# Abstract damped wave equations:
The optimal decay rate

Filippo Dell’Oro, Lorenzo Liverani, Vittorino Pata  Politecnico di Milano -
Dipartimento di Matematica
Via Bonardi 9, 20133 Milano, Italy<EMAIL_ADDRESS>(F. Dell’Oro)
<EMAIL_ADDRESS>(L. Liverani<EMAIL_ADDRESS>(V. Pata)

###### Abstract.

The exponential decay rate of the semigroup $S(t)=e^{t{\mathbb{A}}}$ generated
by the abstract damped wave equation

$\ddot{u}+2f(A)\dot{u}+Au=0$

is here addressed, where $A$ is a strictly positive operator. The continuous
function $f$, defined on the spectrum of $A$, is subject to the constraints

$\inf f(s)>0\qquad\text{and}\qquad\sup f(s)/s<\infty$

which are known to be necessary and sufficient for exponential stability to
occur. We prove that the operator norm of the semigroup fulfills the estimate

$\|S(t)\|\leq Ce^{\sigma_{*}t}$

being $\sigma_{*}<0$ the supremum of the real part of the spectrum of
${\mathbb{A}}$. This estimate always holds except in the resonant cases, where
the negative exponential $e^{\sigma_{*}t}$ turns out to be penalized by a
factor $(1+t)$. The decay rate is the best possible allowed by the theory.

###### Key words and phrases:

Abstract damped wave equations, semigroups, exponential stability, decay rate,
resonance

###### 2010 Mathematics Subject Classification:

35B35, 35P05, 47D06

Work partially supported by the Italian MIUR-PRIN Grant 2020F3NCPX
“Mathematics for industry 4.0 (Math4I4)”

##  1\. Introduction

###  1.1. The equation

Let $H$ be a (separable) real Hilbert space, endowed with the scalar product
and norm $\langle\cdot,\cdot\rangle,$ and $\|\cdot\|$, respectively, and let
$A:{\rm{dom}}(A)\subset H\to H$ be a strictly positive selfadjoint linear
operator. We define the product Hilbert space

$\mathcal{H}={\rm{dom}}(A^{\frac{1}{2}})\times H,$

endowed with the standard Euclidean product norm

$\|(u,v)\|_{\mathcal{H}}^{2}=\|A^{\frac{1}{2}}u\|^{2}+\|v\|^{2}.$

Denoting by $\sigma(A)\subset(0,\infty)$ the spectrum of $A$, let

$f:\sigma(A)\to(0,\infty)$

be a continuous function subject to the structural assumptions

(1.1)
$\inf_{s\in\sigma(A)}f(s)>0\qquad\text{and}\qquad\sup_{s\in\sigma(A)}\frac{f(s)}{s}<\infty.$

For $t>0$, we consider the abstract damped wave equation in the unknown
$u=u(t)$

(1.2) $\ddot{u}+2f(A)\dot{u}+Au=0,$

the dot standing for derivative with respect to time, subject to the initial
conditions

$u(0)=u_{0}\quad\text{and}\qquad\dot{u}(0)=v_{0},$

where the vector $(u_{0},v_{0})\in\mathcal{H}$ is arbitrarily assigned. Here,
$f(A)$ is the selfadjoint linear operator on $H$ constructed via the
functional calculus of $A$, namely,

$f(A)=\int_{\sigma(A)}f(s)\,dE_{A}(s),$

being $E_{A}$ the spectral measure of $A$ (see, e.g., [20]). Calling
$\boldsymbol{u}=(u,v)$, and introducing the linear operator

${\mathbb{A}}(u,v)=(v,-Au-f(A)v),$

with domain

${\rm{dom}}({\mathbb{A}})=\big{\\{}\boldsymbol{u}\in\mathcal{H}:\,{\mathbb{A}}\boldsymbol{u}\in\mathcal{H}\big{\\}},$

we can rewrite (1.2) as the first order ODE in $\mathcal{H}$

$\dot{\boldsymbol{u}}={\mathbb{A}}\boldsymbol{u}.$

The linear operator ${\mathbb{A}}$ is known to be the infinitesimal generator
of a ${\mathcal{C}}_{0}$-semigroup

$S(t)=e^{t{\mathbb{A}}}:\mathcal{H}\to\mathcal{H}$

of linear contractions (see, e.g., [8]). Hence, for any given
$\boldsymbol{u}_{0}=(u_{0},v_{0})\in\mathcal{H}$, the unique mild solution
$\boldsymbol{u}(t)$ to the equation above, subject to the initial condition
$\boldsymbol{u}(0)=\boldsymbol{u}_{0}$, is given by

$\boldsymbol{u}(t)=(u(t),\dot{u}(t))=S(t)\boldsymbol{u}_{0}.$

We denote (twice) the energy of such a solution by

${\mathsf{E}}(t)=\|\boldsymbol{u}(t)\|^{2}_{\mathcal{H}}=\|A^{\frac{1}{2}}u(t)\|^{2}+\|\dot{u}(t)\|^{2}.$

###  1.2. Exponential decay of the semigroup

As far as the sole existence of $S(t)$ is concerned, we could have taken _any_
continuous function $f:\sigma(A)\to[0,\infty)$. However, as shown for instance
in [8], the exponential stability of the semigroup occurs if and only if (1.1)
is in place. Recall that $S(t)$ is said to be _exponentially stable_ if there
exist $\omega>0$ and $C\geq 1$ such that

(1.3) $\|S(t)\|_{L(\mathcal{H})}\leq Ce^{-\omega t},$

where the norm is taken in the Banach space $L(\mathcal{H})$ of bounded linear
operators on $\mathcal{H}$. Once the exponential stability is known, one would
like to find the best possible $\omega>0$ (i.e., the largest) for which (1.3)
holds. This number is the (exponential) _decay rate_ , defined as111The number
$-\omega_{*}$ is usually called the _growth bound_ of the semigroup.

$\omega_{*}=\sup\big{\\{}\omega>0:\,\|S(t)\|_{L(\mathcal{H})}\leq Ce^{-\omega
t}\big{\\}},$

for some $C=C(\omega)\geq 1$. However, for general (exponentially stable)
semigroups, computing $\omega_{*}$ might not be an easy task. Another relevant
quantity, usually much easier to detect, is the _spectral bound_ of
${\mathbb{A}}$

$\sigma_{*}=\sup_{\lambda\in\sigma({\mathbb{A}})}{\mathfrak{Re}}\,\lambda,$

$\sigma({\mathbb{A}})$ being the spectrum of (the complexification of)
${\mathbb{A}}$, which is related to $\omega_{*}$ through the (possibly strict)
inequality $\omega_{*}\leq-\sigma_{*}$ (see [9, 19]).

