Real Analysis

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Real Analysis

Department of Mathematics, Aalto University 2022

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L ^p spaces 1

In this section we study theL^pspaces in order to be able to capture quantitative information on the average size of measurable functions and boundedness of operators on such functions. The cases 0<p<1,p=1,p=2, 1<p< ∞andp= ∞ are different in character, but they all play an important role in in Fourier analysis, harmonic analysis, functional analysis and partial differential equations. The spaceL¹of integrable functions plays a central role in measure and integration theory. The Hilbert spaceL²of square integrable functions is important in the study of Fourier series. Many operators that arise in applications are bounded in L^pfor 1<p< ∞, but the limit casesL¹andL^∞require a special attention.

1.1 ^L ^p functions

Definition 1.1. Letµbe an outer measure onRⁿ,A⊂Rⁿaµ-measurable set and f:A→[−∞,∞] aµ-measurable function. Then f∈L^p(A), 1Ép< ∞, if

kfkp= µZ

A|f|^pdµ

¶¹_p

< ∞.

TH E M O R A L: Forp=1, f ∈L¹(A) if and only if|f|is integrable in A. For 1Ép< ∞,f∈L^p(A) if and only if|f|^pis integrable inA.

Remark 1.2. The measurability assumption on f essential in the definition. For example, letA⊂[0, 1] be a non-measurable set with respect to the one-dimensional Lebesgue measure and considerf: [0, 1]→R,

f(x)=







1, x∈A,

−1, x∈[0, 1] \A.

Thenf²=1 is integrable on [0, 1], butf is not a Lebesgue measurable function.

1

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Example 1.3. Letf:Rⁿ→[0,∞], f(x)= |x|⁻ⁿand assume thatµis the Lebesgue measure. Let A=B(0, 1)={x∈Rⁿ :|x| <1}, Ωn =m(B(0, 1)) and denote A_i= B(0, 2⁻ⁱ⁺¹) \B(0, 2⁻ⁱ),i=1, 2, . . .. Then

Z

B(0,1)|x|^{−n p}dx= X∞ i=1

Z

Ai

|x|^{−n p}dx É

X∞ i=1

Z

A_i

2^{n pi}dx (x∈Ai=⇒ |x| Ê2⁻ⁱ=⇒ |x|^{−n p}É2^{n pi})

= X∞ i=1

2^{n pi}|A_i| É X∞ i=1

2^{n pi}|B(0, 2⁻ⁱ⁺¹)|

=Ωn

X∞ i=1

2^{n pi}(2⁻ⁱ⁺¹)ⁿ (Ωn= |B(0, 1)|)

=Ωn

X∞ i=1

2^{n pi−ni+n}=2ⁿΩn

X∞ i=1

2^in(p−1)< ∞, ifn(p−1)<0⇐⇒p<1. Thusf∈L^p(B(0, 1)) forp<1.

On the other hand, Z

B(0,1)|x|⁻^{n p}dx= X∞ i=1

Z

Ai

|x|⁻^{n p}dx Ê

X∞ i=1

Z

Ai

2^{n p(i−1)}dx (x∈A_i=⇒ |x| <2⁻ⁱ⁺¹=⇒ |x|^{−n p}>2^{n p(i−1)})

= X∞ i=1

2^{n p(i−1)}|Ai| =Ωn(2ⁿ−1)2^{−n p} X∞ i=1

2^{n pi}2⁻ⁱⁿ (|A_i| = |B(0, 2⁻ⁱ⁺¹)| − |B(0, 2⁻ⁱ)|

=Ωn(2⁽⁻ⁱ⁺¹⁾ⁿ−2⁻ⁱⁿ)=Ωn(2ⁿ−1)2⁻ⁱⁿ)

=C(n,p) X∞ i=1

2^in(p−1)= ∞,

ifn(p−1)Ê0⇐⇒pÊ1. Thusf∉L^p(B(0, 1)) forpÊ1. This shows that f∈L^p(B(0, 1))⇐⇒p<1.

If A=Rⁿ\B(0, 1), then we denoteA_i=B(0, 2ⁱ) \B(0, 2ⁱ⁻¹),i=1, 2, . . ., and a similar argument as above shows that

f∈L^p(Rⁿ\B(0, 1))⇐⇒p>1.

Observe that f 6∈L¹(B(0, 1)) and f 6∈L¹(Rⁿ\B(0, 1)). Thus f(x)= |x|⁻ⁿ is a borderline function inRⁿas far as integrability is concerned.

TH E M O R A L: The smaller the parameterpis, the worse local singularities anL^pfunction may have. On the other hand, the larger the parameterpis, the more anL^pfunction may spread out globally.

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Example 1.4. Assume thatf:Rⁿ→[0,∞] is radial. Thusf depends only on|x| and it can be expressed asf(|x|), wheref is a function defined on [0,∞). Then

Z

Rⁿf(|x|)dx=ω_n−1 Z _∞

0

f(r)rⁿ⁻¹dr, (1.1)

where

ωn−1= 2π^n/2 Γ^(n/2)

is the (n−1)-dimensional volume of the unit sphereÇB(0, 1)={x∈Rⁿ:|x| =1}.

