7. The convolution of distributions 25

Note that

∂kϕ(x

− y)

∂xk

=

(−1)k

∂kϕ(x

− y)

∂yk

.

Therefore, by the definition of the derivative of a distribution,

f

∂kϕ(x

− y)

∂xk

=

(−1)kf

∂kϕ(x

− y)

∂yk

=

∂kf

∂xk

(ϕ(x − y)) =

∂kf

∂xk

∗ ϕ.

Thus

(7.4)

∂k

∂xk

(f ∗ ϕ) =

∂kf

∂xk

∗ ϕ = f ∗

∂kϕ

∂xk

, ∀f ∈ D , ∀ϕ ∈ D.

Proposition 7.1. Let f ∈ E , i.e., f has a compact support, supp f ⊂ U1,

and let supp ϕ ⊂ U2, where U1 and U2 are bounded open sets. Then f ∗ ϕ ∈

C0

∞(Rn),

(7.4) holds, and

(7.5) supp(f ∗ ϕ) ⊂ U1 + U2,

where U1 + U2 is the open set of all sums x + y, x ∈ U1, y ∈ U2.

Proof: It has already been proven that f ∗ϕ ∈

C∞(Rn)

and (7.4) holds.

Note that supp ϕ(x − y) in y is contained in x − U2. If x / ∈ U1 + U2, then

x−U2 does not intersect U1 and therefore f(ϕ(x−·)) = 0 since supp f ⊂ U1.

Thus (f ∗ ϕ)(x) ∈ C0

∞(U1

+ U2) if f has a compact support.

Proposition 7.2. For any ψ ∈ D, ϕ ∈ D, and f ∈ D , the following

formula holds:

(7.6) (f ∗ ϕ)(ψ) =

Rn

(f ∗ ϕ)(x)ψ(x)dx = f(ϕ1 ∗ ψ),

where ϕ1(x) = ϕ(−x), and (ϕ1 ∗ ψ)(x) =

Rn

ϕ1(x − t)ψ(t)dt =

Rn

ϕ(t − x)ψ(t)dt.

Proof: We have

(7.7) (f ∗ ϕ)(ψ) =

Rn

f(ϕ(x − ·))ψ(x)dx.

Let SN (y) =

∑N

|k|=0

ϕ(ξk − y)ψ(ξk)Δxk be a Riemann sum of the integral

(ϕ1 ∗ ψ)(y) =

Rn

ϕ(x − y)ψ(x)dx. Since SN (y) converges to (ϕ1 ∗ ψ)(y)

uniformly in y and the same is true for each partial derivative

∂k

∂yk

SN (y), we

have that SN (y) → (ϕ1 ∗ ψ)(y) in D. Therefore

(7.8) f(SN ) =

N

k=0

f(ϕ(ξk − ·))ψ(ξk)Δxk → f(ϕ1 ∗ ψ).

Since

∑N

k=0

f(ϕ(ξk − ·))ψ(ξk)Δxk is the Riemann sum of the integral (7.7),

we get (7.6) from (7.8).