######  Definition 1.1.

The semigroup $S(t)$ satisfies the _spectrum determined growth_ (SDG)
condition if $\omega_{*}=-\sigma_{*}$.

In view of finding the decay rate, it is then of paramount importance to
establish whether or not the SDG condition is satisfied. This is true, for
instance, for eventually norm continuous (such as analytic or differentiable)
semigroups [9]. Still, even if $S(t)$ fulfills the SDG condition, this does
not mean at all that (1.3) holds for $\omega=\omega_{*}$, as the following
simple example shows.

######  Example 1.2.

Consider the damped pendulum equation

$\ddot{u}+2a\dot{u}+u=0,$

a particular instance of (1.2) for $\mathcal{H}=\mathbb{R}^{2}$, $A=1$ and
$f\equiv a>0$. The related semigroup, being norm continuous, always fulfills
the SDG condition. In particular, the decay rate is attained whenever $a\neq
1$. But, when $a=1$, the norm of the solution corresponding to the initial
datum $\boldsymbol{u}_{0}=(u_{0},v_{0})$ reads

$\|S(t)\boldsymbol{u}_{0}\|_{\mathcal{H}}=\sqrt{u_{0}^{2}+v_{0}^{2}+2(u_{0}^{2}-v_{0}^{2})t+2(u_{0}+v_{0})^{2}t^{2}}\,e^{-\frac{t}{2}}.$

The decay rate, equal to $\frac{1}{2}$ in this case, is certainly not
attained, and the reason is that a resonance phenomenon is encountered.

This motivates the next definition.

######  Definition 1.3.

The semigroup $S(t)$ satisfies the _strong spectrum determined growth_ (SSDG)
condition if the decay rate $\omega_{*}=-\sigma_{*}$ is attained, that is, if
(1.3) holds with $\omega=\omega_{*}=-\sigma_{*}$.

###  1.3. Earlier contributions

The equation (1.2) of this paper belongs to a general class of abstract
differential models introduced in [2], and “exhibiting the empirically
observed damping rates in elastic systems”. For this reason, $A$ is usually
called the _elastic operator_ , while $f(A)$ is the _damping operator_. The
prototypical case occurs when $f(s)$ controls and is controlled by
$s^{\theta}$ for some $\theta\in[0,1]$, namely, when $f(A)$ is comparable with
the powers $A^{\theta}$. Here, the semigroup is know to be analytic for
$\theta\in[\frac{1}{2},1]$ and differentiable (actually, of Gevrey type) for
$\theta\in(0,\frac{1}{2})$. See, among others, the classical papers [3, 4, 5,
13, 14, 15]. Further works, dealing with a damping operator not necessarily
comparable with $A^{\theta}$, include [1, 12, 16, 17, 18]. Nevertheless, the
literature concerning the analysis of the decay rate $\omega_{*}$ of $S(t)$,
with particular reference to whether or not it is actually attained, is
exceptionally bare. The full solution to the problem has been found only for
the weakly damped wave equation, sometimes referred to as the _telegrapher’s
equation_ , corresponding to the choice $f(s)=a>0$ (see [6, 10]). The much
more general (and difficult) situation of a nonconstant $f$ has been tackled
in [11]. There, within a certain number of assumptions, which prevent $f(s)$
to grow at infinity faster than $s^{\theta}$ with $\theta<\frac{1}{2}$, the
authors obtain sharp exponential decay estimates for trajectories originating
from sufficiently regular initial data, but they cannot generally exhibit the
decay rate of the semigroup. On the other hand, the necessary and sufficient
condition (1.1) for the exponential stability of $S(t)$ clearly allows $f(s)$
to have up to a linear growth at infinity. And indeed, the special case
$f(s)=s$ corresponds to the widely studied strongly damped wave equation, also
known by the name of _Kelvin-Voigt equation_ , which appears in several areas
of Mathematical Physics.

###  1.4. Our result

The aim of this paper is to provide a complete answer. Within the sole
condition (1.1), we show that the semigroup $S(t)$ fulfills (1.3) with
$\omega=\omega_{*}=-\sigma_{*}$, except in some particular cases, called
resonant, where the term $e^{-\omega_{*}t}$ is penalized by a factor $(1+t)$.
This result is optimal. The key idea of the proof is to decompose $\sigma(A)$
into the disjoint union of suitably chosen subsets $\sigma_{\imath}$, and then
perform energy estimates with _ad hoc_ multipliers on each of the (orthogonal)
subspaces $E_{A}(\sigma_{\imath})H$, making use of the functional calculus of
$A$.

The paper is organized as follows. In the next Section 2, we introduce two
objects that will play a crucial role in our analysis. In Section 3, we recall
the properties of the spectrum of the operator ${\mathbb{A}}$. The main result
is stated and proved in Section 4, and an application is then discussed in
Section 5. The final Sections 6-9 are devoted to the proofs of four lemmas
encountered in the proof of the main theorem.

### A word of warning

Although we work with a real Hilbert space $H$, having in mind the concrete
examples of Mathematical Physics, all the results of this paper hold
_verbatim_ if $H$ is a complex Hilbert space. The proofs are exactly the same,
up to replacing in the calculations any occurrence of a scalar product with
its real part.

##  2\. The Function ${\phi}$ and the Number ${m_{*}}$

A crucial object for our analysis is the continuous function
$\phi:\sigma(A)\to(0,\infty)$ defined as

$\phi(s)=\begin{cases}f(s)&\quad\text{if }f(s)\leq\sqrt{s}\,,\\\ \vskip
4.2679pt\cr f(s)-\sqrt{f^{2}(s)-s}&\quad\text{if }f(s)>\sqrt{s}\,,\end{cases}$

along with the number

$m_{*}=\inf_{s\in\sigma(A)}\phi(s)>0.$

The fact that $m_{*}>0$ merely follows by observing that

$f(s)=\begin{cases}\phi(s)&\quad\text{if }f(s)\leq\sqrt{s}\,,\\\ \vskip
1.42262pt\cr\displaystyle\frac{s}{2\phi(s)}+\frac{\phi(s)}{2}&\quad\text{if
}f(s)>\sqrt{s}\,.\end{cases}$

Hence, $m_{*}=0$ would violate (1.1). Then, in the region $f(s)>\sqrt{s}\,$,
we have that

