Master Dissertation

Proof. Let F : t ↦→ f(xt). By application of Taylor’s formula [10] to F ,
f(x) = F (1) =
=
N−1
k=0
N−1
k=0
=
1
k! ∂k t F (0) +
|α|≤N−1
|α|=k
1
(N − 1)!
xα α! f (α)
(0) + N
xα α! f (α) (0) +
|α|=N
1
(1 − t)
0
N−1 ∂ N t F (t)dt
1
N−1
(1 − t)
xα α! f (α) (xt)dt
0
|α|=N
x α
α! fα(x), (7.18)
where fα(x) = N 1
0 (1 − t)N−1 f (α) (xt)dt and where we have used (7.17).
Now, since |x α | ≤ maxj |xj| N ≤ |x| N when |α| = N, we can make the
following estimate
As wanted.
|α|=N
xα α! fα(x)
N 1
≤ |x|
α!
|α|=N
fα(x)
N
= |x|
|α|=N
N
≤ |x|
|α|=N
N
≤ |x|
|α|=N
1
α! N
1
(1 − t)
0
N−1 f (α) (xt)dt
N
α! sup |(1 − t)
t∈[0,1]
N−1 f (α) (xt)|
N
α! sup |f (α) (x ′ )| |x ′ | < |x| .
Lemma 7.14. Let u ∈ D(R n ) with support supp u = 0, then there exists
an N ≥ 0 such that 〈u, φ〉 = 0 for all φ ∈ C ∞ 0 (Rn ) and ∂ α φ(0) = 0 when
|α| ≤ N.
Proof. Let ψ ∈ C ∞ 0 (Rn ) be defined such that
Let ɛ ∈ (0, 1). Then
⎧
⎪⎨ 1, |x| < 1/2,
ψ(x) = s ∈ [0, 1],
⎪⎩
0,
1/2 ≤ |x| ≤ 1,
|x| > 1.
〈u, φ〉 = 〈u(x), φ(x)ψ(x/ɛ)〉, for all φ ∈ C ∞ 0 (R n ),
since supp u ⊂ {x ∈ R n ||x| < ɛ/2} =: U and ψ(x/ɛ) = 1 on U.
If x > 1, then x/ɛ > x > 1, hence the map x ↦→ φ(x)ψ(x/ɛ) is supported in
the unit-ball. From now on we may therefore assume that |x| ≤ ɛ.
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