Let us show how to use this formula to compute the volume of a ballB(x,r)= {y∈Rⁿ:|y−x| <r},x∈Rⁿ andr>0. By the translation and scaling invariance, we have

rⁿΩn=rⁿm(B(0, 1))=m(B(x,r))=m(B(0,r))

= Z

RⁿχB(0,r)(y)d y= Z

Rⁿχ(0,r)(|y|)d y

=ωn−1

Z r

0 ρⁿ⁻¹dρ=ωn−1

rⁿ n . In particular, it follows thatωn−1=nΩnand

m(B(x,r))= 2π^n/2 Γ^(n/2)

rⁿ

n = π^n/2 Γ⁽ⁿ₂+1)rⁿ. Letr>0. Then

Z

Rⁿ\B(0,r)

1

|x|^αdx= Z

Rⁿ

1

|x|^αχ_Rⁿ\B(0,r)(x)dx

=rⁿ Z

Rⁿ

1

|rx|^αχ_Rⁿ\B(0,r)(rx)dx

=rⁿ^−α Z

Rⁿ

1

|x|^αχ_Rⁿ\B(0,1)(x)dx

=rⁿ^−α Z

Rⁿ\B(0,1)

1

|x|^αdx< ∞, α>n, and, in a similar way,

Z

B(0,r)

1

|x|^αdx=r^n−α Z

B(0,1)

1

|y|^αd y=r^n−α Z

B(0,1)

1

|x|^αdx< ∞, α<n.

Observe, that here we make a change of variablesx=r y.

On the other hand, the integrals can be computer directly by (1.1). This gives Z

Rⁿ\B(0,r)

1

|x|^αdx=ωn−1

Z _∞

r ρ^−αρⁿ⁻¹dρ

= ωn−1

−α+nρ^−α+n

¯

∞ r

=ωn−1

α−nr^−α+n< ∞, α>n and

Z

B(0,r)

1

|x|^αdx=ωn−1

Z r

0 ρ^−αρⁿ⁻¹dρ

= ωn−1

−α+nρ^−α+n

¯

r 0

=ωn−1

α−nr^n−α< ∞, α<n.

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Remarks 1.5:

Formula (1.1) implies following claims:

(1) If|f(x)| Éc|x|^−αin a ballB(0,r),r>0, for someα<n, thenf∈L¹(B(0,r)).

On the other hand, if |f(x)| Êc|x|^−α in B(0,r) for someα>n, then f ∉ L¹(B(0,r)).

(2) If|f(x)| Éc|x|^−α in Rⁿ\B(0,r) for someα>n, then f ∈L¹(Rⁿ\B(0,r)).

On the other hand, if|f(x)| Êc|x|^−αin Rⁿ\B(0,r) for someα<n, then f∉L¹(Rⁿ\B(0,r)).

Remark 1.6. If f∈L^p(A), then|f(x)| < ∞forµ-almost everyx∈A.

Reason. LetA_i={x∈A:|f(x)| Êi},i=1, 2, . . . . Then {x∈A:|f(x)| = ∞}=

\∞ i=1

Ai. By Chebyshev’s inequality

µ({x∈A:|f(x)| = ∞})Éµ(Ai)= Z

A_i

1dµ É

Z

A_i

µ|f| i

¶p

dµ (|f| ÊiinA_i) É 1

i^p Z

A|f|^pdµ

| {z }

<∞

−−−→i→∞ 0. _■

The converse is not true, as the previous example shows.

Remark 1.7. If f∈L^p(A), then {x∈Rⁿ:|f(x)| 6=0} isσ-finite with respect toµ. Reason. LetA_i={x∈A:|f(x)| Ê¹_i},i=1, 2, . . . . Then

{x∈A:|f(x)| 6=0}= [∞ i=1

Ai. By Chebyshev’s inequality

µ(Ai)=µ({x∈A:|f(x)| Ê¹_i})= Z

A_i

1dµ Éi^p

Z

A_i|f|^pdµ< ∞ (|f| Ê¹_i inA_i)

for everyi=1, 2, . . . . _■

1.2 ^L ^p norm

If f∈L^p(A), 1Ép< ∞, the norm off is the number kfkp= kfkL^p(A)=

µZ

A|f|^pdµ

¶¹_p . We shall see that this has the usual properties of the norm:

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(1) (Nonnegativity) 0É kfkp< ∞,

(2) kfkp=0⇐⇒ f=0µ-almost everywhere, (3) (Homogeneity)ka fkp= |a|kfkp,a∈R, (4) (Triangle inequality)kf+gkpÉ kfkp+ kgkp.

The claims (1) and (3) are clear. For p=1, the claim (4) follows from the pointwise triangle inequality|f(x)+g(x)| É |f(x)| + |g(x)|. Forp>1, the claim (4) is not trivial and we shall prove it later in this section.

Let us recall how to prove (2). Recall that if a property holds except on a set of µmeasure zero, we say that it holdsµ-almost everywhere.

⇐= Assume thatf=0µ-almost everywhere inA. Then Z

A|f|^pdµ= Z

A∩{|f|=0}|f|^pdµ

| {z }

+ Z

A∩{|f|>0}|f|^pdµ

| {z }

=0.

=0, =0,

|f| =0µ-a.e. µ(A∩{|f| >0})=0 Thuskfkp=0.

=⇒ Assume thatkfkp=0. LetAi=©

x∈A:|f(x)| Ê¹_iª

,i=1, 2, . . .. Then {x∈A:|f(x)| >0}=

[∞ i=1

A_i. By Chebyshev’s inequality

µ(A_i)= Z

Ai

1dµÉ Z

Ai

|i f|^pdµÉi^p Z

A|f|^pdµ

| {z }

=0

=0. (i|f| Ê1 in A_i)

Thusµ(A_i)=0 for everyi=1, 2, . . . and µ

Ã∞

[

i=1

A_i

! É

X∞ i=1

µ(A_i)=0.

In other words,f=0µ-almost everywhere inA.

Forµ-measurable functionsfandgon aµ-measurable setA, we are interested in the casef(x)=g(x) forµ-almost everyx∈A, which means that

µ({x∈A:f(x)6=g(x)})=0.

In the case f=gµ-almost everywhere, we do not usually distinguishf fromg.