(2.1) $\frac{f(s)}{s}\leq\frac{1}{2m_{*}}+\frac{m_{*}}{2s},$

this fact being equivalent to $\phi(s)\geq m_{*}$. Finally, for every $m>0$
and every sequence $s_{n}\to\infty$, we have the implication

(2.2)
$\frac{f(s_{n})}{s_{n}}\to\frac{1}{2m}\quad\Rightarrow\quad\phi(s_{n})\to m.$

Indeed, the first convergence implies in particular that

$\frac{s_{n}}{f^{2}(s_{n})}\to 0\qquad\text{and}\qquad
f(s_{n})>\sqrt{s_{n}}\,,$

the latter for all $n$ large. Accordingly,

$\phi(s_{n})=f(s_{n})\Big{(}1-\sqrt{1-\frac{s_{n}}{f^{2}(s_{n})}}\,\Big{)}\sim\frac{s_{n}}{2f(s_{n})}\to
m.$

##  3\. The Spectrum of the Infinitesimal Generator

For every fixed $s\in\sigma(A)$, we introduce the pair of complex numbers

$\lambda_{s}^{\pm}=\begin{cases}\displaystyle-f(s)\pm
i\sqrt{s-f^{2}(s)}&\quad\text{if }f(s)\leq\sqrt{s}\,,\\\ \vskip
2.84526pt\cr\displaystyle-f(s)\pm\sqrt{f^{2}(s)-s}&\quad\text{if
}f(s)>\sqrt{s}\,,\end{cases}$

which are nothing but the solutions to the second order equation

$\lambda^{2}+2f(s)\lambda+s=0.$

We define

$\Sigma=\bigcup_{s\in\sigma(A)}\big{\\{}\lambda_{s}^{\pm}\big{\\}}.$

We also consider the (possibly empty) set

$\Lambda=\Big{\\{}\lambda<0:\,\exists\,s_{n}\in\sigma(A)\,\,\,\text{such
that}\,\,\,s_{n}\to\infty\,\,\,\text{and}\,\,\,\lim_{n\to\infty}\frac{f(s_{n})}{s_{n}}=-\frac{1}{2\lambda}\Big{\\}}.$

Then, as shown in [8], the spectrum of (the complexification of) the operator
${\mathbb{A}}$ reads

$\sigma({\mathbb{A}})=\Sigma\cup\Lambda,$

where the union is not necessarily disjoint. In light of (2.2), it is apparent
that $\Lambda$ belongs to the closure of $\Sigma$. This yields the equality

(3.1)
$\sigma_{*}=\sup_{\lambda\in\sigma({\mathbb{A}})}{\mathfrak{Re}}\,\lambda=\sup_{\lambda\in\Sigma}{\mathfrak{Re}}\,\lambda=-\inf_{s\in\sigma(A)}\phi(s)=-m_{*}.$

######  Remark 3.1.

Concerning the structure of the spectrum of ${\mathbb{A}}$, we have that

$\sigma({\mathbb{A}})=\sigma_{\rm p}({\mathbb{A}})\cup\sigma_{\rm
c}({\mathbb{A}}),$

where

$\sigma_{\rm p}({\mathbb{A}})=\bigcup_{s\in\sigma_{\rm
p}(A)}\big{\\{}\lambda_{s}^{\pm}\big{\\}}.$

With standard notation, $\sigma_{\rm p}$ and $\sigma_{\rm c}$ are the point
spectrum and the continuous spectrum, respectively. This characterization,
although not explicitly stated in [8], can be inferred from the calculations
of that paper, by arguing as in [7, Sec. 5].

##  4\. The Theorem

###  4.1. Statement of the result

We begin with the rigorous definition of resonance.

######  Definition 4.1.

The semigroup $S(t)$ is said to be _resonant_ if there exists
$s_{*}\in\sigma(A)$ such that

$m_{*}=\phi(s_{*})\qquad\text{and}\qquad f(s_{*})=\sqrt{s_{*}}\,.$

In other words, the resonance phenomenon appears whenever the real part of the
spectrum of ${\mathbb{A}}$ attains its supremum $-m_{*}$ and the equality

$-m_{*}=\lambda^{+}_{s_{*}}=\lambda^{-}_{s_{*}}$

holds for some $s_{*}\in\sigma(A)$. In fact, if such a point $s_{*}$ exists,
it is easily seen to be unique.

######  Theorem 4.2.

There exists a constant $C\geq 1$ such that

  * I.

$\|S(t)\|_{L(\mathcal{H})}\leq Ce^{-m_{*}t},\,\,\,$ if $S(t)$ not resonant;
and

  * II.

$\|S(t)\|_{L(\mathcal{H})}\leq C(1+t)e^{-m_{*}t},\,\,\,$ if $S(t)$ resonant.

On account of (3.1), the theorem produces a straightforward consequence.

######  Corollary 4.3.

If $S(t)$ is not resonant, then it fulfills the SSDG condition. Instead, if
$S(t)$ is resonant, it fulfills the SDG condition, but not the SSDG one.

######  Remark 4.4.

In view of Example 1.2, the result is optimal.

It is sometimes possible to provide a precise estimate of the constant $C$.

######  Proposition 4.5.

Assume that the semigroup is subdamped, that is,

$\sup_{s\in\sigma(A)}\frac{f(s)}{\sqrt{s}}=\ell<1.$

Then point I of Theorem 4.2 holds with

$C=\sqrt{\frac{1+\ell}{1-\ell}}.$

Also in this case, the result is optimal, as the following example shows.

######  Example 4.6.

Consider the subdamped pendulum equation

$\ddot{u}+2a\dot{u}+u=0,\quad a\in(0,1),$

which meets the hypotheses of the proposition above with $\ell=a$. Taking the
unit vector $\boldsymbol{u}_{0}=(\frac{1}{\sqrt{2}}\,,\frac{1}{\sqrt{2}})$,
the norm of the solution at time $t$ reads

$\|S(t)\boldsymbol{u}_{0}\|_{\mathcal{H}}=\sqrt{\frac{1-a\cos\big{(}2\sqrt{1-a^{2}}\,t\big{)}}{1-a}}\,e^{-\frac{a}{2}t}.$

Choosing

$t=t_{k}=\frac{(2k+1)\pi}{2\sqrt{1-a^{2}}\,},$

for any $k\in{\mathbb{N}}$, we get exactly

$\|S(t_{k})\|_{L(\mathcal{H})}\geq\|S(t_{k})\boldsymbol{u}_{0}\|_{\mathcal{H}}=\sqrt{\frac{1+a}{1-a}}\,e^{-\frac{a}{2}{t_{k}}}.$

