[email protected] Real Analysis Analysis_Folland_Ch6.pdfApr 30, 2018 · 2. Prove Theorem 6.8. a. If...

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Real Analysis

Byeong Ho BanMathematics and Statistics

Texas Tech University

Chapter 6. Lp spaces(Last update : April 30, 2018)

1. When does equality hold in Minkowski’s inequality? (The answer is different for p = 1 and for 1 < p <∞. What about p =∞?)

Proof. (April 7th 2018)

(1)When p = 1

‖f + g‖1 = ‖f‖1 + ‖g‖1 ⇐⇒∫|f + g| =

∫|f |+

∫|g|

⇐⇒∫

(||f + g| − (|f |+ |g|)|)

⇐⇒ |f + g| = |f |+ |g| a.e.⇐⇒ ∀ almost every x,∃α ≥ 0 such that f(x) = αg(x)

(2) When 1 < p 0(When αx = 0 it is clear.) it is equality iff

|f |p

‖f‖pp=|f + g|

pp−1

‖f + g‖pp−1pp−1

=|f |

pp−1

‖f‖pp−1pp−1

and|g|p

‖g‖pp=|f + g|

pp−1

‖f + g‖pp−1pp−1

=|f |

pp−1

‖f‖pp−1pp−1

a.e.

Thus, (‖g‖p‖f‖p

= αx or αx = 0

)a.e. x

1

2

(3) When p =∞

‖f + g‖∞ = ‖f‖∞ + ‖g‖∞ ⇐⇒ ‖f + g‖∞ ≥ |f |+ |g| a.e.⇐⇒ |f + g| ≥ |f |+ |g| a.e.⇐⇒ f = αxg a.e.

where αx ≥ 0. �

2. Prove Theorem 6.8.

a. If f and g are measurable functions on X, then ‖fg‖1 ≤ ‖f‖1 ‖g‖∞. If f ∈ L1 and g ∈ L∞,‖fg‖1 = ‖f‖1 ‖g‖∞ iff |g(x)| = ‖g‖∞ a.e. on the set where f(x) 6= 0.

b. ‖·‖∞ is a norm on L∞

c. ‖fn − f‖∞ → 0 iff there exists E ∈M such that µ(Ec) = 0 and fn → f uniformly on E.

d. L∞ is a Banach space.

e. The simple functions are dense in L∞.


a.Observe that

|fg| = |f ||g| ≤ |f | ‖g‖∞ =⇒ ‖fg‖1 =∫|fg| ≤ ‖g‖∞

∫|f | = ‖f‖1 ‖g‖∞

Suppose that f ∈ L1 and g ∈ L∞.Observe that

‖fg‖1 = ‖f‖1 ‖g‖∞ ⇐⇒∫|fg| = ‖g‖∞

∫|f |

⇐⇒∫|f |(||g| − ‖g‖∞ |) = 0

⇐⇒ |g(x)| = ‖g‖∞ a.e. on E = {x : f(x) 6= 0}

b.(1)

‖f‖∞ = 0 ⇐⇒ |f(x)| ≤ 0 a.e.⇐⇒ f = 0 a.e.

3

(2) if λ = 0, it is obvious, so let’s assume λ 6= 0.

|λf | = |λ||f | ≤ |λ| ‖f‖∞ =⇒ ‖λf‖∞ ≤ |λ| ‖f‖∞|λ||f | = |λf | ≤ ‖λf‖∞ =⇒ |λ| ‖f‖∞ ≤ ‖λf‖∞

(3)

|f + g| ≤ |f |+ |g| ≤ ‖f‖∞ + ‖g‖∞ =⇒ ‖f + g‖∞ ≤ ‖f‖∞ + ‖g‖∞

c. Let � > 0 be given.Suppose that ‖fn − f‖∞ → 0.Let

En = {x : |fn(x)− f(x)| > ‖fn − f‖∞}c E =

∞⋂n=1

En

Note that µ(Ecn) = 0 ∀n. so µ(Ec) = 0.And observe that, there is N ∈ N such that

|fn(x)− f | ≤ ‖fn − f‖∞ < � ∀n ≥ N∀x ∈ E

Therefore, fn → f uniformly on E.

