Analysis Comprehensive Exam Review NotesRevised: August 18, 2018

Topics Index

1 Set theory and measures 21.1 Definitions and key properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Sets and sequences of sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Sample problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Differentiation 62.1 Key results and theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Past exam problems using the LDT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Other differentiation problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Integration 93.1 Absolute Continuity, bounded variation, and related characterizations . . . . . . . . . . 93.2 Practice problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Notions of convergence 124.1 Modes of convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.2 Flowchart of modes of convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.3 Convergence theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

5 Fubini and Tonelli 155.1 Statements of key theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155.2 List of previous exam problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

6 Lp–Spaces 176.1 Key results and inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176.2 Dense class arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186.3 Convex functions and Jensen’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 196.4 Proofs of standard results from Folland . . . . . . . . . . . . . . . . . . . . . . . . . . . 196.5 Previous exam problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206.6 Proofs of results from the comp prep homework . . . . . . . . . . . . . . . . . . . . . . 23

7 Signed measures 257.1 Definitions and key theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257.2 Practice problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267.3 Extra problems from Folland . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

8 A few key examples 28

9 Functional analysis 299.1 Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299.2 Linear transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299.3 Bounded linear operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299.4 Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309.5 Practice problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

10 Miscellaneous other notes 3210.1 Other problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3210.2 Facts and reminder notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1

1 SET THEORY AND MEASURES

1.1 Definitions and key properties

Definition 1.1 (σ-algebras). A sigma algebra is a collection of sets A in some universal set U suchthat

(i) ∅ ∈ A;

(ii) Xn ⊆ A =⇒ ∪iXi ∈ A;

(iii) X ∈ A =⇒ Xc ∈ A.

Note that (ii) and (iii) imply that countable intersections of sets in A are also in A. For example,we can take U = 1, 2 and A = ∅, 1, 2. For every set X the power set 2X is a σ-algebra. Wecan form the Borel σ-algebra on R by considering all open intervals (a, b) and then taking all possiblecombinations of unions, intersections, and complements of these intervals.

Definition 1.2 (Measures). A (positive) measure is a function µ : A→ [0,∞] (or onto R∪∞) whichsatisfies

(i) µ(∅) = 0;

(ii) (Disjoint Unions) µ (∪iXi) =∑

i µ(Xi);

(iii) (Non-Negativity) µ ≥ 0.

Notice that these axioms imply monotonicity. A finite measure is a measure such that µ(U) < ∞. Aσ-finite measure space is such that there exist finite measure sets, Xi, such that U = ∪∞i=1Xi. TheBorel measure is a measure on the Borel σ-algebra. The Lebesgue measure is defined on U = R with`(a, b) = |(a, b)| = b− a. The Lebesgue measure in invariant under translation: |A| = |A+ x|.

Definition 1.3 (Outer measures). For X ⊂ ∪iXi, we define the outer measure by

µ∗(X) = |X|e = inf∞∑i=1

µ(Xi).

Outer measures have the following properties:

(i) (Monotonicity) If X ⊂ Y then µ∗(X) ≤ µ∗(Y );

(ii) (Domain) µ∗ : 2R → R;

(iii) (Non-negativity) µ∗ ≥ 0;

(iv) (Null Sets) µ∗(∅) = 0;

(v) (Countable Subadditivity) µ∗ (∪iXi) ≤∑

i µ∗(Xi).

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By Caratheodory, we see that E is measurable iff for ALL subsets A (i.e., not just the measurable ones)we have that

µ∗(A) = µ∗(E ∩ A) + µ∗(Ec ∩ A).

Every Borel measurable set is Lebesgue measurable, but not vice versa. Open sets are measurable.

Proposition 1.4 (Approximation of outer measure by open sets). For all E ⊆ Rn and ∀ε > 0, thereexists an open set H such that E ⊆ H and µ(H) < µ∗(E) + ε. One can alternately choose a compact(closed) set F such that F ⊆ E with µ(F ) > µ∗(E) − ε. Note that we can form any open set by acountable disjoint union of boxes in Rn.

Definition 1.5 (Borel sets). A Gδ set is a countable intersection of open sets. A Fσ set is a countableunion of closed sets.

Theorem 1.6 (E is measurable iff it differs from a Borel set of zero measure). Let E ⊆ Rn. Then

(a) E is measurable iff we can find a Gδ set H and a set Z of measure zero such that E = H \ Z.

(b) E is measurable iff we can find a Fσ set F and Z with |Z| = 0 such that E = F ∪ Z.

Proof of (a). For sufficiency, suppose that E = H \ Z. Here H is Gδ, so measurable, and |Z| = 0is measurable. By a theorem E is measurable. For necessity, Suppose that E is measurable. By atheorem, there exists a Gδ set H with E ⊆ H and such that |E|e = |H|e. Set Z := H \ E so that Z ismeasurable as above and |Z| = |H| − |E| = 0.

Proof of (b). For sufficiency, suppose that E = F ∪Z which is a union of measurable sets and is henceitself measurable. To show necessity, suppose that E is measurable. Then Ec is measurable, so by part(a) Ec = H \ Z for some H,Z with |Z| = 0 and H = ∩k≥1Ek for Ek all open. Then

E = C(Ec) = C(H \ Z) = C(H ∩ CZ)− CH ∪ Z = ∪k≥1CEk ∪ Z.

Here each CEk is closed, so F = supk≥1CEk is a Fσ set.

1.2 Sets and sequences of sets

Definition 1.7 (Limsups and liminfs of sets). Given a sequence of sets En define

lim supn→∞

En =⋂m≥1

⋃n≥m

En (In ∞-many of the sets En)

lim infn→∞

En =⋃m≥1

⋂n≥m

En (In all but finitely-many of the En)

Observe that for any N ≥ 1:

lim supn→∞

En ⊆∞⋃i=N

Ei.

Definition 1.8 (Monotone sequences of sets). We use the notation Ek E if E1 ⊆ E2 ⊆ · · · andE = ∪k≥1Ek. Similarly, we say that Ek E if E1 ⊇ E2 ⊇ · · · and E = ∩k≥1Ek. We get continuityproperties of limits of sets for monotone sequences of sets (see below).

Theorem 1.9 (Continuity of the Lebesgue measure on monotone sets). Let Ek be a sequence ofmeasurable sets in Rn.

(1) If Ek E, then limk→∞ |Ek| = |E|;

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(2) If Ek E and |Ek| <∞ for some k, then limk→∞ |Ek| = |E|.

Proof of 1. If for some j ≥ 1, |Ej| =∞ then |Ek| =∞ for all k ≥ j, and hence |E| =∞. Now we canassume that |Ek| <∞ for all k ≥ 1. We write

E = E1 ∪ (E2 \ E1) ∪ · · · = E1 ∪⋃k≥1

(Ek+1 \ Ek).

Then

|E| = |E1|+∑k≥1

|Ek+1 \ Ek| = |E1|+ limn→∞

n∑k=1

(|Ek+1| − |Ek|) = limn→∞

|En+1|.

Proof of 2. We can assume that |E1| <∞ since otherwise we can truncate the sequence and relabel toobtain the same limit. Moreover, we can write E1 as a disjoint union of measurable sets:

E1 = E ∪⋃k≥1

(Ek \ Ek+1).

Now we have that

|E1| = |E|+∑k≥1

|Ek \ Ek+1|

= |E|+ limn→∞

n∑k=1

(|Ek| − |Ek+1|)

= |E|+ limn→∞

(|E1| − |En+1|),

which implies the result.

Example 1.10 (Spring 2018, #6). Prove each of the following:

i. Let Ek ⊂ Rn for k ≥ 1 be sets such that E1 ⊆ E2 ⊆ E3 ⊆ · · · , and denote E := ∪k≥1Ek. Assumethat |E|e <∞. Prove that

|E|e = limk→∞|Ek|e.

ii. Let E be a set in Rn with 0 < |E|e < ∞. Let 0 < ϑ < 1. Show that there is a set Eϑ ⊂ E with|Eϑ|e = ϑ · |E|e.

1.3 Sample problems

Example 1.11 (Spring 2017, #6). Given A ⊆ [0, 1], prove that A is Lebesgue measurable iff

|A|e + |[0, 1] \ A|e = 1.

Example 1.12 (Fall 2016, #2). Show that for A ⊂ Rd, A is Lebesgue measurable iff for every ε > 0there is a Lebesgue measurable set E ⊂ Rd such that |A∆E|e < ε.

Example 1.13 (Spring 2016, #4). Assume E ⊂ Rd is measurable such that |E| <∞.

(a) Suppose that f : E → [−∞,∞] is measurable and finite a.e. Given ε > 0, prove that there is aclosed set F ⊆ E such that |E \ F | < ε and f is bounded on F .

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(b) For each n ≥ 1 let fn be a measurable function on E, and suppose that

∀x ∈ E : Mx = supn≥1|fn(x)| <∞.

Prove that for each ε > 0 there is a closed set F ⊆ E and a finite constant M such that |E \F | < εand |fn(x)| ≤M for all x ∈ F and n ≥ 1.

Example 1.14 (Spring 2014, #2). Let A be any subset of Rd. Prove that there exists a measurableset H ⊇ A that satisfies |A ∩ E|e = |H ∩ E| for every measurable set E ⊆ Rd.

Example 1.15 (Fall 2012, #1). Given a set E ⊆ Rd with |E|e < ∞, show that E is Lebesguemeasurable iff for each ε > 0 we can write E = (S ∪ A) \ B where S is a union of finitely manynon-overlapping boxes and |A|e, |B|e < ε.

Example 1.16 (Spring 2012, #3). Suppose that 0 < θ < 1, E ⊂ Rn, and 0 < |E| < ∞. Prove thatthere is a cube Q such that θ · |Q| < |E ∩Q|.

Example 1.17 (Spring 2008, #2 ∗). Let | · |e denote the exterior Lebesgue measure on R. Supposethat E is a subset of R with |E|e <∞. Show that E is Lebesgue measurable iff for every ε > 0 we canwrite E = (S ∪N1) \N2, where S is a finite union of non-overlapping intervals and |N1|e, |N2|e < ε.

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2 DIFFERENTIATION

2.1 Key results and theorems

Theorem 2.1 (LDT). Let f be measurable and defined on the domain A. Then for a.e. x ∈ A wehave that

limr→0

1

2r

∫ x+r

x−rf(y)dy = f(x),

i.e., if the previous equation holds for all x ∈ B, then |A \B| = 0. Lebesgue points satisfy this relation,and almost every x ∈ [a, b] is a Lebesgue point. We can also express the LDT in the special case off := χE, the characteristic function of E, in the form of∣∣∣∣ 1

2r

∫ x+r

x−rχE(y)dy − χE(x)

∣∣∣∣→ 0, a.e. as r → 0.

A few useful facts:

• By the LDT, if |A| > 0, ∃a ∈ A, δ0 > 0 such that |A ∩ (a− δ, a+ δ)| > 3δ2

for all 0 < δ < δ0.

• Consider f(x) := χE(x) for |E| <∞. Then f ∈ L1(R) and so a.e. x ∈ R is a Lebesgue point of f .