That is, we shall regard them as equal. We could be formal and consider the equivalence relation

f∼g⇐⇒f=g µ-almost everywhere inA

but this is hardly necessary. In practice, we are thinking f as the equivalence class of all functions which are equal tof µ-almost everywhere inA. ThusL^p(A)

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actually consists of equivalence classes rather than functions, but we shall not make the distinction. In measure and integration theory we cannot distinguishf fromg, if the functions are equalµ-almost everywhere. In fact, iff=gµ-almost everywhere inA, thenf ∈L^p(A)⇐⇒g∈L^p(A) andkf−gkp=0. In particular, this implies that kfkp= kgkp. On the other hand, ifkf−gkp=0, then f =g µ-almost everywhere inA.

Another situation that frequently arises is that the functionf is defined only almost everywhere. Then we say that f is measurable if and only if its zero extension to the whole space is measurable. Observe, that this does not affect the L^pnorm of f.

Next we show thatL^p(A) is a vector space.

Lemma 1.8. (i) If f∈L^p(A), thena f ∈L^p(A),a∈R. (ii) Iff,g∈L^p(A), thenf+g∈L^p(A).

Proof. (1) Z

A|a f|^pdµ= |a|^p Z

A|f|^pdµ< ∞.

(2) p=1 The triangle inequality|f+g| É |f| + |g|implies Z

A|f+g|dµÉ Z

A|f|dµ+

Z

A|g|dµ< ∞. 1<p< ∞ The elementary inequality

(a+b)^pÉ(2 max{a,b})^p=2^pmax{a^p,b^p}

É2^p(a^p+b^p), a,bÊ0, 0<p< ∞ (1.2) implies

Z

A|f+g|^pdµÉ2^p µZ

A|f|^pdµ+ Z

A|g|^pdµ

¶

< ∞.

ä

Remark 1.9. Note that the proof applies for 0<p< ∞. Thus L^p(A) is a vector space for 0<p< ∞. However, it will be a normed space with theL^pnorm only for pÊ1 as we shall see later.

Remark 1.10. A more careful analysis gives the useful inequality

(a+b)^pÉ2^p⁻¹(a^p+b^p), a,bÊ0, 1Ép< ∞. (1.3) Remarks 1.11:

(1) Iff:A→Cis a complex-valued function, thenf is said to beµ-measurable if and only if Ref and Imf areµ-measurable. We say that f∈L¹(A) if Ref∈L¹(A) and Imf∈L¹(A), and we define

Z

Af dµ= Z

A

Ref dµ+i Z

A

Imf dµ,

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whereiis the imaginary unit. This integral satisfies the usual linearity properties. It also satisfies the important inequality

¯

¯ Z

Af dµ

¯

¯É Z

A|f|dµ.

The definition of theL^p spaces and the norm extends in a natural way to complex-valued functions. Note that the propertyka fkp= |a|kfkpfor everya∈Cand thusL^pis a complex vector space.

(2) The spaceL²(A) is an inner product space with the inner product

〈f,g〉 = Z

A

f g dµ, f,g∈L²(A).

Here gis the complex conjugate which can be neglected if the functions are real-valued. This inner product induces the standardL²norm, since

kfk2= µZ

A|f|²dµ

¶¹₂

= µZ

Af f dµ

¶¹₂

= 〈f,f〉¹².

(3) In the special case thatA=Nandµis the counting measure, theL^p(N) spaces are denoted byl^pand

l^p= (

(x_i) : X∞ i=1

|x_i|^p< ∞ )

, 1Ép< ∞.

Here (x_i) is a sequence of real (or complex) numbers. In this case, Z

Nx dµ= X∞ i=1

x(i) for every nonnegative functionxonN. Thus

kxkp= Ã∞

X

i=1

|x_i|^p

!¹_p .

Note that the theory ofL^pspaces applies to these sequence spaces as well.

Definition 1.12. Let 1<p< ∞. The Hölder conjugate p⁰ of p is the number which satisfies

1 p+ 1

p⁰=1.

Forp=1 we definep⁰= ∞and ifp= ∞, thenp⁰=1.

Remark 1.13. Note that

p⁰= p p−1, p=2=⇒p⁰=2, 1<p<2=⇒p⁰>2, 2<p< ∞ =⇒1<p⁰<2, p→1=⇒p⁰→ ∞, (p⁰)⁰=p.

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Lemma 1.14 (Young’s inequality). Let 1<p< ∞. Then for everyaÊ0,bÊ0, abÉa^p

p +b^p⁰ p⁰, with equality if and only ifa^p=b^p⁰.

TH E M O R A L: Young’s inequality is a very useful tool in splitting a product to a sum. Morever, it shows where the conjugate exponentp⁰comes from.

Proof. The claim is obviously true, ifa=0 orb=0. Thus we may assume that a>0 andb>0. Clearly

abÉa^p p +b^p⁰

p⁰ ⇐⇒1 p

a^p b^p⁰ + 1

p⁰−ab¹⁻^p⁰Ê0⇐⇒ 1 p

Ã a b

p0 p

!p

+ 1 p⁰− a

b

p0 p

Ê0 Lett=a/b^p⁰^/pand defineϕ: (0,∞)→R,

ϕ(t)=1 pt^p+ 1

p⁰−t.

Then

ϕ(0)= 1 p⁰, lim

t→∞ϕ(t)= ∞ and ϕ⁰(t)=t^p⁻¹−1.

Note thatϕ⁰(t)=0⇐⇒t=1, from which we conclude ϕ(t)Êϕ(1)=1

p+ 1

p⁰−1=0 for everyt>0.

Moreover,ϕ(t)>0, ift6=1. It follows thatϕ(t)=0 if and only ifa/b^p⁰^/p=t=1. ä Remarks 1.15:

(1) Young’s inequality forp=2 follows immediately from (a−b)²Ê0⇐⇒a²−2ab+b²Ê0⇐⇒a²

2 +b²

2 ÊabÊ0.

