# The Szász inequality for matrix polynomials and functional calculus

Piotr Pikul and Oskar Jakub Szymański and Michał Wojtylak

###### Abstract.

The Szász inequality is a classical result providing a bound for polynomials
with zeros in the upper half of the complex plane in terms of its low-order
coefficients. Some generalisations of this result to polynomials in several
variables were done by Borcea-Brändén and Knese. In this article the
inequality is discussed in the context of matrix polynomials and matrix
variables. As a byproduct we improve the scalar Szász inequality by relaxing
the assumption of the location of zeros.

##  1\. Introduction

A polynomial is called stable if all its zeros lie outside the open upper
half-plane. O. Szász [16] discovered an inequality bounding a stable
polynomial
$p(\lambda)=a_{d}\lambda^{d}+a_{d-1}\lambda^{d-1}+\ldots+a_{1}\lambda+1\in\mathbb{C}[\lambda]$
in terms of its first few coefficients:

$|p(\lambda)|\leq\exp\bigl{(}|a_{1}||\lambda|+3(|a_{1}|^{2}+|a_{2}|)|\lambda|^{2}\bigr{)}.$

The result was used to characterise local uniform limits of stable
polynomials. See [12, Chapter VIII, Theorem 4].

Later on the inequality was improved by de Branges [5, Lemma 5] to

(1.1)
$|p(\lambda)|\leq\exp\bigl{(}\operatorname{Re}(a_{1}\lambda)+{\textstyle\frac{1}{2}}(|a_{1}|^{2}-2\operatorname{Re}(a_{2}))|\lambda|^{2}\bigr{)},\quad\lambda\in\mathbb{C}.$

Knese in [10, Theorem 1.3] provided a simplifed proof and showed sharpness of
the bound on the imaginary axis. The elegant presentation in [10] actually
shows that stability is not a necessary condition for (1.1) to hold. We will
elaborate on that in Corollary 3.3, Proposition 3.6 and Examples 3.4 and 3.5.

However, the main focus of this paper are analogues of the Szász inequality
for matrix polynomials and matrix variables. One result of this kind was
already proved in [17] as Proposition 4.5. Namely, if the numerical range of
the matrix polynomial $P(\lambda)$ is contained in some half-plane
$H_{\varphi}$, $\varphi\in[0;2\pi)$, then

(1.2) $\|P(\lambda)\|\leq 2\exp\Bigl{[}\lambda_{H}\bigl{(}\lambda
A_{1}-|\lambda|^{2}A_{2}\bigr{)}+\frac{1}{2}|\lambda|^{2}\left\|A_{1}\right\|^{2}\Bigr{]},\quad\lambda\in\mathbb{C}.$

where $\lambda_{H}(X)$ is the largest eigenvalue of the matrix
$\operatorname{Re}X=\frac{A+A^{*}}{2}$ (equivalently: maximal real coordinate
of the numerical range).

As shown by Example 3.1 below, in general (hyper)stability of a matrix
polynomial is not enough to ensure a bound of the norm based exclusively on
the low order coefficients. However, assuming additionally a factorisation of
the polynomial we receive the inequality

$\left\|P(\lambda)\right\|_{F}\leq\sqrt{n^{d}}\exp\Bigl{(}\frac{1}{n}\operatorname{tr}\operatorname{Re}(\lambda
A_{1})+\frac{1}{2n}\big{(}\left\|A_{1}\right\|_{F}^{2}-2\operatorname{tr}\operatorname{Re}A_{2}\big{)}|\lambda|^{2}\Bigr{)},$

see Theorem 3.2 below.

Another topic considered in the current paper are the scalar polynomials in
one matrix variable. In Propositions 4.1, 4.2 and Theorem 4.4 we estimate the
norm of $p(A)$ for a matrix $A$ and a stable scalar polynomial $p$ in various
ways

$\displaystyle\left\|p(A)\right\|_{F}\leq$ $\displaystyle\
\sqrt{n^{d}}\exp\Bigl{(}\frac{1}{n}\operatorname{tr}\operatorname{Re}(a_{1}A)+\frac{1}{2n}(|a_{1}|^{2}-2\operatorname{Re}a_{2})\left\|A\right\|_{F}^{2}\Bigr{)}$
$\displaystyle\left\|p(A)\right\|\leq$ $\displaystyle\
\exp\Bigl{(}\left\|A\right\|\sqrt{d(|a_{1}|^{2}-2\operatorname{Re}a_{2})}\Bigr{)}$
$\displaystyle\left\|p(A)\right\|\leq$ $\displaystyle\
\exp\bigl{(}|a_{1}|\left\|A\right\|+{\textstyle\frac{1}{2}}(|a_{1}|^{2}-2\operatorname{Re}(a_{2}))\left\|A\right\|^{2}\bigr{)}.$

Although the first inequality seems initially to be the weakest one due to the
$\sqrt{n^{d}}$ factor, apparently it is the only one of those three that can
beat the von Neumann inequality, see Example 4.5.

Further, we generalise scalar results on multivariate polynomials to the
matrix variables. We always assume that the involved matrices commute.
Proposition 5.3 provides a Szász type inequality for scalar polynomials of two
matrix variables.

$\left\|p(A_{1},A_{2})\right\|\leq\exp\bigg{(}\\!\left\|A_{1}\right\||p_{1}|+\left\|A_{2}\right\||p_{2}|+\frac{1}{2}\left\|A\right\|^{2}_{\infty}\Big{(}\big{|}p_{1}+p_{2}\big{|}^{2}-\operatorname{Re}\\!\sum_{j,k=1}^{2}p_{j,k}\Big{)}\bigg{)}$

In the above formula, $p_{j}$ and $p_{j,k}$ denote partial derivatives of
order 1 and 2 at $(0,0)$. For more than two variables the analogous
inequalities are provided by Proposition 5.2 and Theorem 5.3.

Finally, in Section 6 we apply the results to matrix polynomial in matrix
variables, understood in two ways. Fist we consider the complete version of
the inequality. Subsequently, we discuss the direct substitution
$P(T):=\sum_{j=0}^{d}A_{j}T^{j}$.

##  2\. Preliminaries

Throughout this article we denote fields of real and complex numbers by
$\mathbb{R}$ and $\mathbb{C}$ respectively. $\mathbb{N}$ stands for positive
and $\mathbb{Z}_{+}$ for nonnegative integers. Symbol ${\mathbb{C}}^{n\times
n}$ stands for the set of square $n\times n$ matrices with complex entries.
The set of matrix polynomials of $k$ variables with square $n\times n$
coefficients will be denoted by ${\mathbb{C}}^{n\times
n}[z_{1},\ldots,z_{k}]$. Matrix polynomials will be denoted by capital letters
to be easily distinguishable from scalar ones.

A scalar polynomial $p\in\mathbb{C}[z_{1},\ldots,z_{k}]$ is called stable with
respect to the set $D\subseteq\mathbb{C}^{k}$ if all its zeros lie outside
$D$. If the set is not specified explicitly, we assume it is a product of
upper half planes (i.e.
$D=H_{0}^{k}=\\{z\in\mathbb{C}^{k}:\operatorname{Im}(z_{j})\geq 0,\
j=1,\ldots,k\\}$). For matrix polynomial $P\in{\mathbb{C}}^{n\times
n}[\lambda]$ stability with respect to $D\subseteq\mathbb{C}$ means that
$P(\lambda)$ is invertible for every $\lambda\in D$. Point
$\lambda\in\mathbb{C}$ that $\ker P(\lambda)\neq\\{0\\}$ is called
_eigenvalue_ of the matrix polynomial $P$. The numerical range of $P(\lambda)$
is defined as

$\\{\lambda\in\mathbb{C}:x^{*}P(\lambda)x=0\text{ for some
}x\in\mathbb{C}^{n}\setminus\\{0\\}\\}.$

For a matrix $A\in{\mathbb{C}}^{n\times n}$ we denote its standard operator
norm by $\|A\|:=\sup\big{\\{}\|Ax\|:x\in\mathbb{C}^{n},\ \|x\|=1\big{\\}}$. We
will also use the Frobenius norm
$\left\|A\right\|_{F}:=\sqrt{\operatorname{tr}(A^{*}A)}$ and the matrix
imaginary part (skew part of the matrix)
$\operatorname{Im}A:=\frac{1}{2i}(A-A^{*})$. For a vector $z\in\mathbb{C}^{n}$
we denote its standard euclidean norm by $\left\|z\right\|$, and
$\left\|z\right\|_{\infty}$ stands for the supremum norm.

We provide now several lemmas, which will be used later on. Note that for
$A\in{\mathbb{C}}^{n\times n}\setminus\\{-I\\}$ one easily obtains

(2.1) $\log\left\|I+A\right\|\leq\left\|A\right\|.$

We present now a more subtle estimate.

