Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 203 / 4 / c / ii

Table of contents
- Solution ii

Solution

For

ϕ \in C_{0}^{\infty} (D)

f_{ϕ} = - 2 π Δ^{- 1} ϕ = \int_{D} G_{D} (\cdot, y) ϕ (y) d y \in H_{0}^{1} (D)

and define the distributional pairing by

(h, ϕ) := (h, f_{ϕ})_{\nabla} .

It is centered Gaussian by the definition of the GFF. Integration by parts gives

Var (h, ϕ) = ∥ f_{ϕ} ∥_{\nabla}^{2} = \frac{1}{2 π} \int_{D} f_{ϕ} (- Δ f_{ϕ}) d x = \int_{D} f_{ϕ} (x) ϕ (x) d x = \iint_{D \times D} ϕ (x) G_{D} (x, y) ϕ (y) d x d y .

This is the Test-function pairing with a Gaussian free field.

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