Definition 15.5.1 Let U be an open subset of ℝ^{n}. C_{c}^{∞}(U) is the vector space of all infinitely differentiable functions which equal zero for all x outside of some compact set contained in U. Similarly, C_{c}^{m}

Then a little work shows ψ ∈ C_{c}^{∞}(U). The following also is easily obtained.
Proof: Pick z ∈ U and let r be small enough that B
Definition 15.5.4 Let U = {x ∈ ℝ^{n} : x < 1}. A sequence {ψ_{m}}⊆ C_{c}^{∞}(U) is called a mollifier (This is sometimes called an approximate identity if the differentiability is not included.) if

and ∫ ψ_{m}(x) = 1. Sometimes it may be written as
As before, ∫ f(x,y)dμ(y) will mean x is fixed and the function y → f(x,y) is being integrated. To make the notation more familiar, dx is written instead of dm_{n}(x).

with ψ(x) ≥ 0 and ∫ ψdm = 1. Let ψ_{m}(x) = c_{m}ψ(mx) where c_{m} is chosen in such a way that ∫ ψ_{m}dm = 1. By the change of variables theorem c_{m} = m^{n}.
Definition 15.5.6 A function, f, is said to be in L_{loc}^{1}(ℝ^{n},μ)if fis μ measurable and if fX_{K} ∈ L^{1}(ℝ^{n},μ)for every compact set, K.Here μis a Radon measure on ℝ^{n}. Usually μ = m_{n}, Lebesgue measure. When this is so, write L_{loc}^{1}(ℝ^{n})or L^{p}(ℝ^{n}), etc. If f ∈ L_{loc}^{1}(ℝ^{n},μ), and g ∈ C_{c}(ℝ^{n}),

The following lemma will be useful in what follows. It says that one of these very unregular functions in L_{loc}^{1}
Lemma 15.5.7 Let f ∈ L_{loc}^{1}(ℝ^{n},μ), and g ∈ C_{c}^{∞}(ℝ^{n}). Then f ∗g is an infinitely differentiable function. Here μ is a Radon measure on ℝ^{n}.
Proof: Consider the difference quotient for calculating a partial derivative of f ∗ g.

Using the fact that g ∈ C_{c}^{∞}

is uniformly bounded. To see this easily, use Theorem 5.5.2 on Page 279 to get the existence of a constant, M depending on

such that

for any choice of x and y. Therefore, there exists a dominating function for the integrand of the above integral which is of the form C

Now letting
Recall also the following partition of unity result. It was proved earlier. See Corollary 10.1.15 on Page 785.
Theorem 15.5.8 Let K be a compact set in ℝ^{n} and let

If K_{1} is a compact subset of U_{1} there exist such functions such that also ψ_{1}
Note that in the last conclusion of above corollary, the set U_{1} could be replaced with U_{i} for any fixed i by simply renumbering.
Theorem 15.5.9 For each p ≥ 1,C_{c}^{∞}(ℝ^{n}) is dense in L^{p}(ℝ^{n}). Here the measure is Lebesgue measure.
Proof: Let f ∈ L^{p}(ℝ^{n}) and let ε > 0 be given. Choose g ∈ C_{c}(ℝ^{n}) such that f − g_{p} <

where {ψ_{m}} is a mollifier. It follows from Lemma 15.5.7 g_{m} ∈ C_{c}^{∞}(ℝ^{n}). It vanishes if x

is continuous. Thus when m is large enough,

This proves the theorem.
This is a very remarkable result. Functions in L^{p}
Corollary 15.5.10 Let U be an open set. For each p ≥ 1,C_{c}^{∞}(U) is dense in L^{p}(U). Here the measure is Lebesgue measure.
Proof: Let f ∈ L^{p}(U) and let ε > 0 be given. Choose g ∈ C_{c}(U) such that f − g_{p} <

where {ψ_{m}} is a mollifier. It follows from Lemma 15.5.7 g_{m} ∈ C_{c}^{∞}(U) for all m sufficiently large. It vanishes if x

is continuous. Thus when m is large enough,

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 having compact support have anything to do with analytic functions which also have infinitely many derivatives. The answer is no! Recall that if an analytic function has a limit point in the set of zeros then it is identically equal to zero. Thus these functions in C_{c}^{∞}