Conversely, suppose that fn → f on E ⊂ X with µ(Ec) = 0. Then, there is N ∈ N such that

|fn(x)− f(x)| < � ∀n ≥ N∀x ∈ E

Then

‖fn − f‖∞ < � ∀n ≥ N

d.Let {fn} ⊂ L∞ such that

∞∑n=1

‖fn‖∞ ‖fn‖∞} E =∞⋂n=1

En

Note that µ(Ecn) = 0, so µ(Ec) = 0 and that

∞∑n=1

|fn(x)| ≤∞∑n=1

‖fn‖∞

4

Now, observe that, for any m ∈ N∣∣∣∣∣∞∑

n=m+1

fn(x)

∣∣∣∣∣ ≤∞∑

n=m+1

|fn(x)| ≤∞∑

n=m+1

‖fn‖∞

Since ∥∥∥∥∥∞∑

n=m+1

fn(x)

∥∥∥∥∥∞

≤∞∑

n=m+1

‖fn‖∞ → 0 as m→∞

the series converges in L∞, so L∞ is a Banach space.

e.Let f ∈ L∞, then there exists a sequence of simple functions {fn} such that

fn → f a.e, |fn| ≤ |f |Thus, there exists E such that fn → f uniformly on E and µ(Ec) = 0. Thus, by c.,

‖fn − f‖∞ → 0 as n→∞therefore, the space of simple function is dense in L∞

�

3. If 1 ≤ p ≤ r ≤ ∞, Lp ∩ Lr is a Banach space with norm‖f‖ = ‖f‖p + ‖f‖r

and if p < q < r, the inclusion map Lp ∩ Lr → Lq is continuous.


First of all, note that Lp ∩ Lr is a vector space. And observe the following argument.‖f‖ = 0 ⇐⇒ ‖f‖p + ‖f‖r = 0

⇐⇒ ‖f‖p = 0 ∧ ‖f‖r = 0⇐⇒ f = 0

‖λf‖ = ‖λf‖p + ‖λf‖r = |λ|(‖f‖p + ‖f‖r

)= |λ| ‖f‖

‖f + g‖ = ‖f + g‖p + ‖f + g‖r ≤ ‖f‖p + ‖g‖p + ‖f‖r + ‖g‖r = ‖f‖+ ‖g‖

Thus, Lp ∩ Lr is a normed vector space with the norm ‖·‖.Let {fn} ⊂ Lp ∩ Lr be a sequence such that

∞∑n=1

‖fn‖

5

Thus, ∥∥∥∥∥∞∑

n=m+1

fn

∥∥∥∥∥p

→ 0

∥∥∥∥∥∞∑

n=m+1

fn

∥∥∥∥∥r

→ 0

as m→∞.Therefore, ∥∥∥∥∥

∞∑n=m+1

fn

∥∥∥∥∥ =∥∥∥∥∥

∞∑n=m+1

fn

∥∥∥∥∥p

+

∥∥∥∥∥∞∑

n=m+1

fn

∥∥∥∥∥r

→ 0

Since∑∞

n=1 fn ∈ Lp ∩ Lr, this space is a Banach space.Furthermore,Observe that, if

‖f − g‖ < �then

‖f − g‖q = ‖f − g‖λp ‖f − g‖

1−λr ≤ ‖f − g‖

λ ‖f − g‖1−λ < � with 1q

=λ

p+

1− λr

Thus, the inclusion map is continuous.�

4. If 1 ≤ p < r ≤ ∞, Lp + Lr is a Banach space with norm‖f‖ = inf{‖g‖p + ‖h‖r : f = g + h}

and, if p < q < r, the inclusion map Lq → Lp + Lr is continuous.


Firstly, note that Lp + Lr is, clearly, a vector space. And the given mapping is a norm due to followingargument.

Let f ∈ Lp + Lr be given function. Then there exist {gn} ⊂ Lp and {hn} ⊂ Lr such that

limn→∞

(‖gn‖p + ‖hn‖r

)= ‖f‖ gn + hn = f ∀n ∈ N

‖f‖ = 0 =⇒ inf{‖g‖p + ‖h‖r : f = g + h} = 0

=⇒ limn→∞

(‖gn‖p + ‖hn‖r

)= 0

=⇒ limn→∞

‖gn‖p = 0 = ‖0‖p ∧ limn→∞ ‖hn‖r = 0 = ‖0‖r=⇒ f = 0 + 0 = 0

f = 0 =⇒ ‖0‖p + ‖0‖r ≥ inf{‖g‖p + ‖h‖r : f = g + h}=⇒ ‖f‖ = 0

‖λf‖ = inf{‖g‖p + ‖h‖r : λf = g + h} = |λ| inf{∥∥∥g

λ

∥∥∥p

+

∥∥∥∥hλ∥∥∥∥r

: f =g

λ+h

λ

}= |λ| ‖f‖

6

Let f1, f2 ∈ Lp + Lr. Note that, for given � > 0, there exists g1, g2 and h1, h2 such that fi = gi + hii = 1, 2 and