(2) Young’s inequality can be also proved geometrically. To see this, consider the curves y=x^p⁻¹and the inversex=y^1/(p⁻¹⁾=y^p⁰⁻¹. Then

Z _a

0

x^p⁻¹dx=a^p p and

Z _b

0

y^p⁰⁻¹d y=b^p⁰ p⁰ .

By comparing the areas under the curves that these integrals measure, we have

abÉ Z a

0

x^p−1dx+ Z b

0

y^p⁰⁻¹d y=a^p p +b^p⁰

p⁰.

Theorem 1.16 (Hölder’s inequality). Let 1<p< ∞and assume thatf∈L^p(A) andg∈L^p⁰(A). Thenf g∈L¹(A) and

Z

A|f g|dµÉ µZ

A|f|^pdµ

¶¹_pµZ

A|g|^p⁰dµ

¶_p0¹ .

Moreover, an equality occurs if and only if there exists a constant csuch that

|f(x)|^p=c|g(x)|^p⁰forµ-almost everyx∈A.

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TH E M O R A L: Hölder’s inequality is very useful tool in estimating a product of functions.

Remark 1.17. Hölder’s inequality states thatkf gk1É kfkpkgkp⁰, 1<p< ∞. Ob- serve that for p=2 this is the Cauchy-Schwarz inequality|〈f,g〉| É kfk2kgk2. Proof. Ifkfkp=0, thenf=0µ-almost everywhere inAand thusf g=0µ-almost everywhere inA. Thus the result is clear, ifkfkp=0 orkgkp⁰=0. We may assume thatkfkp>0 andkgkp⁰>0. Let

fe= f kfkp

and ge= g kgkp⁰

. Then

kfekp=

°

° f kfkp

°

°p

=kfkp

kfkp =1 and kg˜kp⁰=1.

By Young’s inequality 1 kfkpkgkp⁰

Z

A|f g|dµ= Z

A|fe||eg|dµ É

Z

A

µ1

p|fe|^p+ 1 p⁰|ge|^p⁰

¶ dµ É1

p Z

A|fe|^pdµ

| {z }

=1

+1 p⁰ Z

A|ge|^p⁰dµ

| {z }

=1

=1 p+ 1

p⁰ =1.

An equality holds if and only if Z

A

µ1

p|fe|^p+ 1

p⁰|eg|^p⁰− |feeg|

¶

| {z }

Ê0

dµ=0,

which implies that

1

p|fe|^p+ 1

p⁰|ge|^p⁰− |fege| =0

µ-almost everywhere inA. This shows that an equality occurs in Young’s inequality if and only if|fe|^p= |eg|^p⁰ µ-almost everywhere inA. In this case

|f(x)|^p= kfk^pp

kgk^p_p⁰0

|g(x)|^p⁰

forµ-almost everyx∈A. ä

WA R N I N G: f∈L^p(A) andg∈L^p(A) does not imply thatf g∈L^p(A).

Reason. Let

f: (0, 1)→R, f(x)= 1

px, g=f,

and assume thatµis the Lebesgue measure. Then f∈L¹((0, 1)) andg∈L¹((0, 1)), but

(f g)(x)=f(x)g(x)=1

x and f g∉L¹((0, 1)).

■

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Remarks 1.18:

(1) Forp=2 we have the Cauchy-Schwarz inequality Z

A|f g|dµÉ µZ

A|f|²dµ

¶¹₂µZ

A|g|²dµ

¶¹₂ .

(2) Hölder’s inequality holds for arbitrary measurable functions with the interpretation that the integrals may be infinite. (Exercise)

Lemma 1.19 (Jensen’s inequality). Let 1Ép<q< ∞and assume that 0<

µ(A)< ∞. Then

µ 1 µ(A)

Z

A|f|^pdµ

¶¹

p

É µ 1

µ(A) Z

A|f|^qdµ

¶¹

q.

TH E M O R A L: The integral average is an increasing function of the power.

Proof. By Hölder’s inequality Z

A|f|^pdµÉ µZ

A|f|

pq p dµ

¶^p_qµZ

A

1

q q−pdµ

¶^q−p_q

= µZ

A|f|^qdµ

¶^p_q

µ(A)¹⁻^p^q. ä

Remark 1.20. If 1Ép<q< ∞andµ(A)< ∞, thenL^q(A)⊂L^p(A).

WA R N I N G: Let 1Ép<q< ∞. In general,L^q(A)6⊂L^p(A) orL^p(A)6⊂L^q(A).

Reason. Letf: (0,∞)→R,f(x)=x^aand assume thatµis the Lebesgue measure.

Then

f∈L¹((0, 1))⇐⇒a> −1 and f∈L¹((1,∞))⇐⇒a< −1.

Assume that 1Ép<q< ∞. Choosebsuch that 1/qÉb<1/p. Then the functionx^−bχ(0,1)(x) belongs toL^p((0,∞)), but does not belong toL^q((0,∞)). On the other hand, the functionx⁻^bχ(1,∞)(x) belongs toL^q((0,∞)), but does not belong to

L^p((0,∞)). _■

Examples 1.21:

(1) LetA=(0, 1),µbe the Lebesgue measure and 1Ép< ∞. Define f: (0, 1)→ R,

f(x)= 1 x¹^p¡

log²_x¢²_p .

Thenf∈L^p((0, 1)), butf∉L^q((0, 1)) for anyq>p. Thus for everypwith 1Ép< ∞, there exists a function f which belongs toL^p((0, 1)), but does not belong to anyL^q((0, 1)) withq>p. (Exercise)

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(2) Let 1Ép<q< ∞. Assume thatAcontainsµ-measurable sets of arbitrarily small positive measure. Then there are pairwise disjointµ-measurable setsA_i⊂A,i=1, 2, . . . , such thatµ(A_i)>0 andµ(A_i)→0 asi→ ∞. Let

f= X∞ i=1

a_iχA_i,

wherea_iÊ0 witha_i→ ∞asi→ ∞are chosen so that X∞

i=1

a^q_iµ(Ai)= ∞ and X∞ i=1

a^p_iµ(Ai)< ∞.

