412 CHAPTER 15. THE Lp SPACES

Lemma 15.5.3 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.

Definition 15.5.4 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 and ineverything that follows ε → 0 instead of m→ ∞.

As before,∫

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).

Example 15.5.5 Let

ψ ∈C∞c (B(0,1)) (B(0,1) = {x : |x|< 1})

with ψ(x) ≥ 0 and∫

ψdm = 1. Let ψm(x) = cmψ(mx) where cm is chosen in such a waythat

∫ψmdm = 1. By the change of variables theorem cm = mn.

Definition 15.5.6 A function, f , is said to be in L1loc(Rn,µ) if f is µ measurable and if

| f |XK ∈ L1(Rn,µ) for every compact set, K. Here µ is a Radon measure on Rn. Usu-ally µ = mn, Lebesgue measure. When this is so, write L1

loc(Rn) or Lp(Rn), etc. Iff ∈ L1

loc(Rn,µ), and g ∈Cc(Rn),

f ∗g(x)≡∫

f (y)g(x−y)dµ .

The following lemma will be useful in what follows. It says that one of these veryunregular functions in L1

loc (Rn,µ) is smoothed out by convolving with a mollifier.

Lemma 15.5.7 Let f ∈ L1loc(Rn,µ), and g ∈C∞

c (Rn). Then f ∗ g is an infinitely differen-tiable function. Here µ is a Radon measure on Rn.

Proof: Consider the difference quotient for calculating a partial derivative of f ∗g.

f ∗g(x+ te j)− f ∗g(x)t

=∫

f (y)g(x+ te j−y)−g(x−y)

tdµ (y) .