‖g1‖p + ‖h1‖r < ‖f1‖+�

2‖g2‖p + ‖h2‖r < ‖f2‖+

�

2

Then

‖f1 + f2‖ = ‖g1 + g2‖p + ‖h1 + h2‖r ≤ ‖g1‖p + ‖g2‖p + ‖h1‖r + ‖h2‖r < ‖f1‖+ ‖f2‖+ �

Since � > 0 is arbitrary,

‖f1 + f2‖ ≤ ‖f1‖+ ‖f2‖Thus, Lp + Lr is a normed vector space.Let {fn} ⊂ Lp + Lr be a sequence such that

∞∑n=1

‖fn‖

7


(1) Let

a = inf{µ(E) : E ⊂ X ∧ µ(E) > 0}Suppose a > 0 and, letting f ∈ Lp(µ) and An = {x : |f(x)| > n}, observe that

npµ(An) =

∫An

npdµ <

∫An

|f |pdµ ≤ ‖f‖pp =⇒ µ(An) =‖f‖ppnp

Note that

limn→∞

µ(An) = 0 =⇒ ∃N ∈ N such that µ(An) < a ∀n ≥ N

=⇒ µ(An) = 0 ∀n ≥ N, then, by the definition of An, ‖f‖∞

8

Suppose b 1n}), observe that

1

nqµ(Bn) <

∫Bn

|f |qdµ ≤∫|f |qdµ = ‖f‖qq

9

If a = 0, let

f =∞∑n=1

2np+1χEn

Then, by MCT,

‖f‖pp =∫|f |pdµ =

∞∑n=1

2npp+1µ(En) <

∞∑n=1

2n(−1p+1) 0 be given and

E = {x : |f(x)| > ‖f‖∞ − �}

10

And, ∀q > p, observe that

∞ > ‖f‖qq ≥∫E

|f |qdµ >∫E

(‖f‖∞ − �)qdµ = µ(E)(‖f‖∞ − �)

q =⇒ ‖f‖q > µ(E)1q (‖f‖∞ − �)

=⇒ limq→∞‖f‖q ≥ (‖f‖q − �) limq→∞µ(E)

1q

=⇒ limq→∞‖f‖q ≥ ‖f‖q − �

Since � is arbitrary,

limq→∞‖f‖q ≥ ‖f‖∞

On the other hands, observe that

limq→∞‖f‖q ≤ limq→∞ ‖f‖

pqp ‖f‖

1− pq

∞ = ‖f‖∞

Therefore,

limq→∞‖f‖q = ‖f‖∞

�

8. Suppose µ(X) = 1 and f ∈ Lp for some p > 0, so that f ∈ Lq for 0 < q < p.a. log ‖f‖q ≥

∫log |f |. (Use exercise 42d in §3.5, with F (t) = et.)

b. (∫|f |q − 1)/q ≥ log ‖f‖q, and (

∫|f |q − 1)/q →

∫log |f | as q → 0.

c. limq→0 ‖f‖q = exp(∫

log |f |).

Proof. �

9. Suppose 1 ≤ p 0 be given. Then for any n ∈ N, letEn,� = {x : |fn(x)− f(x)| ≥ �}

Then, observe that

�pµ(En,�) ≤∫En,�

|fn − f |pdµ ≤ ‖fn − f‖pp → 0 =⇒ µ(En,�)→ 0 as n→∞

Therefore, fn → f in measure. Hence, by Theorem 2.30, there is a subsequence {fnj} ⊂ {fn} such thatfnj → f a.e.