######  Lemma 2.1.

Let $A\in{\mathbb{C}}^{n\times n}\setminus\\{-n^{-1/2}I\\}$. Then

$\log\left\|\frac{1}{\sqrt{n}}I+A\right\|_{F}\leq\frac{1}{\sqrt{n}}\operatorname{tr}\operatorname{Re}A+\frac{1}{2}\left\|A\right\|_{F}^{2}\text{.}$

###### Proof.

The proof is based on a direct computation:

$\displaystyle\log\left\|\frac{1}{\sqrt{n}}I+A\right\|_{F}$
$\displaystyle=\frac{1}{2}\log\left\|\frac{1}{\sqrt{n}}I+A\right\|_{F}^{2}$
$\displaystyle=\frac{1}{2}\log\operatorname{tr}\Bigl{[}\Bigl{(}\frac{1}{\sqrt{n}}I+A\Bigr{)}^{*}\Bigl{(}\frac{1}{\sqrt{n}}I+A\Bigr{)}\Bigr{]}$
$\displaystyle=\frac{1}{2}\log\operatorname{tr}\Big{(}\frac{1}{n}I+\frac{1}{\sqrt{n}}A+\frac{1}{\sqrt{n}}A^{*}+A^{*}A\Big{)}$
$\displaystyle=\frac{1}{2}\log\Big{(}1+\frac{2}{\sqrt{n}}\operatorname{tr}\operatorname{Re}A+\left\|A\right\|_{F}^{2}\Big{)}$

Now it is enough to use the scalar inequality $\log(1+x)\leq x$,
$x\in(-1;\infty)$.

$\log\left\|\frac{1}{\sqrt{n}}I+A\right\|_{F}\leq\frac{1}{2}\Big{(}\frac{2}{\sqrt{n}}\operatorname{tr}\operatorname{Re}A+\left\|A\right\|_{F}^{2}\Big{)}=\frac{1}{\sqrt{n}}\operatorname{tr}\operatorname{Re}A+\frac{1}{2}\left\|A\right\|_{F}^{2}\text{.}$

∎

Knese showed in [10, Proof of Lemma 2.1] that for
$b_{1},\dots,b_{d}\in\mathbb{C}$ satisfying $\sum_{1\leq j<k\leq
d}\operatorname{Im}b_{j}\operatorname{Im}b_{k}\geq 0$ one has

(2.2)
$\sum_{j=1}^{d}|b_{j}|^{2}\leq\Big{|}\sum_{j=1}^{d}b_{j}\Big{|}^{2}-2\operatorname{Re}\sum_{1\leq
j<k\leq d}b_{j}b_{k}.$

We present a matrix version of this inequality.

######  Lemma 2.2.

Let $B_{1},B_{2},\dots,B_{d}$ be $n\times n$ complex matrices such that

(2.3) $\sum_{1\leq j<k\leq
d}\operatorname{tr}(\operatorname{Im}B_{j}\operatorname{Im}B_{k})\geq 0.$

Then

$\sum_{j=1}^{d}\left\|B_{j}\right\|_{F}^{2}\leq\biggl{\|}\sum_{j=1}^{d}B_{j}\biggr{\|}_{F}^{2}-2\operatorname{tr}\operatorname{Re}\sum_{1\leq
j<k\leq d}B_{j}B_{k}\text{.}$

###### Proof.

Observe that

$\displaystyle\biggl{\|}\sum_{j=1}^{d}B_{j}\biggr{\|}_{F}^{2}$
$\displaystyle=\operatorname{tr}\biggl{(}\sum_{j=1}^{d}B_{j}^{*}\sum_{k=1}^{d}B_{k}\biggr{)}$
$\displaystyle=\operatorname{tr}\biggl{(}\sum_{j=1}^{d}B_{j}^{*}B_{j}+\sum_{1\leq
j\neq k\leq d}B_{j}^{*}B_{k}\biggr{)}$
$\displaystyle=\operatorname{tr}\sum_{j=1}^{d}B_{j}^{*}B_{j}+\operatorname{tr}\sum_{1\leq
j<k\leq d}(B_{j}^{*}B_{k}+B_{k}^{*}B_{j})$
$\displaystyle=\sum_{j=1}^{d}\operatorname{tr}B_{j}^{*}B_{j}+\operatorname{tr}\sum_{1\leq
j<k\leq d}2\operatorname{Re}(B_{j}^{*}B_{k})$
$\displaystyle=\sum_{j=1}^{d}\left\|B_{j}\right\|_{F}^{2}+2\operatorname{tr}\operatorname{Re}\sum_{1\leq
j<k\leq d}B_{j}^{*}B_{k}\text{,}$

therefore it is sufficient to prove the following statement

$2\operatorname{tr}\operatorname{Re}\sum_{1\leq j<k\leq
d}(B_{j}^{*}B_{k}-B_{j}B_{k})\geq 0.$

To do so, note that

$\displaystyle 2\operatorname{tr}\operatorname{Re}$ $\displaystyle\sum_{1\leq
j<k\leq d}(B_{j}^{*}B_{k}-B_{j}B_{k})$ $\displaystyle=\sum_{1\leq j<k\leq
d}\bigl{(}\operatorname{tr}(B_{j}^{*}B_{k})-\operatorname{tr}(B_{j}B_{k})+\operatorname{tr}(B_{k}^{*}B_{j})-\operatorname{tr}(B_{k}^{*}B_{j}^{*})\bigr{)}$
$\displaystyle=\sum_{1\leq j<k\leq
d}\bigl{(}\operatorname{tr}(B_{j}^{*}B_{k})-\operatorname{tr}(B_{j}B_{k})+\operatorname{tr}(B_{j}B_{k}^{*})-\operatorname{tr}(B_{j}^{*}B_{k}^{*})\bigr{)}$
$\displaystyle=\sum_{1\leq j<k\leq
d}\operatorname{tr}\bigl{(}(B_{j}-B_{j}^{*})(B_{k}^{*}-B_{k})\bigr{)}$
$\displaystyle=4\sum_{1\leq j<k\leq
d}\operatorname{tr}(\operatorname{Im}B_{j}\operatorname{Im}B_{k}),$

which is a non-negative number by assumption. ∎

For a better understanding of the assumption (2.3) we present two lemmas.

######  Lemma 2.3.

Let $B_{1},B_{2},\dots,B_{d}$ be $n\times n$ complex matrices such that
$\operatorname{Im}B_{j}\leq 0$ for $j\in\\{1,2,\dots,d\\}$. Then

$\sum_{1\leq j<k\leq
d}\operatorname{tr}(\operatorname{Im}B_{j}\operatorname{Im}B_{k})\geq 0.$

###### Proof.

Observe that

$\operatorname{tr}(\operatorname{Im}B_{j}\operatorname{Im}B_{k})=\operatorname{tr}\bigl{(}\sqrt{-\operatorname{Im}B_{j}}(-\operatorname{Im}B_{k})(\sqrt{-\operatorname{Im}B_{j}})\bigr{)}\geq
0$

as the matrix
$\sqrt{-\operatorname{Im}B_{j}}(-\operatorname{Im}B_{k})(\sqrt{-\operatorname{Im}B_{j}})$
is positive semi-definite due to the assumption that $\operatorname{Im}B_{j}$
and $\operatorname{Im}B_{k}$ are both nonpositive semi-definite matrices.

∎

The following lemma is elementary.

######  Lemma 2.4.

Let $B_{1},B_{2},\dots,B_{d}$ be $n\times n$ complex matrices, then

$\sum_{1\leq j<k\leq
d}\operatorname{tr}(\operatorname{Im}B_{j}\operatorname{Im}B_{k})=\Big{\|}\sum_{j=1}^{d}\operatorname{Im}B_{j}\Big{\|}^{2}_{F}-\sum_{j=1}^{d}\left\|\operatorname{Im}B_{j}\right\|^{2}_{F}.$

In particular, for $b_{1},\dots,b_{d}\in\mathbb{C}$ one has

$\sum_{1\leq j<k\leq
d}(\operatorname{Im}b_{j}\operatorname{Im}b_{k})=d\bigg{(}(d-1)\operatorname{Im}(\operatorname{mean}(b_{1},\dots,b_{d}))-\operatorname{var}(\operatorname{Im}b_{1},\dots,\operatorname{Im}b_{d})\bigg{)}.$

##  3\. Matrix polynomials in one variable

In [17] the authors asked whether in [17, Proposition 4.5.] the assumption
regarding numerical range can be replaced by hyperstability of $P$. Let us
recall that a matrix polynomial $P\in{\mathbb{C}}^{n\times n}[\lambda]$ is
called _hyperstable_ with respect to $D$ if for every
$x\in\mathbb{C}^{n}\setminus\\{0\\}$ there exists
$y\in\mathbb{C}^{n}\setminus\\{0\\}$ such that $y^{*}P(\lambda)x\neq 0$ for
any $\lambda\in D$. A matrix polynomial with numerical range contained in a
half-plane is always hyperstable, but not conversely, cf. [17]. We begin the
paper with the negative answer to the question above.

######  Example 3.1.

Consider the matrix polynomial

$P(\lambda)=\begin{bmatrix}1&\lambda^{d}\\\ 0&1\end{bmatrix},$

with $d\geq 3$. Then $P(0)=I_{2}$, and $P(\lambda)$ is hyperstable with
respect to any set $D\subseteq\mathbb{C}$ (cf. [17, Proposition 3.3]).
However,

$\left\|P(\lambda)\right\|\geq\frac{1}{\sqrt{2}}\left\|P(\lambda)\right\|_{F}=\sqrt{1+\frac{|\lambda|^{2d}}{2}}.$

Hence, as $A_{1}=A_{2}=0$, there can not be any global bound on
$\left\|P(\lambda)\right\|$ of Szász type.

Let us also compute the numerical range of $P(\lambda)$. Let
$x\in\mathbb{C}^{2}\setminus\\{0\\}$. We have
$x^{*}P(\lambda)x=|x_{1}|^{2}+|x_{2}|^{2}+\lambda^{d}\overline{x}_{1}x_{2}$.
In particular, for $x_{1}=x_{2}=1$, we see that the numerical range of
$P(\lambda)$ is not contained in any half-plane.

