On non-local ergodic Jacobi semigroups: spectral theory, convergence-to-equilibrium and contractivity

(1)

Bruce K. Driver

Analysis Tools with Applications,

SPIN Springer’s internal project number, if known

June 9, 2003 File:analpde.tex

Springer

Berlin Heidelberg NewYork Hong Kong London

Milan Paris Tokyo

(2)

Preface

These are lecture notes from Real analysis and PDE, Math 240 and Math 231.

Some sections are in better shape than others. I am sorry for those sections which are still a bit of a mess. These notes are still not polished. Nevertheless, I hope they may be of some use even in this form.

(16)

Part I

Basic Topological, Metric and Banach Space

Notions

(17)

1 Limits, sums, and other basics

1.1 Set Operations

Suppose that X is a set. LetP(X)or 2^X denote the power set ofX, that is elements ofP(X) = 2^X are subsets ofA. ForA∈2^X let

A^c=X\A={x∈X:x /∈A} and more generally ifA, B⊂X let

B\A={x∈B:x /∈A}. We also define the symmetric diﬀerence ofAand B by

A4B= (B\A)∪(A\B).

As usual if {A_α}α∈I is an indexed collection of subsets of X we define the union and the intersection of this collection by

∪^α∈IAα:={x∈X:∃α∈I 3 x∈Aα}and

∩α∈IA_α:={x∈X:x∈A_α∀α∈I}. Notation 1.1 We will also write `

α∈IA_α for ∪α∈IA_α in the case that {A_α}α∈I are pairwise disjoint, i.e.A_α∩A_β =∅if α6=β.

Notice that ∪ is closely related to ∃ and ∩ is closely related to ∀. For example let{A_n}^∞n=1 be a sequence of subsets fromX and define

{A_n i.o.}:={x∈X : #{n:x∈A_n}=∞}and

{An a.a.}:={x∈X :x∈An for allnsuﬃciently large}.

(One should read{A_n i.o.}asA_ninfinitely often and{A_n a.a.}asA_n almost always.) Then x∈ {A_n i.o.}iﬀ ∀N ∈ N ∃ n ≥N 3 x∈ A_n which may be written as

(18)

{A_n i.o.}=∩^∞N=1∪n≥NA_n.

Similarly, x ∈ {An a.a.} iﬀ ∃ N ∈ N 3 ∀ n ≥ N, x ∈ An which may be written as

{A_n a.a.}=∪^∞N=1∩n≥NA_n.

1.2 Limits, Limsups, and Liminfs

Notation 1.2 The Extended real numbers is the setR¯:=R∪{±∞},i.e. it is Rwith two new points called ∞ and−∞.We use the following conventions,

±∞·0 = 0,±∞+a=±∞for anya∈R, ∞+∞=∞ and−∞ − ∞=−∞

while∞ − ∞ is not defined.

If Λ ⊂ R¯ we will let supΛ and infΛ denote the least upper bound and greatest lower bound ofΛrespectively. We will also use the following convention, ifΛ=∅, thensup∅=−∞andinf∅= +∞.

Notation 1.3 Suppose that{xn}^∞n=1⊂R¯ is a sequence of numbers. Then lim inf

n→∞x_n= lim

n→∞inf{x_k:k≥n}and (1.1) lim sup

n→∞

x_n= lim

n→∞sup{x_k:k≥n}. (1.2) We will also writelimforlim inf andlimforlim sup.

Remark 1.4.Notice that if ak := inf{xk : k ≥ n} and bk := sup{xk : k ≥ n},then {ak}is an increasing sequence while {bk}is a decreasing sequence.

Therefore the limits in Eq. (1.1) and Eq. (1.2) always exist and lim inf

n→∞xn= sup

n

inf{xk:k≥n}and lim sup

n→∞

x_n= inf

n sup{x_k:k≥n}.

The following proposition contains some basic properties of liminfs and limsups.

Proposition 1.5.Let{an}^∞n=1and{bn}^∞n=1be two sequences of real numbers.

Then

1.lim infn→∞an ≤lim sup_n_→∞anandlimn→∞anexists inR¯iﬀlim infn→∞an= lim sup_n_→∞an∈R¯.