• By the LDT, a.e. point in [0, 1] is a density point of χA(x) in that for a.e. x ∈ A:

limr→0

|A ∩ [x− r, x+ r]||[x− r, x+ r]|

= 1.

Lemma 2.2 (The Riemann-Lebesgue lemma). For all f ∈ L1, we have that

limλ→0

∫ ∞−∞

f(x)eıλxdx = 0.

Note that the exponential function in the previous integral can effectively be replaced by any otherfunction that integrates to zero to obtain an analogous result.

Proof Sketch. We use a dense class argument to approximate f(x) on R within some small ε > 0.Namely, we know that the functions g ∈ C1

0 , or the compactly upported continuous functions with onecontinuous derivative, are dense in L1. Then we can complete the proof using IBP where we pick up afactor of 1/λ and give ourselves a little room, i.e., show that the limit is less than ε for every ε > 0.

The mean value theorem (MVT for derivatives) states that if f is defined and continous on [a, b] anddifferentiable on (a, b), then ∃c ∈ (a, b) such that (b − a)f ′(c) = f(b) − f(a). The intermediate valuetheorem (IVT) states that if f is continuous on [a, b], then for all c in the range between f(a) and f(b),∃x ∈ (a, b) such that f(x) = c. The MVT for integrals states that if f is continuous on [a, b] then∃c ∈ (a, b) such that

f(c) =1

b− a

∫ b

a

f(t)dt.

Lemma 2.3 (Growth lemma II). Suppose that f : [a, b]→ R and E ⊂ [a, b] be measurable. Then if fis differentiable on E, |f(E)|e ≤

∫E|f ′|.

6

2.2 Past exam problems using the LDT

Example 2.4 (Spring 2018, #1). Let A ⊆ R be a measurable set. For x ∈ R denote A+ x = a+ x :a ∈ A. Prove that if A satisfies

|A \ (A+ x)| = 0, ∀x ∈ R,

then either |A| = 0 or |R \ A| = 0.

Example 2.5 (Spring 2017, #4). Let A be a measurable subset of [0, 1].

(a) Prove that if |A| > 2/3, then A contains an arithmetic progression of length 3, that is, prove thatthere are a, d ∈ R such that a, a+ d, a+ 2d ∈ A;

(b) Use part (a) to prove that if |A| > 0, then A contains an arithmetic progression of length 3.

Example 2.6 (Fall 2016, #1). Let E ⊆ R be a Lebesgue measurable set with 0 < |E| <∞.

(1) For each x ∈ R and r > 0 define Ir(x) := [x − r/2, x + r/2] and hr(x) := |E ∩ Ir(x)|. Prove thatfor a fixed r > 0, the function hr(x) is continuous at every x ∈ R.

(2) Prove that there exists r0 > 0 such that for each 0 < r < r0 there exists a closed interval I ⊂ Rwhich satisfies |I| = r and |E ∩ I| = r/2.

Example 2.7 (Spring 2016, #3). Suppose that E ⊆ [0, 1] is measurable and that there exists δ > 0such that

|E ∩ [x− r, x+ r]| ≥ δr,

for all x ∈ (0, 1) and r > 0 such that (x− r, x+ r) ⊆ [0, 1]. Prove that |E| = 1.

2.3 Other differentiation problems

Example 2.8 (Fall 2016, #4). Prove that if f(x), xf(x) ∈ L1(R) then the function

F (w) :=

∫Rf(x) sin(wx)dx,

is defined, continuous, and differentiable at every point w ∈ R. (HINT: Use the identity that sin(α)−sin(β) = 2 sin(α−β

2) cos(α+β

2).)

∗∗ NOTE: Continuity proof uses the DCT. ∗∗

Example 2.9 (Fall 2007, #4). Prove that if f is integrable on [a, b], and∫ x

a

f(t)dt = 0,

for all x ∈ [a, b], then f(t) = 0 a.e. in [a, b].

Example 2.10 (Spring 2008, #1). Consider a sequence of functions fn ∈ L1(Rd) for n ≥ 0 with

C := supn≥0

∫Rd

|fn|dx <∞.

Suppose that the following assumptions are satisfied:

(i) fn → f0 in measure as n→∞;

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(ii) For all ε > 0, ∃δ > 0 such that

|A| ≤ δ =⇒ supn≥0

∫A

|fn|dx ≤ ε;

(iii) For all ε > 0 there exists a set K ⊆ Rd with |K| <∞ such that

supn≥0

∫Rd\K

|fn|dx ≤ ε.

Prove that fn → f0 strongly (i.e., in norm) in L1(Rd). Also show that all three conditions are necessaryby constructing three counterexamples, each of which satisfies exactly two of the three hypotheses, andfor which fn does not converge strongly to f0.

8

3 INTEGRATION

3.1 Absolute Continuity, bounded variation, and related characterizations

Definition 3.1 (Absolute continuity). Let I be an interval in R. Then we say that f : I → R isabsolutely continuous on I if ∀ε > 0 ∃δ > 0 such that for (xk, yk) ⊂ I whenever

∑k(yk − xk) < δ then∑

k |f(xk)− f(yk)| < ε. Equivalently, we have the characterization that f is absolutely continuous on[a, b] when f has a derivative f ′ a.e., its derivative is Lebesgue integrable, and

f(x) = f(a) +

∫ x

a

f ′(t)dt, ∀x ∈ [a, b].

In other words, if f is absolutely continuous then f ′ exists and is integrable a.e., and absolutelycontinuous functions satisfy the FTC. Note that if fn is absolutely continuous on [0, 1] then ∃f ′n ∈L1([0, 1]) such that fn(x) =

∫ x0f ′n(t)dt.

NOTE: Ordinarily, we only can show that∫ baf ′(t)dt ≤ f(b) − f(a) (See previous exam problem on

[0, 1].).

Proof strategies: To prove absolute continuity of a function: Use Banach-Zaretsky, can show f isLipschitz, use the FTC, etc.

Theorem 3.2 (Banach-Zaretsky). Let f : [a, b]→ R. Then TFAE:

(1) f is absolutely continuous.

(2) f is continuous, f ∈ BV[a, b], and satisfies Luzin’s condition: if |A| = 0 then |f(A)| = 0.

(3) f is continuous, f is differentiable a.e., f satisfies Luzin’s condition, f ′ ∈ L1[a, b].

Definition 3.3 (Functions of bounded variation). Let Γ = a = x0 < x1 < . . . < xm = b be apartition of [a, b] and define SΓ :=

∑mi=1 |f(xi)− f(xi−1)|. Let the variation of f over [a, b] be defined

asV := sup

ΓSΓ : Γ a partition of [a, b] .

If V < ∞, then we say that f is of bounded variation on [a, b]. If f has bounded variation on [a, b],then we can express f = g−h where g, h are both bounded and monotone increasing on [a, b] (Jordandecomposition theorem). Note that these functions can be extended to be monotone increasing onall of R as well.

Proof strategies: A typical trick used whenever sup is used in problems is applied in this specialcase. Namely, V (f) = supΓ SΓ should be interpreted as: ∀ε > 0, there exists a partition Γε := a =x0 < x1 < · · · < xn = b such that

V (f) ≤n∑k=1

|f(xk)− f(xk−1)|+ ε.

9

This version of the definition is much easier to manipulate. Also, if we let V (x) = V [f ; a, x] denote thetotal variation of f on [a, a + x], then we see that V is monotone increasing and hence differentiablea.e. Moreover, the total variation of f on [x, x+ h] dominates |f(x+ h)− f(x)|:

|f(x+ h)− f(x)| ≤ V [f ;x, x+ h] = V (x+ h)− V (x).

Then it follows that |f ′| ≤ V ′ so that, for example,∫ b

a

|f ′(t)|dt ≤∫ b

a

V ′ ≤ V (a)− V (b) = V [f ; a, b].

Notice that f monotone increasing on [a, b] =⇒ f ∈ BV[a, b]. Also, f monotone increasing implies

that f is differentiable a.e., f ′ ≥ 0, and∫ baf ′(t)dt ≤ f(b)− f(a) for all a < b.

Proposition 3.4 (Monotone increasing function properties). If f is monotone increasing, then

(a) f has at most countably many points of discontinuity.

(b) f is differentiable a.e.

(c) f ′ is a measurable function.

(d) f ′ ∈ L1 and we have the FTC as an upper bound: 0 ≤∫ baf ′ ≤ f(b)− f(a).

Nested characterizations: We say that f is Lipschitz on [a, b] if there exists a K > 0 (where Kis independent of all x, y ∈ [a, b]) such that |f(x) − f(y)| ≤ K · |x − y| for all x, y ∈ [a, b]. We havethe following important nested inclusion of function types (C1 denotes the set of functions with onecontinuous derivative):

C1[a, b] ( Lip[a, b] ( AC[a, b] ( BV[a, b] ( L∞[a, b].

Typical counter examples to these types of functions are√x or are of the form f(x) = xa sin(1/xb).

3.2 Practice problems

Example 3.5 (Spring 2018, #4). Let f : [0, 1]→ [0, 1] be defined by f(0) = 0 and f(x) = x2| sin(1/x)|for x ∈ (0, 1]. Show that f is absolutely continuous on [0, 1]. Give an example of a function φ : [0, 1]→[0, 1] that is of bounded variation, and such that φ′ exists in (0, 1], but such that φ f is not absolutelycontinuous in [0, 1].NOTE:

√x is increasing, and hence of bounded variation.

Example 3.6 (Spring 2014, #4). Suppose that f ∈ L1(R) is such that f ′ ∈ L1(R) and f is absolutelycontinuous on every finite interval [a, b]. Show that limx→∞ f(x) = 0.NOTE: Suppose not, and then show that f is not integrable.

Example 3.7 (Spring 2017, #1). Assume f is real-valued and has bounded variation on [a, b]. Extendf to R by setting f(x) = f(a) for x < a and f(x) = f(b) for x > b. Prove that there exists a constantC > 0 such that ||Ttf − f ||1 ≤ C · |t| when t ∈ R where Ttf(x) = f(x− t) denotes the translation of fby t.

Example 3.8 (Fall 2017, #3). Assume that f : R → R is monotone increasing, and that we havelimx→−∞ f(x) = 0 and limx→∞ f(x) = 1. Prove that f is absolutely continuous on every finite interval[a, b] iff

∫∞−∞ f

′(x)dx = 1.

10

Example 3.9 (Spring 2016, #8). Show that f : [a, b] → R is Lipschitz iff f is absolutely continuousand f ′ ∈ L∞[a, b].

Example 3.10 (Fall 2015, #2). Suppose that fn are absolutely continuous functions on [0, 1] suchthat fn(0) = 0 and ∑

n≥1

∫ 1

0

|f ′n(x)|dx <∞.

Show that

•∑

n≥1 fn(x) converges for every x. Call this limit f(x);(absolutely convergent series are convergent)

• f is absolutely continuous;

• For a.e. x ∈ [0, 1] we have that

f ′(x) =∑n≥1

f ′n(x).

Example 3.11 (Spring 2015, #7). Assume that f has bounded variation on [a, b]. Letting V [f ; a, b]denote the total variation of f on [a, b], prove that∫ b

a

|f ′| ≤ V [f ; a, b].