The example shows that hyperstability is not enough to obtain any bound
depending on $A_{1}$ and $A_{2}$ only. On the other hand, ordering that the
numerical range of the polynomial lies in a halfplane in (1.2) is a strong
assumption. Mimicking the proof in the scalar case from [10] we provide a
version that requires a factorisation instead of the numerical range
condition. We refer the reader to [13, Section 3] for a relation between
factorisation and the location of the numerical range.

######  Theorem 3.2.

Let $P(\lambda)=I+\sum_{j=1}^{d}\lambda^{j}A_{j}\in{\mathbb{C}}^{n\times
n}[\lambda]$ be a matrix polynomial with the following factorisation
$P(\lambda)=\prod_{j=1}^{d}(I+\lambda B_{j})$, where

(3.1) $\sum_{1\leq j<k\leq
d}\operatorname{tr}(\operatorname{Im}B_{j}\operatorname{Im}B_{k})\geq 0.$

Then the Frobenius norm of $P(\lambda)$ can be estimated from above as

$\left\|P(\lambda)\right\|_{F}\leq\sqrt{n^{d}}\exp\Bigl{(}\frac{1}{n}\operatorname{tr}\operatorname{Re}(\lambda
A_{1})+\frac{1}{2n}\big{(}\left\|A_{1}\right\|_{F}^{2}-2\operatorname{tr}\operatorname{Re}A_{2}\big{)}|\lambda|^{2}\Bigr{)},\quad\lambda\in\mathbb{C}.$

According to Lemma 2.3, if $\operatorname{Im}B_{j}\leq 0$ for all
$j=1,\ldots,d$ then the assumption (3.1) is satisfied. Moreover, the
eigenvalues of the polynomial $P$ are then precisely of the form
$-\frac{1}{\lambda}$ for $\lambda\in\sigma(B_{j})$. Hence, $P$ is stable (with
respect to the open half-plane).

###### Proof.

First, we use Lemma 2.1:

(3.2) $\displaystyle\log\left\|P(\lambda)\right\|_{F}$
$\displaystyle\leq\sum_{j=1}^{d}\log\|I+\lambda B_{j}\|_{F}$
$\displaystyle=d\log\sqrt{n}+\sum_{j=1}^{d}\log\|\frac{1}{\sqrt{n}}I+\lambda\tilde{B}_{j}\|_{F}$
$\displaystyle\leq\log\sqrt{n^{d}}+\frac{1}{\sqrt{n}}\sum_{j=1}^{d}\operatorname{tr}\operatorname{Re}(\lambda\tilde{B}_{j})+\frac{1}{2}\sum_{j=1}^{d}\|\lambda\tilde{B}_{j}\|_{F}^{2}\text{,}$

where $\tilde{B}_{j}:=n^{-1/2}B_{j}$ for $j\in\\{1,2,\dots,d\\}$. Since
$I+\sum_{j=1}^{d}\lambda^{j}A_{j}=\prod_{j=1}^{d}(I+\lambda B_{j})$, we have
$\sum_{j=1}^{d}\tilde{B}_{j}=n^{-1/2}A_{1}$ and $\sum_{1\leq j<k\leq
d}\tilde{B}_{j}\tilde{B}_{k}=n^{-1}A_{2}$. Then, the right side of the last
inequality can be rewritten and estimated by Lemma 2.2 as follows:

$\displaystyle\log\sqrt{n^{d}}+$
$\displaystyle\frac{1}{n}\operatorname{tr}\operatorname{Re}(\lambda
A_{1})+\frac{1}{2}|\lambda|^{2}\sum_{j=1}^{d}\left\|\tilde{B}_{j}\right\|_{F}^{2}$
$\displaystyle\leq$
$\displaystyle\log\sqrt{n^{d}}+\frac{1}{n}\operatorname{tr}\operatorname{Re}(\lambda
A_{1})+\frac{1}{2n}(\left\|A_{1}\right\|_{F}^{2}-2\operatorname{tr}\operatorname{Re}A_{2})|\lambda|^{2}.$

Therefore, we obtain

$\log\left\|P(\lambda)\right\|_{F}\leq\log\sqrt{n^{d}}+\frac{1}{n}\operatorname{tr}\operatorname{Re}(\lambda
A_{1})+\frac{1}{2n}(\left\|A_{1}\right\|_{F}^{2}-2\operatorname{tr}\operatorname{Re}A_{2})|\lambda|^{2}.$

Taking exponent of both sides of the above inequality ends the proof. ∎

Let us discuss condition (3.1) in the scalar case. Let $p$ be a scalar
polynomial with nonzero roots $\lambda_{1},\ldots,\lambda_{d}$. Condition
(3.1) reads as

(3.3) $\sum_{1\leq j<k\leq
d}\operatorname{Im}\frac{1}{\lambda_{j}}\operatorname{Im}\frac{1}{\lambda_{k}}\geq
0.$

Observe that this is equivalent, by Lemma 2.4, to ordering that

(3.4)
$\operatorname{var}\Lambda\leq(d-1)(\operatorname{mean}\Lambda)^{2},\quad\Lambda=\left(\operatorname{Im}\frac{1}{\lambda_{1}},\dots,\operatorname{Im}\frac{1}{\lambda_{d}}\right).$

Clearly, any polynomial stable with respect to the lower or upper half plane
satisfies (3.4). For $p$ of degree not greater than $2$, (3.4) implies
stability with respect to the upper or lower half-plane. For a higher degree
we discuss the condition below.

######  Corollary 3.3.

For a polynomial $p(\lambda)=1+\sum_{j=1}^{d}a_{j}\lambda^{d}$ satisfying
(3.4) the inequality (1.1) holds, i.e.

$|p(\lambda)|\leq\exp\bigl{(}\operatorname{Re}(a_{1}\lambda)+(|a_{1}|^{2}-2\operatorname{Re}a_{2})|\lambda|^{2}\bigr{)},\quad\lambda\in\mathbb{C}.$

For $d\geq 3$ this is a generalisation of Szász inequality onto a broader
class of polynomials. Below we provide an example of a polynomial satisfying
(3.4) which is stable with respect to no half-plane.

######  Example 3.4.

Consider polynomials of the form

$p_{c}(\lambda):=\Big{(}1+\frac{i}{c+1}\lambda\Big{)}\Big{(}1-\frac{i+1}{c}\lambda\Big{)}^{c}\Big{(}1-\frac{i-1}{c}\lambda\Big{)}^{c},$

where $c\in{\mathbb{N}}$. The roots are clearly
$\\{(c+1)i,\frac{c(1-i)}{2},\frac{c(-1-i)}{2}\\}$ and are contained in no
half-plane $\\{\operatorname{Re}e^{\operatorname{i}\theta}\lambda>0\\}$,
$\theta\in[0,2\pi)$. Condition (3.4) can be readily verified, since among the
$2c+1$ imaginary parts involved there are $2c$ values $-\frac{1}{c}$ and a
single $+\frac{1}{c+1}$. The sum in (3.3) equals
$\frac{2c-1}{c}-\frac{2}{c+1}=\frac{2c^{2}-c-1}{c^{2}+c}>0$. Expanding
$p_{c}(\lambda)=1+a_{1}\lambda+a_{2}\lambda+\ldots+a_{2c+1}\lambda^{2c+1}$ we
have $a_{1}=(\frac{1}{c+1}-2)i$ and $a_{2}=\frac{-2c}{c+1}$.

Note that (3.3) can be satisfied even if the numbers of zeros on each side of
the real axis are equal.

######  Example 3.5.

Choosing $k$ roots satisfying
$\operatorname{Im}\frac{1}{\lambda_{1}}=\ldots=\operatorname{Im}\frac{1}{\lambda_{k}}=1$
and the other $k$, with
$\operatorname{Im}\frac{1}{\lambda_{k+1}}=\ldots=\operatorname{Im}\frac{1}{\lambda_{2k}}=-\frac{k+\sqrt{2k-1}}{k-1}$
results in fulfilling (3.3). Indeed, the left-hand-side is equal

$\displaystyle\binom{k}{2}1^{2}+$
$\displaystyle\binom{k}{2}\bigg{(}\frac{k+\sqrt{2k-1}}{k-1}\bigg{)}^{2}-k^{2}\frac{k+\sqrt{2k-1}}{k-1}$
$\displaystyle=\frac{k}{2}\bigg{(}(k-1)+\frac{(k+\sqrt{2k-1})^{2}}{k-1}\bigg{)}-\frac{k^{3}+k^{2}\sqrt{2k-1}}{k-1}$
$\displaystyle=\frac{k}{2}\frac{2k^{2}+2k\sqrt{2k-1}}{k-1}-\frac{k^{3}+k^{2}\sqrt{2k-1}}{k-1}=0.$

We conclude the comment on the condition (3.4) with the following property.

######  Proposition 3.6.

Assume that we draw an infinite sequence $\\{b_{j}\\}_{j=1}^{\infty}$ of
complex numbers independently from a probability distribution on the complex
plane with a non-real mean and a finite variance. Then with probability $1$
there exists $d\in{\mathbb{N}}$ such that the polynomial
$p(\lambda)=\prod_{j=1}^{d}(1+\lambda b_{j})$ satisfies the global Szász
inequality (1.1).

