A friend and I are attempting to answer part 3) of the exercise quoted below (from Continuous Martingales and Brownian Motion) regarding Brownian Motion (BM). We have some questions apropos thereof.
To show that $(X^{\mu}, X^{\nu})$ is Gaussian, we argued that $X^{\mu}$ and $X^{\nu}$ are individually Gaussian because their defining integrals are limits of convex combinations of simple functions, each of which is Gaussian, and since the Gaussian subspace is closed, the result is Gaussian. Is that reasoning correct? Is there another (better) way to conclude that the pair is Gaussian?
May I please have a hint for showing that $E[X^{\mu}X^{\nu}] = \int_{0}^{1}\int_{0}^{1}\inf(s,t) \, d\mu(s) \, d\nu(t) = (h,g)$? Here's what we tried in attempt to show the 2nd equality, that $\int_{0}^{1}\int_{0}^{1}\inf(s,t) \, d\mu(s) \, d\nu(t) = (h,g)$ (we didn't have a clue for the first equality): Split the integral up according to whether $s<t$ or $s>t$: $$ \begin{align*} \int_{0}^{1}\int_{0}^{1}\inf(s,t) \, d\mu(s) \, d\nu(t) &= \int_{0}^{1}\int_{0}^{t}s \, d\mu(s) \, d\nu(t) + \int_{0}^{1}\int_{t}^{1}t \, d\mu(s) \, d\nu(t) \\ &= \int_{0}^{1} \int_{0}^{t} 1\cdot h'(s) \, ds \, d\nu(t) + \int_{0}^{1} t \mu(]t,1])\,d\nu(t) \\ &= \int_{0}^{1} h(t) \, d\mu(t) + \int_{0}^{1} t \mu(]t,1])\,d\nu(t)\\ &= (h,g) + \int_{0}^{1} t \mu(]t,1])\,d\nu(t), \end{align*} $$ but we can't show that the second term is zero (and I don't think it is). Is what we've done here correct?
(1.12) Exercise. We denote by $H$ the subspace of $C[0,1]$ of functions $h$ such that $h(0)=0$, $h$ is absolutely continuous and its derivative $h'$ (which exists a.e.) satisfies $$\int_{0}^{1} h'(s)^{2}\, ds < +\infty. $$ 1) Prove that $H$ is a Hilbert space for the scalar product $$ (g,h) = \int_{0}^{1} g'(s)h'(s)\, ds. $$
2) For any bounded measure $\mu$ on $[0,1]$, show that there exists an element $h$ in $H$ such that for every $f \in H$ $$ \int_{0}^{1} f(x)\, d\mu(x) = (f,h) $$ and $h'(s) = \mu(]s,1])$.
3) Let $B$ be a standard linear BM, $\mu$ and $\nu$ two bounded measures associated as in 2) with $h$ and $g$. Prove that $$X^{\mu}(\omega) = \int_{0}^{1} B_{s}(\omega)\,d\mu(s) \quad \text{and} \quad X^{\nu}(\omega) = \int_{0}^{1} B_{s}(\omega)\,d\nu(s) $$ are random variables, that the pair $(X^{\mu}, X^{\nu})$ is Gaussian and that $$ E[X^{\mu}X^{\nu}] = \int_{0}^{1}\int_{0}^{1}\inf(s,t) \, > d\mu(s) \, d\nu(t) = (h,g). $$
Forgot to address other questions.
The reasoning for Gaussians seems right, would do the same.
The first equality you can use linearity of expectation and the fact that for $s<t$, $B_t = (B_t - B_s) + B_s$ where this decomposition is a sum of independent mean 0 Gaussians. Putting the expectation inside the integral , only $E(B_s^2) = s$ survives.
Your solution is almost correct for the second equality: It appears you are applying the inner product $(s,h)$ to the second line, but really you are using $s \chi_{s < t}$, which is not in $C[0,1]$, so you should be careful. If you replace with $f(s) = s \chi_{s < t} + t \chi_{s > t}$, which doesn't change your integral when you convert to the inner product, then you would subtract the integral of $\int_{t}^1 t d\mu(s)$ which cancels the term on the other side.