On the other hand, suppose that fn → f in measure and |fn| ≤ g ∈ Lp for all n.Note that there is a subsequence {fnk}k such that

fnk → f a.e.so |f | ≤ g and |fn − f | ≤ 2g ∈ Lp and lim infn→∞ |fn − f | = 0.Observe that, by Fatou’s lemma,∫

2pgp =

∫2pgp +

∫|fn − f |p ≤ lim inf

n→∞

∫{2pgp + |fn − f |p} =

∫2pgp + lim inf

n→∞

∫|fn − f |p∫

2pgp =

∫2pgp −

∫|fn − f |p ≤ lim inf

n→∞

∫{2pgp − |fn − f |p} =

∫2pgp − lim sup

n→∞

∫|fn − f |p

11

Thus,

lim supn→∞

∫|fn − f |p ≤ 0 ≤ lim inf

n→∞

∫|fn − f |p =⇒ ‖fn − f‖p → 0 as n→∞

�

10. Suppose 1 ≤ p < ∞. If fn, f ∈ Lp and fn → f a.e., then ‖fn − f‖p → 0 iff ‖fn‖p → ‖f‖p. (UseExercise 20 in §2.3. )

Proof. (April 11st 2018)

Suppose ‖fn − f‖p → 0 and, by Minkowski’s Inequality, note that

| ‖fn‖p − ‖f‖p | ≤ ‖fn − f‖p → 0 =⇒ ‖fn‖p → ‖f‖pConversely, suppose ‖fn‖p → ‖f‖p and note that |fn − f |, |fn|+ |f |, 2|f | ∈ Lp and that

|fn|+ |f | → 2|f | a.e. limn→∞

∫(|fn|+ |f |)p → lim

n→∞‖|fn|+ |f |‖pp = 2

p ‖f‖pp

Therefore, by the Exercise 20 in §2.3, since |fn − f | ≤ |fn|+ |f |,

limn→∞

∫|fn − f |p =

∫limn→∞

|fn − f |p = 0

Thus,

‖fn − f‖p → 0 as n→∞�

11.

Proof. �

12.

Proof. �

13. Lp(Rn,m) is separable for 1 ≤ p

12

And observe that ∫|f(x)− p(x)χRk |pdm ≤ �pµ(Rk) = �p(2k)n

Thus,

P ′ = {pχRk : p ∈ P and k ∈ N}

is dense in Cc(Rn) in p−norm.Since, if PQ is a set of polynomials of n variables with rational coefficients,

P ′Q = {pχRk : p ∈ PQ and k ∈ N}is countable and for any p ∈ P , there is p′ ∈ PQ such that

|p′(x)− p(x)| < � ∀x ∈ Rn,observe that ∫

|p′(x)− p(x)|pχRkdm ≤ �pm(Rk) = �p(2k)n.

Thus, P ′Q is dense in P′, so is dense in Cc(Rn) and so Lp.

(2) L∞ is not separable.

( =⇒ )For each r ∈ (0,∞), let fr = χBr(0). Then note that

‖fr − fs‖∞ = 1 ∀r, s ∈ (0,∞) with r 6= sSuppose that F be a dense subset of L∞, then

B 12(fr) ∩ F 6= ∅ ∀r ∈ (0,∞)

Since

B 12(fr) ∩B 1

2(fs) = ∅,

r 7→ B 12(fr)

is an injective map, so F should be uncountable. �

14. If g ∈ L∞, the operator T defined by Tf = fg is bounded on Lp for 1 ≤ p ≤ ∞. Its operator norm isat most ‖g‖∞, with equality if µ is semifinite.


If g = 0,

‖Tf‖p = ‖0‖p = 0 = ‖0‖∞ = ‖g‖∞ ∀f ∈ Lp

When p

13

|fg| = |f ||g| ≤ ‖f‖∞ ‖g‖∞ =⇒ ‖fg‖∞ ≤ ‖f‖∞ ‖g‖∞thus,

‖T‖ ≤ ‖g‖∞Conversely, when 1 ≤ p ≤ ∞, if µ is semifinite, then ∀� > 0 there is a set A ⊂ X, such that

0 < µ(A) ‖g‖∞ − �Then, note that χA ∈ Lp and

‖TχA‖p =(∫|gχA|pdµ

) 1p

>

((‖g‖∞ − �)

p

∫|χA|pdµ

) 1p

= (‖g‖∞ − �) ‖χA‖p

or

‖TχA‖∞ = ‖gχA‖∞ > (‖g‖∞ − �) ‖χA‖∞Thus,

‖T‖ ≥‖TχA‖p‖χA‖p

> ‖g‖∞ − �

Since � > 0 is arbitrary,

‖T‖ ≥ ‖g‖∞Therefore,

‖T‖ = ‖g‖∞�

15.

Proof. �

16.