###### Proof.

We may assume without loss of generality that $b_{j}\neq 0$ for $j=1,\dots,d$
as the zero $b_{j}$’s do not contribute to the formula for $p(\lambda)$ and
the mean of the underlying distribution is nonzero. With probability $1$ the
value
$\operatorname{var}(-\operatorname{Im}b_{1},\ldots,-\operatorname{Im}b_{d})$
tends to the variance $\sigma^{2}>0$ of the imaginary part of the random
variable. Analogously, $-\frac{1}{d}\operatorname{Im}\sum_{j=1}^{d}b_{j}$
tends to a nonzero value $\beta$ (imaginary part of the mean of the random
variable). There exists $d_{0}\in{\mathbb{N}}$ such that for $d\geq d_{0}$ we
have simultaneously

$\begin{array}[]{r@{\
}l}\operatorname{var}(-\operatorname{Im}b_{1},\ldots,-\operatorname{Im}b_{d})\leq&\sigma^{2}+|\beta|/2,\\\
\big{|}-\frac{1}{d}\operatorname{Im}\sum_{j=1}^{d}b_{j}|\geq&|\beta|-|\beta|/2=|\beta|/2,\\\
\sigma^{2}+|\beta|/2\leq&(d-1)(|\beta|/2)^{2}.\end{array}$

Combining the above inequalities, we obtain

$\displaystyle\operatorname{var}(-\operatorname{Im}b_{1},\ldots,-\operatorname{Im}b_{d})$
$\displaystyle\leq\sigma^{2}+|\beta|/2$
$\displaystyle\leq(d-1)(|\beta|/2)^{2}$
$\displaystyle\leq(d-1)\Big{(}-\frac{1}{d}\operatorname{Im}\sum_{j=1}^{d}b_{j}\Big{)}^{2}.$

This shows that
$\Lambda=(-\operatorname{Im}b_{1},\ldots,-\operatorname{Im}b_{d})$ satisfies
(3.4), where the roots of $p$ are clearly $\lambda_{j}=-\frac{1}{b_{j}}$ for
$j=1,\ldots,d$. Application of Corollary 3.3 finishes the proof. ∎

Below we present an example of a sequence of matrix polynomials
$P_{k}(\lambda)$ of the same size $n\times n$ but increasing degree, based on
the scalar polynomials in [10, Proof of Theorem 1.3]. The aim of this lengthy
example is the following. Note that the outcome inequality in Theorem 3.2 has
a factor $\sqrt{n^{d}}$ in front of the exponent. We will show that this
factor needs to depend on $n$.

######  Example 3.7.

Take two matrices of size $n\times n$: $C_{1}=[1]_{s,t=1}^{n}$ and
$C_{2}:=-I_{n}$. Keeping the analogy with [10] we have:

$\Gamma:=(C_{1}^{2}-2C_{2})/2=I+\frac{n}{2}C_{1}$

and

$\displaystyle D_{k}:=\Gamma-C_{1}^{2}/(2k)=I+\frac{n(k-1)}{2k}C_{1}\text{.}$

Now, let us define

$P_{k}(\lambda):=\biggl{(}I+\frac{C_{1}\lambda}{k}\biggr{)}^{k}\biggl{(}I+\frac{\sqrt{D_{k}}\lambda}{\sqrt{k}}\biggr{)}^{k}\biggl{(}I-\frac{\sqrt{D_{k}}\lambda}{\sqrt{k}}\biggr{)}^{k}$

and observe it has the form as in Theorem 3.2. In the following computations
we will use the equality:

$(I+\alpha
C_{1})^{k}=I+\sum_{j=1}^{k}\binom{k}{j}\alpha^{j}n^{j-1}C_{1}=I+\frac{1}{n}\bigl{(}(n\alpha+1)^{k}-1\bigr{)}C_{1}.$

Then

$\displaystyle P_{k}(\operatorname{i}y)=$
$\displaystyle\Bigl{(}I+\frac{\operatorname{i}yC_{1}}{k}\Bigr{)}^{k}\Bigl{(}I+\frac{y^{2}D_{k}}{k}\Bigr{)}^{k}$
$\displaystyle=$
$\displaystyle\biggl{(}I+\frac{\operatorname{i}y}{k}C_{1}\biggr{)}\Bigl{(}\frac{y^{2}+k}{k}\Bigr{)}^{k}\biggl{(}I+\frac{y^{2}n(k-1)}{2k(y^{2}+k)}C_{1}\biggr{)}$
$\displaystyle=$
$\displaystyle\Bigl{(}\frac{y^{2}+k}{k}\Bigr{)}^{k}\Biggl{[}I+\frac{1}{n}\biggl{[}\Bigl{(}\frac{ny\operatorname{i}}{k}+1\Bigr{)}^{k}-1\biggr{]}C_{1}\Biggr{]}$
$\displaystyle\
\cdot\Biggl{[}I+\frac{1}{n}\biggl{[}\Bigl{(}\frac{n^{2}y^{2}(k-1)}{2k(y^{2}+k)}+1\Bigr{)}^{k}-1\biggr{]}C_{1}\Biggr{]}.$

Denoting $a_{k}:=\bigl{(}\frac{ny\operatorname{i}}{k}+1\bigr{)}^{k}$ and
$b_{k}:=\bigl{(}\frac{n^{2}y^{2}(k-1)}{2k(y^{2}+k)}+1\bigr{)}^{k}$ we obtain

$\displaystyle P_{k}(\operatorname{i}y)=$
$\displaystyle\Bigl{(}\frac{y^{2}+k}{k}\Bigr{)}^{k}\Bigl{(}I+\frac{1}{n}(a_{k}-1)C_{1}\Bigr{)}\Bigl{(}I+\frac{1}{n}(b_{k}-1)C_{1}\Bigr{)}$
$\displaystyle=$
$\displaystyle\Bigl{(}\frac{y^{2}+k}{k}\Bigr{)}^{k}\biggl{[}I+\frac{1}{n}(a_{k}+b_{k}-2)C_{1}+\frac{1}{n^{2}}(a_{k}b_{k}-a_{k}-b_{k}+1)nC_{1}\biggr{]}$
$\displaystyle=$
$\displaystyle\Bigl{(}\frac{y^{2}+k}{k}\Bigr{)}^{k}\biggl{[}I+\frac{1}{n}(a_{k}b_{k}-1)C_{1}\biggr{]}.$

Finally, we can calculate the Frobenius norm of $P_{k}(\operatorname{i}y)$

$\displaystyle\left\|P_{k}(\operatorname{i}y)\right\|_{F}^{2}$
$\displaystyle=\Bigl{(}\frac{y^{2}+k}{k}\Bigr{)}^{2k}\operatorname{tr}\Bigl{[}I+\frac{a_{k}b_{k}-1}{n}C_{1}\Bigr{]}^{*}\Bigl{[}I+\frac{a_{k}b_{k}-1}{n}C_{1}\Bigr{]}$
$\displaystyle=\Bigl{(}\frac{y^{2}+k}{k}\Bigr{)}^{2k}\operatorname{tr}\Bigl{[}I+2\operatorname{Re}\frac{a_{k}b_{k}-1}{n}C_{1}+\frac{|a_{k}b_{k}-1|^{2}}{n^{2}}C_{1}^{2}\Bigr{]}$
$\displaystyle=\Bigl{(}\frac{y^{2}+k}{k}\Bigr{)}^{2k}\operatorname{tr}\Bigl{[}I+\frac{|a_{k}b_{k}|^{2}-1}{n}C_{1}\Bigr{]}=\Bigl{(}\frac{y^{2}+k}{k}\Bigr{)}^{2k}(n-1+|a_{k}b_{k}|^{2}).$

The last step is taking limit of $\left\|P_{k}(\operatorname{i}y)\right\|_{F}$
with $k\to\infty$. To do this, we make use of a classical property of the
exponent

$\lim_{n\to\infty}\bigl{(}1+f(n)\bigr{)}^{g(n)}=\exp\bigl{[}\
\lim_{n\to\infty}f(n)g(n)\ \bigr{]}\text{,}$

which holds whenever $\lim_{n\to\infty}f(n)=0$ and
$\lim_{n\to\infty}g(n)=\infty$. The straightforward application of this fact
leads us to the following identities:

$\lim_{k\to\infty}a_{k}=e^{\operatorname{i}ny},\qquad\lim_{k\to\infty}b_{k}=e^{n^{2}y^{2}/2},\qquad\lim_{k\to\infty}\Bigl{(}\frac{y^{2}+k}{k}\Bigr{)}^{2k}=e^{2y^{2}}.$

Combining these equalities we obtain:

(3.5)
$\lim_{k\to\infty}\left\|P_{k}(\operatorname{i}y)\right\|_{F}=\Bigl{[}e^{2y^{2}}\Big{(}n-1+\big{|}e^{\operatorname{i}yn}e^{y^{2}n^{2}/2}\big{|}^{2}\Big{)}\Bigr{]}^{1/2}=e^{y^{2}}(e^{n^{2}y^{2}}+n-1)^{1/2}.$

It can be readily verified that
$P_{k}(\lambda)=I+C_{1}\lambda+C_{2}\lambda+\ldots+C_{3k}\lambda^{3k}$. Hence
the exponential part of the estimate in Theorem 3.2 equals

(3.6)
$\exp\Big{(}\frac{1}{n}\operatorname{Re}(n\operatorname{i}y)+\frac{1}{2n}(n^{2}+2n)|\operatorname{i}y|^{2})\Big{)}=e^{(\frac{n}{2}-1)y^{2}},\quad
y>0.$

Suppose that we are able to show the following inequality

$\left\|P(\lambda)\right\|_{F}\leq
c(n)\exp\Bigl{(}\frac{1}{n}\operatorname{tr}\operatorname{Re}(\lambda
A_{1})+\frac{1}{2n}\big{(}\left\|A_{1}\right\|_{F}^{2}-2\operatorname{tr}\operatorname{Re}A_{2}\big{)}|\lambda|^{2}\Bigr{)},\quad\lambda\in\mathbb{C},$

with some $c(n)$. Then from (3.6) and (3.5) we receive

$c(n)\geq\frac{e^{y^{2}}(e^{n^{2}y^{2}}+n-1)^{1/2}}{e^{(\frac{n}{2}-1)y^{2}}},\quad
y>0.$

Setting $y=1$ we obtain

$c(n)\geq e^{n/2}.$

##  4\. Functional calculus in one variable

In this section we prove inequalities (4.2), (4.3) and (4.4) providing bounds
for the norm of the scalar polynomial evaluated on the matrix variable. In
Proposition 4.1 the estimation depends on both the size $n$ of a matrix and
the degree $d$ of a polynomial $p\in\mathbb{C}[\lambda]$. The next result
(Proposition 4.2) provides the estimation dependent on the parameter $d$ only,
while the last Theorem 4.4 contains the estimation independent on both
parameters $n$ and $d$. At the end of the section we provide examples
comparing the estimates and the von Neumann inequlity.

