Kenneth Kuttler

414 CHAPTER 15. THE Lp SPACES

Corollary 15.5.9 Let K be a compact set in Rn and let {Ui}∞

i=1 be an open cover of K.Then there exist functions, ψk ∈C∞

c (Ui) such that ψ i ≺Ui and for all x ∈ K,∞

∑i=1

ψ i (x) = 1.

If K1 is a compact subset of U1 there exist such functions such that also ψ1 (x) = 1 for allx ∈ K1.

Proof: This follows from a repeat of the proof of Theorem 12.2.11 on Page 287, re-placing the lemma used in that proof with Theorem 15.5.8.

Note that in the last conclusion of above corollary, the set U1 could be replaced with Uifor any fixed i by simply renumbering.

Theorem 15.5.10 For each p ≥ 1, C∞c (Rn) is dense in Lp(Rn). Here the measure is

Lebesgue measure.

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

2 . This can be done by using 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 (Rn). It vanishes if

x /∈ spt(g)+B(0, 1m ).

||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 the uniform continuity of g. Theorem 15.1.12 wasused to 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 theorem.This is a very remarkable result. Functions in Lp (Rn) don’t need to be continuous

anywhere 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.

414 CHAPTER 15. THE L? SPACESCorollary 15.5.9 Let K be a compact set in R" and let {U;};_, be an open cover of K.Then there exist functions, W,, € Ce (U;) such that w, ~< U; and for all x € K,If K, is a compact subset of U, there exist such functions such that also Wy, (x) = 1 for allxeE€ Kj.Proof: This follows from a repeat of the proof of Theorem 12.2.11 on Page 287, re-placing the lemma used in that proof with Theorem 15.5.8.Note that in the last conclusion of above corollary, the set U; could be replaced with U;for any fixed i by simply renumbering.Theorem 15.5.10 For each p > 1, C2(IR") is dense in L?(IR"). Here the measure isLebesgue measure.Proof: Let f € L?(R") and let € > 0 be given. Choose g € C,(IR”) such that || f — g||) <5. This can be done by using Theorem 15.2.4. Now let8m (8) = 8% Wy (8) = f 8% —Y) Vo (¥) dm (9) = [ 8(Y) Wy (XY) dota (9)where {y,,} is a mollifier. It follows from Lemma 15.5.7 gm, € C2(R"). It vanishes ifx ¢ spt(g) + B(0, 7).e—enle = (fabs) [x -y)vl dint yy Pd) )(Ji [ 1) —8—¥) Wal )domny iamuts))/ (/ Ia(x) — s(x —y) Pirin (x 1) vals)E= fou leBulloYnladrmty) <5IAIAwhenever m is large enough thanks to the uniform continuity of g. Theorem 15.1.12 wasused to obtain the third inequality. There is no measurability problem because the function(x,y) > |g(x) —g(x—y)|Vn(y)is continuous. Thus when m is large enough,E EIf — 8ml|p < If — ello + lle —amllp < 515 =€EThis proves the theorem.This 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.