Example 3.12 (Fall 2012, #2). (a) Let E ⊆ Rd be a measurable set such that |E| <∞. Let fnn≥1

be a sequence of measurable functions on E, and suppose that fn is finite a.e. for each n. Show thatif fn → f a.e. on E, then fn

m−→ f .(b) Show by example that part (a) can fail if |E| =∞.

Example 3.13 (Fall 2012, #3). Suppose that f is a bounded, real-valued, measurable function on

[0, 1] such that∫ 1

0xnf(x)dx = 0 for all n ≥ 0. Show that f(x) = 0 a.e.

Example 3.14 (Spring 2008, #4). Suppose that fn ∈ C1[0, 1] for n ≥ 1, and we have:

(a) fn(0) = 0;

(b) |f ′n(x)| ≤ 1√x

a.e.; and

(c) There exists a measurable function h such that f ′n(x)→ h(x) for every x ∈ [0, 1].

Prove that there exists an absolutely continuous function f such that fn → f uniformly as n→∞.

11

4 NOTIONS OF CONVERGENCE

4.1 Modes of convergence

Pointwise a.e. Convergence: We say that fn(x) → f(x) pointwise a.e. on E if fn → f for allx ∈ E \ Z where |Z| = 0. Equvalently, µ (x : fn(x) 9 f(x)) = 0.

Convergence in Measure: We write fnµ−→ f if ∀ε > 0 ∃N such that µ (|fn − f | > ε) < ε (or < η)

for all n ≥ N . Note that a sequence being convergent in measure is the same as it being Cauchy inmeasure .

Theorem 4.1 (Pointwise convergence of a subsequence). In general, convergence in measure does not

imply pointwise convergence a.e. However, if fnµ−→ f what we can say is that there exists a subsequence

fnk which does converge pointwise to f .

Counter Example. We will demonstrate a sequence that converges in measure to zero on [0, 1], butwhich does not converge pointwise a.e. to zero. For n ≥ 1 and 1 ≤ j ≤ n, define

Sn,j :=

[j − 1

n,j

n

]⊆ [0, 1],

and set fn(x) := χSn,j(x). Let ε > 0 and observe that

`(x ∈ [0, 1] : fn(x) > ε) = `(x ∈ [0, 1] : fn(x) = 1) = |Sn,j| =1

n→ 0,

as n → ∞. So fnm−→ 0. However, given any x ∈ [0, 1] there are infinitely-many n ∈ N such that

x ∈ Sn,j. This implies that there is a subsequence fnk such that fnk

→ 1 and hence fn 9 0 pointwisea.e. on [0, 1].

Proof. Suppose that fnµ−→ f . We need to find a subsequence fnk

k≥1 such that fnk→ f a.e. on X as

k →∞. For j ≥ 1, choose Lj such that for all k ≥ Lj

µ (x : |fk − f |(x) ≥ 1/j) < 1

2j.

We can just as well assume that L1 < L2 < L3 < · · · . For j ≥ 1, we define

Ej := x : |fLj− f |(x) ≥ 1/j,

and setZ := lim sup

j→∞Ej =

⋂m≥1

⋃j≥m

Ej.

Now for m ≥ 1, we can see that

µ(Z) ≤ µ

(⋃j≥m

Ej

)≤∑j≥m

µ(Ej) <∑j≥m

2−j = 21−m → 0,

12

as m → ∞. So we conclude that µ(Z) = 0. Next, if x ∈ X \ Z, then x /∈ ∪j≥mEj for some m ≥ 1.This implies that x /∈ Ej for all j ≥ m. Thus |fLj

− f |(x) ≤ 1/j for j ≥ m, which implies thatlimj→∞ fLj

(x) = f(x) for all x ∈ X \Z. So it suffices to take fnj:= fLj

so that fnk→ f pointwise a.e.

in X.

Almost Uniform Convergence: We have almost uniform convergence when ∀ε > 0 ∃E ⊂ X such

that µ(E) < ε and fnunif−−→ f for all x ∈ X \ E, or write fn

∣∣∣Ec→ f

∣∣∣Ec

. Almost uniform convergence

implies both pointwise a.e. convergence and convergence in measure: ∀ε > 0 there is En ⊂ X suchthat µ(En) < ε/2n.

Theorem 4.2 (Egorov). Let E ⊆ Rn be measurable with |E| < ∞ and let fk, f : E → R for k ≥ 1.Furthermore, assume that limk→∞ fk(x) = f(x) a.e. in E and that f is finite a.e. in E. Let ε > 0.Then there exists a closed F ⊂ E such that |E \F | < ε and fk converges uniformly to f on F . Thatis,

supx∈F|fk(x)− f(x)| → 0,

as k →∞. This convergence is almost uniform on E.

Note that to apply Egorov on the exam, one should state all of the necessary conditions which must holdbefore we can apply it: µ(X) < ∞, fn are all measurable, and fn → f a.e.; The following statementof Lusin’s theorem follows from Egorov and a dense classs argument for L1 functions: If f : [a, b]→ C,

or equivalently if f is bounded, then ∀ε > 0 ∃ a compact set E ⊂ [a, b] such that µ(E) < ε and f∣∣∣Ec

is

continuous.

Lp-Norm Convergence: fn → f in Lp if ||fn − f ||p → 0 as n → ∞. L1, Lp-norm convergence of asequence imply convergence in measure of the same sequence: By Chebyshev we have that

µ (|fn − f | ≥ ε) ≤ 1

εp

∫|fn − f |p =

1

εp||fn − f ||pp.

Weak Convergence in a Hilbert Space: We say that xn converges weakly to x ∈ H, writtenxn

w−→ x, if 〈xn, y〉H → 〈x, y〉H for all y ∈ H.

4.2 Flowchart of modes of convergence

The following diagram is a useful summary of modes of convergence (From Dr. Heil’s notes):

13

4.3 Convergence theorems

Theorem 4.3 (Lebesgue’s BCT). Let fn be integrable ∀n ≥ 1 such that (a) fnm−→ f ; or (b) fn → f

a.e. for some measurable function f . If |fn(x)| ≤ g(x) a.e. for all n where g is integrable, then f isintegrable and

limn→∞

∫|fn − f |dµ = 0.

Theorem 4.4. If f, g are measurable such that |f | ≤ g and g is integrable, then f is integrable.

Theorem 4.5 (Monotone convergence theorem). Let fkk≥1 be a sequence of measurable functionson E. Then

(1) If fk f a.e. on E and ∃ϕ integrable such that fk ≥ ϕ a.e. in E for all k ≥ 1, then

limk→∞

∫E

fk =

∫E

f.

(2) If fk f a.e. on E and ∃ϕ integrable such that fk ≤ ϕ ∀k ≥ 1, then

limk→∞

∫E

fk =

∫E

f.

Theorem 4.6 (General form of LDCT). Let fk be measurable for all k ≥ 1. Suppose that (a)limk→∞ fk = f a.e. in E; and (b) ∃ϕ integrable such that for all k ≥ 1, |fk| ≤ ϕ a.e. in E. Then

limk→∞

∫fk =

∫f.

Sketch. For a proof sketch of the DCT, we require that the fk ≥ 0 and then use Fatou’s lemma.

Lemma 4.7 (Fatou). Suppose that E ⊆ Rn is measurable and let fk ≥ 0 be measurable for all k ≥ 1.Then

lim infk→∞

∫E

fk ≥∫E

(lim infk→∞

fk

).

We can obtain the same conclusion if we instead assume that the fk ≥ ϕ for all k ≥ 1 when ϕ isintegrable on E. Notice that the statement of Fatou’s lemma makes no a priori assumptions on theconvergence of the sequence fkk≥1.

Theorem 4.8 (Uniform convergence theorems). We have the following variants of uniform convergencetheorems:

• fn → f uniformly on [a, b] with fn all continuous =⇒ f is continuous.

• If fn is differentiable on [a, b] and limn→∞ fn(x0) exists for some x0 ∈ [a, b] and f ′n convergeuniformly on [a, b], then fn → f uniformly on [a, b] and f ′(x) = limn→∞ f

′n(x) for all x ∈ [a, b].

∗∗ If fn is integrable on [a, b] and fn → f uniformly, then f is integrable and∫[a,b]

f = limn→∞

∫[a,b]

fn.

14

5 FUBINI AND TONELLI

5.1 Statements of key theorems

Theorem 5.1 (Fubini). Let f(x, y) be integrable over I = I1 × I2. Then

i. For a.e. x ∈ I1, fx is measurable and integrable on I1. That is,∫I2

|fx(y)|dy =

∫I2

|f(x, y)|dy <∞.

ii. The function F (x) =∫I2f(x, y)dy is measurable and integrable on I1. That is,∫

I1

|F (x)|dx =

∫I1

∣∣∣∣∫I2

f(x, y)dy

∣∣∣∣ dx <∞.Moreover,

∫ ∫If(x, y)dxdy =

∫I1

[∫I2f(x, y)dy

]dx.

The necessary conditions to apply Fubini’s theorem can often be obtained by first applying Tonelli’stheorem stated below. Note that to apply Tonelli, we must require that f ≥ 0 is non-negative!

Theorem 5.2 (Tonelli). Let E = A × B where A,B are respectively measurable sets in Rn and Rm.Let f : E → [0,∞] be measurable. Then

i. For a.e. x ∈ A, fx : B → [0,∞] is measaurable.

ii. F (x) =∫Bf(x, y)dy is a measurable function of x.

iii.∫ ∫

Ef =

∫A

[∫Bf(x, y)dy

]dx.

Proof strategy: Generally speaking, one applies Tonelli on |f | to prove that f ∈ L1, then we canapply the Fubini theorem to help with reversing the order of integration. Checking the conditions ofFubini’s theorem, we see that this step is actually necessary, unless it is trivial that f ∈ L1.

5.2 List of previous exam problems

Example 5.3 (Spring 2017, #2). Let W (x) := max(1 − |x|, 0) be the hat function on [−1, 1]. Givenf ∈ L1(R), let

g(y) =

∫ ∞−∞

f(t)e−2πıytdt.

Prove that g is bounded on R and that for a.e. x we have that∫ ∞−∞

f(y)

(sinπ(x− y)

π(x− y)

)2

dy =

∫ 1

−1

g(t)(1− |t|)e2πıtxdt.

HINT: Use the fact that∫∞−∞W (t)e2πıytdt =

(sinπyπy

)2

.

15

Example 5.4 (Fall 2016, #5). Let f ∈ Lp(R) for 1 ≤ p < ∞. Given y > 0 denote Ay := x ∈ R :|f(x)| > y. Prove that ∫

R|f(x)|pdx = p

∫ ∞0

yp−1|Ay|dy.

Example 5.5 (Spring 2016, #6). Given f ∈ L1(R), define

g(x) =

∫ x

−∞f(t)dt, x ∈ R.

Given c > 0, prove that g(x+ c)− g(x) is an integrable function of x, and show that∫ ∞−∞

(g(x+ c)− g(x)) dx = c

∫ ∞−∞

f(t)dt.

