10.5. APPROXIMATION WITH SMOOTH FUNCTIONS 253

Proof: If f ≥ 0 then extend it to be 0 off U . In the above argument, let all functionsinvolved, the simple functions and the continuous functions be zero off U . Simply intersectall Vi with U and no harm is done. Now to extend to Lp (U) , simply apply Theorem 10.5.1to the positive and negative parts of real and imaginary parts of f . ■

Example 10.5.4 Let U = B(z,2r)

ψ (x) =

 exp[(|x− z|2− r2

)−1]

if |x− z|< r,

0 if |x− z| ≥ r.

Then a little work shows ψ ∈C∞c (U). The following also is easily obtained.

Lemma 10.5.5 Let U be any open set. Then C∞c (U) ̸= /0.

Proof: Pick z ∈U and let r be small enough that B(z,2r)⊆U . Then let

ψ ∈C∞c (B(z,2r))⊆C∞

c (U)

be the function of the above example. ■For a different approach see Problem 13 on Page 213.This leads to a really remarkable result about approximation with smooth functions.

Definition 10.5.6 Let U = {x∈Rn : |x|< 1}. A sequence {ψm} ⊆C∞c (U) is called

a mollifier (This is sometimes called an approximate identity if the differentiability is notincluded.) if

ψm(x)≥ 0, ψm(x) = 0, if |x| ≥ 1m,

and∫

ψm(x) = 1. Sometimes it may be written as {ψε} where ψε satisfies the aboveconditions except ψε (x) = 0 if |x| ≥ ε . In other words, ε takes the place of 1/m. Therecertainly exist mollifiers. Let ψ ∈C∞

c (B(0,1)) , ψ (x)≥ 0,∫

ψ (x)dmn = 1. Then let

ψm (x)≡ cmψ (mx)

where cm is chosen to make∫

cmψ (mx)dmn = 1. Thus ψm is 0 off B(0, 1

m

).

The notation∫

f (x,y)dµ(y) will mean x is fixed and the function y→ f (x,y) is beingintegrated. To make the notation more familiar, dx is written instead of dmn(x).

Lemma 10.5.7 Let g ∈Cc (U) then there exists h ∈C∞c (U) such that∫

|g−h|p dmn < ε.

Proof: Let ψm be a mollifier. Consider

hm (x)≡∫

g(x−y)ψm (y)dmn (y)

Then since the integral of ψm is 1, it follows that

hm (x)−g(x) =∫

(g(x−y)−g(x))ψm (y)dmn (y)

10.5. APPROXIMATION WITH SMOOTH FUNCTIONS 253Proof: If f > 0 then extend it to be 0 off U. In the above argument, let all functionsinvolved, the simple functions and the continuous functions be zero off U. Simply intersectall V; with U and no harm is done. Now to extend to L? (U) , simply apply Theorem 10.5.1to the positive and negative parts of real and imaginary parts of f. HlExample 10.5.4 Let U = B(z,2r)-1exp (x2?) | if |x-z| <1,Oif |x—z| >r.y(x) =Then a little work shows w € C?(U). The following also is easily obtained.Lemma 10.5.5 Let U be any open set. Then C?(U) 4 0.Proof: Pick z € U and let r be small enough that B(z,2r) C U. Then letwe Ce (B(z,2r)) CCC (U)be the function of the above example. MlFor a different approach see Problem 13 on Page 213.This leads to a really remarkable result about approximation with smooth functions.Definition 10.5.6 Ler U = {x ER": |x| < 1}. A sequence {y,,} C C2(U) is calleda mollifier (This is sometimes called an approximate identity if the differentiability is notincluded.) if. 1Vin(X) 2 0, Win(X) = 0, if |x| 2 m’and [ W,,(x) = 1. Sometimes it may be written as {y,} where W, satisfies the aboveconditions except W, (x) = 0 if |x| > €. In other words, € takes the place of 1/m. Therecertainly exist mollifiers. Let y € Ce (B(0,1)), w(x) > 0, f w(x) dm, = 1. Then letVin (x) = Cm (mx)where Cm is chosen to make f Cm (mx) dmn = 1. Thus YW, is 0 off B (0, +).The notation { f(x,y)du(y) will mean x is fixed and the function y > f(x,y) is beingintegrated. To make the notation more familiar, dx is written instead of dm, (x).Lemma 10.5.7 Let g € C.(U) then there exists h € C2 (U) such thatJig-HlPam, <€é.Proof: Let y,,, be a mollifier. ConsiderInn (8) = [ 8(%—y) Vy (¥) dma (9)Then since the integral of y,,, is 1, it follows thattin (8) ~ (8) = | (@(%—y) ~8(8)) Yn (Vda (9)