2. There is a subsequence {ank}^∞k=1 of {an}^∞n=1 such that limk→∞ank = lim sup_n_→∞an.

3.

lim sup

n→∞

(an+bn)≤lim sup

n→∞

an+ lim sup

n→∞

bn (1.3)

whenever the right side of this equation is not of the form∞ − ∞.

(19)

1.2 Limits, Limsups, and Liminfs 5 4. Ifa_n≥0andb_n ≥0for all n∈N,then

lim sup

n→∞

(anbn)≤lim sup

n→∞

an·lim sup

n→∞

bn, (1.4)

provided the right hand side of (1.4) is not of the form0·∞ or∞·0.

Proof.We will only prove part 1. and leave the rest as an exercise to the reader. We begin by noticing that

inf{a_k:k≥n}≤sup{a_k:k≥n}∀n so that

lim inf

n→∞an≤lim sup

n→∞

an.

Now suppose thatlim infn→∞an = lim sup_n_→∞an =a∈R.Then for all

>0,there is an integerN such that

a− ≤inf{ak:k≥N}≤sup{ak:k≥N}≤a+ , i.e.

a− ≤a_k≤a+ for allk≥N.

Hence by the definition of the limit,limk→∞ak =a.

Iflim infn→∞an=∞,then we know for allM∈(0,∞)there is an integer N such that

M≤inf{ak :k≥N}

and hencelim_n_→∞a_n=∞.The case wherelim sup_n_→∞a_n=−∞is handled similarly.

Conversely, suppose thatlim_n_→∞a_n =A ∈R¯ exists. If A∈ R, then for every >0there existsN( )∈Nsuch that|A−a_n|≤ for alln≥N( ),i.e.

A− ≤a_n≤A+ for alln≥N( ).

From this we learn that

A− ≤lim inf

n→∞a_n≤lim sup

n→∞

a_n≤A+ . Since >0is arbitrary, it follows that

A≤lim inf

n→∞a_n≤lim sup

n→∞

a_n≤A, i.e. that A= lim inf_n_→∞a_n= lim sup_n_→∞a_n.

IfA=∞,then for allM >0there existsN(M)such thatan≥M for all n≥N(M). This show that

lim inf

n→∞a_n≥M and sinceM is arbitrary it follows that

∞ ≤lim inf

n→∞a_n ≤lim sup

n→∞

a_n. The proof is similar ifA=−∞as well.

(20)

1.3 Sums of positive functions

In this and the next few sections, let X and Y be two sets. We will write α⊂⊂X to denote thatαis afinitesubset ofX.

Definition 1.6.Suppose thata: X →[0,∞] is a function and F ⊂X is a subset, then

X

F

a=X

x∈F

a(x) = sup (X

x∈α

a(x) :α⊂⊂F )

. Remark 1.7.Suppose thatX =N={1,2,3, . . .},then

X

N

a= X∞ n=1

a(n) := lim

N→∞

XN n=1

a(n).

Indeed for allN, PN

n=1a(n)≤P

Na,and thus passing to the limit we learn

that X∞

n=1

a(n)≤X

N

a.

Conversely, ifα⊂⊂N,then for allN large enough so thatα⊂{1,2, . . . , N}, we have P

αa≤PN

n=1a(n)which upon passing to the limit implies that X

α

a≤ X∞ n=1

a(n)

and hence by taking the supremum overαwe learn that X

N

a≤ X∞ n=1

a(n).

Remark 1.8.Suppose that P

Xa < ∞, then {x∈X :a(x)>0} is at most countable. To see this first notice that for any >0, the set {x:a(x)≥ } must befinite for otherwiseP

Xa=∞. Thus {x∈X:a(x)>0}=[_∞

k=1{x:a(x)≥1/k}

which shows that {x∈X :a(x)>0}is a countable union of finite sets and thus countable.

Lemma 1.9.Suppose that a, b:X →[0,∞] are two functions, then X

X

(a+b) =X

X

a+X

X

b and X

X

λa=λX

X

a for allλ≥0.