Example 5.6 (Spring 2015, #5). Show that∫ ∞0

sin2 x

xe−sxdx =

ln(1 + 4/s2)

4,

for s > 0 by applying Fubini’s theorem to the function f(x, y) = e−sx sin(2xy) on E = (0,∞)× (0, 1).

Example 5.7 (Comp Prep Exam #6). Given f ∈ L1[0, 1], define g(x) =∫ 1

xf(t)tdt for 0 < x ≤ 1. Show

that g is defined a.e. on [0, 1], that g ∈ L1[0, 1], and that∫ 1

0g(x)dx =

∫ 1

0f(x)dx.

Example 5.8 (Fall 2008, #3). Let f, g be absolutely continuous functions on [0, 1]. Show that forx ∈ [0, 1] we have ∫ x

0

f(t)g′(t)dt = f(x)g(x)− f(0)g(0)−∫ x

0

f ′(t)g(t)dt.

(HINT: Consider the integral∫ ∫

Ef ′(t)g′(t)dt over the set E := (s, t) ∈ [0, x]2 : s ≤ t.)

16

6LP–SPACES

6.1 Key results and inequalities

If p <≤ ∞ then we define the Lp-norm to be

||f ||p :=

(∫|f |pdµ

)1/p

,

and when p = ∞ we set ||f ||∞ = supx∈E |f(x)|. Alternately we could define it as the essential supwhich takes x ∈ E \ Z for some |Z| = 0:

essupx∈E |f(x)| = supα : µ(|f | ≤ α) > 0.

The sequence analog to this space is

`p =

(ai)i≥1 :

(∑i

|ai|p)1/p

<∞

, p <∞,

and where we set ||(ai)||`∞ = supn |an|.Fact: If µ(X) < ∞, then Lq(µ) ( Lp(µ) for any 0 < p < q ≤ ∞. This can be proved by applying

Holder with the conjugate exponents p0 := q/p and q0 := q/(q − p) .

Proposition 6.1 (Cauchy-Schwarz and Holder inequalities). Note that we assume that the functionsf, g are both square integrable, i.e., |f |2, |g|2 are both integrable. If this is the case then,∫ 1

0

|f | ≤(∫ 1

0

|f |2)1/2

∣∣∣∣∫ fg

∣∣∣∣ ≤ (∫ |f |2)1/2(∫|g|2)1/2

.

Note that the latter equation above is the special case of the more general cases in Holder’s inequalityfor 1

p+ 1

q= 1 ⇐⇒ p+ q = pq when p = q = 2:∣∣∣∣∫ fg

∣∣∣∣ ≤ ||f ||p · ||g||q.Dual (conjugate) exponents: If p+ q = pq, then p = q/(q − 1).

Note that Minkowski’s theorem states that ||f + g||p ≤ ||f ||p + ||g||p. Minkowski also implies that∥∥∥∥∫ f

∥∥∥∥p

≤∫||f ||p.

17

Proof of Minkowski (Know This Proof Technique). We let p be arbitrary with f, g ∈ Lp and ob-serve that:

||f + g||pp =

∫|f + g|p =

∫|f + g|p−1|f + g|

≤∫|f + g|p−1(|f |+ |g|)dµ, then we apply Holder

≤ (||f ||p + ||g||p)(∫|f + g|(p−1) p

p−1

)1−1/p

= (||f ||p + ||g||p)||f + g||pp||f + g||p

.

If α > 0, we can obtain that

µ (x : |f(x)| > α) ≤(||f ||pα

)p.

Another important result which is non-trivial to prove and due to Riesz is that for 1 ≤ q < ∞ andg ∈ L1:

||g||q = sup

∣∣∣∣∫ fg

∣∣∣∣ : ||f ||p = 1

, where

1

p+

1

q= 1.

Proposition 6.2 (Completeness of normed vector spaces). A normed vector space is complete iff anyCauchy sequence converges (in E). Also, if any absolutely convergent series converges. In other words,if ||fn−fm|| → 0 as n,m→∞, then ∃f ∈ E such that ||fn−f || → 0 as n→∞. Also,

∑n≥1 ||fn|| <∞

=⇒ ∃f ∈ E such that∑

n≥1 fn = f , or equivalently∥∥∥∥∥N∑n=1

fn − f

∥∥∥∥∥→ 0, as N →∞.

To prove these completeness properties, we can appeal to the completeness of Rn or other known completespaces. For example, Lp is complete for 1 ≤ p ≤ ∞.

6.2 Dense class arguments

Compactly supported continous functions are dense in L1: Cc[a, b] = L1[a, b]. Other dense functionsin Lp are the simple (i.e., staircase) functions, and polynomials. If f ∈ Lp and C is dense in Lp then∀ε > 0, ∃g ∈ C such that ||f − g||p < ε.

Example 6.3 (Invariance under translation in L1). Given f ∈ L1, let Taf(x) = f(x − a) denotetranslation of f by a. Prove that Taf → f in the L1-norm as a→ 0, i.e., lima→0 ||Taf − f ||1 = 0.

Proof. We use a density argument noting that compactly supported continuous functions on R aredense in L1: Cc(R) = L1(R). Then for all ε > 0, ∃g ∈ Cc(R) such that ||f − g||1 < ε/2. So

||Taf − f ||1 ≤ ||Taf − Tag||1 + ||Tag − g||1 + ||g − f ||1 →ε

2+ 0 +

ε

2,

where ||Tag − g||1 → 0 since g is continuous.

18

6.3 Convex functions and Jensen’s inequality

Definition 6.4 (Convexity). The function ϕ : R→ R is convex if

ϕ(ϑt+ (1− ϑ)u) ≤ ϑϕ(t) + (1− ϑ)ϕ(u),

for all t, u ∈ R and ϑ ∈ [0, 1]. We note an important special case which is that the function ϕ(t) := tp

is convex for all p > 1.

Theorem 6.5 (Jensen’s inequality). Suppose that ϕ is convex on X. Then for all functions f : X → Rwhich are measurable and integrable on X:∫

X

ϕ(f(x))dm(x) ≥ ϕ

(∫f(x)dm(x)

).

If ϕ is convex and increasing, then

ϕ−1

(∫(ϕ f)dm

)≥∫fdm.

For simple functions of the form f :=∑n

i=1 ciχEi, i.e., where ϕ f =

∑ni=1 ϕ(ci)χEi

, we obtain thecorollary that ∫

φ f =n∑i=1

ϕ(ci)m(Ei) ≥ ϕ

(n∑i=1

cim(Ei)

).

6.4 Proofs of standard results from Folland

Theorem 6.6. If 0 < p < q < r ≤ ∞, then Lq ⊂ Lp + Lr.

Proof. We first note that f = fχf≤1 + fχf>1. Let f ∈ Lq. Define E := x : |f(x)| > 1 and setg := fχE, h := fχEc . Then

|g|p = |f |pχE ≤ |f |qχE =⇒ g ∈ Lp

|h|r = |f |rχEc ≤ |f |qχEc =⇒ h ∈ Lr,

with f = g + h.

Theorem 6.7. If 0 < p < q < r ≤ ∞, then Lp ∩ Lr ⊂ Lq and

||f ||q ≤ ||f ||λp ||f ||1−λr ,

where λ = (q−1 − r−1)/(p−1 − r−1), i.e., where λ ∈ (0, 1) is defined by 1q

= λ · 1p

+ (1− λ)1r.

Proof. If r =∞, |f |q ≤ ||f ||q−p∞ |f |p with λ = p/q so that

||f ||q ≤ ||f ||p/qp · ||f ||1−p/q∞ = ||f ||λp ||f ||1−λ∞ .

If r <∞, then we can use Holder’s inequality with the conjugate exponents pqλ

and rq1−λ :∫

|f |q =

∫|f |λq|f |(1−λ)q ≤ || |f |λq ||p/λq||f (1−λ)q||rq/(1−λ)

=

(∫|f |p)λq/p(∫

|f |r)(1−λ)q/r

= ||f ||λqp · ||f ||(1−λ)qr .

Taking qth roots implies the result in this case as well.

19

Theorem 6.8. If 0 < p < q ≤ ∞ and A is any set, then `p(A) ⊂ `q(A) and ||f ||q ≤ ||f ||p (in `∗)

Proof. First, ||f ||p∞ = supα |f(α)|p ≤∑

α |f(α)|p =⇒ ||f ||∞ ≤ ||f ||p (in `∗). The case of q <∞ followsfrom the previous if λ = p/q:

||f ||q ≤ ||f ||λp ||f ||1−λ∞ ≤ ||f ||p.

Theorem 6.9. If µ(X) <∞ and 0 < p < q ≤ ∞, then Lq ⊂ Lp and ||f ||p ≤ ||f ||qµ(X)1/p−1/q.

Proof. If q =∞ then the result is obvious:

||f ||pp =

∫|f |p ≤ ||f ||p∞

∫1 = ||f ||p∞µ(X).

If q <∞, then we can use Holder’s inequality with the conjugate exponents q/p and q/(q − p):

||f ||pp =

∫|f |p · 1 ≤ || |f |p ||q/p · ||1||q/(q−p) = ||f ||pqµ(X)(q−p)/q.

6.5 Previous exam problems

Example 6.10 (Fall 2016, # 4). Let 1 ≤ p < q ≤ ∞ and f ∈ Lp ∩ Lq. Show that f ∈ Lr for allp ≤ r ≤ q.

Proof. First, suppose that 1 ≤ p < r < q < ∞ and write r = tp + (1 − t)q for 0 < t < 1. Set theconjugate exponents u = 1/t and u′ = u/(u− 1) = 1/(1− t). Then this implies that∫

|f |r =

∫|f |tp|f |(1−t)q ≤

(∫|f |tpu

)1/u(∫|f |(1−t)qu′

)1/u′

=

(∫|f |p)t(∫

|f |q)1−t

<∞.

Next, for 1 ≤ p < r < q =∞:∫|f |r =

∫|f |p|f |r−p ≤ ||f ||r−p∞

∫|f |p <∞.

The cases where r = p or r = q are trivial.

Example 6.11 (Fall 2007, #2). Let (X,M, µ) be a measure space, let µ be a positive measure, andlet fn, f ∈ L1(X,M, µ) for n ≥ 1. Assume that:

(1) fn(x)→ f(x) for a.e. x ∈ X;

(2) ||fn||1 → ||f ||1.

Prove that ||fn − f ||1 → 0 as n→∞.(HINT: Fatou’s lemma.)

Example 6.12 (Spring 2010, #5). Let (X,M, µ) be a measure space and let f, fn ∈ Lp where 1 ≤p <∞. Prove that if fn → f a.e., then ||fn − f ||p → 0 iff ||fn||p → ||f ||p.

20

Slick Proof. The forward direction is easy by Minkowski: 0 ≤ | ||fn||p − ||f ||p | ≤ ||fn − f ||p. For thereverse direction, we must show that ||fn||p → ||f ||p =⇒ ||fn − f ||p → 0. We note that the function| · |p is always convex when p ≥ 1. Then

|fn − f |p = |fn + (−f)|p = 2p ·∣∣∣∣12fn +

1

2(−f)

∣∣∣∣ , then by convexity

≤ 2p(|fn|p

2+|f |p

2

).

