Kenneth Kuttler

368 CHAPTER 12. THE Lp SPACES

is continuous. Thus when m is large enough,

∥ f −gm∥p ≤ ∥ f −g∥p +∥g−gm∥p <ε

2= ε . ■

This is a very remarkable result. Functions in Lp (Rn) don’t need to be continuousanywhere and yet every such function is very close in the Lp norm to one which is infinitelydifferentiable having compact support. The same result holds for Lp (U) for U an open set.This is the next corollary.

Corollary 12.5.9 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 12.2.4. 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 12.5.7 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)

) 1p

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 12.1.11 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 . ■

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 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.

368 CHAPTER 12. THE L? SPACESis continuous. Thus when m is large enough,—€ €Il f — 8mllp < lf —gllp + \lg—gmllp < a5 =e. HEThis is a very remarkable result. Functions in L? (R") don’t need to be continuousanywhere and yet every such function is very close in the L” norm to one which is infinitelydifferentiable having compact support. The same result holds for L? (U) for U an open set.This is the next corollary.Corollary 12.5.9 Let U be an open set. For each p > 1, C2(U) is dense in L?(U).Here the 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 12.2.4. Now letBin (@) = 8% Wy (@) = | (@— 4) Vy) ditn y) = [ 8(4) Yn (@—y) dim (9)where {y,,,} is a mollifier. It follows from Lemma 12.5.7 g, € C2(U) for all m sufficientlylarge. It vanishes if x ¢ spt(g) + B(0, +). ThenLle—enlle = (flet@)— [ ele —u)y(wdin(y)|"a(0))iPpIA(fo Is(a) —g(@—y)| Vn(U}dg (yarn)IA/ (/ Ig(@) — 8(@— y)|Pdiny (x)) Winly)dmn(y)N/Mhoo le — SyllpWin(y)dnmn(y) <whenever m is large enough thanks to uniform continuity of g. Theorem 12.1.11 was usedto obtain the third inequality. There is no measurability problem because the function(x,y) > |g(x) —9(@—y)|W,,(y)is continuous. Thus when m is large enough,If —gnllp <IIf-8llp +llg—Snllp <5 +5 = 8Another 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 C> (R”) are notanalytic. This is a strictly real analysis phenomenon and has absolutely nothing to do withthe theory of functions of a complex variable.