######  Proposition 4.1.

Let $p(\lambda)=1+\sum_{j=1}^{d}a_{j}\lambda^{j}\in\mathbb{C}[\lambda]$ be a
scalar polynomial satisfying (3.4). Then the Frobenius norm of the polynomial
$p(A)$ can be estimated from above as follows:

$\left\|p(A)\right\|_{F}\leq\sqrt{n^{d}}\exp\Bigl{(}\frac{1}{n}\operatorname{tr}\operatorname{Re}(a_{1}A)+\frac{1}{2n}(|a_{1}|^{2}-2\operatorname{Re}a_{2})\left\|A\right\|_{F}^{2}\Bigr{)},\;A\in{\mathbb{C}}^{n\times
n}.$

###### Proof.

Since $p(0)=1$ we can write $p(\lambda)=\prod_{j=1}^{d}(1+\alpha_{j}\lambda)$,
and by stability of $p$ we know that $\operatorname{Im}\alpha_{j}\leq 0$ for
$j\in\\{1,2,\dots,d\\}$. Then obviously, we have
$p(A)=\prod_{j=1}^{d}(I+\alpha_{j}A)$ and $\sum_{j=1}^{d}\alpha_{j}=a_{1}$,
and $\sum_{1\leq j<k\leq d}\alpha_{j}\alpha_{k}=a_{2}$. We will use Lemma 2.1,
assuming that $A\neq-1/\alpha_{j}I$, $j=1,\dots,d$. We have

$\displaystyle\log\left\|p(A)\right\|_{F}$
$\displaystyle\leq\log\prod_{j=1}^{d}\left\|I+\alpha_{j}A\right\|_{F}$
$\displaystyle=\sum_{j=1}^{d}\log\left\|I+\alpha_{j}A\right\|_{F}$
$\displaystyle=d\log\sqrt{n}+\sum_{j=1}^{d}\log\left\|\frac{1}{\sqrt{n}}I+\tilde{\alpha}_{j}A\right\|_{F}$
$\displaystyle\leq\log\sqrt{n^{d}}+\frac{1}{\sqrt{n}}\sum_{j=1}^{d}\operatorname{tr}\operatorname{Re}(\tilde{\alpha}_{j}A)+\frac{1}{2}\sum_{j=1}^{d}\left\|\tilde{\alpha}_{j}A\right\|_{F}^{2},$

where $\tilde{\alpha}_{j}:=n^{-1/2}\alpha_{j}$ for $j\in\\{1,2,\dots,d\\}$.
Now, we obviously have $\sum_{j=1}^{d}\tilde{\alpha}_{j}=n^{-1/2}a_{1}$ and
$\sum_{1\leq j<k\leq d}\tilde{\alpha}_{j}\tilde{\alpha}_{k}=n^{-1}a_{2}$, so
we can rewrite the right side of the last inequality and then, using (2.2),
estimate it further as follows:

$\displaystyle\log\sqrt{n^{d}}$
$\displaystyle+\frac{1}{n}\operatorname{tr}\operatorname{Re}(a_{1}A)+\frac{1}{2}\left\|A\right\|_{F}^{2}\sum_{j=1}^{d}|\tilde{\alpha}_{j}|^{2}$
$\displaystyle\leq\log\sqrt{n^{d}}+\frac{1}{n}\operatorname{tr}\operatorname{Re}(a_{1}A)+\frac{1}{2n}(|a_{1}|^{2}-2\operatorname{Re}a_{2})\left\|A\right\|_{F}^{2}$

Therefore, we obtain

$\log\left\|p(A)\right\|_{F}\leq\log\sqrt{n^{d}}+\frac{1}{n}\operatorname{tr}\operatorname{Re}(a_{1}A)+\frac{1}{2n}(|a_{1}|^{2}-2\operatorname{Re}a_{2})\left\|A\right\|_{F}^{2}$

Taking exponent on the both side of the above inequality ends the proof for
$A\neq-1/\alpha_{j}I$, $j=1,\dots,d$. By continuity the inequality holds for
any $A$. ∎

In the next results we move on to estimating the operator norm of $p(A)$.

######  Proposition 4.2.

Let $p(\lambda)=1+\sum_{j=1}^{d}a_{j}\lambda^{j}\in\mathbb{C}[\lambda]$ be a
scalar polynomial satisfying (3.4). Then the operator norm of $p(A)$ can be
estimated from above as follows:

$\left\|p(A)\right\|\leq\exp\Bigl{(}\left\|A\right\|\sqrt{d(|a_{1}|^{2}-2\operatorname{Re}a_{2})}\Bigr{)},\;A\in{\mathbb{C}}^{n\times
n}.$

###### Proof.

As before, because of stability of polynomial $p(\lambda)$ we can write
$p(\lambda)=\prod_{j=1}^{d}(1+\alpha_{j}\lambda)$, where
$\operatorname{Im}\alpha_{j}\leq 0$ for $j\in\\{1,2,\cdots,d\\}$. Then we have
$p(A)=\prod_{j=1}^{d}(I+\alpha_{j}A)$ and $\sum_{j=1}^{d}\alpha_{j}=a_{1}$,
and $\sum_{1\leq j<k\leq d}\alpha_{j}\alpha_{k}=a_{2}$. Applying (2.1) $d$
times and using (2.2), we obtain

$\displaystyle\log\left\|p(A)\right\|$
$\displaystyle\leq\log\prod_{j=1}^{d}\left\|I+\alpha_{j}A\right\|=\sum_{j=1}^{d}\log\left\|I+\alpha_{j}A\right\|$
$\displaystyle\leq\sum_{j=1}^{d}\left\|\alpha_{j}A\right\|=\left\|A\right\|\sum_{j=1}^{d}|\alpha_{j}|$
$\displaystyle\leq\left\|A\right\|\Bigl{(}d\sum_{j=1}^{d}|\alpha_{j}|^{2}\Bigr{)}^{1/2}$
$\displaystyle\leq\left\|A\right\|\sqrt{d(|a_{1}|^{2}-2\operatorname{Re}a_{2})}\text{,}$

where the last but one inequality comes from the inequality of arithmetic and
square means. Taking the exponent finishes the proof. ∎

######  Remark 4.3.

From the proof of the previous proposition, we can see that for a polynomial
$p(\lambda)$ satisfying (3.4) and $p(0)=1$ the expression
$|a_{1}|^{2}-2\operatorname{Re}a_{2}=\sum_{j=1}^{d}|\alpha_{j}|^{2}\geq 0$ has
always a non-negative value.

As one may observe, in Proposition 4.1 the estimate contained $d$ and $n$
while in Proposition 4.2 the estimate contained $d$. We present now an
estimate that is independent on both $d$ and $n$. Its proof is based on the
von Neumann inequality in the form

(4.1) $\left\|p(A)\right\|\leq\sup_{|z|\leq\left\|A\right\|}|p(z)|,$

where $p$ is a polynomial and $A$ is an arbitrary square matrix. The
inequality above can be easily derived from the usual von Neumann inequality
by considering $\tilde{p}(z)=p(z\cdot\left\|A\right\|)$ and
$\tilde{A}=A/\left\|A\right\|$.

