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).
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