Thenf∈L^p(A) \L^q(A). It can be shown, thatL^p(A) is not contained in L^q(A) if and only ifAcontains measurable sets of arbitrarily small positive measure. (Exercise)

(3) Let 1Ép<q< ∞. Assume thatAcontainsµ-measurable sets of arbitrarily large measure. Then there are pairwise disjointµ-measurable setsA_i⊂A, i=1, 2, . . . , such thatµ(A_i)>0 andµ(A_i)→ ∞asi→ ∞. Let

f= X∞ i=1

aiχA_i,

wherea_iÊ0 witha_i→ ∞asi→ ∞are chosen so that X∞

i=1

a^q_iµ(Ai)< ∞ and X∞ i=1

a^p_iµ(Ai)= ∞.

Then f ∈L^q(A) \L^p(A). It can be shown, thatL^q(A) is not contained in L^p(A) if and only if A contains measurable sets of arbitrarily large measure. (Exercise)

Remark 1.22. There is a more general version of Jensen’s inequality. Assume that 0<µ(A)< ∞. Let f∈L¹(A) such thata<f(x)<bfor everyx∈A. Ifϕis a convex function on (a,b), then

ϕ µ 1

µ(A) Z

A

f dµ

¶

É 1

µ(A) Z

Aϕ◦f dµ.

The casesa= −∞andb= ∞are not excluded. Observe, that in this case may happen thatϕ◦f is not integrable. We leave the proof as an exercise.

Theorem 1.23 (Minkowski’s inequality). Assume 1Ép< ∞andf,g∈L^p(A).

Thenf+g∈L^p(A) and

kf+gkpÉ kfkp+ kgkp.

Moreover, an equality occurs if and only if there exists a positive constantcsuch that f(x)=c g(x) forµ-almost everyx∈A.

TH E M O R A L: Minkowski’s inequality is the triangle inequality for theL^p norm. It implies that theL^pnorm, with 1Ép< ∞, is a norm in the usual sense and thatL^p(A) is a normed space if the functions that coincide almost everywhere are identified.

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Remark 1.24. Elementary inequalities (1.3) and (1.4) imply that kf+gkp=

µZ

A|f+g|^pdµ

¶¹_p

É2^p−1^p µZ

A

(|f|^p+ |g|^p)dµ

¶¹_p

É2

p−1 p

ÃµZ

A|f|^pdµ

¶¹_p +

µZ

A|g|^pdµ

¶¹_p!

=2^p−1^p (kfkp+ kgkp), 1Ép< ∞.

Observe that the factor 2^(p⁻^1)/^pis strictly greater than one forp>1 and Minkowski’s inequality does not follow from this.

Proof. p=1 : The triangle inequality, as in the proof of Lemma1.8, shows that kf+gk1É kfk1+ kgk1.

1<p< ∞ : Ifkf+gkp=0, there is nothing to prove. Thus we may assume thatkf+gkp>0. By Hölder’s inequality, we have

Z

A|f+g|^pdµÉ Z

A|f+g|^p⁻¹|f|dµ+

Z

A|f+g|^p⁻¹|g|dµ

(|f+g|^p= |f+g|^p⁻¹|f+g| É |f+g|^p⁻¹(|f| + |g|)) É

µZ

A|f+g|^(p−1)p⁰dµ

¶_p0¹ µZ

A|f|^pdµ

¶¹_p

+ µZ

A|f+g|^(p⁻^1)p⁰dµ

¶¹

p0µZ

A|g|^pdµ

¶¹_p . Since (p−1)p⁰=pand 0< kf+gkp< ∞, we have

µZ

A|f+g|^pdµ

¶1−^p−1_p

É µZ

A|f|^pdµ

¶¹_p +

µZ

A|g|^pdµ

¶¹_p .

It remains to consider when the equality can occur. This happens if there is an equality in the pointwise inequality

|f(x)+g(x)|^p= |f(x)+g(x)|^p−1|f(x)+g(x)| É |f(x)+g(x)|^p−1(|f(x)| + |g(x)|) for µ-almost every x∈A as well as an equality in the application of Hölder’s inequality. An equality occurs in Hölder’s inequality if

c₁|f(x)|^p= |f(x)+g(x)|^p=c₂|g(x)|^p

forµ-almost everyx∈A. Equality occurs in in the pointwise inequality above if f(x) andg(x) have the same sign. This completes the proof. ä Remark 1.25. It is possible to prove Minkowski’s inequality directly by Young’s inequality instead of Hölder’s inequality (exercise).

Note that the normed spaceL^p(A), 1Ép< ∞, is a metric space with the metric d(f,g)= kf−gkp.

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1.3 ^L ^p spaces for ⁰ _< ^p _< ¹

It is sometimes useful to considerL^pspaces for 0<p< ∞. Observe that Definition 1.1makes sense also when 0<p<1 and the space is a vector space by the same argument as in the proof of Lemma1.8. However,kfkpis not a norm for 0<p<1.

Reason. Let f,g:R→R, f =χ_[0,¹

2) and g=χ_[¹

2,1]. Then f+g=χ[0,1] so that kf+gkp=1. On the other hand,

kfkp=2⁻¹^p and kgkp=2⁻¹^p. Thus

kfkp+ kgkp=2·2⁻¹^p=2¹⁻¹^p<1,

when 0<p<1. This shows thatkfkp+ kgkp< kf+gkp. _■ Thus the triangle inequality does not hold true when 0<p<1, but we have the following result.

