected to $0$ by a path lying entirely in $\mathrm{Ber}(V_g)$ . Let $z\in \mathrm{Ber}(V_g)$ . If $z=0$ , we may take the constant path. If $z\neq 0$ , then there exist $r_0>0$ and $\theta _0\in [0,2\pi ]$ such that $ z=\Phi (r_0,\theta _0). $ Consider the curve $ \gamma _z(t) = \Phi (\mathrm{tr}_0,\theta _0), \ t\in [0,1]. $ The continuity of $\Phi $ shows that $\gamma _z$ is continuous. In addition, $ \gamma _z(0)=0 \ \text {and}\ \gamma _z(1)=z. $ For every $t\in [0,1]$ , we also have $ \gamma _z(t)\in \mathrm{Ber}(V_g). $ Thus, every point of $\mathrm{Ber}(V_g)$ can be joined to $0$ by a continuous path contained entirely in $\mathrm{Ber}(V_g)$ .

Now, let $z_1,z_2\in \mathrm{Ber}(V_g)$ . Choose $r_1,r_2\geq 0$ and $\theta _1,\theta _2\in [0,2\pi ]$ such that

$$ \begin{align*} z_1=\Phi(r_1,\theta_1) \ \text{and}\ z_2=\Phi(r_2,\theta_2). \end{align*} $$

Define

$$ \begin{align*} \gamma(t) = \begin{cases} \Phi((1-2t)r_1,\theta_1), & 0\leq t\leq\frac12,\\ \Phi((2t-1)r_2,\theta_2), & \frac12\leq t\leq1. \end{cases} \end{align*} $$

Then,

$$ \begin{align*} \gamma(0)=z_1, \ \gamma\left(\frac12\right)=0, \ \gamma(1)=z_2. \end{align*} $$

The two parts of $\gamma $ are continuous and agree at $t=1/2$ because $ \Phi (0,\theta _1) = 0 = \Phi (0,\theta _2). $ Hence, $\gamma $ is a continuous path in $\mathrm{Ber}(V_g)$ joining $z_1$ to $z_2$ . Therefore, $\mathrm{Ber}(V_g)$ is path connected and, consequently, connected.

(ii) If $ab=0$ , then, by Theorem 2.2, $ \mathrm{Ber}(V_g) = \{ 0\} $ , $ \mathrm{Ber}(V_g) = \{z\in \mathbb C:|z|<2|a|\}, $ or $ \mathrm{Ber}(V_g) = \{z\in \mathbb C:|z|\leq |b|\rho _0\} $ all of which are convex.

(iii) If $a=0$ , the assertion is immediate since $ \{z\in \mathbb C:|z|<2|a|\} = \varnothing. $ Suppose, therefore, that $a\neq 0$ . Using $A(r)$ and $B(r)$ as in (2.2) and since

$$ \begin{align*} \lim_{r\to\infty}A(r)=2a \ \text{and}\ \lim_{r\to\infty}B(r)=0, \end{align*} $$

we obtain $ \widetilde {V_g}(re^{i\theta }) \to 2ae^{2i\theta } $ as $r\to \infty $ for every $\theta \in [0,2\pi )$ . Therefore,

$$ \begin{align*} \left\{ 2ae^{2i\theta}: \theta\in[0,2\pi) \right\} \subseteq \overline{\mathrm{Ber}(V_g)}. \end{align*} $$

Thus, the closure of $\mathrm{Ber}(V_g)$ contains the circle centered at the origin with radius $2|a|$ .

Suppose now that $\mathrm{Ber}(V_g)$ is convex. Its closure is then also convex. Since a convex set containing a circle must contain its closed convex hull, it follows that

$$ \begin{align*} \left\{ z\in\mathbb C:|z|\leq2|a| \right\} \subseteq \overline{\mathrm{Ber}(V_g)}. \end{align*} $$

We next use the standard fact that a convex subset of a finite-dimensional vector space contains the interior of its closure. Since $a\neq 0$ , the disk above has nonempty interior, and hence $ \left \{ z\in \mathbb C:|z|<2|a| \right \} \subseteq \mathrm{Ber}(V_g). $

(iv) Writing $w=re^{i\theta }$ , we have

$$ \begin{align*} \widetilde{V_g}(re^{i\theta}) = 2a\Big( 1-\frac{1-e^{-r^2}}{r^2} \Big)e^{2i\theta} + b\frac{1-e^{-r^2}}{r}e^{i\theta}. \end{align*} $$

For each fixed $r>0$ , the triangle inequality gives

$$ \begin{align*} \left| \widetilde{V_g}(re^{i\theta}) \right| \leq M_{(a,b)}(r). \end{align*} $$