(21)

1.3 Sums of positive functions 7 I will only prove thefirst assertion, the second being easy. Letα⊂⊂X be afinite set, then

X

α

(a+b) =X

α

a+X

α

b≤X

X

a+X

X

b which after taking sups overαshows that

X

(a+b)≤X

X

a+X

X

b.

Similarly, ifα, β⊂⊂X,then X

α

a+X

β

b≤X

α∪β

a+X

α∪β

b=X

α∪β

(a+b)≤X

X

(a+b).

Taking sups overαandβ then shows that X

X

a+X

X

b≤X

X

(a+b).

Lemma 1.10.LetX andY be sets,R⊂X×Y and suppose thata:R→R¯ is a function. LetxR:={y∈Y : (x, y)∈R}andRy :={x∈X: (x, y)∈R}. Then

sup

(x,y)∈R

a(x, y) = sup

x∈X

sup

y∈xR

a(x, y) = sup

y∈Y

sup

x∈Ry

a(x, y)and

(x,y)inf∈Ra(x, y) = inf

x∈X inf

y∈xRa(x, y) = inf

y∈Y inf

x∈Ry

a(x, y).

(Recall the conventions: sup∅=−∞andinf∅= +∞.)

Proof.LetM = sup_(x,y)_∈_Ra(x, y), Nx:= sup_y_∈_x_Ra(x, y).Thena(x, y)≤ M for all(x, y)∈RimpliesNx= sup_y_∈_x_Ra(x, y)≤M and therefore that

sup

x∈X

sup

y∈xR

a(x, y) = sup

x∈X

N_x≤M. (1.5)

Similarly for any(x, y)∈R,

a(x, y)≤Nx≤sup

x∈X

Nx= sup

x∈X

sup

y∈xR

a(x, y) and therefore

sup

(x,y)∈R

a(x, y)≤sup

x∈X

sup

y∈xR

a(x, y) =M (1.6)

Equations (1.5) and (1.6) show that sup

(x,y)∈R

a(x, y) = sup

x∈X

sup

y∈xR

a(x, y).

The assertions involving infinums are proved analogously or follow from what we have just proved applied to the function−a.

(22)

Fig. 1.1.Thexandy— slices of a setR⊂X×Y.

Theorem 1.11 (Monotone Convergence Theorem for Sums).Suppose that f_n:X→[0,∞]is an increasing sequence of functions and

f(x) := lim

n→∞fn(x) = sup

n

fn(x).

Then

nlim→∞

X

fn =X

X

f

Proof.We will give two proves. For thefirst proof, letP^f(X) ={A⊂X : A⊂⊂X}.Then

nlim→∞

X

fn= sup

n

X

fn= sup

n

sup

α∈Pf(X)

X

α

fn= sup

α∈Pf(X)

sup

n

X

α

fn

= sup

α∈Pf(X) nlim→∞

X

α

fn= sup

α∈Pf(X)

X

α

nlim→∞fn

= sup

α∈P^f(X)

X

α

f =X

X

f.

(Second Proof.) LetSn =P

Xfn andS =P

Xf.Since fn ≤fm≤f for alln≤m,it follows that

S_n≤S_m≤S

which shows thatlim_n_→∞S_n exists and is less thatS,i.e.

A:= lim

n→∞

X

fn≤X

X

f. (1.7)

Noting thatP

αf_n≤P

Xf_n=S_n ≤Afor allα⊂⊂X and in particular, X

α

fn ≤A for allnandα⊂⊂X.

(23)

1.3 Sums of positive functions 9 Lettingntend to infinity in this equation shows that

X

α

f ≤A for allα⊂⊂X and then taking the sup over allα⊂⊂X gives

X

f ≤A= lim

n→∞

X

fn (1.8)

which combined with Eq. (1.7) proves the theorem.

Lemma 1.12 (Fatou’s Lemma for Sums). Suppose that f_n :X →[0,∞] is a sequence of functions, then

X

lim inf

n→∞f_n≤lim inf

n→∞

X

f_n. Proof.Define g_k ≡ inf

n≥kf_n so thatg_k ↑lim inf_n_→∞f_n as k → ∞. Since g_k ≤f_n for allk≤n,

X

g_k ≤X

X

f_n for alln≥k

and therefore X

X

g_k ≤lim inf

n→∞

X

f_n for allk.