######  Theorem 4.4.

Let $p(\lambda)\in\mathbb{C}[\lambda]$ be a polynomial satisfying (3.4) with
$p(0)=1$ and let $A\in{\mathbb{C}}^{n\times n}$ be a complex square matrix.
Then the following inequality holds:

$\left\|p(A)\right\|\leq\exp\bigl{(}|a_{1}|\left\|A\right\|+{\textstyle\frac{1}{2}}(|a_{1}|^{2}-2\operatorname{Re}(a_{2}))\left\|A\right\|^{2}\bigr{)}\text{.}$

###### Proof.

As mentioned before, we use the general von Neumann inequality (4.1) first.
The next step is to use the scalar Szász inequality (1.1), obtaining

$\displaystyle\left\|p(A)\right\|$
$\displaystyle\leq\sup_{|z|\leq\left\|A\right\|}|p(z)|$
$\displaystyle\leq\sup_{|z|\leq\left\|A\right\|}\exp\bigl{(}\operatorname{Re}(a_{1}z)+\frac{1}{2}(|a_{1}|^{2}-2\operatorname{Re}(a_{2}))|z|^{2}\bigr{)}$
$\displaystyle\leq\exp\bigl{(}|a_{1}|\left\|A\right\|+\frac{1}{2}(|a_{1}|^{2}-2\operatorname{Re}(a_{2}))\left\|A\right\|^{2}\bigr{)}.$

∎

Let us recall that for any matrix $A\in{\mathbb{C}}^{n\times n}$ the relation
$\left\|A\right\|_{F}\leq\sqrt{n}\left\|A\right\|$ between the operator two-
norm and the Frobenius norm holds. Below, we make a comparison of the global
bounds presented in this Section. The inequalities resulting from Proposition
4.1, Proposition 4.2 and Theorem 4.4 are listed below

(4.2) $\displaystyle\left\|p(A)\right\|_{F}$
$\displaystyle\leq\sqrt{n^{d}}\exp\Bigl{(}\frac{1}{n}\operatorname{tr}\operatorname{Re}(a_{1}A)+\frac{1}{2n}(|a_{1}|^{2}-2\operatorname{Re}a_{2})\left\|A\right\|_{F}^{2}\Bigr{)}=:E_{1},$
(4.3) $\displaystyle\left\|p(A)\right\|_{F}$
$\displaystyle\leq\sqrt{n}\exp\Bigl{(}\left\|A\right\|\sqrt{d(|a_{1}|^{2}-2\operatorname{Re}a_{2})}\Bigr{)}=:E_{2},$
(4.4) $\displaystyle\left\|p(A)\right\|_{F}$
$\displaystyle\leq\sqrt{n}\exp\bigl{(}|a_{1}|\left\|A\right\|+\frac{1}{2}(|a_{1}|^{2}-2\operatorname{Re}a_{2})\left\|A\right\|^{2}\bigr{)}=:E_{3}.$

Below we provide three examples where each of these inequalities gives the
best bound. The drawback of (4.2) lies in the dependence on $n$ and $d$.
However, it appears that (4.2) can beat the von Neumann inequality in some
cases, while the other two inequalities can not.

######  Example 4.5.

Set $n=2$ and $d=3$. As the first example take
$p(\lambda)=-(\lambda-1)^{3}=-\lambda^{3}+3\lambda^{2}-3\lambda+1$, which is a
stable polynomial with respect to the open upper half-plane $H_{0}$ with
$p(0)=1$. We have $a_{1}=-3$ and $a_{2}=3$. For matrix

$A=\begin{bmatrix}1&1\\\
1&1\end{bmatrix},\quad\left\|A\right\|_{F}=2,\quad\left\|A\right\|=2$

we obtain $E_{1}=2\sqrt{2}$, $E_{2}=e^{6}\sqrt{2}$, $E_{3}=e^{12}\sqrt{2}$.
Obviously, in this case first inequality (4.2) gives the best estimation of
$\left\|p(A)\right\|_{F}$. It is worth noting that a direct use of von Neumann
inequality (4.1) in this case gives a worse estimation, since
$\sup_{|\lambda|\leq 2}|p(\lambda)|=3^{3}=27$ and the corresponding estimation
of the Frobenius norm equals $27\sqrt{2}$.

Next, for

$A=\begin{bmatrix}-1&1\\\
1&-1\end{bmatrix},\quad\left\|A\right\|_{F}=2,\quad\left\|A\right\|=2$

the estimations from (4.2) – (4.4) attain $E_{1}=2e^{6}\sqrt{2}$,
$E_{2}=e^{6}\sqrt{2}$ and $E_{3}=e^{12}\sqrt{2}$. This time, the best
estimation is given by the second inequality (4.3). It is worse than
$27\sqrt{2}$ obtained from von Neumann inequality, though. Actually it is
evident from the proofs of Proposition 4.2 and Theorem 4.4 that the estimation
(4.1) is never worse than (4.3) and (4.4).

Now take
$p(\lambda)=(\lambda-1)^{2}(\lambda+1)=\lambda^{3}-\lambda^{2}-\lambda+1$. As
before, $p(\lambda)$ is stable with respect to the open upper half-plane
$H_{0}$ and $p(0)=1$. We see that $a_{1}=a_{2}=-1$. Consider matrix
$A=-I_{2}$. Then $\left\|A\right\|_{F}=\sqrt{2}$, $\left\|A\right\|=1$ and
(4.2) - (4.4) lead to $E_{1}=2e^{5/2}\sqrt{2}$, $E_{2}=e^{3}\sqrt{2}$ and
$E_{3}=e^{5/2}\sqrt{2}$. Clearly, in this case the best upper bound for
$\left\|p(A)\right\|_{F}$ is given by the third inequality (4.4).

##  5\. Multivariate functional calculus

A Szász-type inequality for multivariate scalar polynomials can be found in
the work of Borcea and Branden [3, Theorem 6.5]. The constants in estimation
$|p(\boldsymbol{z})|\leq B\exp(C\left\|\boldsymbol{z}\right\|_{\infty})$
depend only on the coefficients of terms with degree at most $2$ and the
number of variables. In [10] the result was improved and in this section we
will continue the topic. For a multivariate polynomial
$p\in\mathbb{C}[z_{1},z_{2},\ldots,z_{d}]$ we express the low-degree
coefficients in terms of partial derivatives and denote them by
$p_{j}:=\frac{\partial p}{\partial z_{j}}(0,0)$ and
$p_{j,k}:=\frac{\partial^{2}p}{\partial z_{j}\partial z_{k}}(0,0)$ ($1\leq
j,k\leq d$).

If a polynomial $p\in\mathbb{C}[z_{1},z_{2}]$ is stable with respect to the
open upper half plane and satisfies $p(0,0)=1$, then the following inequality
holds [10, Theorem 1.4]:

(5.1)
$|p(z_{1},z_{2})|\leq\exp\bigg{(}\operatorname{Re}\big{(}z_{1}p_{1}+z_{2}p_{2}\big{)}+\frac{1}{2}\left\|z\right\|^{2}_{\infty}\Big{(}\big{|}p_{1}+p_{2}\big{|}^{2}-\operatorname{Re}\Big{(}\sum_{j,k=1}^{2}p_{j,k}\Big{)}\Big{)}\bigg{)}.$

The main tool below will be the two-variable von Neumann inequality, a result
due to Tsuyoshi Ando [1]. For a polynomial $p\in\mathbb{C}[z_{1},z_{2}]$ and
two commuting contractive matrices (result is valid also for operators on an
infinite-dimensional Hilbert space) $T_{1}$, $T_{2}$ the following inequality
holds:

$\left\|p(T_{1},T_{2})\right\|\leq\sup\big{\\{}|p(z_{1},z_{2})|:z\in\mathbb{C}^{2},\
|z_{1}|,|z_{2}|\leq 1\big{\\}}$

General operators $T_{j}$ can be easily reduced to the case of contractions by
taking $\tilde{T}_{j}=T_{j}/\left\|T_{j}\right\|$ and
$\tilde{p}(z_{1},z_{2}):=p(\left\|T_{1}\right\|z_{1},\left\|T_{2}\right\|z_{2})$,
leading to the inequality

(5.2)
$\left\|p(T_{1},T_{2})\right\|\leq\sup\big{\\{}|p(z_{1},z_{2})|:z\in\mathbb{C}^{2},\
|z_{1}|\leq\left\|T_{1}\right\|,\ |z_{2}|\leq\left\|T_{2}\right\|\big{\\}}$

valid for arbitrary commuting pair of operators (the case when one of the
operators is zero naturally reduces to the univariate setting).

######  Proposition 5.1.

Let $p\in\mathbb{C}[z_{1},z_{2}]$ be a polynomial stable with respect to the
open upper half plane such that $p(0,0)=1$. If
$A_{1},A_{2}\in{\mathbb{C}}^{n\times n}$ and $A_{1}A_{2}=A_{2}A_{1}$ then

$\left\|p(A_{1},A_{2})\right\|\leq\exp\bigg{(}\\!\left\|A_{1}\right\||p_{1}|+\left\|A_{2}\right\||p_{2}|+\frac{1}{2}\left\|A\right\|^{2}_{\infty}\Big{(}\big{|}p_{1}+p_{2}\big{|}^{2}-\operatorname{Re}\\!\sum_{j,k=1}^{2}p_{j,k}\Big{)}\bigg{)},$

where
$\left\|A\right\|_{\infty}:=\max\\{\left\|A_{1}\right\|,\left\|A_{2}\right\|\\}$.

