6. Supports of distributions 23
Therefore, by the Cauchy-Schwartz inequality,
(6.14) |[ϕ]|m1,BR
2
C1 ϕ
2
m1+n, ∀ϕ C0
∞(BR),
i.e.,
(6.15) |f(ϕ)| C2 ϕ m1+n, ∀ϕ C0
∞(BR).
Denote by L2(BR) the space of vector-valued functions g(x) = {gk(x)
L2(BR), k = (k1,...,kn), 0 |k| m1 + n}, with the norm g
2
0
=

m1+n
|k|=0
gk
2.
0
By the Riesz theorem (see, for example, [R]), any bounded
linear functional Φ on L2(BR) has the form Φ(g) =
∑m1+n(gk,hk),
|k|=0
where
hk L2(BR) and (gk,hk) is the scalar product in L2(BR). Let L be a linear
(not closed) subspace of L2(BR) consisting of all vector-valued functions of
the form {
∂kϕ
∂xk
, ϕ C0
∞(Bk),
0 |k| m1 + n}. It follows from (6.15) that
f(ϕ) defines a bounded linear functional Ψ on L L(BR). Extending Ψ as
a bounded linear functional to L2(BR) (the Hahn-Banach theorem; see, for
example, [R]) we get {ψk(x), ψk(x) L2(BR), 0 |k| m1 + n} so that
Ψ(g) =
∑m1+n(gk,ψk).
|k|=0
In particular,
(6.16) f(ϕ) =
m1+n
|k|=0
∂kϕ
∂xk
, ψk =
m1+n
|k|=0
BR
∂kϕ(x)
∂xk
ψk(x)dx.
Let
ψk1(x) =
x1
a−1
· · ·
xn
an
ψk(y)dy1 · · · dyn,
where (a1,...,an) is a point in BR. Then ψk1(x) are continuous in BR and
ψk(x) =
∂nψk1(x)
∂x1 · · · ∂xn
in BR.
Therefore we have
(6.17) f(ϕ) =
m
|k|=0
BR
∂kϕ(x)
∂xk
fk1(x)dx, ∀ϕ C0
∞(BR),
where fk1(x) are continuous in BR, m = m1 +2n. Since supp f BR, there
exists χ(x) C0
∞(BR)
equal to 1 in a neighborhood of supp f. Therefore,
for any ϕ C0
∞(Rn),
we have from (6.17):
(6.18)
f(ϕ) = f(χϕ) =
m
|k|=0
fk1
∂k(χϕ)
∂xk
=
m
|k|=0
∂fk
∂xk
(ϕ), ∀ϕ C0
∞(Rn),
where fk are regular functionals corresponding to continuous functions fk(x)
in
Rn
with supp fk BR.
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