We may now use the monotone convergence theorem to letk→ ∞tofind X

X

lim inf

n→∞f_n=X

X

klim→∞g_k^{M CT}= lim

k→∞

X

g_k ≤lim inf

n→∞

X

f_n.

Remark 1.13.IfA=P

Xa <∞,then for all >0there existsα ⊂⊂X such that

A≥X

α

a≥A− for allα⊂⊂X containingα or equivalently,

¯¯

¯A−X

α

a

¯¯

¯≤ (1.9)

for allα⊂⊂X containingα. Indeed, choose α so thatP

α a≥A− .

(24)

1.4 Sums of complex functions

Definition 1.14.Suppose that a:X →Cis a function, we say that X

X

a= X

x∈X

a(x)

exists and is equal to A∈C, if for all >0 there is a finite subsetα ⊂X such that for all α⊂⊂X containingα we have

¯¯

¯A−X

α

a

¯¯

¯≤ .

The following lemma is left as an exercise to the reader.

Lemma 1.15.Suppose thata, b:X →C are two functions such that P

Xa andP

Xb exist, thenP

X(a+λb) exists for allλ∈Cand X

X

(a+λb) =X

X

a+λX

X

b.

Definition 1.16 (Summable). We call a functiona:X →C summable

if X

X

|a|<∞.

Proposition 1.17.Let a : X → C be a function, then P

Xa exists iﬀ P

X|a|<∞,i.e. iﬀa is summable.

Proof. If P

X|a| < ∞, then P

X(Rea)^± < ∞ and P

X(Ima)^± < ∞ and hence by Remark 1.13 these sums exists in the sense of Definition 1.14.

Therefore by Lemma 1.15,P

Xaexists and X

X

a=X

X

(Rea)⁺−X

X

(Rea)⁻+i ÃX

X

(Ima)⁺−X

X

(Ima)⁻

! . Conversely, if P

X|a| =∞ then, because |a| ≤|Rea|+|Ima|, we must

have X

X

|Rea|=∞or X

X

|Ima|=∞.

Thus it suﬃces to consider the case wherea:X→Ris a real function. Write a=a⁺−a⁻ where

a⁺(x) = max(a(x),0)anda⁻(x) = max(−a(x),0). (1.10) Then|a|=a⁺+a⁻ and

(25)

1.4 Sums of complex functions 11

∞=X

X

|a|=X

X

a⁺+X

X

a⁻ which shows that eitherP

Xa⁺=∞orP

Xa⁻ =∞.Suppose, with out loss of generality, thatP

Xa⁺=∞.LetX⁰ :={x∈X:a(x)≥0},then we know that P

X⁰a =∞ which means there are finite subsets αn ⊂ X⁰ ⊂ X such that P

αna≥ n for alln. Thus if α⊂⊂ X is any finite set, it follows that limn→∞P

αn∪αa=∞,and therefore P

Xacan not exist as a number inR. Remark 1.18.Suppose that X =N and a : N→C is a sequence, then it is not necessarily true that

X∞ n=1

a(n) =X

n∈N

a(n). (1.11)

This is because

X∞ n=1

a(n) = lim

N→∞

XN n=1

a(n) depends on the ordering of the sequenceawhere asP

n∈Na(n)does not. For example, take a(n) = (−1)ⁿ/n then P

n∈N|a(n)| = ∞ i.e. P

n∈Na(n) does notexist whileP∞

n=1a(n)does exist. On the other hand, if X

n∈N

|a(n)|= X∞ n=1

|a(n)|<∞ then Eq. (1.11) is valid.

Theorem 1.19 (Dominated Convergence Theorem for Sums). Sup- pose that f_n : X → C is a sequence of functions on X such that f(x) = limn→∞fn(x)∈Cexists for allx∈X. Further assume there is adominat- ing function g:X →[0,∞)such that

|fn(x)|≤g(x)for allx∈X andn∈N (1.12) and that g is summable. Then

nlim→∞

X

x∈X

f_n(x) = X

x∈X

f(x). (1.13)

Proof. Notice that |f| = lim|f_n| ≤ g so that f is summable. By con- sidering the real and imaginary parts off separately, it suﬃces to prove the theorem in the case wheref is real. By Fatou’s Lemma,

X

(g±f) =X

X

lim inf

n→∞(g±f_n)≤lim inf

n→∞

X

(g±f_n)

=X

X

g+ lim inf

n→∞

Ã

±X

X

fn

! .