###### Proof.

According to (5.2):

$\left\|p(A_{1},A_{2})\right\|\leq\sup\big{\\{}|p(z_{1},z_{2})|:|z_{1}|\leq\left\|A_{1}\right\|,\
|z_{2}|\leq\left\|A_{2}\right\|\big{\\}}.$

Estimating the right hand side using (5.1) gives us

$\displaystyle\left\|p(A_{1},A_{2})\right\|$
$\displaystyle\leq\sup\bigg{\\{}\exp\\!\Big{[}\operatorname{Re}\big{(}z_{1}p_{1}+z_{2}p_{2}\big{)}+\frac{1}{2}\left\|z\right\|^{2}_{\infty}\Big{(}\big{|}p_{1}+p_{2}\big{|}$
$\displaystyle\quad-\operatorname{Re}\\!\sum_{j,k=1}^{2}p_{j,k}\Big{)}\Big{]}:|z_{1}|\leq\left\|A_{1}\right\|,\
|z_{2}|\leq\left\|A_{2}\right\|\bigg{\\}}$
$\displaystyle\leq\exp\\!\bigg{[}\sup\\!\Big{\\{}\operatorname{Re}\big{(}z_{1}p_{1}+z_{2}p_{2}\big{)}:|z_{1}|\leq\left\|A_{1}\right\|,\
|z_{2}|\leq\left\|A_{2}\right\|\Big{\\}}$
$\displaystyle\quad+\frac{1}{2}\left\|A\right\|^{2}_{\infty}\Big{(}\big{|}p_{1}+p_{2}\big{|}-\operatorname{Re}\\!\sum_{j,k=1}^{2}p_{j,k}\Big{)}\bigg{]}$
$\displaystyle=\exp\\!\bigg{(}\\!\left\|A_{1}\right\||p_{1}|+\left\|A_{2}\right\||p_{2}|+\frac{1}{2}\left\|A\right\|^{2}_{\infty}\Big{(}\big{|}p_{1}+p_{2}\big{|}-\operatorname{Re}\\!\sum_{j,k=1}^{2}p_{j,k}\Big{)}\bigg{)}.$

∎

Knese used the two variable case to prove the following inequality for scalar
polynomials of arbitrarily many variables [10, Theorem 1.5]: If
$p\in\mathbb{C}[z_{1},\ldots,z_{k}]$ is a stable multivariate polynomial
satisfying $p(0,\ldots,0)=1$, then the following inequality holds:

(5.3) $|p(\boldsymbol{z})|\leq\exp\Big{(}\sqrt{2}\left\|\nabla
p(\mathbf{0})\right\|\left\|\boldsymbol{z}\right\|+\big{(}\left\|\nabla
p(\mathbf{0})\right\|^{2}+\left\|\operatorname{Re}H_{p}(\mathbf{0})\right\|\big{)}\left\|\boldsymbol{z}\right\|^{2}\Big{)},$

where $H_{p}$ stands for the Hessian matrix of $p$.

Due to the counterexamples by Kaijser and Varopoulos [18] (see also [8]) the
scalar inequality (5.3) for more than two variables cannot be applied in the
matrix case independently on the number of variables and size of the matrix.
However, there exist applicable multivariate versions of the von Neumann
inequality.

In 1961 Brehmer [4] proven that for a commuting system of operators
$(T_{1},\ldots,T_{k})$ such that $\sum_{j=1}^{k}\left\|T_{j}\right\|^{2}\leq
1$ the inequality
$\left\|p(T_{1},\ldots,T_{k})\right\|\leq\sup\\{|p(z)|:\|\boldsymbol{z}\|_{\infty}\leq
1\\}$ holds for any polynomial $p\in\mathbb{C}[z_{1},\ldots,z_{k}]$. We can
drop the assumption on the norm of the operators in the similar way as for the
one-variable inequality, obtaining:

(5.4)
$\left\|p(T_{1},\ldots,T_{k})\right\|\leq\sup\Big{\\{}|p(\boldsymbol{z})|\colon\left\|\boldsymbol{z}\right\|_{\infty}^{2}\leq\sum_{j=1}^{k}\left\|T_{j}\right\|^{2}\Big{\\}}.$

######  Proposition 5.2.

Let $p\in\mathbb{C}[z_{1},\ldots,z_{k}]$ be a polynomial stable with respect
to $H_{0}^{k}$ such that $p(\mathbf{0})=1$. If
$\mathbf{A}=(A_{1},\ldots,A_{k})\in({\mathbb{C}}^{n\times n})^{k}$ is a
commuting $k$-tuple then

$\left\|p(\mathbf{A})\right\|\leq\exp\Big{[}\sqrt{2k}\left\|\nabla
p(\mathbf{0})\right\|\left\|\mathbf{A}\right\|_{2}+k\big{(}\left\|\nabla
p(\mathbf{0})\right\|^{2}+\left\|\operatorname{Re}H_{p}(\mathbf{0})\right\|\big{)}\left\|\mathbf{A}\right\|_{2}^{2}\Big{]},$

where
$\left\|\mathbf{A}\right\|_{2}=\sqrt{\sum_{j=1}^{k}\left\|A_{j}\right\|^{2}}$.

###### Proof.

According to the aforementioned result of Brehmer:

$\left\|p(\mathbf{A})\right\|\leq\sup\Big{\\{}|p(\boldsymbol{z})|:\|\boldsymbol{z}\|_{\infty}\leq\left\|\mathbf{A}\right\|_{2}\Big{\\}}.$

We estimate the right hand side by (5.3), keeping in mind that
$\left\|\boldsymbol{z}\right\|\leq\sqrt{k}\left\|\boldsymbol{z}\right\|_{\infty}$.

$\displaystyle\left\|p(\mathbf{A})\right\|$
$\displaystyle\leq\sup_{\|\boldsymbol{z}\|_{\infty}\leq\left\|\mathbf{A}\right\|_{2}}\exp\Big{(}\sqrt{2}\left\|\nabla
p(\mathbf{0})\right\|\left\|\boldsymbol{z}\right\|+\big{(}\left\|\nabla
p(\mathbf{0})\right\|^{2}+\left\|\operatorname{Re}H_{p}(\mathbf{0})\right\|\big{)}\left\|\boldsymbol{z}\right\|^{2}\Big{)}$
$\displaystyle\leq\exp\Big{[}\sqrt{2}\left\|\nabla
p(\mathbf{0})\right\|\sqrt{k}\left\|\mathbf{A}\right\|_{2}+\big{(}\left\|\nabla
p(\mathbf{0})\right\|^{2}+\left\|\operatorname{Re}H_{p}(\mathbf{0})\right\|\big{)}\big{(}\sqrt{k}\left\|\mathbf{A}\right\|_{2}\big{)}^{2}\Big{]}$

∎

Apart from the Brehmer’s generalisation of the von Neumann inequality
involving inflation of the polydisc, a constant factor depending on dimension
of the matrices can be included. Existence and estimation of appropriate
constants was proven in [6].

There exist constants $C_{k,n}\geq 1$ such that for a commuting row
contraction $\mathbf{T}=(T_{j})_{j=1}^{k}\in({\mathbb{C}}^{n\times n})^{k}$
(i.e. $\|\sum_{j=1}^{n}T_{j}T_{j}^{*}\|\leq 1$) the following inequality holds
(see [6, Theorem 4.7])

(5.5) $|p(T_{1},\ldots,T_{k})|\leq
C_{k,n}\sup\\{|p(\boldsymbol{z})|:\left\|\boldsymbol{z}\right\|\leq
1,j=1,\ldots,k\\}.$

The assumption on $\mathbf{T}$ used above is slightly weaker than
$\left\|\mathbf{T}\right\|_{2}\leq 1$. Namely it is required that

$\left\|\bf
A\right\|_{R}:=\sqrt{\Big{\|}\sum_{j=1}^{k}A_{j}A_{j}^{*}\Big{\|}}\leq 1,$

while $\left\|\bf A\right\|_{R}\leq\left\|\bf A\right\|_{2}$. It is worth
noting that $\sup_{k\geq 1}C_{k,n}<\infty$ for any $n\geq 1$, but in general
the constants tend to infinity. Note also that in (5.5) we take the supremum
on a ball instead of a polydisc.

######  Theorem 5.3.

Let $p\in\mathbb{C}[z_{1},\ldots,z_{k}]$ be a polynomial stable with respect
to $H_{0}^{k}$ and satisfying $p(\mathbf{0})=1$ and
$\mathbf{T}\in({\mathbb{C}}^{n\times n})^{k}$ be a commuting tuple of
matrices. Then

(5.6) $\left\|p(\mathbf{A})\right\|\leq
C_{k,n}\exp\Big{[}\sqrt{2}\left\|\nabla
p(\mathbf{0})\right\|\left\|\mathbf{A}\right\|_{R}+\big{(}\left\|\nabla
p(\mathbf{0})\right\|^{2}+\left\|\operatorname{Re}H_{p}(\mathbf{0})\right\|\big{)}\left\|\mathbf{A}\right\|_{R}^{2}\Big{]}.$

Moreover, for $n\leq 3$ we have:

$\left\|p(\mathbf{A})\right\|\leq\exp\Big{[}\sqrt{2}\left\|\nabla
p(\mathbf{0})\right\|\left\|\mathbf{A}\right\|_{2}+\big{(}\left\|\nabla
p(\mathbf{0})\right\|^{2}+\left\|\operatorname{Re}H_{p}(\mathbf{0})\right\|\big{)}\left\|\mathbf{A}\right\|_{2}^{2}\Big{]}$

what is a better estimation than in Proposition 5.2.

###### Proof.

We proceed analogously as in the proofs of Propositions 5.1 and 5.2, starting
from the inequality (5.5). An arbitrary commuting tuple $\bf A$ can be
transformed into a row contraction through dividing by $\left\|\bf
A\right\|_{R}$

$\Big{\|}\sum_{j=1}^{k}\frac{A_{j}}{\left\|\bf
A\right\|_{R}}\frac{A_{j}^{*}}{\left\|\bf
A\right\|_{R}}\Big{\|}=\Big{\|}{\sum_{j=1}^{k}A_{j}A_{j}^{*}}\Big{\|}/\left\|\bf
A\right\|_{R}^{2}=1.$

On the right-hand-side of (5.5) we consider $\tilde{p}\big{(}{\bf
A}/\left\|\bf A\right\|_{R}\big{)}$, where
$\tilde{p}(z_{1},\ldots,z_{k}):=p\big{(}z_{1}\left\|\bf
A\right\|_{R},\ldots,z_{k}\left\|\bf A\right\|_{R}\big{)}$ for
$\left\|\boldsymbol{z}\right\|\leq 1$, which is equivalent to
$p(\boldsymbol{z})$ for $\left\|\boldsymbol{z}\right\|\leq\left\|\bf
A\right\|_{R}$. Then we estimate
$\sup\\{|p(\boldsymbol{z})|\colon\left\|\boldsymbol{z}\right\|\leq\left\|\bf
A\right\|_{R}\\}$ by a straightforward use of (5.3) to obtain the inequality
(5.6).