Now noticing that || · ||p ≥ 0 where we have that ||fn||p → ||f ||p a.e., we will immediately try Fatou’slemma:∫

X

limn→∞

2p(|fn|p

2+|f |p

2− |fn − f |p

)dµ ≤ lim inf

n→∞

∫X

2p(|fn|p

2+|f |p

2− |fn − f |p

)dµ

= lim infn→∞

(∫X

2p−1|fn|pdµ+

∫X

2p−1|f |pdµ)

− lim supn→∞

∫X

|fn − f |pdµ

= lim infn→∞

(2p−1||fn||pp + 2p−1||f ||pp

)− lim sup

n→∞2p · ||fn − f ||pp.

This implies by subtraction that

0 ≤ lim supn→∞

∫X

|fn − f |pdµ ≤ 0.

Alternate “Obtainable” Proof. The forward direction is easy and uses the (reverse) triangle inequality.The reverse direction uses Egorov which can be applied in this case since informally f ∈ Lp impliesthat “f lives on a compact set”. In particular, let ε > 0 be fixed. We can choose a bounded set E suchthat ||fχEc ||p < ε. So by Egorov, we can choose a measurable F ⊆ E so that fn(x)→ f(x) uniformlyon F and where ||fχF c||p < 2ε. Then we have that∫

E

|fn|pdx→∫E

|f |pdx ≥ ||f ||pp − 2ε.

Lastly, by hypothesis we must have that

lim supn→∞

∫Ec

|fn|pdx < 2ε,

so that

lim supn→∞

∫Ec

|fn|p ≤∫|fn|p −

∫Ec

|fn|p

→ ||f ||pp − ||fχEc ||pp ≤ ||f ||pp − 2ε.

Example 6.13 (Fall 2015, #7). Fix a finite measure space (X,M, µ) and 1 ≤ p < q ≤ ∞. Show thatLp 6 ⊂Lq iff X contains sets of arbitrarily small positive measure. (HINT: For the “if” implication:construct a disjoint sequence En with 0 < µ(En) < 2−n, and consider f =

∑n anχEn for suitable

constants an.)

Proof. [ =⇒ ]: Suppose first that there exists f ∈ Lp \ Lq. Next, we construct the following sets forn ≥ 1: En := x ∈ X : |f(x)| > n. Then by Chebyshev we see that

µ(En) ≤(||f ||pn

)p→ 0,

21

as n → ∞. So it suffices to prove that µ(En) > 0 for all n ≥ 1. Suppose that µ(En) = 0 for somen. Then if |f | ≤ n a.e., i.e., f ∈ L∞ ⊆ Lq: ||f ||qq =

∫|f |q ≤ nq · µ(X) < ∞. But this contradicts the

assumption that f /∈ Lq.

[⇐=]: We inductively construct measurable sets Fn such that 0 < µ(F1) < 1/2 and 0 < µ(Fn) <µ(Fn−1)/4 for n ≥ 2. If we then set En := Fn \∪n−1

j=1Fj then the En are disjoint and 0 < µ(En) < 2−n.

Suppose first that q <∞ and consider f =∑

n anχEn for an := 1/µ(En)1/q. Then if we set λ = 1− p/qwe have that

||f ||pp =∑n

|an|pµ(En) =∑n

µ(En)λ <∑n

1

2λn<∞,

which shows that f ∈ Lp. On the other hand,

||f ||qq =∑n

|an|qµ(En) =∑n

1 = +∞,

which shows that f /∈ Lq. If q = ∞, we can define f =∑

n anχEn with an :=(

1µ(En)

) 12p → ∞. Then

f ∈ Lp as before, while f = an on En. Since µ(En) > 0, we see that ||f ||∞ =∞.

Example 6.14 (Spring 2011, #7). (a) Assume that µ is a finite measure on a measurable space(X,M). With q ∈ (1,∞), let fkk≥1 ⊂ Lq(X) and f ∈ Lq(X) be given. Suppose also that

• supk≥1 ||fk||Lq(X) <∞; and

• fk → f for µ-a.e. on X.

Prove that fk → f in Lp(X) for all p ∈ [1, q).(b) Is the statement in (a) still true if µ is only assumed to be σ-finite? Justify your answer.

Proof of (a). First, note that by Fatou we have that

0 ≤ ||f ||q =

(∫lim infn→∞

|fn|q)1/q

≤ lim infn→∞

||fn||q ≤ supn≥1||fn||q <∞,

which implies (again) that f ∈ Lq(X). So we consider the functions gk := fk − f , which are uniformlybounded and converge to zero a.e. in X. We need to show that ||gk||p → 0 for all p ∈ [1, q). Now wetake the conjugate exponents p0 := q/p and q0 := q/(q− p), i.e., so that 1/p0 + 1/q0 = 1, in Holder andapply Chebyshev in the forllowing forms for m ≥ 1:∫X

χ|gk|>m|gk|pdµ ≤ µ(|gk| > m) · || |gk|p ||q/p ≤

1

mq−p

(∫X

|gk|q) q−p

q

·(∫

X

|gk|q)p/q

=1

mp−q ||gk||qq.

Next, we define

gk :=

gk, |gk| ≤ m;

0, otherwise.

Then |gk| ≤ m pointwise a.e. in X and we are given that gk → 0 a.e. in X. Now by the DCT andMinkowski’s inequality, we obtain that

limk→∞||gk||p = lim

k→∞

(∫X

χ|gk|>m|gk|p +

∫X

|gk|p)1/p

≤ limk→∞||χ|gk|>m|gk|

p||p +∥∥∥ limk→∞

gk

∥∥∥≤ 1

mp−q ||gk||qq −→ 0,

as each of k,m→∞.

22

Example 6.15 (Spring 2014, #3). Let fn be a sequence in Lp(Ω) for 1 ≤ p ≤ ∞, and assume thatf ∈ Lp(Ω) is such that ||fn − f ||p → 0. Show that there exists a subsequence fnk

and a functionh ∈ Lp(Ω) such that

(i) fnk→ f a.e. on Ω; and

(ii) For every k ≥ 1, we have that |fnk(x)| ≤ h(x) a.e. on Ω.

Proof. The conclusion is apparently obvious when p = ∞, so we assume that 1 ≤ p < ∞. Since fnis Cauchy in Lp(Ω) by hypothesis, we have a subsequence fnk

such that

||fnk+1− fnk

|| ≤ 2−k, ∀k ≥ 1.

From this point on, we simplify notation by calling fnkinstead by just fk. Let

gn(x) :=n∑k=1

|fk+1(x)− fk(x)|,

and observe that by computation ||gn||p ≤ 1. Now since gn(x) ≤ gn+1(x), the MCT implies that gn(x)tends to a finite limit g(x) a.e. on Ω. We also know that g ∈ Lp(Ω). On the other hand, for anym ≥ n ≥ 2, we have that

|fm(x)− fn(x)| ≤ |fm − fm−1|(x) + · · ·+ |fn+1 − fn|(x) ≤ g(x)− gn−1(x) −→ 0,

as n → ∞. Thus a.e. on Ω, fn(x) is Cauchy and therefore converges to some finite limiting f ∗(x).We notice that a.e. on Ω:

|f ∗(x)− fn(x)| ≤ g(x), for n ≥ 2.

In particular, this shows that f ∗ ∈ Lp(Ω). Finally, since

|fn(x)| ≤ g(x) + |f ∗(x)| =: h(x),

where h ∈ Lp(Ω) since Lp is closed under addition, the DCT implies that fk → f ∗ in Lp(Ω). Thisimplies that f = f ∗ a.e., completing the proof.

Example 6.16 (Spring 2015, #3). Let f ∈ Lp[0,∞]. Show that for 1 < p <∞ we have

limx→∞

1

x1−1/p

∫ x

0

f(t)dt = 0.

6.6 Proofs of results from the comp prep homework

Example 6.17 (Problem #3). Given a measurable set E ⊆ Rd with |E| < ∞, prove that for eachmeasurable function f on E we have that ||f ||p → ||f ||∞ as p→∞.

Proof. Suppose that f is bounded. Then

||f ||p ≤ ||f ||∞|E|1/p → ||f ||∞,

as p→∞. Secondly, we notice that for any ε > 0

||f ||p ≥ ||fχ|f |>||f ||∞−ε||p ≥ (||f ||∞ − ε)||f | > ||f ||∞ − ε|1/p → ||f ||∞ − ε.

Note that somewhat in contrast to the result proved above, we have that if f ∈ `p and 1 < p < ∞,then ||f ||∞ ≤ ||f ||p.

23

Example 6.18 (Problem #4). Let 1 ≤ p ≤ ∞ be given. Suppose that ϕ is a measurable function on Rsuch that fϕ ∈ Lp(R) for every f ∈ Lp(R). Prove that ϕ ∈ L∞(R), i.e., we must prove that ϕLp ⊆ Lp.

Proof. We argue by contradiction. Suppose that ϕ /∈ L∞. Fix a sequence of integers kj so that the sets

Ej := x : 2kj < ϕ(x) < 2kj+1,

have positive measure. Set fj := χEj/|Ej|1/p. Then ||fj||p = 1 and ||ϕfj||p ≥ 2kj . Take f :=

∑j≥1 fj/j

2

where f ∈ Lp by the triangle inequality. But we can compute that ϕf /∈ Lp. This yields our desiredcontradiction.

Example 6.19 (Problem #5). Let E ⊆ Rd be measurable and fix 1 < p <∞. Assume that fn ∈ Lp(E),fn → f a.e., and sup ||fn||p <∞. Prove that f ∈ Lp(E), and if g ∈ Lp′(E) (1/p+ 1/p′ = 1) then

limn→∞

∫E

fng =

∫E

fg.

Does the same result hold if p = 1?

Counter example when p = 1. Let fn := χ[n,n+1) so that fn → 0 a.e., but where ||fn||p = 1 for all p.Let g ≡ 1 so that g ∈ L∞ when |E| < ∞. Then

∫fng = 1, but

∫fg = 0. Thus the hypothesis fails

when p = 1.

Proof. Guided by two examples, namely,

(a) fn = χ[n,n+1) → 0 a.e., but ||fn||p = 1 ∀p ∀n;

(b) fn = n1/pχ[0,1/n) → 0 a.e., but ||fn||p = 1 for all n,

we argue that f ∈ Lp. By Fatou’s lemma:(∫|f |p)1/p

=

(∫| lim inf fn|p

)1/p

≤ lim inf

(∫|fn|p

)1/p

≤ sup ||fn||p < +∞.

Take 1 < p < ∞ and g ∈ Lp′ . We argue that∫fng →

∫fg. The key point in our argument is the

reduction to an integral over a finite-measure set. In particular, let ε > 0 and pick |E0| <∞ such thatg is bounded on E0 so that ∫

Ec0

|g|p′ <(

ε

sup ||fn||p

)p′,

and on which fn → f uniformly (by Egorov). By uniform convergence and the boundedness of g onE0,

∫E0fng →

∫E0fg, and by Holder∣∣∣∣∣

∫Ec

0

fng

∣∣∣∣∣ ≤ ||fn||p||gχEc0||p′ < ε.

