Master Dissertation

Theorem 8.1. Identify ˆg(k) with the distribution it induces, then
Proof. For all φ ∈ S(R 4 )
lim〈ˆg,
φ〉 = lim
ɛ→0 ɛ→0 〈g, ˆ φ〉 = lim
ɛ→0
lim
ɛ→0 ˆg(k) = (2π)4δ(k).
g(k) ˆ φ(k)d 4
k = lim
ɛ→0
g0(ɛk) ˆ φ(k)d 4
k
= ˆφ(k)d 4 k, (8.5)
by the Dominated Convergence Theorem.
Now note that generally
〈 ˆ δ, φ〉 = 〈δ, ˆ φ〉 = ˆ
φ(0) = φ(x)dx = 〈1, φ〉. (8.6)
Therefore ˆ δ = 1. Similarly
〈 ˇ δ, φ〉 = 〈δ, ˇ φ〉 = ˇ φ(0) = (2π) −n
hence
φ(x)dx = (2π) −n 〈1, φ〉,
〈ˆ1, φ〉 = (2π) −n 〈F( ˇ δ), φ〉 = (2π) −n 〈δ, φ〉. (8.7)
That is, ˆ1 = (2π) nδ. It then follows from (8.5), that
lim〈ˆg,
φ〉 =
ɛ→0
as wanted.
Now by (8.4)
= 1
ɛ 4n
= 1
ɛ 4n
ˆφ(k)d 4 k = 〈1, ˆ φ〉 = 〈ˆ1, φ〉 = (2π) 4 〈δ, φ〉,
d 4 k1 · · · d 4 kn ˆ T (k1, . . . , kn)ˆg(k1) · · · ˆg(kn)
d 4 k1 · · · d 4 kn ˆ T (k1, . . . , kn)ˆg0(k1/ɛ) · · · ˆg0(kn/ɛ)
d 4 k1 · · · d 4 kn ˆ T (ɛk1, . . . , ɛkn)ˆg0(k1) · · · ˆg0(kn),
by substituting ki by ɛki. Finally, we may calculate in the usual way and
in the end take the limit ɛ → 0 without problems.
70