Lemma 1.26. Iff,g∈L^p(A) and 0<p<1, thenf+g∈L^p(A) and kf+gk^ppÉ kfk^pp+ kgk^pp.

Proof. The elementary inequality

(a+b)^pÉa^p+b^p, a,bÊ0, 0<p<1, (1.4) implies

kf+gk^pp= Z

A|f+g|^pdµÉ Z

A|f|^pdµ+

Z

A|g|^pdµ= kfk^pp+ kgk^pp.

ä However,L^p(A) is a metric space with the metric

d(f,g)= kf−gk^pp= Z

A|f−g|^pdµ

This metric is not induced by a norm, sincekfk^ppdoes not satisfy the homogeneity required by the norm. On the other hand,kfkpsatisfies the homogeneity, but not satisfy the triangle inequality.

Remarks 1.27:

(1) By (1.2), we have kf+gkpÉ¡

kfk^pp+ kgk^pp

¢¹_p

É2¹^p(kfkp+ kgkp), 0<p<1.

Thus a quasi triangle inequality holds with a multiplicative constant.

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(2) Iff,g∈L^p(A), fÊ0,gÊ0, then

kf+gkpÊ kfkp+ kgkp, 0<p<1.

This is the triangle inequality in the wrong direction (exercise).

Remark 1.28. It is possible to define theL^pspaces also whenp<0. Aµ-measurable function is inL^p(A) for p<0, if

0<

Z

A|f|^pdµ< ∞.

If f ∈L^p(A) for p<0, then f 6=0 µ-almost everywhere and|f| < ∞µ-almost everywhere inA. However, this is not a vector space.

1.4 Completeness of L ^p

Next we prove a famous theorem, which is not only important in the theory ofL^p spaces, but has a historical interest as well. The result was found independently by F. Riesz and E. Fisher in 1907, primarily in connection with the Fourier series which culminates in showing completeness ofL².

Recall that a sequence (f_i) of functions f_i∈L^p(A), i=1, 2, . . ., converges in L^p(A) to a function f∈L^p(A), if for everyε>0 there existsi_εsuch that

kf_i−fkp<ε when iÊi_ε. Equivalently,

ilim→∞kfi−fkp=0.

A sequence (f_i) is a Cauchy sequence inL^p(A), if for everyε>0 there existsi_ε such that

kf_i−f_jkp<ε when i,jÊi_ε. WA R N I N G: This is not the same condition as

kfi+1−fikp<ε when iÊi_ε.

Indeed, the Cauchy sequence condition implies this, but the converse is not true (exercise).

CL A I M: If f_i→f inL^p(A) asi→ ∞, then (f_i) is a Cauchy sequence inL^p(A).

Reason. Letε>0. Since f_i→f in L^p(A) as i→ ∞, there exists i_ε such that kf_i−fkp<^ε₂ wheniÊi_ε. Minkowski’s inequality implies

kf_i−f_jkpÉ kf_i−fkp+ kf−f_jkp<ε 2+ε

2=ε

wheni,jÊi_ε. _■

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Theorem 1.29 (Riesz-Fischer). For every Cauchy sequence (f_i) inL^p(A), 1É p< ∞, there existsf∈L^p(A) such thatf_i→f inL^p(A) asi→ ∞.

TH E M O R A L: L^p(A), 1Ép< ∞, is a Banach space with the normk · kp. In particular,L²(A) is a Hilbert space.

Proof. Assume that (f_i) is a Cauchy sequence inL^p(A). We construct a subsequence as follows. Choosei₁such that

kf_i−f_jkp<1

2 when i,jÊi₁.

We continue recursively. Suppose thati₁,i₂, . . . ,i_khave been chosen such that kf_i−f_jkp< 1

2^k when i,jÊi_k. Then choosei_k₊₁>i_ksuch that

kf_i−f_jkp< 1

2^k⁺¹ when i,jÊi_k+1. For the subsequence (f_i_k), we have

kf_i_k−f_i_k+1kp< 1

2^k, k=1, 2, . . . . Let

g_l=

l

X

k=1

|f_i_k+1−f_i_k| and g= X∞ k=1

|f_i_k+1−f_i_k|. Then

lim

l→∞g_l=lim

l→∞

l

X

k=1

|f_i_k+1−f_i_k| = X∞ k=1

|f_i_k+1−f_i_k| =g

and as a limit ofµ-measurable functions gis aµ-measurable function. Fatou’s lemma and Minkowski’s inequality imply

µZ

A

g^pdµ

¶¹_p

= µZ

A

lim inf

l→∞ g_l^pdµ

¶¹_p

Élim inf

l→∞

µZ

A

g_l^pdµ

¶¹_p

Élim inf

l→∞

l

X

k=1

µZ

A|fi_k₊₁−fi_k|^pdµ

¶¹_p

É X∞ k=1

1 2^k=1.

Thus g∈L^p(A) and consequently g(x)< ∞forµ-almost every x∈A. It follows that the telescoping series

f_i₁(x)+ X∞ k=1

(f_i_k+1(x)−f_i_k(x))

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converges absolutely forµ-almost everyx∈A. Denote the sum of the series by f(x) for thosex∈Aat which it converges and setf(x)=0 in the remaining set of measure zero. Then

f(x)=f_i₁(x)+ X∞ k=1

(f_i_k+1(x)−f_i_k(x))

=lim

l→∞

Ã f_i₁(x)+

l−1

X

k=1

(f_i_k₊₁(x)−f_i_k(x))

!

=lim

l→∞f_i_l(x)= lim

k→∞f_i_k(x)

for µ-almost every x∈A. Thus there is a subsequence (f_i_k) which converges µ-almost everywhere inA. Next we show that the original sequence converges to

f inL^p(A).