According to [6, Corollary 3.4] we have $C_{k,2}=1$, what resolves the special
case of $n=2$ (since $\left\|\bf A\right\|_{R}\leq\left\|\bf A\right\|_{2}$).
Knese [9], basing on Kosiński’s work on three point Pick problem [11], showed
that

$\left\|p(T_{1},\ldots,T_{k})\right\|\leq\sup\Big{\\{}|p(\boldsymbol{z})|:\|\boldsymbol{z}\|_{\infty}\leq
1\Big{\\}}$

for commuting contractive $3\times 3$ matrices $(T_{1},\ldots,T_{k})$.
Analogously as before, we can obtain inequality valid for arbitrary commuting
$k$-tuple. Therefore

$\displaystyle\left\|p(\mathbf{A})\right\|$
$\displaystyle\leq\sup\Big{\\{}|p(\boldsymbol{z})|:|z_{j}|\leq\left\|A_{j}\right\|,\
j=1,\ldots,k\Big{\\}}$
$\displaystyle\leq\sup_{|z_{j}|\leq\left\|\mathbf{A}_{j}\right\|}\exp\Big{(}\sqrt{2}\left\|\nabla
p(\mathbf{0})\right\|\left\|\boldsymbol{z}\right\|+\big{(}\left\|\nabla
p(\mathbf{0})\right\|^{2}+\left\|\operatorname{Re}H_{p}(\mathbf{0})\right\|\big{)}\left\|\boldsymbol{z}\right\|^{2}\Big{)}$
$\displaystyle\leq\exp\Big{[}\sqrt{2}\left\|\nabla
p(\mathbf{0})\right\|\left\|\mathbf{A}\right\|_{2}+\big{(}\left\|\nabla
p(\mathbf{0})\right\|^{2}+\left\|\operatorname{Re}H_{p}(\mathbf{0})\right\|\big{)}\left\|\mathbf{A}\right\|_{2}^{2}\Big{]}.$

∎

Since up to our knowledge the exact values of $C_{k,n}$ are not known for
$n\geq 4$, it is not possible to precisely decide whether the estimation from
Theorem 5.3 is lower than the one from Proposition 5.2.

##  6\. Matrix polynomials of matrix variables

First we present a complete version of the Szász inequality. It is known (see
e.g. [15, p.54], [2]) that the unit disc is not only a spectral set for
contractions but a complete spectral set, i.e. the von Neumann inequality
(4.1) extends on matrices of polynomials (even more generally, on matrices
whose entries are rational functions with no poles in the disc). Combining
this fact with Theorem 3.2 leads to the following complete version of the
Szász inequality.

######  Proposition 6.1.

Let $P(\lambda)=I_{n}+\sum_{j=1}^{d}\lambda^{j}A_{j}\in{\mathbb{C}}^{n\times
n}[\lambda]$ be a matrix polynomial with the following factorisation
$P(\lambda)=\prod_{j=1}^{d}(I_{n}+\lambda B_{j})$ satisfying (3.1). Then, for
$T\in{\mathbb{C}}^{m\times m}$, the following inequality holds

$\biggl{\|}I_{n}\otimes I_{m}+\sum_{j=1}^{d}A_{j}\otimes
T^{j}\biggr{\|}\leq\sqrt{n^{d}}\exp\Bigl{(}\frac{1}{n}|\operatorname{tr}A_{1}|\left\|T\right\|+\frac{1}{2n}\big{(}\left\|A_{1}\right\|_{F}^{2}-2\operatorname{tr}\operatorname{Re}A_{2}\big{)}\left\|T\right\|^{2}\Bigr{)}.$

###### Proof.

By the fact that the unit disc is a complete spectral set we have

$\displaystyle\bigg{\|}I_{n}\otimes
I_{m}+\sum_{j=1}^{d}A_{j}\left\|T\right\|\otimes\frac{T^{j}}{\left\|T\right\|}\bigg{\|}$
$\displaystyle\leq\sup\left\\{\bigl{\|}P(\left\|T\right\|z)\bigr{\|}\colon|z|\leq
1\right\\}$
$\displaystyle\leq\sup\left\\{\left\|P(z)\right\|_{F}\colon|z|\leq\left\|T\right\|\right\\}.$

Now we estimate the right-hand-side by Theorem 3.2:

$\displaystyle\sup_{|\lambda|\leq\left\|T\right\|}\left\|P(\lambda)\right\|_{F}$
$\displaystyle\leq\sqrt{n^{d}}\exp\sup_{|\lambda|\leq\left\|T\right\|}\Bigl{(}\frac{1}{n}\operatorname{tr}\operatorname{Re}(\lambda
A_{1})+\frac{1}{2n}\big{(}\left\|A_{1}\right\|_{F}^{2}-2\operatorname{tr}\operatorname{Re}A_{2}\big{)}|\lambda|^{2}\Bigr{)}$
$\displaystyle\leq\sqrt{n^{d}}\exp\Bigl{(}\frac{1}{n}|\operatorname{tr}A_{1}|\left\|T\right\|+\frac{1}{2n}\big{(}\left\|A_{1}\right\|_{F}^{2}-2\operatorname{tr}\operatorname{Re}A_{2}\big{)}\left\|T\right\|^{2}\Bigr{)}.$

∎

Another inequality was shown by Mlak [14] (cf. [2]). If
$P(\lambda)=A_{0}+A_{1}\lambda+\ldots+A_{d}\lambda^{d}\in{\mathbb{C}}^{n\times
n}[\lambda]$ is a matrix polynomial and a contraction
$T\in{\mathbb{C}}^{n\times n}$ doubly commutes with coefficients of $P$, i.e.
$TA_{j}=A_{j}T$, $TA_{j}^{*}=A_{j}^{*}T$ for $j=0,\ldots,d$, then defining

$P(T):=A_{0}+A_{1}T+A_{2}T^{2}+\ldots+A_{d}T^{d}$

we have

$\left\|P(T)\right\|\leq\sup\\{\left\|P(z)\right\|:|z|\leq 1\\}.$

######  Proposition 6.2.

Let $P(\lambda)=I_{n}+\sum_{j=1}^{d}\lambda^{j}A_{j}\in{\mathbb{C}}^{n\times
n}[\lambda]$ be a matrix polynomial with the following factorisation
$P(\lambda)=\prod_{j=1}^{d}(I_{n}+\lambda B_{j})$ satisfying (3.1). For
$T\in{\mathbb{C}}^{n\times n}$ doubly commuting with every $A_{j}$ the norm of
$P(T)$ can be estimated from above as

$\left\|P(T)\right\|\leq\sqrt{n^{d}}\exp\Bigl{(}\frac{1}{n}|\operatorname{tr}A_{1}|\left\|T\right\|+\frac{1}{2n}\big{(}\left\|A_{1}\right\|_{F}^{2}-2\operatorname{tr}\operatorname{Re}A_{2}\big{)}\left\|T\right\|^{2}\Bigr{)}.$

Note that if $T$ doubly commutes with $B_{j}$ for $j=1,\ldots,d$ then it
doubly commutes with all the $A_{j}$’s as well.

###### Proof.

According to the Mlak’s generalisation of von Neumann inequality we obtain:

$\left\|P(T)\right\|\leq\sup\\{\left\|P(\lambda)\right\|:|\lambda|\leq\left\|T\right\|\\}\leq\sup\\{\left\|P(\lambda)\right\|_{F}:|\lambda|\leq\left\|T\right\|\\}$

The right hand side can be estimated by Theorem 3.2 in the same way as in the
proof of Proposition 6.1. ∎

In the recent preprint [7], Hartz provides a matrix polynomial version of von
Neumann inequality, namely for $P(\lambda)=A_{0}+A_{1}\lambda+\ldots
A_{d}\lambda^{d}\in{\mathbb{C}}^{n\times n}[\lambda]$ and a contraction
$T\in{\mathbb{C}}^{n\times n}$ the following inequality holds

$\left\|P(T)\right\|\leq\sqrt{d+1}\sup\\{\left\|P(\lambda)\right\|\colon|\lambda|\leq
1\\}.$

Actually, the result is valid also when we replace matrices by operators on an
infinite dimensional Hilbert space. Repeating the same reasoning as in the
proofs of Propositions 6.1 and 6.2 we obtain the following.

######  Corollary 6.3.

Let $P(\lambda)=I_{n}+\sum_{j=1}^{d}A_{j}\lambda^{j}\in{\mathbb{C}}^{n\times
n}[\lambda]$ be a matrix polynomial admitting factorisation
$P(\lambda)=\prod_{j=1}^{d}(I_{n}+\lambda B_{j})$ and satisfying the condition
(3.1). Then for $T\in{\mathbb{C}}^{n\times n}$ we have the following
estimation:

$\left\|P(T)\right\|\leq\sqrt{n^{d}(d+1)}\exp\Bigl{(}\frac{1}{n}|\operatorname{tr}A_{1}|\left\|T\right\|+\frac{1}{2n}\big{(}\left\|A_{1}\right\|_{F}^{2}-2\operatorname{tr}\operatorname{Re}A_{2}\big{)}\left\|T\right\|^{2}\Bigr{)}.$

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