Proof. �

17. With the notation as in Theorem 6.14, if µ is semifinite, q 0 and hence Sg is σ−finite.

Proof. .

If not, ∃δ > 0 such thatµ(E�) =∞ ∀0 < � ≤ δ

where

E� = {x : |g(x)| > �}.Let

λ = sup{α > 0 : µ(Eα) =∞}.

(λ exists since µ is semifinite, so {α > 0 : µ(Eα) =∞} is bounded.)

14

Note that, when f = 1µ(E

λ+ 1n)1pχE

λ+ 1n

sgn(g),

‖f‖p = 1So

Mq(g) ≥∣∣∣∣∫ fg∣∣∣∣ = 1

µ(Eλ+ 1n)1p

∫Eλ+ 1n

|g|dµ > µ(Eλ+ 1n)1−

1p

(λ+

1

n

)then

∞ > Mq(g) ≥ limn→∞

µ(Eλ+ 1n)1−

1p

(λ+

1

n

)= λµ(Eλ)

1− 1p =∞.

And it is a contradiction.�

18.

Proof. �

19.

Proof. �

20. Suppose supn ‖fn‖p 0 such that

∫E|g|p < � whenever µ(E) < δ, (ii)A ⊂ X such that µ(A) 0 be given and let q be the conjugate exponent of p. And let A = supn ‖fn‖p + 1.

Note that

ν(E) =

∫E

|g|q

is a finite measure and is absolutely continuous with respect to µ. Thus, there is δ > 0 such that

∀E with µ(E) < δ =⇒∫E

|g|q =∣∣∣∣∫E

|g|q∣∣∣∣ < 12 ( �4A)q

Now, let

Bn = {x : |g(x)|q ≥1

n} ∀n

then ⋃n

Bn = {x : |g(x)|q > 0} =let= B.

15

therefore,

limn→∞

∫Bcn

|g|q = limn→∞

(∫|g|q −

∫Bn

|g|q)

=

∫|g|q − lim

n→∞

∫Bn

|g|q =∫|g|q −

∫B

|g|q = 0

Thus, there exists N ∈ N such that ∫BN

|g|q < 12

( �4A

)qNote that, since

µ(BN)

N=

∫BN

1

N≤∫BN

|g|q ≤∫|g|q

17

Proof. .

a. Recall that L2 space is a Hilbert space with respect to the inner product

〈f, g〉 =∫[0,1]

fgdm.

And also note that

〈cos 2πnx, cos 2πmx〉 =∫[0,1]

cos 2πnx cos 2πmxdm = 0 if n 6= m.

Therefore { cos 2πnx‖cos 2πnx‖}n is an orthonormal sequence.Then, by Exercise 5.63,

cos 2πnx

‖cos 2πnx‖→ 0 weakly in L2.

Assume that cos 2πnx→ 0 in measure. Then there exists a subsequence (cos 2πnjx)2 → 0 a.e. Observethat, since

|cos 2πnjx|2 ≤ 1 ∀j on [0, 1]and 1 ∈ L2([0, 1]), by DCT,

limj→∞

∫[0,1]

(cos 2πnjx)2dm =

∫[0,1]

0dm = 0.

However,

limj→∞

∫[0,1]

(cos 2πnjx)2dm = lim

n→∞

1

8

(sin 4πnjπnj

+ 4

)=

1

2

Thus, it is a contradiction.

b.Let � > 0 and x ∈ (0, 1] be given and note that ∃N ∈ N such that 1

N< x, then

|fn(x)| = |nχ(0, 1n)(x)| = 0 < � ∀n ≥ N.

Since {0} has measure 0, fn → 0 a.e.Now, let En(�) = {x ∈ [0, 1] : |fn(x)| > �}. Note that

limn→∞

µ({x ∈ [0, 1] : |fn(x)| > �}) = limn→∞

µ({x ∈ [0, 1] : |nχ(0, 1n)(x)| > �}) = 0.

Thus, fn → 0 in measure.

However, for any p, note that 1 ∈ Lq([0, 1]) (q is a conjugate exponent of p) and

limn→∞

∫fndm = lim

n→∞

∫nχ(0, 1

n)dm = 1.

Therefore, fn 6→ 0 weakly in Lp. �

23.

Proof. �

24.

18

Proof. �

25.

Proof. �

26.

Proof. �

27.

Proof. �

28.

Proof. �

29.

Proof. �

30.