###  4.2. Some preparatory lemmas

For any given $K\geq 2$ and $\varepsilon\in(0,1)$, we decompose the spectrum
of $A$ into the disjoint union

$\sigma(A)=\sigma_{0}\cup\sigma_{1}\cup\sigma_{2}\cup\sigma_{3},$

where the four regions $\sigma_{\imath}$ (some of which possibly empty) are
defined as

$\displaystyle\sigma_{0}$
$\displaystyle=\Big{\\{}s\in\sigma(A):\,\frac{f(s)}{\sqrt{s}}>K\Big{\\}},$
$\displaystyle\sigma_{1}$
$\displaystyle=\Big{\\{}s\in\sigma(A):\,\frac{f(s)}{\sqrt{s}}\leq
1-\varepsilon\Big{\\}},$ $\displaystyle\sigma_{2}$
$\displaystyle=\Big{\\{}s\in\sigma(A):\,1+\varepsilon\leq\frac{f(s)}{\sqrt{s}}\leq
K\Big{\\}},$ $\displaystyle\sigma_{3}$
$\displaystyle=\Big{\\{}s\in\sigma(A):\,1-\varepsilon<\frac{f(s)}{\sqrt{s}}<1+\varepsilon\Big{\\}}.$

The proofs of Theorem 4.2 and the subsequent Proposition 4.5 are based on the
following four lemmas, which hold for all regular initial data, that is, for
all $(u_{0},v_{0})$ in the domain of ${\mathbb{A}}$. Given the trajectory

$(u(t),\dot{u}(t))=S(t)(u_{0},v_{0})\in{\rm{dom}}({\mathbb{A}}),$

we split the corresponding energy at time $t\geq 0$ into the sum

${\mathsf{E}}(t)=\sum_{\imath=0}^{3}{\mathsf{E}}_{\imath}(t),$

where

${\mathsf{E}}_{\imath}(t)=\|A^{\frac{1}{2}}E_{A}(\sigma_{\imath})u\|^{2}+\|E_{A}(\sigma_{\imath})\dot{u}\|^{2}.$

Note that ${\mathsf{E}}_{\imath}\equiv 0$ if $\sigma_{\imath}=\emptyset$. The
constants $K$ and $\varepsilon$ in the statements of the lemmas are understood
to be the ones appearing in the definitions of $\sigma_{\imath}$.

######  Lemma 4.7.

For every $K\geq 2$ large enough we have the inequality

${\mathsf{E}}_{0}(t)\leq 3{\mathsf{E}}_{0}(0)e^{-2m_{*}t}.$

######  Lemma 4.8.

For every $\varepsilon\in(0,1)$ we have the inequality

${\mathsf{E}}_{1}(t)\leq\frac{2-\varepsilon}{\varepsilon}{\mathsf{E}}_{1}(0)e^{-2m_{*}t}.$

######  Lemma 4.9.

For every $\varepsilon\in(0,1)$ and every $K\geq 2$, we have the inequality

${\mathsf{E}}_{2}(t)\leq\frac{9K^{2}}{\varepsilon}{\mathsf{E}}_{2}(0)e^{-2m_{*}t}.$

######  Lemma 4.10.

For every $\varepsilon\in(0,\frac{1}{16})$ such that
$\sigma_{3}\neq\emptyset$, we have the inequality

${\mathsf{E}}_{3}(t)\leq\frac{8}{\varepsilon}{\mathsf{E}}_{3}(0)e^{-2m_{3}(1-4\sqrt{\varepsilon}\,)t},$

where $m_{3}=\inf_{s\in\sigma_{3}}\phi(s)$.

The rather technical proofs of the four lemmas, heavily based on the
functional calculus of $A$, will be carried out in the final Sections 6, 7, 8
and 9 of the paper.

###  4.3. Proof of Theorem 4.2

We consider the two cases separately.

I. Let us choose $K\geq 2$ sufficiently large in order for Lemma 4.7 to hold.
We claim that, up to fixing $\varepsilon\in(0,\frac{1}{16})$ small enough,

${\mathsf{E}}_{3}(t)\leq\frac{8}{\varepsilon}{\mathsf{E}}_{3}(0)e^{-2m_{*}t}.$

This is trivially true if $\sigma_{3}=\emptyset$ for some $\varepsilon$. Let
us then assume that $\sigma_{3}\neq\emptyset$ for all
$\varepsilon\in(0,\frac{1}{16})$. We preliminarily observe that, for all
$\varepsilon$ small,

$m_{3}>m_{*}.$

If not, there would exist a sequence $s_{n}\in\sigma(A)$ such that

$\frac{f(s_{n})}{\sqrt{s_{n}}}\to 1\qquad\text{and}\qquad\phi(s_{n})\to
m_{*}=m_{3}.$

If $s_{n}$ is unbounded, then (up to a subsequence) $s_{n}\to\infty$.
Therefore,

$m_{*}\sim\phi(s_{n})\sim f(s_{n})\sim\sqrt{s_{n}}\,\to\infty,$

a contradiction. This tells that $s_{n}$ must be bounded. But then (up to a
subsequence) $s_{n}\to s_{*}\in\sigma(A)$, as $\sigma(A)$ is closed. By
continuity, we deduce that $f(s_{*})=\sqrt{s_{*}}$ and $\phi(s_{*})=m_{*}$,
against the assumption that $S(t)$ is not resonant. At this point, the claim
follows by fixing $\varepsilon$ in Lemma 4.10 small enough that
$m_{3}(1-4\sqrt{\varepsilon}\,)\geq m_{*}$ (just note that $m_{3}$ is a
decreasing function of $\varepsilon$). Once the claim is proven, setting

$M=\frac{9K^{2}}{\varepsilon},$

and taking advantage of Lemmas 4.7-4.9, we conclude that

${\mathsf{E}}(t)=\sum_{\imath=0}^{3}{\mathsf{E}}_{\imath}(t)\leq
M{\mathsf{E}}(0)e^{-2m_{*}t}.$

Due to the continuity of the semigroup, the latter inequality remains valid by
density for all initial data $(u_{0},v_{0})$ belonging to the phase space
$\mathcal{H}$. This finishes the proof. ∎

II. In this case, we have the equality $m_{3}=m_{*}$ for every
$\varepsilon\in(0,\frac{1}{16})$. Select then any $K\geq 2$ sufficiently large
in order for Lemma 4.7 to hold. From Lemmas 4.7-4.10 we learn that the
inequality

