Master Dissertation

Let n, m ∈ N then
and
Hence
mf( n n
n
) = f( ) + · · · + f(
m m m ) = f(m
m terms
n
) = f(n)
m
f(n) = nf(1).
f( n n
) =
m m f(1).
Since ρ0 is continuous f(x) = xf(1) for all x > 0.
Thus
ρ0(e x ) = e f(x) = e xf(1) ,
that is, substituting x = ln a,
Let ω := f(1).
ρ0(a) = e f(1) ln(a) ln af(1)
= e = a f(1) .
Note that (7.5) is the condition of a regularly varying function.
We regard ω as an indication of how singular a causal distribution is near
{0}. This leads us to define:
Definition 7.4. The distribution d ∈ S ′ (R m ) is called singular of order ω,
if it has a quasi-asymptotics d0(x) at x = 0, or its Fourier transform has
quasi-asymptotics ˆ d0(p) at p = ∞, respectively, with power-counting
function ρ(δ) satisfying
for each a > 0.
ρ(aδ)
lim
δ→0 ρ(δ) = aω ,
Note that d0 = 0 in (7.1) and the corresponding requirement in (7.2),
respectively, are required to make sure that the singular order is uniquely
defined. If we skipped the requirement any ω ′ ≥ ω could be chosen as
singular order.
Fixme: It this explained well enough?
Lemma 7.5. The distributions d0 and ˆ d0 are homogeneous of degree
−(m + ω) and ω respectively.
Proof. From (7.6) and (7.5),
Hence by Proposition 1.12
a −ω 〈 ˆ d0(p), ˇ φ(p)〉 = a m 〈 ˆ d0(p), ˇ φ(ap)〉.
〈 ˆ d0( p
a ), ˇ φ(p)〉 = a m 〈 ˆ d0(p), ˇ φ(ap)〉 = a −ω 〈 ˆ d0(p), ˇ φ(p)〉. (7.7)
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