Proof. �

31. (A Generalized Höder Inequality) Suppose that 1 ≤ pj ≤ ∞ and∑n

1 p−1j = r

−1 ≤ 1. If fj ∈ Lpjfor j = 1, . . . , n, then

∏n1 fj ∈ Lr and ‖

∏n1 fj‖r ≤

∏n1 ‖fj‖pj . (First do the case n = 2.)

Proof. .

When n = 1, it is obvious. When n = 2, then with 1 < p1, p2

19

Now assume that ∥∥∥∥∥n∏1

fj

∥∥∥∥∥r

≤n∏1

‖fj‖pj withn∑j=1

1

pj=

1

r′≤ 1.

And let fn+1 ∈ Lpn+1 and∑n+1

11pj

= 1r≤ 1, so 1

r′+ 1

pn+1= 1

r≤ 1. Observe that, by the case when n = 2,∥∥∥∥∥

n+1∏1

fj

∥∥∥∥∥r

≤

∥∥∥∥∥n∏1

fj

∥∥∥∥∥r

‖fn+1‖pn+1 ≤n+1∏1

‖fj‖pj

Therefore, ∥∥∥∥∥n∏1

fj

∥∥∥∥∥r

≤n∏1

‖fj‖pj ∀n ∈ N

�

32. Suppose that (X,M, µ) and (Y,N , ν) are σ−finite measure spaces and K ∈ L2(µ× ν). If f ∈ L2(ν),the integral Tf(x) =

∫K(x, y)f(y)dν(y) converge absolutely for a.e. x ∈ X; moreover, Tf ∈ L2(µ) and

‖Tf‖2 ≤ ‖K‖2 ‖f‖2.

Proof. .

Observe that K ∈ L2(µ× ν), so by Toneli,∫|K(x, y)|2d(µ× ν) =

∫ (∫|K(x, y)|2dν(y)

)dµ(x)

20

33. (Exercise 8.5) If s : Rn × Rn → Rn is defined by s(x, y) = x− y, then s−1(E) is Lebesgue measurablewhenever E is Lebesgue measurable. (For n = 1, draw a picture of s−1(E) ⊂ R2. It should be clear thatafter rotation through an angle π

4, s−1(E) becomes F × R where F = {x :

√2x ∈ E}, and Theorem 2.44

can be applied. The same idea works in higher dimensions.)

Proof. .

Observe that, when n = 1,

s−1((a, b)) = {(x, y) ∈ R2 : y = x− c ∀c ∈ (a, b)}.Rotating the set counterclockwise centered at (0, 0) through π

4, you get

(

√2a

2,

√2b

2)× R

With same idea, if E ⊂ R is Lebesgue measurable set, after rotating s−1(E) counterclockwise centeredat the origin, through π

4, we have F × R where F = {x :

√2x ∈ E}. Since F × R is Lebesgue measurable

and measurability is independent of scaling and rotation, s−1(E) is also Lebesgue measurable.

In general, when n is arbitrary natural number, let E ⊂ Rn be a Lebesgue measurable set. Then, withappropriate rotational operator, we rotate s−1(E) and get

(Fn × Rn) =let {x ∈ Rn :√

2nx ∈ E} × Rn

Since Lebesgue measuability is independent of scaling and rotating, Fn is Lebesgue measurable, and sos−1(E) is a Lebesgue measurable set. �

34. (Exercise 8.7) If f is locally integrable on Rn and g ∈ Ck has compact support, then f ∗ g ∈ Ck.

Proof. .

Recall

f ∗ g =∫f(y)g(x− y)dy.

Let Sg = {x : g(x) 6= 0} = supp(g) and let S = {x− y : y ∈ Sg}.Now, observe that, for all 0 ≤ j ≤ k,

∂j

∂xj(f ∗ g) =

∫f(y)

∂jg(x− y)∂xj

dy =

∫S

f(y)∂jg(x− y)

∂xjdy

Since S is reflection and translation of a Lebesgue measurable set Sg, S is also Lebesgue measurable, sof is integrable over S. In addition, since g ∈ Ck, ∂jg is continuous, so is integrable over compact set S.Therefore, fτx(

∂jg∂xj

) is integrable over S. Therefore, ∂j(f∗g)∂xj

is continuous for all 0 ≤ j ≤ k, so

f ∗ g ∈ Ck

�

35.

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[email protected] Real Analysis Analysis_Folland_Ch6.pdfApr 30, 2018 · 2. Prove Theorem 6.8. a. If...

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