${\mathsf{E}}(t)\leq\frac{9K^{2}}{\varepsilon}{\mathsf{E}}(0)e^{-2m_{*}(1-4\sqrt{\varepsilon}\,)t}$

holds for every $\varepsilon\in(0,\frac{1}{16})$. Fixing an arbitrary
$\varepsilon_{*}\in(0,\frac{1}{16})$, we choose for every $t\geq 0$

$\varepsilon=\varepsilon(t)=\frac{\varepsilon_{*}}{(1+t)^{2}}.$

Then, calling

$M=\frac{9K^{2}}{\varepsilon_{*}}e^{8m_{*}\sqrt{\varepsilon_{*}}\,},$

we finally obtain

${\mathsf{E}}(t)\leq M(1+t)^{2}{\mathsf{E}}(0)e^{-2m_{*}t}.$

As before, the sought inequality is valid by density for all initial data. ∎

###  4.4. Proof of Proposition 4.5

If the semigroup is subdamped, then $\sigma(A)=\sigma_{1}$ with
$\varepsilon=1-\ell$. Accordingly, the energy ${\mathsf{E}}$ reduces to
${\mathsf{E}}_{1}$, and point I of Theorem 4.2 is nothing but Lemma 4.8. ∎

##  5\. An Application: Wave Equations with Fractional Damping

For $a>0$ and $\theta\in[0,1]$, we consider equation (1.2) with
$f(s)=as^{\theta}$, namely,

$\ddot{u}+2aA^{\theta}\dot{u}+Au=0.$

The function $f$ clearly complies with (1.1), so that Theorem 4.2 and
Proposition 4.5 apply. We denote by $s_{0}>0$ the minimum of the spectrum of
$A$.

######  Remark 5.1.

Given a bounded domain $\Omega\subset\mathbb{R}^{N}$ with smooth boundary
$\partial\Omega$, a concrete instance of this model is obtained by taking
$H=L^{2}(\Omega)$ and the Laplace-Dirichlet operator $A=-\Delta$ with domain

${\rm{dom}}(-\Delta)=H^{2}(\Omega)\cap H_{0}^{1}(\Omega).$

In particular, for $\theta=0$ we have the weakly damped wave equation, whereas
for $\theta=1$ we have the strongly damped wave equation. Another physically
relevant model is the beam (for $N=1$) or plate (for $N=2$) equation with
fractional damping, subject to the hinged boundary conditions. In this case,
the operator $A$ is the Bilaplacian $\Delta^{2}$ with domain

${\rm{dom}}(\Delta^{2})=\big{\\{}u\in H^{2}(\Omega)\cap
H_{0}^{1}(\Omega):\,\Delta u\in H^{2}(\Omega)\cap H_{0}^{1}(\Omega)\big{\\}}.$

For this choice of $f$, the function $\phi$ reads

$\phi(s)=\begin{cases}as^{\theta}&\quad\text{if }s\geq
a^{\frac{2}{1-2\theta}},\\\ \vskip 4.2679pt\cr
as^{\theta}-\sqrt{a^{2}s^{2\theta}-s}&\quad\text{if
}s<a^{\frac{2}{1-2\theta}},\end{cases}\qquad\text{if }\theta<\frac{1}{2},$

and

$\phi(s)=\begin{cases}as^{\theta}&\quad\text{if }s\leq
a^{\frac{2}{1-2\theta}},\\\ \vskip 4.2679pt\cr
as^{\theta}-\sqrt{a^{2}s^{2\theta}-s}&\quad\text{if
}s>a^{\frac{2}{1-2\theta}},\end{cases}\qquad\text{if }\theta>\frac{1}{2},$

whereas

$\phi(s)=\begin{cases}a\sqrt{s}\,&\quad\text{if }a\leq 1,\\\ \vskip
4.2679pt\cr\big{(}a-\sqrt{a^{2}-1}\,\big{)}\sqrt{s}\,&\quad\text{if
}a>1,\end{cases}\qquad\text{if }\theta=\frac{1}{2}.$

Note that $\phi(s)\leq\sqrt{s}\,$. Besides, for $\theta\neq\frac{1}{2}$,
equality occurs only for $s=a^{\frac{2}{1-2\theta}}$. Accordingly, whenever
$\theta\neq\frac{1}{2}$ and the semigroup is resonant,

$m_{*}=a^{\frac{1}{1-2\theta}}.$

In general, the picture strongly depends on the exponent $\theta$. To this
end, we first observe that

$\frac{f(s)}{\sqrt{s}}\,=as^{\theta-\frac{1}{2}}$

is decreasing for $\theta<\frac{1}{2}$, constant and equal to $a$ for
$\theta=\frac{1}{2}$, increasing and diverging to infinity for
$\theta>\frac{1}{2}$. In particular, this tells that, if $\theta>\frac{1}{2}$
and $A$ is an unbounded operator, the semigroup $S(t)$ is never subdamped.

$\boldsymbol{\theta=0}$

$\boldsymbol{\theta\in(0,\frac{1}{2})}$

$\boldsymbol{\theta\in(\frac{1}{2},1)}$

$\boldsymbol{\theta=1}$

Figure 1. Profile of $\phi(s)$ versus $\sqrt{s}$ (dashed) for different values
of $\theta$.

###  $\bullet$ Case $\boldsymbol{\theta\in[0,\tfrac{1}{2})}$

The function $\phi$ is increasing (actually, strictly increasing if
$\theta>0$). Therefore, Theorem 4.2 holds with

$m_{*}=\phi(s_{0}).$

The semigroup $S(t)$ is resonant if and only if
$s_{0}=a^{\frac{2}{1-2\theta}}$. Besides, the semigroup is subdamped whenever
$s_{0}>a^{\frac{2}{1-2\theta}}$. In which case, we infer from Proposition 4.5
that the constant $C$ of the theorem reads

$C=\sqrt{\frac{{s_{0}}^{\frac{1}{2}-\theta}+a}{{s_{0}}^{\frac{1}{2}-\theta}-a}}\,.$

###  $\bullet$ Case $\boldsymbol{\theta=\frac{1}{2}}$

Again, $m_{*}=\phi(s_{0})$. The semigroup is resonant if and only if $a=1$,
and is subdamped if and only if $a<1$, with

$C=\sqrt{\frac{1+a}{1-a}}\,.$

Note that $C$ is now independent of $s_{0}$.

###  $\bullet$ Case $\boldsymbol{\theta\in(\tfrac{1}{2},1)}$

This is the most intriguing situation, since:

  * -

$\phi$ is increasing for $s<a^{\frac{2}{1-2\theta}}$.