The same inequality holds for f .

24

7 SIGNED MEASURES

7.1 Definitions and key theorems

Definition 7.1 (Signed measures). Let (X,M) be a measurable space. A signed measure on (X,M)is a function v : M → [−∞,∞] such that

(1) v(∅) = 0;

(2) v assumes at most one of the values ±∞;

(3) If Ej is a sequence of disjoint sets in M , then v (∪jEj) =∑

j v(Ej), where the latter sumconverges absolutely if v (∪jEj) is finite.

Notation: (Mutual signularity of measures) We write u ⊥ v if ∃E,F ∈ M such that E ∩ F = ∅,E ∪ F = X and E is null for u, F is null for v. Note that we say a set E is NULL for u if u(E) = 0.

Examples of signed measures:

• If u1, u2 are measures onM and at least one of them is finite, then v = u1−u2 is a signed measure.

• If u is a measure on M and f : X → [−∞,∞] is a measurable function such that at least one of∫f±du is finite, then the set function defined by v(E) =

∫Efdu is a signed measure.

Theorem 7.2 (Hahn decomposition). If v is a signed measure on (X,M), then ∃P,N ⊂ X such that

(1) P ∪N = X and P ∩N = ∅;

(2) P is positive for v: ∀E ∈M such that E ⊆ P , v(E) ≥ 0;

(3) N is negative for v: ∀E ∈ N such that E ⊆ N , v(E) ≤ 0.

Moreover, this composition is essentially unique in that if (P ′, N ′) are another such pair, then everyM-measurable subset of P∆P ′, N∆N ′ has measure zero.

Theorem 7.3 (Jordan decomposition). Let (P,N) be a Hahn decomposition for the signed measureu. Then u has a unique decomposition into u+ − u− of positive measures u±, at least one of which isfinite, such that u+(E) = 0 for all measurable E ⊆ N and for all measurable E ⊆ P , u−(E) = 0.

Definition 7.4. The total variation of the signed measure u is given by |u| = u+ + u−. The signedmeasure u is absolutely continuous with respect to v, written u v if for every measurable A, v(A) =0 =⇒ u(A) = 0.

Theorem 7.5 (Radon-Nikodym). Let (X,Σ) be a measurable space on which two σ-finite measuresu, v are defined (i.e., u(X) < ∞). If u v then ∃ measurable f : X → [0,∞) such that for allmeasurable A ⊆ X:

u(A) =

∫A

fdv.

25

Theorem 7.6 (Lebesgue-Radon-Nikodym). Let v be a σ-finite signed measure and u a σ-finite positivemeasure on (X,M). Then there exists unique σ-finite signed measures λ, ρ on (X,M) such that λ ⊥ ρ,ρ u, and v = λ + ρ. Moreover, there is an extended u-integrable function f : X → R such thatdρ = fdu, and any two such functions are equal u-a.e.

Consequences:

• v u =⇒ dv = fdu for f a real extended u-integrable function;

• d|v| = |f |du and∫|f |du =

∫d|v| = |v|(X) = ||v|| <∞;

• u u+ v;

• v+(E) = v(E ∩ P ), v−(E) = −v(E ∩N) (the sign here is important);

• Good sets to take to find quick examples include: E1 := E ∩ P , E2 := E ∩N .

7.2 Practice problems

Example 7.7 (Fall 2017,#1). Let v be a signed Borel measure on R. Prove that

|v|(E) = sup

∣∣∣∣∫E

fdv

∣∣∣∣ : |f | ≤ 1

.

Example 7.8 (Spring 2017,#8). Let X be a set and M be a sigma algebra of subsets of X. Supposethat (X,M, u) and (X,M, v) are two finite measure spaces with u a positive measure and v a signedmeasure. Prove that the following two statements are equivalent:

(a) v u, i.e., if E ∈M satisfies u(E) = 0 then v(E) = 0;

(b) For every ε > 0 there exists some δ > 0 such that if E ∈M satisfies u(E) < δ, then v(E) < ε.

Notes on (b) =⇒ (a). Fix ε > 0. If this were not true then ∀δ > 0, u(E) < δ =⇒ v(E) ≥ ε. Nowfor n ≥ 1, choose En so that u(En) < 2−n and define E := ∩m≥1 supj≥mEj. Then for any fixed m ≥ 1,

u(E) ≤ u(∪j≥mEj) ≤ 21−m → 0,

as m → ∞. So we conclude that u(E) = 0. On the other hand, if we define Fm := ∪n≥mEn, thenFm ⊇ Fm+1 for all m, which implies by a continuity theorem that

v(∩m≥1Fm) = limm→∞

v(Fm) ≥ ε.

Indeed, since for all m ≥ 1, we have that

v(Fm) = v(∪n≥mEn) ≥ v(Em) ≥ ε,

by assumption. We thus arrive at a contradiction.

Example 7.9 (Fall 2016,#6). Let u and v be two σ-finite positive measures on a measure space (X,M).Show that there exists a measurable function f : X → R such that for each E ∈M,∫

E

(1− f)du =

∫E

fdv.

Does the above statement hold for every pair of signed finite measures u, v?

26

Example 7.10 (Spring 2014, #7). Let v be a bounded signed Borel measure on R, and assume thatf ∈ L1(R). Prove that

g(x) :=

∫ ∞−∞

f(x− y)dv(y),

is defined at almost every x ∈ R, and that

||g||1 ≤ ||f ||1 · |v|(R),

where |v| is the total variation of v. Show further that if f is uniformly continuous on R, then so is g.

Example 7.11 (Spring 2008, #6). Let v be a signed Borel measure on I = [0, 1] such that |v|(I) = 1and v(I) = 0. Suppose that there exists a continuous function f : I → R such that ||f ||∞ ≤ 1 and∫ 1

0

fdv = 1.

Show that the Lebesgue measure on I is not absolutely continuous with respect to |v|.

7.3 Extra problems from Folland

Example 7.12. Suppose that v is a signed measure and that λ, µ are positive measures such thatv = λ− µ. Prove that λ ≥ v+ and µ ≥ v−.

Proof. Let X = P ∪N be the Hahn decomposition for v on X and suppose that E ⊆ X is v-measurable.Then

v+(E) = v(E ∩ P ) = λ(E ∩ P )− µ(E ∩ P ) ≤ λ(E ∩ P ) ≤ λ(E)

v−(E) = −v(E ∩N) = µ(E ∩N)− λ(E ∩N) ≤ µ(E).

Example 7.13. Suppose that v is a signed measure on (X,M) and that E ∈M. Prove that

|v|(E) = sup

n∑j=1

|v(Ej)| : E1, . . . , En are disjoint, and E =n⋃j=1

Ej

.

Example 7.14. Let u, v be signed measures. Prove the following:

(A) v ⊥ u iff |v| ⊥ u iff v+, v− ⊥ u;

(B) v u iff |v| u iff v+, v− u.

27

8 A FEW KEY EXAMPLES

Note: See Dr. Lacey’s handout and key notes. This section is a summary of the key examples fromthat write-up.

∗∗ Pointwise convergence to zero, but the norm is bounded: Take fn := χ[n,n+1). Then fn → 0pointwise a.e., but ||fn||p = 1 for any 0 < p ≤ ∞.

∗∗ Pointwise convergence to zero, but the norm is bounded (Compact case): Fix 1 ≤ p <∞.Take fn := n1/pχ[0,1/n). Then fn → 0 pointwise a.e., but ||fn||p = 1.

∗∗ Convergence to zero in Lp, |fn| ≤ g ∈ Lp, but do not have pointwise convergence to

zero: For an integer j such that 2j ≤ n < 2j+1, define fn := χ[n−2j

2j,n−2j+1

2j

] so that fnµ−→ 0, but

fn 9 0 since lim sup fn = 1 and lim inf fn = 0.

• Sequence of functions that converge weakly to zero, but do not themselves converge:Here, for example, converging weakly to zero means∫

fngdx→ 0, g ∈ L1.

The basic example on [0, 1] is fn := sin(nx)w−→ 0 ⇐⇒ 〈fn, g〉 → 0 ∀g ∈ L1 (or for all g ∈ L2, etc.).

• A sequence of sets in An ⊂ [0, 1] with∑

n |An| = ∞, but where lim supnAn has measurezero: Take An := [0, 1/n) and apply the Borel-Cantelli lemma.

• A measurable E ⊂ R with positive measure that does not contain an interval: Any “fat”Cantor set will work here. (Any fat Cantor set is nowhere dense.)

• A uniformly continuous function on [0,∞) that is unbounded: Take f(x) :=√x+ 1.

• An element of `2(N) that is not in `1(N): Select an := n−3/4, let’s say. Then ||an||`1 =∑n n−3/4 = ∞, but ||an||`2 =

∑n n−3/2 < ∞. Likewise, for any 0 < p < q ≤ ∞, there are

sequences in `q \ `p.

• A function in L1[0, 1) that is not in L2[0, 1): Choose f(x) := 1/√x. Then ||f ||L1 =

∫ 1

0dx√x

= 2,

but ||f ||2L2 =∫ 1

0dxx

=∞. Likewise, for any 0 < p < q ≤ ∞, there are functions in Lp[0, 1)\Lq[0, 1).

• Sequence of continuous functions which converge to a non-continuous function: Con-sider the sequence fn(x) := xn on [0, 1]. Then fn → χ1, which is not a continuous function onthe full compact interval.

• A continuous function on (0, 1] that is not uniformly continuous: Take f(x) := sin(1/x)as it will work well here.

• A continuous function on [0, 1] that is not Lipschitz on any subinterval: These standardexamples of so-called “pathological” continuous functions are shown to exist by Baire, but are noteasy to explicitly construct by hand.

• fn := n · χ[0, 1n

], and fn := 1nχ[0,n].

28

9 FUNCTIONAL ANALYSIS

9.1 Banach spaces

A Banach space is a complete normed vector space over some field F. A Hilbert space is a particulartype of Banach space endowed with an inner product.

9.2 Linear transformations

The set Y ⊂ X is a linear subspace (over F) if ∀λ, µ ∈ F and ∀x, y ∈ Y , we have that λx+µy ∈ Y . Theintersection of linear subspaces is always a subspace. A linear transformation satisfies T (λ1x1 +λ2x2) =λ1T (x1) + λ2T (x2).

Proposition 9.1 (Continuity of a linear transformation). Let T : X → Y be a linear transformation.Then TFAE:

(1) T is continuous;

(2) T is continuous at zero;

(3) T is continuous at one point;

(4) T is bounded;

(5) T is Lipschitz.

Proof Sketch. For (1)–(3): T continuous at zero ⇐⇒ ∀xn → 0, T (xn) → 0 (where T (0) = 0 since Tis a linear transformation). Then ∀x∗ ∈ X such that xn → x∗, xn − x∗ → 0 =⇒ T (xn − x∗)→ 0.

For (4): If T is continuous, we have that ∀ε = 1 ∃δ > 0 such that ||x|| < δ =⇒ ||Tx|| < 1=⇒ ||T || < 1/δ.