Equality holds when the two terms in the expression for $\widetilde {V_g}(re^{i\theta })$ have the same argument. This occurs when $ \theta _0 = \arg b-\arg a. $ Indeed, $ \arg a+2\theta _0 = 2\arg b-\arg a $ and $ \arg b+\theta _0 = 2\arg b-\arg a. $ Consequently,

$$ \begin{align*} \widetilde{V_g}(r_{(a,b)}e^{i\theta_0}) = \mathrm{ber}(V_g)e^{i(2\arg b-\arg a)}. \end{align*} $$

Thus, the extremal point

$$ \begin{align*} z_0 = \mathrm{ber}(V_g)e^{i(2\arg b-\arg a)} \end{align*} $$

belongs to $\mathrm{Ber}(V_g)$ . Moreover, since $a\neq 0$ , the limiting behavior $ \widetilde {V_g}(re^{i\theta }) \longrightarrow 2ae^{2i\theta } $ as $r\to \infty $ shows that $ \left \{ z\in \mathbb C:|z|<2|a| \right \} \subseteq \mathrm{Ber}(V_g) $ whenever $\mathrm{Ber}(V_g)$ is convex.

Suppose now that $\mathrm{Ber}(V_g)$ is convex. Since it contains both the open disk of radius $2|a|$ and the point $z_0$ , it must contain their convex hull. Therefore,

$$ \begin{align*} \mathrm{conv} \left( \left\{ z\in\mathbb C:|z|<2|a| \right\} \cup \{z_0\} \right) \subseteq \mathrm{Ber}(V_g), \end{align*} $$

which is the desired conclusion. In particular, since $0$ and $z_0$ belong to $\mathrm{Ber}(V_g)$ , convexity requires $ tz_0\in \mathrm{Ber}(V_g) $ for every $t\in [0,1]$ .

Proof (i) Since $g'(z)=2az+b$ , we already have that

$$ \begin{align*} V_ge_n = \frac{b}{\sqrt{n+1}}e_{n+1} + 2a\sqrt{\frac{n+1}{n+2}}e_{n+2}. \end{align*} $$

Let $f=\sum _{n=0}^{\infty }c_ne_n$ be a unit vector in $\mathcal F^2$ . Then, using the preceding formula and the orthonormality of $(e_n)$ , we obtain

$$ \begin{align*} \langle V_gf,f\rangle = \sum_{n=0}^{\infty} \frac{b}{\sqrt{n+1}} c_n\overline{c_{n+1}} + \sum_{n=0}^{\infty} 2a\sqrt{\frac{n+1}{n+2}} c_n\overline{c_{n+2}}. \end{align*} $$

Conversely, every sequence $(c_n)\in \ell ^2(\mathbb N_0)$ satisfying $\sum _{n=0}^{\infty }|c_n|^2=1$ determines a unit vector $f=\sum _{n=0}^{\infty }c_ne_n$ in $\mathcal F^2$ . This proves part (i).

(ii) We next establish the spectral formula for the numerical radius. Recall that for every bounded operator T on a Hilbert space,

$$ \begin{align*} w(T) = \max_{\phi\in[0,2\pi)} \mathop{\mathrm{sup}}_{\|f\|=1} \mathrm{Re} \left( e^{-i\phi}\langle Tf,f\rangle \right). \end{align*} $$

Equivalently, if

$$ \begin{align*} H_\phi = \mathrm{Re}(e^{-i\phi}T) = \frac{e^{-i\phi}T+e^{i\phi}T^*}{2}, \end{align*} $$

then

$$ \begin{align*} w(T) = \max_{\phi\in[0,2\pi)} \mathrm{sup}\sigma(H_\phi). \end{align*} $$

Applying this identity to $T=V_g$ , we have

$$ \begin{align*} w(V_g) = \max_{\phi\in[0,2\pi)} \mathop{\mathrm{sup}}_{\|f\|=1} \mathrm{Re} \left( e^{-i\phi}\langle V_gf,f\rangle \right). \end{align*} $$

Let $f=\sum _{n=0}^{\infty }c_ne_n$ be a unit vector. Part (i) and the triangle inequality give

$$ \begin{align*} \mathrm{Re} \left( e^{-i\phi}\langle V_gf,f\rangle \right) \leq {} |b| \sum_{n=0}^{\infty} \frac{|c_n||c_{n+1}|}{\sqrt{n+1}} + 2|a| \sum_{n=0}^{\infty} \sqrt{\frac{n+1}{n+2}} |c_n||c_{n+2}|. \end{align*} $$

