20.10. EXERCISES 641

and soP([| f | ≥ λ ])≤ 1

λ

∫[| f |≥λ ]

| f |dµ ≤ 1λ

∫Ω

| f |dµ ≤ Mλ

and so, by the assumption of uniform integrability,∫[| f |≥λ ]

| f |dµ < ε

for all f ∈ K provided λ is large enough. This proves the proposition.

20.10 Exercises1. Suppose µ is a vector measure having values inRn orCn. Can you show that |µ|must

be finite? Hint: You might define for each ei, one of the standard basis vectors, thereal or complex measure, µei

given by µei(E)≡ ei ·µ (E) . Why would this approach

not yield anything for an infinite dimensional normed linear space in place of Rn?

2. The Riesz representation theorem of the Lp spaces can be used to prove a very inter-esting inequality. Let r, p,q ∈ (1,∞) satisfy

1r=

1p+

1q−1.

Then1q= 1+

1r− 1

p>

1r

and so r > q. Let θ ∈ (0,1) be chosen so that θr = q. Then also we have

1r=

1/p+1/p′=1︷ ︸︸ ︷

1− 1p′

+1q−1 =

1q− 1

p′

and soθ

q=

1q− 1

p′

which implies p′ (1−θ) = q. Now let f ∈ Lp (Rn) , g ∈ Lq (Rn) , f ,g ≥ 0. Jus-tify the steps in the following argument using what was just shown that θr = q andp′ (1−θ) = q. Let

h ∈ Lr′ (Rn) .

(1r+

1r′

= 1)

∫f ∗g(x) |h(x)|dx =

∫ ∫f (y)g(x−y) |h(x)|dxdy.

≤∫ ∫

| f (y)| |g(x−y)|θ |g(x−y)|1−θ |h(x)|dydx

20.10. EXERCISES 641and so 1 \ MP >Al\)<= d <;/ du<—(Il2Nsq flaws sy f\flaws 7and so, by the assumption of uniform integrability,fldu<enea |for all f € K provided A is large enough. This proves the proposition.20.10 Exercises1. Suppose pl is a vector measure having values in R” or C”. Can you show that || mustbe finite? Hint: You might define for each e;, one of the standard basis vectors, thereal or complex measure, [,, given by M,, (E) = ej: (E). Why would this approachnot yield anything for an infinite dimensional normed linear space in place of R”?2. The Riesz representation theorem of the L’ spaces can be used to prove a very inter-esting inequality. Let r, p,q € (1,9) satisfy1a |rp 4qThen1 1 1 1-=1+4+--->-q rop frand so r > q. Let 6 € (0,1) be chosen so that 9r = q. Then also we have1/p+1/p'=I—1 1 1r Pp q q Ppand so6 1 1qq pwhich implies p’(1—@) = q. Now let f € L?(R"), g € L4(R"), fig >0. Jus-tify the steps in the following argument using what was just shown that @r = q andp' (1-6) =q. LetheL” (R"). (+45 = 1)[fxscolncolar= | [ f(y) ¢x—y) nx) |dxay.< [ [Fr ovlle—y) I" lex —y)I* no) lava