(26)

Sincelim inf_n_→∞(−a_n) =−lim sup_n_→∞a_n,we have shown, X

X

g±X

X

f ≤X

X

g+

½lim infn→∞P

Xfn

−lim sup_n_→∞P

Xfn

and therefore

lim sup

n→∞

X

f_n≤X

X

f ≤lim inf

n→∞

X

f_n. This shows that lim

n→∞

P

Xfnexists and is equal toP

Xf.

Proof.(Second Proof.) Passing to the limit in Eq. (1.12) shows that|f|≤ g and in particular thatf is summable. Given >0,letα⊂⊂X such that

X

X\α

g≤ . Then forβ⊂⊂X such thatα⊂β,

¯¯

¯¯ X

β

f −X

β

f_n

¯¯

¯¯=

¯¯

¯¯ X

β

(f−f_n)

¯¯

≤X

β

|f−f_n|=X

α

|f−f_n|+X

β\α

|f −f_n|

≤X

α

|f−fn|+ 2X

β\α

g

≤X

α

|f−fn|+ 2 . and hence that ¯¯¯¯¯¯

X

β

f −X

β

f_n

¯¯

¯¯≤X

α

|f −f_n|+ 2 .

Since this last equation is true for all suchβ ⊂⊂X,we learn that

¯¯

¯ X

X

f−X

X

fn

¯¯

¯≤X

α

|f −fn|+ 2 which then implies that

lim sup

n→∞

¯¯

¯ X

X

f−X

X

fn

¯¯

¯≤lim sup

n→∞

X

α

|f −fn|+ 2

= 2 . Because >0is arbitrary we conclude that

lim sup

n→∞

¯¯

¯ X

X

f−X

X

fn

¯¯

¯= 0.

which is the same as Eq. (1.13).

(27)

1.5 Iterated sums 13

1.5 Iterated sums

Let X and Y be two sets. The proof of the following lemma is left to the reader.

Lemma 1.20.Suppose that a : X → C is function and F ⊂ X is a subset such that a(x) = 0for all x /∈F.Show that P

Fa exists iﬀ P

Xaexists, and if the sums exist then X

X

a=X

F

a.

Theorem 1.21 (Tonelli’s Theorem for Sums).Suppose thata:X×Y →

[0,∞], then X

X×Y

a=X

X

Y

a=X

Y

X

a.

Proof.It suﬃces to show, by symmetry, that X

X×Y

a=X

X

Y

a

LetΛ⊂⊂X×Y.The for anyα⊂⊂X andβ⊂⊂Y such thatΛ⊂α×β,we

have X

Λ

a≤X

α×β

a=X

α

X

β

a≤X

α

X

Y

a≤X

X

Y

a, i.e.P

Λa≤P

X

P

Y a.Taking the sup overΛin this last equation shows X

X×Y

a≤X

X

Y

a.

We must now show the opposite inequality. IfP

X×Y a=∞we are done so we now assume thata is summable. By Remark 1.8, there is a countable set{(x⁰_n, y_n⁰)}^∞n=1⊂X×Y oﬀof whichais identically0.

Let {yn}^∞n=1 be an enumeration of {y⁰_n}^∞n=1, then since a(x, y) = 0 if y /∈{y_n}^∞n=1, P

y∈Y a(x, y) =P_∞

n=1a(x, y_n)for allx∈X.Hence X

x∈X

X

y∈Y

a(x, y) = X

x∈X

X∞ n=1

a(x, yn) = X

x∈X Nlim→∞

XN n=1

a(x, yn)

= lim

N→∞

X

x∈X

XN n=1

a(x, y_n), (1.14)

wherein the last inequality we have used the monotone convergence theorem withFN(x) :=PN

n=1a(x, yn).Ifα⊂⊂X,then X

x∈α

XN

n=1

a(x, yn) = X

α×{yn}^Nn=1

a≤ X

X×Y

a

(28)

and therefore,

Nlim→∞

X

x∈X

XN

n=1

a(x, yn)≤ X

X×Y

a. (1.15)

Hence it follows from Eqs. (1.14) and (1.15) that X

x∈X

X

y∈Y

a(x, y)≤ X

X×Y

a (1.16)

as desired.