If

$$ \begin{align*} |f| = \sum_{n=0}^{\infty}|c_n|e_n, \end{align*} $$

then $\|f\|=1$ , and the right-hand side of the preceding inequality is precisely $ \left \langle J_{(|a|,|b|)}|f|,|f| \right \rangle. $ Consequently, $ \mathrm{sup}~\sigma (H_\phi ) \leq \mathrm{sup}~\sigma (J_{(|a|,|b|)}) $ for every $\phi \in [0,2\pi )$ , and hence $ w(V_g) \leq \mathrm{sup}~\sigma (J_{(|a|,|b|)}). $

To prove the reverse inequality, suppose first that $ab\neq 0$ , and write $ a=|a|e^{i\alpha }, b=|b|e^{i\beta }. $ Set $ \phi _0=2\beta -\alpha \ \text {and}\ \tau =\alpha -\beta. $ For a nonnegative sequence $(x_n)\in \ell ^2(\mathbb N_0)$ , define $ c_n=e^{in\tau }x_n. $ Then, $ c_n\overline {c_{n+1}} = e^{-i\tau }x_nx_{n+1} $ and $ c_n\overline {c_{n+2}} = e^{-2i\tau }x_nx_{n+2}. $ The definitions of $\phi _0$ and $\tau $ imply that $ e^{-i\phi _0}be^{-i\tau }=|b| $ and $ e^{-i\phi _0}ae^{-2i\tau }=|a|. $ Therefore,

$$ \begin{align*} \mathrm{Re} \left( e^{-i\phi_0}\langle V_gf,f\rangle \right) ={} |b| \sum_{n=0}^{\infty} \frac{x_nx_{n+1}}{\sqrt{n+1}} + 2|a| \sum_{n=0}^{\infty} \sqrt{\frac{n+1}{n+2}} x_nx_{n+2} ={} \langle J_{(|a|,|b|)}x,x\rangle. \end{align*} $$

Taking the supremum over all nonnegative unit vectors x gives $ \mathrm{sup}~\sigma (H_{\phi _0}) \geq \mathrm{sup}~\sigma (J_{(|a|,|b|)}). $ Here, there is no loss in restricting to nonnegative vectors because the matrix entries of $J_{(|a|,|b|)}$ are nonnegative and $ \langle J_{(|a|,|b|)}x,x\rangle \leq |\langle J_{(|a|,|b|)}x,x\rangle | \leq \langle J_{(|a|,|b|)}|x|,|x|\rangle $ for every complex unit vector x.

If $a=0$ or $b=0$ , the same argument applies after choosing the phase so that the single nonzero coefficient becomes positive. We conclude that $ w(V_g) = \mathrm{sup}~\sigma (J_{(|a|,|b|)}). $

(iii) By (1.3), we already have $ \mathrm{ber}(V_g) \leq w(V_g) \leq \|V_g\| = \sqrt {4|a|^2+|b|^2}. $ Suppose first that $ w(V_g)=\|V_g\|. $ Then, there exists a sequence of unit vectors $(f_k)$ such that $ \left | \langle V_gf_k,f_k\rangle \right | \longrightarrow \|V_g\|. $ For each k, choose $\lambda _k\in \mathbb C$ such that $ |\lambda _k|=\|V_g\| $ and $ \overline {\lambda _k}\langle V_gf_k,f_k\rangle = \|V_g\| \left | \langle V_gf_k,f_k\rangle \right |. $ The sequence $(\lambda _k)$ lies on the compact circle of radius $\|V_g\|$ . After passing to a subsequence, we may therefore assume that $ \lambda _k\longrightarrow \lambda, \ |\lambda |=\|V_g\|. $ Furthermore,

$$ \begin{align*} \|\!V_gf_k-\lambda_kf_k\|^2 ={} \|\!V_gf_k\|^2+\|\!V_g\|^2 - 2\mathrm{Re} \left( \overline{\lambda_k} \langle V_gf_k,f_k\rangle \!\right) \leq{} 2\|\!V_g\|^2 - 2\|\!V_g\| \left| \langle V_gf_k,f_k\rangle \right|. \end{align*} $$

The right-hand side tends to zero, and hence $ \|V_gf_k-\lambda _kf_k\|\longrightarrow 0. $ Since $\lambda _k\to \lambda $ , we also have

$$ \begin{align*} \|V_gf_k-\lambda f_k\| \leq \|V_gf_k-\lambda_kf_k\| + |\lambda_k-\lambda| \longrightarrow0. \end{align*} $$

Thus, $\lambda $ belongs to the approximate point spectrum of $V_g$ and, therefore, to $\sigma (V_g)$ . It follows that $ \|V_g\| = |\lambda | \leq r(V_g). $ Since the spectral radius never exceeds the operator norm, we obtain $ r(V_g)=\|V_g\|. $ Using $ r(V_g)=2|a| \ \text {and}\ \|V_g\|=\sqrt {4|a|^2+|b|^2}, $ we conclude that $b=0$ .

