Kenneth Kuttler

15.6. EXERCISES 415

Corollary 15.5.11 Let U be an open set. For each p≥ 1, C∞c (U) is dense in Lp(U). Here

the measure is Lebesgue measure.

Proof: Let f ∈ Lp(U) and let ε > 0 be given. Choose g∈Cc(U) such that || f −g||p < ε

2 .This is possible because Lebesgue measure restricted to the open set, U is regular. Thusthe existence of such a g follows from Theorem 15.2.4. Now let

gm (x) = g∗ψm (x)≡∫

g(x−y)ψm (y)dmn (y) =∫

g(y)ψm (x−y)dmn (y)

where {ψm} is a mollifier. It follows from Lemma 15.5.7 gm ∈C∞c (U) for all m sufficiently

large. It vanishes if x /∈ spt(g)+B(0, 1m ). Then

||g−gm||p =

(∫|g(x)−

∫g(x−y)ψm(y)dmn(y)|pdmn(x)

) 1p

≤(∫

(∫|g(x)−g(x−y)|ψm(y)dmn(y))pdmn(x)

) 1p

≤∫ (∫

|g(x)−g(x−y)|pdmn(x)) 1

ψm(y)dmn(y)

=∫

B(0, 1m )||g−gy||pψm(y)dmn(y)<

whenever m is large enough thanks to uniform continuity of g. Theorem 15.1.12 was usedto obtain the third inequality. There is no measurability problem because the function

(x,y)→ |g(x)−g(x−y)|ψm(y)

is continuous. Thus when m is large enough,

|| f −gm||p ≤ || f −g||p + ||g−gm||p <ε

2= ε .

This proves the corollary.Another thing should probably be mentioned. If you have had a course in complex

analysis, you may be wondering whether these infinitely differentiable functions havingcompact support have anything to do with analytic functions which also have infinitelymany derivatives. The answer is no! Recall that if an analytic function has a limit point inthe set of zeros then it is identically equal to zero. Thus these functions in C∞

c (Rn) are notanalytic. This is a strictly real analysis phenomenon and has absolutely nothing to do withthe theory of functions of a complex variable.

15.6 Exercises1. Let E be a Lebesgue measurable set in R. Suppose m(E)> 0. Consider the set

E−E = {x− y : x ∈ E,y ∈ E}.

15.6. EXERCISES 415Corollary 15.5.11 Let U be an open set. For each p > 1, C2(U) is dense in L?(U). Herethe measure is Lebesgue measure.Proof: Let f € L?(U) and let € > 0 be given. Choose g € C,(U) such that || f — g||p < 5.This is possible because Lebesgue measure restricted to the open set, U is regular. Thusthe existence of such a g follows from Theorem 15.2.4. Now letBin (8) = 8* Vin (8) = [ g(8—¥) We (9) dim») = [8 (¥) Wn (Ry) dm ()where {y,,,} is a mollifier. It follows from Lemma 15.5.7 gn € C2(U) for all m sufficientlylarge. It vanishes if x ¢ spt(g) + B(0,+). Then7mle—enle = (feta) J ebx—v) valida (ying)1PIA( [6 ies) ~ 20—y)lvlp)daty))Pn (x)IA=[ (fis0»)-ats—v)Pamcs))’ (y)dimn(y)wim= fos, l!8~ Salle Vnlv)ama(9) <whenever m is large enough thanks to uniform continuity of g. Theorem 15.1.12 was usedto obtain the third inequality. There is no measurability problem because the function(x,y) > |g(x) —9(x—y)|¥,,(y)is continuous. Thus when m is large enough,E EIf —8nllp SII —8llp+llg—anllp <5 +5 =€This proves the corollary.Another thing should probably be mentioned. If you have had a course in complexanalysis, you may be wondering whether these infinitely differentiable functions havingcompact support have anything to do with analytic functions which also have infinitelymany derivatives. The answer is no! Recall that if an analytic function has a limit point inthe set of zeros then it is identically equal to zero. Thus these functions in Ce (IR”) are notanalytic. This is a strictly real analysis phenomenon and has absolutely nothing to do withthe theory of functions of a complex variable.15.6 Exercises1. Let E be a Lebesgue measurable set in R. Suppose m(E) > 0. Consider the setE-E={x-y:x€E,yeE}.