CL A I M: fi→f inL^p(A) asi→ ∞.

Reason. Letε>0 and let (f_i_k) be a subsequence which converges to f µ-almost everywhere inA. Since (f_i) is a Cauchy sequence inL^p(A), there existsi_εsuch thatkf_i_k−f_ikp<εwhen i,i_kÊi_ε. For a fixed iÊi_ε, we have f_i_k−f_i→f−f_i µ-almost everywhere inAasi_k→ ∞. By Fatou’s lemma

µZ

A|f−f_i|^pdµ

¶¹_p

= µZ

A

lim inf

k→∞ |f_i_k−f_i|^pdµ

¶¹_p

Élim inf

k→∞

µZ

A|f_i_k−f_i|^pdµ

¶¹_p

Éε. _■

This shows that f−f_i∈L^p(A) and thus f =(f−f_i)+f_i∈L^p(A). Moreover, for everyε>0 there existsi_εsuch thatkfi−fkpÉεwheniÊi_ε. This completes the

proof. ä

WA R N I N G: In general, if a sequence has a converging subsequence, the original sequence need not converge. In the proof above, we used the fact that we have a Cauchy sequence.

We shall often use a part of the proof of the Riesz-Fisher theorem, which we now state.

Corollary 1.30. Iff_i→f inL^p(A), then there exist a subsequence (f_i_k) such that lim

k→∞f_i_k(x)=f(x) µ-almost every x∈A.

Proof. The proof of the Riesz-Fischer theorem gives a subsequence (f_i_k) and a functiong∈L^p(A) such that

lim

k→∞f_i_k(x)=g(x) µ-almost every x∈A

and fi_k→g inL^p(A). On the other hand, fi→f inL^p(A), which implies that f_i_k→f inL^p(A). By the uniqueness of the limit, we conclude that f=gµ-almost

everywhere inA. ä

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Let us compare the various modes of convergence of a sequence (f_i) of functions inL^p.

Remarks 1.31:

(1) Iff_i→f inL^p(A) asi→ ∞, then lim

i→∞kf_ikp= kfkp. Reason. By Minkowski’s inequality

kf_ikp= kf_i−f+fkpÉ kf_i−fkp+ kfkp,

which implieskfikp− kfkpÉ kfi−fkp. By switching the roles off and fi, we havekfkp− kf_ikpÉ kf_i−fkp. Thus

¯

¯kf_ikp− kfkp

¯

¯É kf_i−fkp→0, from which it follows that

lim

i→∞kf_ikp= kfkp.

■

(2) Iff_i→f inL^p(A) asi→ ∞, thenf_i→f in measure.

Reason. By Chebyshev’s inequality µ({x∈A:|f_i(x)−f(x)| Êε})É 1

ε^p Z

A|f_i−f|^pdµ

= 1

ε^pkf_i−fk^pp i→∞

−−−→0. _■

(3) Iff_i→f inL^p(A) asi→ ∞, then there exist a subsequence (f_i_k) such that lim

k→∞f_i_k(x)=f(x) µ-almost every x∈A.

Reason. The convergence in measure implies the existence of an almost everywhere converging subsequence. This gives another proof of the previous

corollary. _■

(4) In the casep=1,f_i→f inL¹(A) asi→ ∞implies not only that

ilim→∞

Z

A|f_i|dµ= Z

A|f|dµ but also that

lim

i→∞

Z

A

f_idµ= Z

A

f dµ. Reason.

¯

¯ Z

A

(fi−f)dµ

¯

¯É Z

A|fi−f|dµ= kfi−fk1 i→∞

−−−→0.

■

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The following example shows that imply pointwise convergence almost eerywhere does not implyL^pconvergence pointwise convergence andL^pconvergence does not imply pointwise convergence almost eerywhere.

Example 1.32. In the following examples we assume thatµis the Lebesgue measure.

(1) f_i→f in L^p as i→ ∞does not imply f_i→f almost everywhere. Let f_i:R→R, f_i(x)=χ[i−1,i)(x), i=1, 2, . . ., and f =0. Assume thatµis the Lebesgue measure. Thenf_i(x)→0 for everyx∈R. However,kf_ikp=1 for everyi=1, 2, . . . andkfkp=0. Thus the sequence (f_i) does not converge to

f inL^p(R), 1Ép< ∞.

(2) f_i→f almost everywhere as i→ ∞does not imply f_i→f in L^p. Let f_i:R→R,

f_i(x)=i²χ^¡_0,¹

i

¢(x), i=1, 2, . . . .

Then Z

R|f_i(x)|^pdx=i^2p Z

Rχ^p

(0,¹_i)(x)dx=i^2p¹_i=i^2p−1< ∞. Thusf_i∈L^p(R), 1Ép< ∞,f_i(x)→0 for everyx∈R, but

kf_ikp=i²⁻¹^pÊi−−−→ ∞ⁱ^→∞ . Thus (f_i) does not converge inL^p(R).

(3) f_i→f inL^pasi→ ∞does not implyf_i→f almost everywhere. Consider the sliding sequence of functionsf_i:R→R,

f₂k+j(x)=kχ^h j 2k,^j⁺¹

2k

i(x), k=0, 1, 2, . . . , j=0, 1, 2, . . . , 2^k−1.

Then

kf₂k+jkp=k2⁻^k^p−−−−→^k^→∞ 0,

which implies that f_i→0 inL^p(R), 1Ép< ∞, as i→ ∞. However, the sequence (f_i(x)) fails to converge for everyx∈[0, 1], since

lim sup

i→∞

f_i(x)= ∞ and lim inf

i→∞ f_i(x)=0

for everyx∈[0, 1]. Note that there are many converging subsequences. For example, f₂k+1(x)→0 for everyx∈[0, 1] ask→ ∞.