  * -

$\phi$ is decreasing for $s\in(a^{\frac{2}{1-2\theta}},s_{\rm m})$, where
$s_{\rm m}>a^{\frac{2}{1-2\theta}}$ is the minimum of $\phi$, here thought as
defined on $(0,\infty)$ and not only on $\sigma(A)$, on the interval
$(a^{\frac{2}{1-2\theta}},\infty)$.

  * -

$\phi$ is increasing and diverging to infinity for $s>s_{\rm m}$.

The value $m_{*}$ is now less explicit, since it might not be equal to
$\phi(s_{0})$. In particular, resonance occurs if and only if

$s_{0}=a^{\frac{2}{1-2\theta}}\qquad\text{and}\qquad\sigma(A)\cap(a^{\frac{2}{1-2\theta}},s_{\rm
b})=\emptyset,$

where $s_{\rm b}$ is the bigger of the two (distinct) solutions to the
equation

$s-2a^{\frac{2-2\theta}{1-2\theta}}s^{\theta}+a^{\frac{2}{1-2\theta}}=0,\quad
s>0,$

the other one being $a^{\frac{2}{1-2\theta}}$. In this case, $s_{0}$ is an
eigenvalue of $A$, being an isolated point of $\sigma(A)$, and in turn
$-m_{*}$ is an eigenvalue of ${\mathbb{A}}$.

###  $\bullet$ Case $\boldsymbol{\theta=1}$

We have the equality $\phi(s)=as$ for $s<\frac{1}{a^{2}}$, where $\phi$,
thought again as defined on $(0,\infty)$, reaches its maximum value, equal to
$\frac{1}{a}$. After that, $\phi$ is decreasing, and converges to
$\frac{1}{2a}$ as $s\to\infty$. Then, if $A$ is unbounded,

$m_{*}=\min\Big{\\{}as_{0},\frac{1}{2a}\Big{\\}},$

whereas if $A$ is bounded,

$m_{*}=\min\big{\\{}as_{0},\phi(s_{\rm M})\big{\\}},$

where $s_{\rm M}$ is the maximum of $\sigma(A)$. Quite interestingly, when
$\theta=1$ resonance cannot occur, except in the trivial situation where
$\sigma(A)=\\{\frac{1}{a^{2}}\\}$.

##  6\. Proof of Lemma 4.7

We choose $K$ sufficiently large in order to have the set inclusion

(6.1) $\sigma_{0}\subset(K,\infty).$

Indeed, recalling the second constraint in (1.1), if $s\leq K$ with $K$ large
then

$\frac{f(s)}{\sqrt{s}}\leq\sup_{s\in\sigma(A)}\frac{f(s)}{s}\,\sqrt{K}\,\leq
K.$

Defining the function

$g(s)=f(s)-\frac{m_{*}}{2},$

we rewrite equation (1.2) in the equivalent form

$\ddot{u}+2g(A)\dot{u}+m_{*}\dot{u}+Au=0.$

Note that, due to (2.1),

(6.2) $s\geq 2m_{*}g(s),\quad\forall s\in\sigma_{0}.$

Next, setting $w=E_{A}(\sigma_{0})u$, we introduce the functional

${\mathsf{F}}_{0}={\mathsf{E}}_{0}+2m_{*}^{2}\|w\|^{2}+4m_{*}\langle
w,\dot{w}\rangle.$

In what follows, as well as in the proofs of the remaining three lemmas, all
the calculations are rigorous, since we are working with regular solutions. In
particular, we use the fact that the selfadjoint projection
$E_{A}(\sigma_{0})$ commutes with all the functions of $A$. Multiplying the
equation by the test function $2\dot{w}+4m_{*}w$, we are led to the
differential identity

$\frac{d}{dt}{\mathsf{F}}_{0}+2m_{*}{\mathsf{F}}_{0}+2{\mathsf{G}}_{0}=0,$

where

${\mathsf{G}}_{0}=m_{*}\big{[}\|A^{\frac{1}{2}}w\|^{2}-\|(2m_{*}g)^{\frac{1}{2}}(A)w\|^{2}\big{]}+2\big{[}\|g^{\frac{1}{2}}(A)(m_{*}w+\dot{w})\|^{2}-m_{*}\|m_{*}w+\dot{w}\|^{2}\big{]}.$

In light of (6.2), we see at once that

$\|A^{\frac{1}{2}}w\|^{2}-\|(2m_{*}g)^{\frac{1}{2}}(A)w\|^{2}\geq 0.$

Besides, from the very definition of $g$ and $\sigma_{0}$,

$g(s)\geq K\sqrt{s}-\frac{m_{*}}{2},\quad\forall s\in\sigma_{0}.$

Thus, we infer from (6.1) that $g(s)\geq m_{*}$ provided that $K$ is large
enough. Accordingly,

$\|g^{\frac{1}{2}}(A)(m_{*}w+\dot{w})\|^{2}-m_{*}\|m_{*}w+\dot{w}\|^{2}\geq
0.$

In summary, we have proved that ${\mathsf{G}}_{0}\geq 0$. Therefore,

$\frac{d}{dt}{\mathsf{F}}_{0}+2m_{*}{\mathsf{F}}_{0}\leq 0,$

and by the Gronwall Lemma we conclude that

${\mathsf{F}}_{0}(t)\leq{\mathsf{F}}_{0}(0)e^{-2m_{*}t}.$

The proof is finished once we show the double inequality

$\frac{1}{2}{\mathsf{E}}_{0}\leq{\mathsf{F}}_{0}\leq\frac{3}{2}{\mathsf{E}}_{0}.$

Indeed, from the Cauchy-Schwarz and the Young inequalities, and exploiting
(6.1),

$2m_{*}^{2}\|w\|^{2}+4m_{*}|\langle w,\dot{w}\rangle|\leq
10m_{*}^{2}\|w\|^{2}+\frac{1}{2}\|\dot{w}\|^{2}\leq\frac{10m_{*}^{2}}{K}\|A^{\frac{1}{2}}w\|^{2}+\frac{1}{2}\|\dot{w}\|^{2}\leq\frac{1}{2}{\mathsf{E}}_{0},$

as soon as $K$ is sufficiently large. ∎

##  7\. Proof of Lemma 4.8

Setting $w=E_{A}(\sigma_{1})u$, we introduce the functional

${\mathsf{F}}_{1}={\mathsf{E}}_{1}+2\langle f(A)w,\dot{w}\rangle.$

Multiplying (1.2) by the test function $2\dot{w}+2f(A)w$, we get the
differential identity