Alternative proof of Eq. (1.16). Let A={x⁰_n:n∈N}and let{x_n}^∞n=1

be an enumeration ofA. Then forx /∈A, a(x, y) = 0 for ally∈Y.

Given >0, letδ:X →[0,∞)be the function such that P

Xδ= and δ(x)>0forx∈A. (For example we may defineδ byδ(x_n) = /2ⁿ for alln andδ(x) = 0ifx /∈A.)For eachx∈X,letβ_x⊂⊂X be afinite set such that

X

y∈Y

a(x, y)≤ X

y∈βx

a(x, y) +δ(x).

Then X

X

Y

a≤ X

x∈X

X

y∈βx

a(x, y) +X

x∈X

δ(x)

= X

x∈X

X

y∈βx

a(x, y) + = sup

α⊂⊂X

X

x∈α

X

y∈βx

a(x, y) +

≤ X

X×Y

a+ , (1.17)

wherein the last inequality we have used X

x∈α

X

y∈βx

a(x, y) =X

Λα

a≤ X

X×Y

a with

Λα:={(x, y)∈X×Y :x∈αandy∈βx}⊂X×Y.

Since >0is arbitrary in Eq. (1.17), the proof is complete.

Theorem 1.22 (Fubini’s Theorem for Sums).Now suppose that a:X× Y →Cis a summable function, i.e. by Theorem 1.21 any one of the following equivalent conditions hold:

1.P

X×Y |a|<∞, 2.P

X

P

Y |a|<∞or 3.P

Y

P

X|a|<∞.

Then X

X×Y

a=X

X

Y

a=X

Y

X

a.

(29)

1.6 ^p— spaces, Minkowski and Holder Inequalities 15 Proof. If a : X → R is real valued the theorem follows by applying Theorem 1.21 toa^±— the positive and negative parts ofa.The general result holds for complex valued functionsaby applying the real version just proved to the real and imaginary parts ofa.

1.6 `

^p

— spaces, Minkowski and Holder Inequalities

In this subsection, letµ:X →(0,∞]be a given function. LetFdenote either CorR.Forp∈(0,∞)andf :X →F,let

kfkp≡(X

x∈X

|f(x)|^pµ(x))^1/p and forp=∞let

kfk∞= sup{|f(x)|:x∈X}. Also, forp >0,let

p(µ) ={f :X→F:kfk^p<∞}.

In the case whereµ(x) = 1for allx∈X we will simply write ^p(X)for ^p(µ).

Definition 1.23.Anormon a vector spaceLis a functionk·k:L→[0,∞) such that

1. (Homogeneity)kλfk=|λ| kfkfor allλ∈F andf ∈L.

2. (Triangle inequality) kf+gk≤kfk+kgkfor allf, g∈L.

3. (Positive definite)kfk= 0 impliesf = 0.

A pair(L,k·k)whereL is a vector space and k·kis a norm onLis called anormed vector space.

The rest of this section is devoted to the proof of the following theorem.

Theorem 1.24.Forp∈[1,∞],( ^p(µ),k · kp)is a normed vector space.

Proof.The only diﬃculty is the proof of the triangle inequality which is the content of Minkowski’s Inequality proved in Theorem 1.30 below.

1.6.1 Some inequalities

Proposition 1.25.Letf : [0,∞)→[0,∞)be a continuous strictly increasing function such that f(0) = 0(for simplicity) and lim

s→∞f(s) =∞.Let g=f⁻¹ and fors, t≥0let

F(s) = Z s

0

f(s⁰)ds⁰ andG(t) = Z t

0

g(t⁰)dt⁰. Then for alls, t≥0,

st≤F(s) +G(t) and equality holds iﬀ t=f(s).