(4) A sequence can converge inL^pwithout converging inL^q. Considerfi:R→ R,

fi(x)=¹_iχ(i,2i)(x), i=1, 2, . . . .

Thenkfikp=i⁻¹⁺¹^p,i=1, 2 . . . . Thus f→0 inL^p(R), 1<p< ∞, asi→ ∞, butkf_ik1=1 for everyi=1, 2 . . . , so that the sequence (f_i) does not converge inL¹(R).

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The following discussion clarifies the difference between the pointwise conven- gence andL^pconvergence.

Theorem 1.33. Let 1Ép< ∞. Assume that fi∈L^p(A), i=1, 2, . . . , fi→f µ- almost everywhere in Aas i→ ∞. If there exists g∈L^p(A), gÊ0, such that

|fi| Égµ-almost everywhere in Afor everyi=1, 2, . . . , thenf∈L^p(A) and fi→f inL^p(A) asi→ ∞.

TH E M O R A L: Pointwise convergence implies the norm convergence inL^p if the sequence is uniformly bounded by a function inL^p. This is a dominated convergence theorem inL^p.

Proof. Since|f_i−f| É |f_i|+|f| Ég+|f|µ-almost everywhere inAandg+|f| ∈L^p(A), the dominated convergence theorem implies

ilim→∞

Z

A|f_i−f|dµ= Z

A

ilim→∞|f_i−f|dµ=0.

ä Theorem 1.34. Assume that fi∈L^p(A),i=1, 2, . . . and f∈L^p(A), 1Ép< ∞. If

f_i→f µ-almost everywhere in Aand lim

i→∞kf_ikp= kfkp, thenf_i→f inL^p(A) as i→ ∞.

Proof. Since|f_i| < ∞and|f| < ∞µ-almost everywhere inA, by (1.2), we have 2^p(|fi|^p+ |f|^p)− |fi−f|^pÊ0

µ-almost everywhere in A. The assumption f_i→f µ-almost everywhere inA implies

ilim→∞(2^p(|f_i|^p+ |f|^p)− |f_i−f|^p)=2^p⁺¹|f|^p µ-almost everywhere inA. Applying Fatou’s lemma, we obtain

Z

A

2^p+1|f|^pdµÉlim inf

i→∞

Z

A

¡2^p(|f_i|^p+ |f|^p)− |f_i−f|^p¢ dµ Élim inf

i→∞

µZ

A

2^p|fi|^pdµ+ Z

A

2^p|f|^pdµ− Z

A|fi−f|^pdµ

¶

=lim

i→∞

Z

A

2^p|f_i|^pdµ+

Z

A

2^p|f|^pdµ−lim sup

i→∞

Z

A|f_i−f|^pdµ

= Z

A

2^p|f|^pdµ+ Z

A

2^p|f|^pdµ−lim sup

i→∞

Z

A|fi−f|^pdµ.

Here we used the facts that if (a_i) is a converging sequence of real numbers and (b_i) is an arbitrary sequence of real numbers, then

lim inf

i→∞ (ai+bi)=lim

i→∞ai+lim inf

i→∞ bi and lim inf

i→∞ (−bi)= −lim sup

i→∞

bi. SubtractingR

A2^p+1|f|^pdµfrom both sides, we have lim sup

i→∞

Z

A|f_i−f|^pdµÉ0.

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On the other hand, since the integrands are nonnegative lim inf

i→∞

Z

A|f_i−f|^pdµÊ0.

Thus

ilim→∞

Z

A|f_i−f|^pdµ=0.

ä Remark 1.35. Let 1Ép< ∞. Assume that f_i∈L^p(A), 0Éf_iÉ f_i+1 µ-almost everywhere in A, i=1, 2, . . . . Then the pointwise limit f =lim_i_→∞f_i existsµ- almost everywhere inA. The monotone convergence theorem implies

lim

i→∞

Z

A

f_i^pdµ= Z

A

lim

i→∞f_i^pdµ= Z

A

f^pdµ. Theorem1.34(or Theorem1.33) implies f_i→f inL^p(A) asi→ ∞.

TH E M O R A L: An increasing sequence of nonnegative functions inL^pcon- verges inL^p, if the limit function belongs toL^p. This is a monotone convergence theorem inL^p.

1.5 L ^∞ space

The definition of the L^∞ space differs substantially from the definition of the L^p space for 1Ép< ∞. The main difference is that instead of the integration the definition is based on the almost everywhere concept. The classL^∞consists of bounded measurable functions with the interpretation that we neglect the behaviour of functions on a set of measure zero.

Definition 1.36. Let A⊂Rⁿ be a µ-measurable set and f :A→[−∞,∞] a µ- measurable function. Thenf∈L^∞(A), if there existsM, 0ÉM< ∞, such that

|f(x)| ÉM forµ-almost everyx∈A.

Functions inL^∞are sometimes called essentially bounded functions. The essential supremum off is

ess sup

x∈A

f(x)=inf{M:f(x)ÉMforµ-almost everyx∈A}

=inf©

M:µ({x∈A:f(x)>M})=0ª and the essential infimum off is

ess inf

x∈A f(x)=sup{m:f(x)Êmforµ-almost everyx∈A}

=sup©

m:µ({x∈A:f(x)<m})=0ª . TheL^∞norm of f is

kfk∞=ess sup

x∈A |f(x)|.

Real Analysis

Real Analysis

Contents

L p spaces 1

1.1 L p functions

1.2 L p norm

1.3 L p spaces for 0 < p < 1

1.4 Completeness of L p

1.5 L ∞ space

L ^p spaces 1

1.1 ^L ^p functions

1.2 ^L ^p norm

1.3 ^L ^p spaces for ⁰ _< ^p _< ¹

1.4 Completeness of L ^p

1.5 L ^∞ space