COMPLEX GEOMETRY Course notes

Claim 2.15.3. For every α ∈ N 2 , we have
D α (ρ ɛ ∗ ρ) = ρ ɛ ∗ (D α f)
This result follows similarly applying a change of variables z ↦→ z + ɛ.
Claim 2.15.4. If z ∈ U (ɛ) and f is harmonic on D(z, ɛ), then
(ρ ɛ ∗ f)(z) = f(z).
Proof:
∫
(ρ ɛ ∗ f)(z) = ρ ɛ (ɛ)f(z + ɛ)dVol(ɛ) =
∫ 2π ∫ ɛ
ρ ɛ (r)f(z + rɛ iθ )rdrdθ = 2πf(z)
D(0,z)
0 0
0
∫ ɛ
ρ ɛ (r)rdr = f(z).
Proof of Weyl’s Lemma: Since ɛ −→ ρ ɛ (ɛ − z) has compact support in U, for every z ∈ U (ɛ) ,
h(z) := T ɛ [ρ ɛ (ɛ − z)]
is defined and belongs to A 0 (U (ɛ) ) by Claim 6.2. By Claim 6.4, it suffices to show that for every
f ∈ A 0 c(U (ɛ) ) we have
∫
T [f] = hf,
U (ɛ)
where h is harmonic since
∆h = T ɛ [∆ z ρ ɛ (ɛ − z)] = ∆T ɛ [ρ(ɛ − z)] = 0.
We want to show that T = T h . We have
Hence it suffices to show
∫
T [ρ ɛ ∗ f] = hf.
U (ɛ)
T [f] = T [ρ ɛ ∗ f].
By Claim 6.1, there exists ψ ∈ A 0 such that ∆ψ = f, where ψ is harmonic on V = C − supp(f). Hence
ψ = ρ ɛ ∗ ψ on V ɛ by Claim 6.4. Therefore, O = ψ − ρ ɛ ∗ ρ ∈ A 0 c(U) satisfies
since ∆T = 0, and T (∆ψ) = 0. Hence
∆O = ∆ψ − ρ ɛ ∗ ∆ψ = f − ρ ∗ f
T [f] = T [ρ ∗ f + ∆O] = T [ρ ɛ ∗ f].
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