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Choose an orthonormal basis $(e_n)$ of the Dirichlet energy space and independent variables $({\xi }_n)$ with the standard normal distribution. The zero-boundary Gaussian free field is the formal seriesIts precise energy-indexed interpretation is the isonormal Gaussian processFor every fixed $f$, this sum converges in $L^2$ of the probability space by Parseval identity, and it is a centered Gaussian random variable. For $f,g\in H_0^1(D)$,This covariance characterizes its law independently of the chosen basis. The series does not define an $H_0^1$-valued random element: ${\sum }_n{\xi }_n^2=\infty$ almost surely. It is realized as a random distribution, and its pairing with ordinary test functions is constructed in part (b).

Suppose inductively thatThe observation innovation isso its conditional variance is $V_{t-1}+1$ and its conditional covariance with $x_{t-1}$ is $V_{t-1}$. The scalar static-state Kalman update therefore givesAdding the known control $u_{t-1}$ shifts the conditional mean but not the variance, giving exactlySince $V_0=1$, induction yieldsEvery update is an affine conditioning operation on jointly Gaussian random variables, so

Choose an even positive $J(q)$. Each independent complex Fourier mode of the real field has density $(J(q)/\pi )e^{-J(q)|{\varphi }_q{|}^2}$, soThe quadratic part therefore has expectation ${\sum }_q^{+}G(q)/J(q)$. For the quartic part, apply Isserlis theorem to its four jointly centered Gaussian random variables, or to their real and imaginary components. Pairing the two differentiated fields together imposes $q_2=-q_1$, and its contribution isEach of the two other pairings instead gives $-(B/V)|{\sum }_{q}q/J(q){|}^2=0$, because the summand is odd under $q\mapsto -q$. Thus only the displayed contribution remains. Each full sum is twice its positive-wavevector sum for a real field; putting $S_0={\sum }_q^{+}1/J(q)$ and $S_2={\sum }_q^{+}{q}^2/J(q)$ gives $⟨H_4{⟩}_0=4B{S}_2{S}_0/V$.

Substitute these expectations, $F_0$ and $⟨H_0{⟩}_0$ into the Feynman-Bogoliubov inequality:The zero mode is absent by the composition constraint. The counting uses one representative of every nonzero $\{q,-q\}$ pair, so no extra factor of two is assigned to an independent complex amplitude.

A finite linear combination of independent Gaussian random variables is Gaussian, soSince $v_n\to v:={\sum }_{k\ge 1}{a}_k^2<\infty$, the characteristic functions converge pointwise:The Lévy continuity theorem yields convergence in distribution toOn the given common probability space, the same variance calculation actually proves convergence in $L^2$ as well.

Because independent Brownian motions have zero quadratic covariation, the Itô product rule givesAfter integration, the random variable in the question is $B_t^1{B}_t^2$. Writing $B_t^j=\sqrt{t}{Z}_j$ for independent standard Gaussian random variables $Z_1,Z_2$, its distribution isthe scaled product of two independent standard normal random variables.

Apply the realized absolute covariation theorem to each dyadic partition of an interval. More explicitly, use the continuous increasing clock $C=[M]+[N]$ and the Radon-Nikodym theorem to writeFor each time, let $(U,V)$ have the centered bivariate normal distribution with covariance matrix $\left(\begin{array}{cc}a& c\\ c& b\end{array}\right)$, and defineThis process is continuous and increasing. To prove convergence, localize $M,N$, represent the pair as stochastic integrals against a two-dimensional Brownian motion, and approximate the integrands in $L^2(dC)$ by bounded step previsible processes. For step integrands, the result is the weak law of large numbers applied on each block to independent Gaussian random variables. The Burkholder-Davis-Gundy inequality and the Cauchy-Schwarz inequality make the error uniform on each compact interval in probability. Consequentlyin the sense of uniform convergence on compacts in probability.

The explicit stochastic exponential solutions satisfyConditionally on the path generated by $W$, the last stochastic integral is a centered Gaussian random variable with variance $Y_T$, because $W^{\perp }$ is independent of $W$. The conditional expectation of $g(S_T)$ is thereforeTaking expectations and using part b proves the result.

Translational invariance makes the pointwise variance independent of $\mathbf{r}$. Applying Parseval identity to the stated Fourier convention givesIn the thermodynamic limit, the reciprocal-lattice sum becomesAt each point $\varphi (\mathbf{r})$ is a centered Gaussian random variable, so the supplied absolute third moment gives

For a test function $\varphi$ on $D$, let $f_{\varphi }=-2\pi {\Delta }^{-1}\varphi \in H_0^1(D)$. The pairingis a centered Gaussian random variable with variance