$\frac{d}{dt}{\mathsf{F}}_{1}+2m_{*}{\mathsf{F}}_{1}+2{\mathsf{G}}_{1}=0,$

having set (recall that in this region $s-f^{2}(s)>0$)

$\displaystyle{\mathsf{G}}_{1}$
$\displaystyle=\|f^{\frac{1}{2}}(A)(A-f^{2}(A))^{\frac{1}{2}}w\|^{2}-m_{*}\|(A-f^{2}(A))^{\frac{1}{2}}w\|^{2}$
$\displaystyle\quad+\|f^{\frac{1}{2}}(A)(f(A)w+\dot{w})\|^{2}-m_{*}\|f(A)w+\dot{w}\|^{2}.$

Since

$m_{*}\leq\phi(s)=f(s),\quad\forall s\in\sigma_{1},$

it is apparent that ${\mathsf{G}}_{1}\geq 0$. Therefore, we arrive at

(7.1) $\frac{d}{dt}{\mathsf{F}}_{1}+2m_{*}{\mathsf{F}}_{1}\leq 0.$

Finally, we claim that

(7.2)
$\varepsilon{\mathsf{E}}_{1}\leq{\mathsf{F}}_{1}\leq(2-\varepsilon){\mathsf{E}}_{1}.$

Indeed, recalling that $f(s)\leq(1-\varepsilon)\sqrt{s}$ for every
$s\in\sigma_{1}$, from the Cauchy-Schwarz and the Young inequalities, we get

$2|\langle
f(A)w,\dot{w}\rangle|\leq(1-\varepsilon)\|A^{\frac{1}{2}}w\|^{2}+(1-\varepsilon)\|\dot{w}\|^{2}.$

Collecting (7.1)-(7.2), and exploiting the Gronwall Lemma, the conclusion
follows. ∎

##  8\. Proof of Lemma 4.9

In this proof, as well as in the following one, the symbol
$\langle\cdot,\cdot\rangle$ will also be used to denote the duality product.
Setting $w=E_{A}(\sigma_{2})u$, we introduce the functional

${\mathsf{F}}_{2}={\mathsf{E}}_{2}+2\langle(f^{2}(A)-A)w,w\rangle+2\langle
f(A)w,\dot{w}\rangle.$

Multiplying (1.2) by the test function $2\dot{w}+2f(A)w$ we get the
differential identity

$\frac{d}{dt}{\mathsf{F}}_{2}+2m_{*}{\mathsf{F}}_{2}+2{\mathsf{G}}_{2}=0,$

having set

$\displaystyle{\mathsf{G}}_{2}$
$\displaystyle=\langle(Af(A)+m_{*}A-2m_{*}f^{2}(A))w,w\rangle+\langle(f(A)-m_{*})\dot{w},\dot{w}\rangle$
$\displaystyle\quad+2\langle(A-m_{*}f(A))w,\dot{w}\rangle.$

We claim that ${\mathsf{G}}_{2}\geq 0$. Indeed, since

$m_{*}\leq\phi(s)=f(s)-\sqrt{f^{2}(s)-s}\,<f(s),\quad\forall s\in\sigma_{2},$

exploiting the Cauchy-Schwarz and the Young inequalities,

$\displaystyle 2\langle(A-m_{*}f(A))w,\dot{w}\rangle$
$\displaystyle=2\langle(f(A)-m_{*})^{-\frac{1}{2}}(A-m_{*}f(A))w,(f(A)-m_{*})^{\frac{1}{2}}\dot{w}\rangle$
$\displaystyle\geq-\langle(f(A)-m_{*})^{-1}(A-m_{*}f(A))^{2}w,w\rangle-\langle(f(A)-m_{*})\dot{w},\dot{w}\rangle.$

Therefore, since

$sf(s)+m_{*}s-2m_{*}f^{2}(s)-\frac{(s-m_{*}f(s))^{2}}{f(s)-m_{*}}=\frac{f^{2}(s)-s}{f(s)-m_{*}}(s+m_{*}^{2}-2m_{*}f(s)),$

we get

${\mathsf{G}}_{2}\geq\langle(f(A)-m_{*})^{-1}(f^{2}(A)-A)(A+m_{*}^{2}-2m_{*}f(A))w,w\rangle.$

The claim follows by noting that, for every $s\in\sigma_{2}$,

$s+m_{*}^{2}-2m_{*}f(s)=\big{(}f(s)-\sqrt{f^{2}(s)-s}\,-m_{*}\big{)}\big{(}f(s)+\sqrt{f^{2}(s)-s}\,-m_{*}\big{)}\geq
0.$

Accordingly, we end up with the differential inequality

(8.1) $\frac{d}{dt}{\mathsf{F}}_{2}+2m_{*}{\mathsf{F}}_{2}\leq 0.$

Next, we show that

(8.2) $\frac{\varepsilon}{3}{\mathsf{E}}_{2}\leq{\mathsf{F}}_{2}\leq
3K^{2}{\mathsf{E}}_{2}.$

To this end, we further apply the Cauchy-Schwarz and the Young inequalities,
to get

$2|\langle
f(A)w,\dot{w}\rangle|\leq\frac{1}{\tau}\|f(A)w\|^{2}+\tau\|\dot{w}\|^{2},$

for every $\tau>0$. Since $f^{2}(s)\leq K^{2}s$ for every $s\in\sigma_{2}$, by
setting $\tau=1$ we obtain

${\mathsf{F}}_{2}\leq{\mathsf{E}}_{2}+2(K^{2}-1)\|A^{\frac{1}{2}}w\|^{2}+K^{2}\|A^{\frac{1}{2}}w\|^{2}+\|\dot{w}\|^{2}\leq
3K^{2}{\mathsf{E}}_{2}.$

As far as the other inequality is concerned, we have

${\mathsf{F}}_{2}\geq(1-\tau){\mathsf{E}}_{2}+\langle((2-\tau^{-1})f^{2}(A)+(\tau-2)A)w,w\rangle.$

Recalling that $f^{2}(s)\geq(1+\varepsilon)^{2}s$ for every $s\in\sigma_{2}$,
and selecting $\tau=1-\frac{\varepsilon}{3}$, we end up with

${\mathsf{F}}_{2}\geq\frac{\varepsilon}{3}{\mathsf{E}}_{2}+\varrho(\varepsilon)\|A^{\frac{1}{2}}w\|^{2},$

where, for every $\varepsilon\in(0,1)$,

$\varrho(\varepsilon)=\frac{(3-2\varepsilon)(1+\varepsilon)^{2}}{3-\varepsilon}-1-\frac{\varepsilon}{3}=\frac{2\varepsilon(6-\varepsilon-3\varepsilon^{2})}{3(3-\varepsilon)}>0.$

