Master Dissertation

It is obvious for k = 0 and k = 1. Now, assume it holds for k = m. Then
1
n
(m + 1)!
= 1
n
m + 1
j=1
= 1
n
m + 1
= 1
m + 1
= 1
m + 1
= 1
m + 1
= 1
m + 1
=
j=1
n
i=1
xj
xj
xi
∂
∂yj
∂
∂yj
∂
∂yi
m+1
f(y)
1
n
m!
x
xj
j=1 |α|=m
α
α!
n
j=1 |α|=m
|α|=m j=1
|β|=m+1 j=1
n
|β|=m+1 j=1
|α|=m
i=1
xi
∂
∂yi
x α
α! ∂α f(y)
∂
∂
∂yj
α f(y)
m
f(y)
xα+(0,...,0,1,0,...,0) ∂
α!
α+(0,...,0,1,0,...,0) f(y)
n
x
(αj + 1)
α+(0,...,0,1,0,...,0)
(α + (0, . . . , 0, 1, 0, . . . , 0))! ∂α+(0,...,0,1,0,...,0) f(y)
n
x β
β! ∂β f(y)
x
βj
β
β! ∂βf(y), where β = α + (0, ..., 0, 1, 0, ...0)
where only the jth entry of (0, . . . , 0, 1, 0, . . . , 0) is non-zero and where we
have used
(α + (0, . . . , 0, 1, 0, . . . , 0))! = (α1, α2, . . . , αj + 1, . . . , αn)!
Finally, rename β = α.
= α1!α2! · · · (αj + 1)! · · · αn!
= (αj + 1)α!.
Lemma 7.13. Let f ∈ C ∞ 0 (Rn ) and t ∈ R, then
f(x) ≤
for any N ≥ 1.
|α|≤N−1
xα α! f (α)
N N
(0) + |x|
α!
|α|=N
sup |f (α) (x ′ )|
′
|x | < |x| ,
60