Conversely, if $b=0$ , then $ r(V_g)=2|a|=\|V_g\| $ and the inequalities $ r(V_g)\leq w(V_g)\leq \|V_g\| $ give $ w(V_g)=\|V_g\|=2|a|. $

(iv) Assume $b=0$ and $a\neq 0$ . Then,

$$ \begin{align*} V_ge_n = 2a\sqrt{\frac{n+1}{n+2}}e_{n+2}, \end{align*} $$

so $V_g$ is a weighted shift of step two. For $\theta \in \mathbb R$ , define the diagonal unitary operator $ U_\theta e_n=e^{in\theta }e_n. $ A direct computation gives

$$ \begin{align*} U_\theta^*V_gU_\theta=e^{-2i\theta}V_g. \end{align*} $$

Since $e^{-2i\theta }$ ranges over the unit circle, the numerical range $W(V_g)$ is rotationally invariant. By the Toeplitz–Hausdorff theorem, $W(V_g)$ is convex and is therefore a disk centered at the origin. Its radius is $ w(V_g)=2|a|. $

It remains to determine whether the boundary is attained. Suppose that a unit vector f satisfies $ |\langle V_gf,f\rangle |=2|a|=\|V_g\|. $ Equality in $ |\langle V_gf,f\rangle | \leq \|V_gf\| \leq \|V_g\| $ implies that $ V_gf=\lambda f $ for some $\lambda $ with $|\lambda |=2|a|$ . This is impossible. Indeed, if $ f=\sum _{n=0}^{\infty }c_ne_n\neq~0$ and $n_0$ is the smallest index for which $c_{n_0}\neq 0$ , then the smallest nonzero coefficient of $V_gf$ occurs at the index $n_0+2$ , whereas the smallest nonzero coefficient of $\lambda f$ occurs at $n_0$ . Thus, $V_g$ has no nonzero eigenvalues. The boundary of the disk is consequently not contained in the numerical range, and $ W(V_g) = \left \{ z\in \mathbb C:|z|<2|a| \right \}. $

(v) Suppose next that $a=0$ and $b\neq 0$ . Then,

$$ \begin{align*} V_ge_n = \frac{b}{\sqrt{n+1}}e_{n+1}, \end{align*} $$

so $V_g$ is a compact unilateral weighted shift. As above, diagonal unitary conjugation shows that $W(V_g)$ is rotationally invariant. Hence, it is a disk centered at the origin.

By part (ii), $ w(V_g) = \mathrm{sup}~\sigma (J_{(0,|b|)}) = |b|\mathrm{sup}~\sigma (J_{(0,1)}) = |b|\omega _0. $ The operator $J_{(0,1)}$ is compact and self-adjoint. Moreover, $\mathrm{sup}~\sigma (J_{(0,1)})>0$ since, for example, its quadratic form is positive on a suitable linear combination of $e_0$ and $e_1$ . Therefore, $\omega _0$ is a positive eigenvalue of $J_{(0,1)}$ . It follows from the unitary equivalence used in part (ii) that there exists a unit vector f such that $ \langle V_gf,f\rangle =|b|\omega _0 $ after a suitable rotation. Thus, the numerical radius is attained. Rotational invariance then shows that the entire circle $ \left \{ z\in \mathbb C:|z|=|b|\omega _0 \right \} $ is contained in $W(V_g)$ . Since $W(V_g)$ is convex, it contains the closed disk bounded by this circle. The reverse inclusion follows from the definition of the numerical radius. Consequently, $ W(V_g) = \left \{ z\in \mathbb C:|z|\leq |b|\omega _0 \right \}, $ which proves (v).

(vi) Suppose that $b\neq 0$ . From the formula for the Berezin number,

$$ \begin{align*} \mathrm{ber}(V_g)=\mathop{\mathrm{sup}}_{r>0}M_{(a,b)}(r), \end{align*} $$

where

$$ \begin{align*} M_{(a,b)}(r) = 2|a|\left( 1-\frac{1-e^{-r^2}}{r^2} \right) + |b|\frac{1-e^{-r^2}}{r}. \end{align*} $$

As $r\to \infty $ ,

$$ \begin{align*} M_{(a,b)}(r) = 2|a| + \frac{|b|}{r} - \frac{2|a|}{r^2} + o\left(\frac{1}{r^2}\right). \end{align*} $$

Since $b\neq 0$ , this implies $ M_{(a,b)}(r)>2|a| $ for all sufficiently large r. Hence, $ \mathrm{ber}(V_g)>2|a|.$ Combining this with $ \mathrm{ber}(V_g)\leq w(V_g) $ and the strict form of part (iii) when $b\neq 0$ , we obtain $ 2|a| < \mathrm{ber}(V_g) \leq w(V_g) < \sqrt {4|a|^2+|b|^2}. $