At this point, the conclusion is drawn from (8.1)-(8.2) and the Gronwall
Lemma. ∎

##  9\. Proof of Lemma 4.10

Along the proof, besides the constraint $\varepsilon\in(0,\frac{1}{16})$, we
will use the fact that

$m_{3}\leq\phi(s)\leq f(s)<(1+\varepsilon)\sqrt{s}\,,\quad\forall
s\in\sigma_{3}.$

Setting $w=E_{A}(\sigma_{3})u$, we introduce the functional

${\mathsf{F}}_{3}={\mathsf{E}}_{3}+2(1+4\sqrt{\varepsilon}\,)\langle
f(A)w,\dot{w}\rangle+8\sqrt{\varepsilon}\,(1+4\sqrt{\varepsilon}\,)\|f(A)w\|^{2}.$

Multiplying (1.2) by the test function
$2\dot{w}+2(1+4\sqrt{\varepsilon}\,)f(A)w$, we get the differential identity

$\frac{d}{dt}{\mathsf{F}}_{3}+2m_{3}(1-4\sqrt{\varepsilon}\,){\mathsf{F}}_{3}+2{\mathsf{G}}_{3}=0.$

Here,

$\displaystyle{\mathsf{G}}_{3}=\langle
p(A)w,w\rangle+(1-4\sqrt{\varepsilon}\,)\langle(f(A)-m_{3})\dot{w},\dot{w}\rangle+2(1-16\varepsilon)\langle
f(A)(f(A)-m_{3})w,\dot{w}\rangle,$

having set

$\displaystyle p(s)$
$\displaystyle=(1+4\sqrt{\varepsilon}\,)sf(s)-m_{3}(1-4\sqrt{\varepsilon}\,)s-8m_{3}\sqrt{\varepsilon}\,(1-16\varepsilon)f^{2}(s)$
$\displaystyle\geq(1+4\sqrt{\varepsilon}\,)sf(s)-m_{3}(1-4\sqrt{\varepsilon}\,)s-8m_{3}\sqrt{\varepsilon}\,(1-16\varepsilon)(1+\varepsilon)^{2}s,$
$\displaystyle=(1+4\sqrt{\varepsilon}\,)s(f(s)-m_{3})+8m_{3}\sqrt{\varepsilon}\,\big{[}1-(1-16\varepsilon)(1+\varepsilon)^{2}\big{]}s$
$\displaystyle\geq(1+4\sqrt{\varepsilon}\,)s(f(s)-m_{3}).$

Accordingly,

$\displaystyle{\mathsf{G}}_{3}$
$\displaystyle\geq(1+4\sqrt{\varepsilon}\,)\langle
A(f(A)-m_{3})w,w\rangle+(1-4\sqrt{\varepsilon}\,)\langle(f(A)-m_{3})\dot{w},\dot{w}\rangle$
$\displaystyle\quad+2(1-16\varepsilon)\langle
f(A)(f(A)-m_{3})w,\dot{w}\rangle.$

By an application of the Cauchy-Schwarz and the Young inequalities, we find

$\displaystyle 2(1-16\varepsilon)\langle f(A)(f(A)-m_{3})w,\dot{w}\rangle$
$\displaystyle=2\langle(1+4\sqrt{\varepsilon}\,)(1-4\sqrt{\varepsilon}\,)^{\frac{1}{2}}f(A)(f(A)-m_{3})^{\frac{1}{2}}w,(1-4\sqrt{\varepsilon}\,)^{\frac{1}{2}}(f(A)-m_{3})^{\frac{1}{2}}\dot{w}\rangle$
$\displaystyle\geq-(1-16\varepsilon)(1+4\sqrt{\varepsilon}\,)\langle
f^{2}(A)(f(A)-m_{3})w,w\rangle-(1-4\sqrt{\varepsilon}\,)\langle(f(A)-m_{3})\dot{w},\dot{w}\rangle,$

implying in turn

$\displaystyle{\mathsf{G}}_{3}$
$\displaystyle\geq(1+4\sqrt{\varepsilon}\,)\langle(f(A)-m_{3})(A-(1-16\varepsilon)f^{2}(A))w,w\rangle.$

Observing that

$s-(1-16\varepsilon)f^{2}(s)\geq\big{[}1-(1-16\varepsilon)(1+\varepsilon)^{2}\big{]}s\geq
0,$

we conclude that ${\mathsf{G}}_{3}\geq 0$, and we arrive at

$\frac{d}{dt}{\mathsf{F}}_{3}+2m_{3}(1-4\sqrt{\varepsilon}\,){\mathsf{F}}_{3}\leq
0.$

After an application of the Gronwall Lemma, the proof is finished once we show
that

$\varepsilon{\mathsf{E}}_{3}\leq{\mathsf{F}}_{3}\leq 8{\mathsf{E}}_{3}.$

Indeed, the second inequality is a straightforward consequence of the Cauchy-
Schwarz and the Young inequalities, and is left to the reader. Concerning the
first one, we have

$2(1+4\sqrt{\varepsilon}\,)\langle
f(A)w,\dot{w}\rangle\geq-(1-\varepsilon)\|\dot{w}\|^{2}-\frac{(1+4\sqrt{\varepsilon}\,)^{2}}{1-\varepsilon}\|f(A)w\|^{2}.$

Therefore,

$\displaystyle{\mathsf{F}}_{3}$
$\displaystyle\geq\varepsilon{\mathsf{E}}_{3}+(1-\varepsilon)\|A^{\frac{1}{2}}w\|^{2}+\Big{[}8\sqrt{\varepsilon}\,(1+4\sqrt{\varepsilon}\,)-\frac{(1+4\sqrt{\varepsilon}\,)^{2}}{1-\varepsilon}\Big{]}\|f(A)w\|^{2}$
$\displaystyle\geq\varepsilon{\mathsf{E}}_{3}+\Big{[}\frac{1-\varepsilon}{(1+\varepsilon)^{2}}+8\sqrt{\varepsilon}\,(1+4\sqrt{\varepsilon}\,)-\frac{(1+4\sqrt{\varepsilon}\,)^{2}}{1-\varepsilon}\Big{]}\|f(A)w\|^{2},$

and the quantity in the square brackets is positive for all
$\varepsilon\in(0,\frac{1}{16})$. ∎

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