ψ

⟨ ρ δ_{0}, ψ ⟩ = ⟨ δ_{0}, ρ ψ ⟩ = ρ (0) ψ (0) = ψ (0) = ⟨ δ_{0}, ψ ⟩ .

Hence

ρ δ_{0} = δ_{0}

ii

Solution

H^{'} = δ_{0}

x δ_{0} = 0

therefore give

(x H)^{'} = H + x δ_{0} = H, (x H)^{''} = H^{'} = δ_{0} .

Applying the distributional Leibniz rule twice gives

(φ u)^{''} = φ^{''} u + 2 φ^{'} u^{'} + φ u^{''}, (φ^{'} u)^{'} = φ^{''} u + φ^{'} u^{'} .

Eliminating the middle term proves

φ u^{''} = φ^{''} u - 2 (φ^{'} u)^{'} + (φ u)^{''} .

Choose

ρ \in D (R)

with

ρ (0) = 1

. Parts i and ii and the identity just proved, with

φ = ρ

and

u = x H

, yield

δ_{0} = ρ (x H)^{''} = ρ^{''} x H - 2 (ρ^{'} x H)^{'} + (ρ x H)^{''} .

f_{0} = x H ρ^{''}, f_{1} = - 2 x H ρ^{'}, f_{2} = x H ρ .

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