For (5): Consider that ||Tx− Ty|| ≤ ||t|| · ||x− y||.

9.3 Bounded linear operators

Definition 9.2 (Bounded linear operators). We say that T : X → Y is a bounded linear operator if∃K such that ∀x ∈ X: ||Tx|| ≤ K · ||x|| (where we note that ||Tx|| ≤ ||T || · ||x|| always). In this casewe write T ∈ B(X, Y ). Note that

||T || = lubx 6=0||Tx||||x||

= lub||x||=1 ||Tx||.

Proposition 9.3 (Completeness of B(X, Y )). The space B(X, Y ) is a Banach space endowed with thepointwise operations: (T1 + T2)(x) = T1(x) + T2(x) and (λT )(x) = λT (x).

29

Proof. We need to show that B(X, Y ) is complete when Y is itself a Banach space. To show this we usethe completeness of Y . Suppose that Tj is Cauchy, which implies that ||Tj − Tk||B → 0 as j, k →∞.By the completeness of Y , Tj(x)→ T (x). Now we see that

||(Tj − T )(x)|| = limk→∞||Tjx− Tkx|| ≤ lim sup

k→∞||Tj − Tk|| · ||x|| −→ ∞.

The following theorem called the Banach-Steinhaus theorem is also refered to as the principle of uniformboundedness.

Theorem 9.4 (Banach-Steinhaus). Let X be a Banach space and let Y be a normed linear space.Let Tα be a family of bounded linear operators from X into Y . If for each x ∈ X the set Tαx isbounded, then the set ||Tα|| is bounded.

Theorem 9.5 (Open mapping theorem). Let X and Y be Banach spaces, and let T : X → Y be abounded linear map. Then T maps open sets of X onto open sets of Y .

Proposition 9.6. Let X and Y be Banach spaces and let T ∈ B(X, Y ) be a bijective linear map. ThenT−1 is a bounded linear map.

9.4 Hilbert spaces

Definition 9.7 (Hilbert spaces). A Hilbert space is a set X satisfying:

(a) X is a vector space over some field F (typically C or R);

(b) X possesses an inner product, 〈·, ·〉 : X ×X → F, that satisfies:

i. 〈x, y〉 = 〈y, x〉, ∀x, y ∈ X. Note that this implies that 〈x, x〉 ∈ R ∀x.

ii. 〈x, x〉 ≥ 0 and 〈x, x〉 = 0 ⇐⇒ x = 0.

iii. For all x, y ∈ X and ∀λ ∈ F, 〈λx, y〉 = λ〈x, y〉. Note that this implies 〈x, λy〉 = λ〈x, y〉.iv. For all z, y, z, u ∈ X: 〈x, y + z〉 = 〈x, y〉+ 〈x, z〉 and 〈x+ u, y〉 = 〈x, y〉+ 〈u, y〉.

(c) X is complete in the sense of normed spaces.

One consequence of the above is that X is normed in the sense that ||x|| :=√〈x, x〉 is a norm.

Examples of Hilbert spaces include: L2(X) IS a Hilbert space as with 〈f, g〉 :=∫Xf(x)g(x)dx, 〈f, f〉 =∫

X|f |2 (and linearity is easy to prove). However, Lp(X) for p 6= 2 is NOT a Hilbert space as it does

not satisfy the polarization identity.

Lemma 9.8 (Cauchy-Schwarz). Let X be a Hilbert space. For all x, y ∈ X, |〈x, y〉| ≤ ||x|| · ||y||.

Proof (Real Case). First, we observe that

||x+ λy||2 = ||x||2 + 2λ〈x, y〉+ λ2||y||2.

Then we complete the square as

||x+ λy||2 =

(λ||y||+ 〈x, y〉

||y||

)2

+ ||x||2 − 〈x, y〉2

||y||2,

which cannot have two roots which then implies that

||x||2 − 〈x, y〉2

||y||2≥ 0,

from which the claimed result follows.

30

Note that we also have Cauchy-Schwarz in Rn:(n∑i=1

uivi

)2

≤

(n∑i=1

u2i

)(n∑i=1

v2i

).

Proposition 9.9 (Polarization identities). We are concerned with the real case here. In particular,for H any Hilbert space over R we have that ∀x, y ∈ H: ||x + y||2 + ||x− y||2 = 2(||x||2 + ||y||2). Thepolarization identity also defines an inner product: 〈x, y〉 = 1

4(||x+ y||2 − ||x− y||2), ∀x, y ∈ H.

Definition 9.10 (Orthogonal sets and completeness). Let X be a Hilbert space. We say that x, y ∈ Xare orthogonal if 〈x, y〉 = 0. The family xαα∈I is an orthogonal set ⇐⇒ 〈xα, xβ〉 = 0 ∀α 6= β. Anorthonormal set is an orthogonal set such that 〈xα, xα〉 = 1 for all α ∈ I.

An orthogonal set is called complete if 〈xα, y〉 = 0 ∀α ∈ I =⇒ y = 0. An example of a completeorthogonal set is the basis exp(2πıkϑ)k∈Z in L2

per(0, 1).

Lemma 9.11 (Least squares minimization). Let xiNi=1 be an orthonormal set. Then the difference∥∥∥x−∑Ni=1 λixi

∥∥∥ is minimized by taking λi := 〈y, xi〉.

Corollary 9.12 (Bessel’s inequality). Let xi1≤i≤N be an orthonormal sequence. Then

n∑i=1

|〈xi, y〉|2 ≤ ||y||2.

Theorem 9.13 (Orthogonal complement theorem). Let X be a Hilbert space and E ⊂ X be a closed(i.e., complete) subspace. Then ∀x ∈ X ∃x∗ ∈ E, x⊥ such that x = x∗ + x⊥ and 〈x⊥, y〉 = 0 for ally ∈ E. Moreover, we can prove that

x∗ = argminy∈E ||x− y||.

Example 9.14 (A linear space that is not closed). In `2(N) consider E := x : ∃N such that xj =0∀j > N. Then E is a linear space, but is NOT closed as there is no minimum to the followingexample: for x := (1, 1/2, 1/3, 1/4, . . .) the norm ||x− (1, 1/2, . . . , 1/N, 0, 0, . . .)|| can be made as smallas you wish by taking N large enough.

Proposition 9.15 (Parseval’s identity). If ei is complete and orthonormal, then∥∥∥x−∑N

i=1 λiei

∥∥∥can be made as small as possible by taking λi := 〈ei, x〉ei and letting N →∞. In particular,

||x||2 = limN→∞

N∑i=1

|〈ei, x〉|2.

9.5 Practice problems

Note that this section is intentionally underdeveloped. See the previous comprehensive exams for par-ticular examples.

31

10 MISCELLANEOUS OTHER NOTES

10.1 Other problems

Example 10.1 (Dr. Lacey). Let ej be the unit basis in `2(N). Fix v ∈ `2(N). Consider the linearoperator T defined such that Tej = v for all j ≥ 1 extended linearly. Note that since the ej’s spab `2,this requirement completely determines T . Prove that T is bounded iff v = 0.

Proof 1. The norm of e1 + · · ·+en is√n in `2 whereas the norm of T (e1 + · · ·+en) is n · ||v||. Therefore

||T || ≥√n · ||v|| for all n. This implies that if v 6= 0 then the norm of T is +∞.

Proof 2. Let x :=∑

j≥1 ej/j ∈ `2. Then Tx = v∑

j≥1 1/j. This implies that Tx = +∞ wheneverv 6= 0 where Tx = 0 when v = 0.

Example 10.2 (Spring 2016, #5). Let f be monotone increasing on [0, 1]. Prove that∫ 1

0f ′(x)dx ≤

f(1)− f(0).

Proof. Let gn(x) := (f(x+ 1/n)− f(x))/(1/n). Then for ε ∈ (0, 1/4):∫ 1−ε

ε

f ′(x)dx =

∫ 1−ε

ε

limn→∞

gn(x)dx, and∫ 1−ε

ε

f(x+ 1/n)−∫ 1−ε

ε

f =

∫ 1−ε+1/n

ε+1/n

f −∫ 1−ε

ε

f

=

∫ 1−ε+1/n

ε+1/n

f −∫ ε+1/n

ε

f,

which implies that ∫ 1−ε

ε

f(x+ 1/n)−∫ 1−ε

ε

f ≤ 1

n(f(1)− f(0)).

Now since f is monotone and non-decreasing gn ≥ 0, so by Fatou:

f(1)− f(0) ≥ lim infn→∞

∫ 1−ε

ε

f(x+ 1/n)− f(x)

1/n≥∫ 1−ε

ε

f ′.

Example 10.3 (Fall 2016, #3). Let E1, . . . , En be Lebesgue measurable subsets of [0, 1] and define

Sq := x ∈ [0, 1] : x belongs to at least q of the sets Ei.

Show that for each 1 ≤ q ≤ n, Sq is Lebesgue measurable, and there exists k such that

q|Sq|n≤ |Ek|.

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Proof. Let f :=∑n

k=1 χEk, which is a finite sum of measurable functions and hence is measurable.

ThenSq = x ∈ [0, 1] : f(x) ≥ q,

and so since f is measurable, it follows that Sq is measurable for any 1 ≤ q ≤ n. Moreoverm byChebyshev’s inequality, we see that

q · |Sq| ≤∫

[0,1]

f(x)dx =

∫[0,1]

n∑k=1

χEk(x)dx =

n∑k=1

|Ek| ≤ n ·(

max1≤k≤n

|Ek|).

This implies thatq|Sq|n≤ max

1≤k≤n|Ek| =: |EK |,

for some K ∈ 1, 2, . . . , n. This inequality is true for any 1 ≤ q ≤ n.

Example 10.4 (Fall 2015, #1). Let f ∈ L1(R) and let α > 0. Show that

limn→∞

f(nx)n−α = 0, for a.e. x ∈ R.

Proof. Let fn(x) := f(nx)n−α for n ≥ 1, Then by a substitution, we have that∫R|fn(x)|dx =

1

nα+1

∫|f(x)|dx.

It follows that∞∑n=1

∫R|fn(x)|dx =

∫R|f(x)|dx×

∑n≥1

1

nα+1= ||f ||1 · ζ(α + 1) <∞, (∗)

since f ∈ L1 and α + 1 > 1. Now since |fn(x)| ≥ 0 for all n and for all x ∈ R, by the monotoneconvergence of the sequence of partial sums on the LHS of (∗), we see that also∫

R

∑n≥1

|fn(x)| <∞ =⇒∑n≥1

|fn(x)| <∞,

and so necessarily we must have that limn→∞ |fn(x)| = 0.

Example 10.5 (Spring 2013, #8). Let (X,M, µ) be a measure space, and for n ≥ 1 let fn, f, gn, g be

measurable complex-valued functions on X such that fnµ−→ f and gn

µ−→ g.

(a) Show that fn + gnµ−→ f + g; and

(b) Show that fngnµ−→ fg if µ(X) <∞, but not necessarily if µ(X) =∞.

Proof of (b). Suppose that µ(X) <∞ and for the sake of contradiction that fngn 6µ−→fg. This implies

that there is some ε > 0 and some subsequence fnkgnk such that

µ(x : |fnkgnk− fg|(x) ≥ ε) ≥ ε, ∀k ≥ 1.