(30)

Proof.Let

A_s:={(σ, τ) : 0≤τ≤f(σ)for0≤σ≤s}and Bt:={(σ, τ) : 0≤σ≤g(τ)for0≤τ≤t}

then as one sees from Figure 1.2,[0, s]×[0, t]⊂As∪Bt.(In thefigure:s= 3, t= 1, A3 is the region under t=f(s)for0≤s≤3and B1 is the region to the left of the curves=g(t)for0≤t≤1.)Hence ifm denotes the area of a region in the plane, then

st=m([0, s]×[0, t])≤m(As) +m(Bt) =F(s) +G(t).

As it stands, this proof is a bit on the intuitive side. However, it will become rigorous if one takes mto be Lebesgue measure on the plane which will be introduced later.

We can also give a calculus proof of this theorem under the additional assumption thatf isC¹.(This restricted version of the theorem is all we need in this section.) To do thisfixt≥0and let

h(s) =st−F(s) = Z s

0

(t−f(σ))dσ.

Ifσ > g(t) =f⁻¹(t),thent−f(σ)<0and hence ifs > g(t), we have h(s) =

Z s 0

(t−f(σ))dσ= Z g(t)

0

(t−f(σ))dσ+ Z s

g(t)

(t−f(σ))dσ

≤ Z g(t)

0

(t−f(σ))dσ=h(g(t)).

Combining this with h(0) = 0 we see that h(s)takes its maximum at some point s ∈ (0, t] and hence at a point where 0 = h⁰(s) = t−f(s). The only solution to this equation iss=g(t)and we have thus shown

st−F(s) =h(s)≤ Z g(t)

0

(t−f(σ))dσ=h(g(t))

with equality when s = g(t). To finish the proof we must show Rg(t) 0 (t − f(σ))dσ=G(t). This is verified by making the change of variablesσ=g(τ) and then integrating by parts as follows:

Z g(t) 0

(t−f(σ))dσ= Z t

0

(t−f(g(τ)))g⁰(τ)dτ = Z t

0

(t−τ)g⁰(τ)dτ

= Z t

0

g(τ)dτ =G(t).

(31)

1.6 ^p— spaces, Minkowski and Holder Inequalities 17

4 3

2 1

0 4

3

2

1

0

x y

Fig. 1.2.A picture proof of Proposition 1.25.

Definition 1.26.The conjugate exponentq∈[1,∞]top∈[1,∞]isq:= _p₋^p₁ with the convention thatq=∞ifp= 1.Notice that qis characterized by any of the following identities:

1 p+1

q = 1, 1 + q

p=q, p−p

q = 1 andq(p−1) =p. (1.18) Lemma 1.27.Let p∈(1,∞) and q:= _p^p

−1 ∈ (1,∞) be the conjugate expo- nent. Then

st≤ s^q q +t^p

p for alls, t≥0 with equality if and only if s^q=t^p.

Proof.LetF(s) = ^s_p^p forp >1.Thenf(s) =s^p⁻¹=tandg(t) =t^p−1¹ = t^q⁻¹, wherein we have used q−1 = p/(p−1)−1 = 1/(p−1). Therefore G(t) =t^q/q and hence by Proposition 1.25,

st≤ s^p p +t^q

q with equality iﬀt=s^p⁻¹.

Theorem 1.28 (Hölder’s inequality). Let p, q∈[1,∞] be conjugate expo- nents. For allf, g:X →F,

kf gk¹≤kfk^p· kgk^q. (1.19) If p∈(1,∞), then equality holds in Eq. (1.19) iﬀ

( |f|

kfk^p)^p= ( |g| kgk^q)^q.

(32)

Proof.The proof of Eq. (1.19) forp∈{1,∞}is easy and will be left to the reader. The cases wherekfkq= 0or∞orkgkp= 0or∞are easily dealt with and are also left to the reader. So we will assume thatp∈ (1,∞)and 0<kfk^q,kgk^p <∞. Lettings=|f|/kfk^p andt =|g|/kgk^q in Lemma 1.27 implies

|f g| kfk^pkgk^q ≤ 1

p

|f|^p kfk^p +1

q

|g|^q kgk^q. Multiplying this equation byµand then summing gives

kf gk1

kfk^pkgk^q ≤ 1 p+1

q = 1 with equality iﬀ

|g|

kgk^q = |f|^p⁻¹

kfk^(p^p⁻¹⁾ ⇐⇒ |g|

kgk^q = |f|^p/q

kfk^p/q^p ⇐⇒ |g|^qkfk^pp=kgk^qq|f|^p.

Definition 1.29.For a complex number λ∈C,let sgn(λ) =

½ _λ

|λ| if λ6= 0 0 ifλ= 0.

Theorem 1.30 (Minkowski’s Inequality). If 1≤p≤ ∞andf, g∈ ^p(µ) then

kf+gk^p≤kfk^p+kgk^p, with equality iﬀ

sgn(f) = sgn(g)whenp= 1and

f =cgfor somec >0whenp∈(1,∞).

Proof.Forp= 1, kf+gk1=X

X

|f+g|µ≤X

X

(|f|µ+|g|µ) =X

X

|f|µ+X

X

|g|µ with equality iﬀ

|f|+|g|=|f +g| ⇐⇒ sgn(f) = sgn(g).

Forp=∞,

kf+gk∞= sup

X |f+g|≤sup

X

(|f|+|g|)

≤sup

X |f|+ sup

X |g|=kfk∞+kgk∞.

(33)

1.7 Exercises 19 Now assume thatp∈(1,∞).Since

|f+g|^p≤(2 max (|f|,|g|))^p= 2^pmax (|f|^p,|g|^p)≤2^p(|f|^p+|g|^p) it follows that

kf +gk^pp≤2^p¡

kfk^pp+kgk^pp

¢<∞.

The theorem is easily verified ifkf+gkp= 0,so we may assumekf+gkp>

0.Now

|f+g|^p=|f+g||f+g|^p⁻¹≤(|f|+|g|)|f +g|^p⁻¹ (1.20) with equality iﬀsgn(f) = sgn(g).Multiplying Eq. (1.20) byµand then summing and applying Holder’s inequality gives

X

|f+g|^pµ≤X

X

|f| |f+g|^p⁻¹µ+X

X

|g| |f+g|^p⁻¹µ

≤(kfkp+kgkp)k |f +g|^p⁻¹kq (1.21) with equality iﬀ

µ |f| kfk^p

¶p

=

µ |f+g|^p⁻¹ k|f+g|^p⁻¹k^q

¶^q

= µ |g|

kgk^p

¶p

andsgn(f) = sgn(g).

By Eq. (1.18),q(p−1) =p,and hence k|f+g|^p⁻¹k^qq=X

X

(|f +g|^p⁻¹)^qµ=X

X

|f +g|^pµ. (1.22) Combining Eqs. (1.21) and (1.22) implies

kf +gk^pp≤kfkpkf+gk^p/qp +kgkpkf +gk^p/qp (1.23) with equality iﬀ

sgn(f) = sgn(g)and µ |f|

kfkp

¶p

= |f+g|^p kf+gk^p^p =

µ |g| kgkp

¶p

. (1.24)

Solving for kf +gk^p in Eq. (1.23) with the aid of Eq. (1.18) shows that kf+gk^p ≤kfk^p+kgk^p with equality iﬀEq. (1.24) holds which happens iﬀ f =cgwithc >0.

1.7 Exercises

1.7.1 Set Theory

Letf :X →Y be a function and{A_i}i∈I be an indexed family of subsets of Y,verify the following assertions.

On non-local ergodic Jacobi semigroups: spectral theory, convergence-to-equilibrium and contractivity

Bruce K. Driver

Analysis Tools with Applications,

June 9, 2003 File:analpde.tex

Springer

Berlin Heidelberg NewYork Hong Kong London

Milan Paris Tokyo

Contents

Preface

Part I

Basic Topological, Metric and Banach Space

Notions

1

Limits, sums, and other basics

1.1 Set Operations

1.2 Limits, Limsups, and Liminfs

1.3 Sums of positive functions

1.4 Sums of complex functions

1.5 Iterated sums

1.6 `

— spaces, Minkowski and Holder Inequalities

1.7 Exercises