Now fnµ−→ f and gn

µ−→ g implies that fnk

µ−→ f and gnk

µ−→ g for any subsequence of a convergentsequence converges to the same limit. This implies further subsequences fnkj

→ f a.e. and gnkj→ g

a.e. on X, which implies that fnkjgnkj→ fg a.e. on X. Now together with the fact that µ(X) < ∞,

and that convergence in measure implies finiteness a.e. on X, we can apply Egorov to seethat fnkj

gnkj→ fg almost uniformly on X, which in turn implies convergence in measure of this

subsequence. We have arrived at our desired contradiction.

As a counter example when µ(X) = ∞, consider either of the following definitions for n ≥ 1: (i)fn(x) = gn(x) = x+ 1/n, f(x) = g(x) = x; or (ii) fn(x) = 1

nχ[0,n)(x), f(x) = 0, g(x) = gn(x) = x.

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Example 10.6 (From the Analysis II homework). Prove that if E ⊂ [0, 1] has positive Lebesguemesaure, then there exist x, y ∈ E such that x− y ∈ Q.

Claim: Let G ⊂ R be selected such that |G| > 0. Then there exists an open interval I ⊂ R such that|G ∩ I| ≥ 3/4|I|.

Proof. This result follows from the Vitali covering lemma.

Proof. We prove that given E1, E2 ⊆ R such that |E1|, |E2| > 0, the difference set

D12 := x− y : x ∈ E1, y ∈ E2 ,

contains an open interval. Here we will specify that E1 = E2 and notice that since the rationals aredense in R, this implies the claim in the exam problem.

From the claim, we have open intervals I, J ⊆ R such that |E1∩ I| ≥ 3/4|I| and |E2∩J | ≥ 3/4|J |, andwhere WLOG we can assume that |I| ≥ |J |. It follows that there exists a α ∈ R such that J + α ⊆ I.Let γ := 1/4|J |. Then for any 0 < |β| < γ, we see that the intervals I, J+α+β intersect in an interval

M where the length of M satisfies |M | > 3/4|J |. Next, let E1 := E1 ∩M , E2 := (E2 + α+ β) ∩M forthe fixed parameters α, β and interval M identitified above. Then we see that (1) we have

|E1 ∩ I| = |E1 ∪ E1 ∩ (I \M)|

= |E1|+ |E1 ∩ (I \M)| < |E1|+|J |4

=⇒ |E1| >3

4|I| − 1

4|J | > 3

4|J | − 1

4|J | = |J |

2, (*)

and that (2) we have that

|E2 ∩ J | = |(E2 + α + β) ∩ (J + α + β)|= |E2|+ |(E2 + α + β) ∩ (J + α + β) \M |

< |E2|+|J |4

=⇒ |E2| >3

4|J | − |J |

4=|J |2, (**)

by similar reasoning to the above. Finally, we notice that E1 ∪ E2 ⊆M , which shows that |E1 ∪ E2| ≤|M | ≤ |J |, where by (*) and (**) we also have that

|E1 ∪ E2| ≤ |E1|+ |E2| <|J |2

+|J |2

= |J |.

Hence, we must have that E1∩E2 6= ∅, which in turn implies that E1∩(E2+α+β) 6= ∅, or equivalentlythat there exists a y ∈ E2 such that y + α + β ∈ E1. Thus we can see that α + β ∈ D12. Just to beclear, but by no means concise, as pointed out in the previous discussion we have that α + β ∈ D12

whenever 0 < |β| < γ = |J |/4. So we obtain that

(α− γ, α + γ) =

(α− |J |

4, α +

|J |4

)⊆ D12.

In other words, D12 defined as above contains an interval – and we are done here.

Example 10.7 (Fall 2017, #4). Let φ be a function in C1(Rd) with compact support. Show that forany f ∈ L1(Rd) the function φ ∗ f(x) :=

∫φ(x − y)f(y)dy is differentiable. (HINT: Be careful about

moving limits.)

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Example 10.8 (Sprint 2017, #7). Let fn : [0, 1] → R be non-negative measurable functions thatconverge a.e. to a measurable function f ∈ L1[0, 1].

(a) Prove that the integrals∫ 1

0min(fn(x), f(x))dx are defined for each n, and that

limn→∞

∫ 1

0

min(fn(x), f(x))dx =

∫ 1

0

f(x)dx.

(b) Assume that limn→∞∫ 1

0fn(x)dx =

∫ 1

0f(x)dx. Use part (a) to prove that ||fn−f ||1 → 0 as n→∞.

Example 10.9 (Fall 2015, #8). Let 1 < p <∞. Show that the operator Tf(x) =∫∞

0f(y)x+y

dy satisfies

||Tf ||p ≤ Cp · ||f ||p where Cp =∫∞

01

(x+1)x1/p.

10.2 Facts and reminder notes

The following reminder list is intended to quickly refresh myself on the little facts and/or less standardtricks I have come across during the comp preparation:

• If a question contains multiple parts, it’s a good guess that the later parts will use the results fromthe earlier ones.

• |a− b| = max(a,b)−min(a,b)2

.

• x+ y = min(x, y) + max(x, y).

• 1−cos(2x)2x

= sin2(x)x

for all x ∈ R.

• For all u ∈ R, | sin(u)| ≤ |u| (follows from a truncated Taylor series expansion of the sine function).

• χ[x,x+c](t) = χ[t−c,t](x), which can be combined with an application of Fubini or Tonelli to swap theorders of integration in a double or multiple integral.

• Whenever see convergence in measure, try to extract a subsequence which converges pointwise.

• Whenever see (i) positive gn; and (ii) gn → g a.e., immediately try Fatou’s lemma.

• Whenever see |E| <∞ and/or fn → f a.e., think Egorov applications.

• If cannot prove convergence in measure directly, think proof by contradiction.

• Let ∆ denote the symmetric difference of sets : A∆B = (A\B)∪ (B \A). Then if Ir is an interval,E is measurable, and 0 < |E| < ∞: ||E ∩ Ir(x)| − |E ∩ Ir(y)|| ≤ |(E ∩ Ir(x))∆(E ∩ Ir(y))| ≤|Ir(x)∆Ir(y)|.

• As a heuristic, convolutions tend to smooth out functions. Convolution problems may involveconvergence theorems.

• All aspects of the convergence theorems on the exam are sharp.

• E \ F = (E \ A)∪(A \ F ).

• |A \ E|e < |A∆E|e.

• f(x) =∑n

j=1 χEj(x) =⇒

∫ 1

0f =

∑nj=1 |Ej|.

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• Try the following special functions: h(x, y) = pyp−1χ|f(x)|>y, f(x) = supn |fn(x)|, F (x, t) =f(t)χ[x,x+c](t) = f(t)χ[t−c,t](x). Try gn = f −min(fn, f) and hn = max(fn, f)− f .

• An eigenvalue of U : U(f) = λf =⇒ ||U(f)|| = |λ|||f ||.

• An := |f | > n.

• CF \ CE = E \ F and Ek = Fk ∪ (Ek \ Fk).

• Let kn := min(fn, f). To show that∫kn is well-defined: Note that kn is measurable and non-

negative a.e. provided that the fn, f are. Therefore,∫kn ∈ [0,+∞] exists.

• Fn(x) :=∫ x

0fn(t)dt =⇒ |Fn(x)− Fn(y)| = |

∫χ[x,y](t)fn(t)dt|.

• Let E(j)k := x ∈ Ek : j − 1 ≤ |x| < j = (Ek ∩ x : |x| < j) \ x : |x| < j − 1. Then E

(j)k is

bounded and measurable. May need to assume that |Ek| <∞ for some k (the other case is easier).

• E \ ∩m≥1Fm = ∪m≥1E \ Fn

• If |E| <∞, then χE(x) ∈ L1(R).

• If 0 < r < s < 1: E ∩Q(s) ⊂ (E ∩Q(r)) ∪ (Q(s) \Q(r)).

• As E ∩Q(r) ⊂ Q(r) and Q(r) are measurable sets:

limn→∞

|E ∩Q(1/n)| ≤ limn→∞

|Q(1/n)| = |∅| = 0.

• From Fall 2016 #2: A ⊆ Rd is Lebesgue measurable ⇐⇒ ∀ε > 0 ∃ a Lebesgue measurable E ⊆ Rd

such that |A∆E| < ε. (here: A \ E| < |A∆E| < ε/3)

• Fall 2016 #4: Uses the DCT to show a function is continuous and differentiable.

• Continuous on E =⇒ measurable on E.

• µ(x : |fn − f |+ |gn − g| ≥ ε) ≤ µ(x : |fn − f | ≥ ε/2) + µ(x : |gn − g| ≥ ε/2)

• Counter examples: fn(x) = 1nχ[0,n)(x).

• infn xn ≤ lim infn→∞ xn ≤ lim supn→∞ xn ≤ supn xn where lim infn→∞ xn = limn→∞ infm≥n xm =supn≥0 infm≥n xm and lim supn→∞ xn = limn→∞ supm≥n xm = infn≥0 supm≥n xm.

• f(x) = 1(x log x)1/q

∈ Lq /∈ Lp for p < q.

• Note that µ∗(E ∩ (∪nj=1Aj)) ≤ µ∗(E ∩ (∪j≥1Aj)) ≤∑

j≥1 µ∗(E ∩ Aj).

• Let φ(s) := |x : g(x) < s|. Then since x : sk+1 ≤ g(x) < s ⊆ x : sl ≤ g(x) < s and since|[0,√s| <∞, we have that

limk→∞|φ(s)− φ(sk)| = | ∩k x : sk ≤ g(x) < s| = 0.

• φ continuously differentiable with compact support =⇒ ∃C constant such that

suph6=0

∣∣∣∣φ(x+ ejh− y)− φ(x− y)

h

∣∣∣∣ ≤ C.

• If α is an irrational multiple of 2π, the numbers nα (mod 2π) are dense on the unit circle. Moreover,any continuous function that vanishes on a dense set is identically zero.

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• |A \ E|e| ≤ |A∆E|e for any A,E.

• Let An := |f | > n and A := |f | =∞. Then An+1 ⊆ An and A =∫n≥1

An so that An A. By

continuity, limn→∞ |An| = |A| = 0 when f is finite a.e.

• H ⊇ [0, 1] \ A open such that |H|e = |[0, 1] \ A|e =⇒ [0, 1] \H ⊆ A ⊆ G.

• f ′ integrable =⇒ for all ε > 0, ∃δ > 0 such that for any measurable A ⊆ R: |A| < δ =⇒∫A|f ′| < ε.

• Handle the cases of |A|e <∞, |A|e =∞ separately (typically, reduce the second to the first).

• First problems on Fall 2012 exam;

•∫E|fn|p = ||(fn − f + f)χE||pp ≤ 2p||(fn − f)χE||pp + 2p||fχE||pp.

• lim inf(sn) + lim inf(tn) ≤ lim inf(sn + tn) ≤ lim sup(sn + tn) ≤ lim sup(sn) + lim sup(tn), for anysequences sn and tn.

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