Proof of Integration by Parts for Wiener Integrals and Mean-Square Riemann Integrals

Throughout, $\lVert\cdot\rVert_{2}$, the triangle inequality, and Cauchy-Schwarz are from Square-Integrable Random Variables and the Mean-Square Inner Product and Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm; uniqueness of mean-square limits (if $\lVert Z_n-V\rVert_2\to0$ and $\lVert Z_n-V'\rVert_2\to0$ then $V=V'$ almost surely) follows from these and the null-equivalence of Square-Integrable Random Variables and the Mean-Square Inner Product. Bilinearity of the covariance in each argument follows from its formula $\operatorname{Cov}(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$ there together with the bilinearity of the mean-square inner product recorded in Square-Integrable Random Variables and the Mean-Square Inner Product; for centered random variables, $\operatorname{Cov}(X,Y)=\mathbb{E}[XY]$. Fix $t\in[0,T]$; for $t=0$ every asserted identity reads $0=0$ by the degenerate-interval conventions of Mean-Square Riemann Integral of a Family of Random Variables, so assume $t>0$. For $n\ge1$ put $x_i=it/n$ ($0\le i\le n$), and let $\omega_n$ be a bound for $|f(u)-f(v)|$ over $u,v\in[0,T]$ with $|u-v|\le t/n$; by Continuity on a Closed Interval Implies Uniform Continuity we may choose $\omega_n\to0$. Both parts rest on the Abel summation identity: for any random variables $U_{x_0},U_{x_1},\dots,U_{x_n}$,

which holds pointwise on $\Omega$ because the left-hand side telescopes to $\sum_i\bigl(f(x_i)U_{x_i}-f(x_{i-1})U_{x_{i-1}}\bigr)$.

Part 1. All the random variables below lie in the centered jointly Gaussian family of claims 2 and 3 of Wiener Integrals of Continuous Functions are Jointly Gaussian (Wiener integrals and values of $B$; expectations vanish and covariances are given by claim 2 there), so all second moments of their linear combinations are computed by bilinearity of the covariance from those covariance formulas, together with Additivity of the Riemann Integral on Adjacent Intervals for splitting Riemann integrals over adjacent intervals.

Mean-square continuity of $V$. For $0\le s\le s'\le T$: $\mathbb{E}[(V_{s'}-V_s)^2]=\int_0^{s'}k^2-2\int_0^{s}k^2+\int_0^{s}k^2=\int_s^{s'}k(u)^2\,du\le(s'-s)\max_{[0,T]}k^2$ (the maximum exists by Extreme Value Theorem on a Compact Interval; monotonicity of the integral via Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval and Linearity and Monotonicity of the Lebesgue Integral). This also covers $s=0$ since $V_0=0$, and degenerate cases by the conventions of Mean-Square Riemann Integral of a Family of Random Variables. Mean-square continuity follows; the existence assertions in the statement then follow from claim 2 of Basic Properties of the Mean-Square Riemann Integral and Existence and Uniqueness of the Mean-Square Riemann Integral for Mean-Square Continuous Families, as indicated there.

Step (a). Let $D_n=\sum_{i=1}^{n}f(x_{i-1})(V_{x_i}-V_{x_{i-1}})-\int_0^t f(u)k(u)\,dB_u$. Expanding $\mathbb{E}[D_n^2]$ by bilinearity and the covariance formulas: for $i\ne j$ the increments are uncorrelated, $\operatorname{Cov}(V_{x_i}-V_{x_{i-1}},V_{x_j}-V_{x_{j-1}})=0$ (evaluate the four $\min$ terms); $\operatorname{Cov}(V_{x_i}-V_{x_{i-1}},V_{x_i}-V_{x_{i-1}})=\int_{x_{i-1}}^{x_i}k^2$; and $\operatorname{Cov}\bigl(V_{x_i}-V_{x_{i-1}},\int_0^tfk\,dB\bigr)=\int_0^{x_i}fk^2-\int_0^{x_{i-1}}fk^2=\int_{x_{i-1}}^{x_i}fk^2$. Hence

Step (b). We show $\bigl\lVert\sum_{i=1}^{n}(f(x_i)-f(x_{i-1}))V_{x_i}-\int_0^tg(u)V_u\,du\bigr\rVert_2\to0$. Using $f(x_i)-f(x_{i-1})=\int_{x_{i-1}}^{x_i}g(u)\,du$ (hypothesis on $f,g$ and Additivity of the Riemann Integral on Adjacent Intervals) and claim 2 of Basic Properties of the Mean-Square Riemann Integral, the $i$-th summand equals $\int_{x_{i-1}}^{x_i}g(u)V_{x_i}\,du$ (a mean-square Riemann integral of the family $(g(u)V_{x_i})_{u}$), while claim 5 of Basic Properties of the Mean-Square Riemann Integral, applied inductively over the division points $x_1,\dots,x_{n-1}$, splits $\int_0^tg(u)V_u\,du=\sum_i\int_{x_{i-1}}^{x_i}g(u)V_u\,du$. By linearity (claim 1 there; the family $(g(u)(V_{x_i}-V_u))_u$ is mean-square continuous by claims 1 and 2 there) the difference is $\sum_i\int_{x_{i-1}}^{x_i}g(u)(V_{x_i}-V_u)\,du$, and by the norm bound (claim 4 there),

where $\max_{[0,T]}|g|$ exists by Extreme Value Theorem on a Compact Interval and

by the mean-square continuity estimate above.

By the Abel identity with $U=V$ (note $V_{x_0}=V_0=0$), $\sum_if(x_{i-1})(V_{x_i}-V_{x_{i-1}})=f(t)V_t-\sum_i(f(x_i)-f(x_{i-1}))V_{x_i}$. The left side converges in mean square to $\int_0^tfk\,dB$ by step (a); the right side converges to $f(t)V_t-\int_0^tgV_u\,du$ by step (b). Uniqueness of mean-square limits gives the identity of part 1, for every version of the integrals involved (each version is a limit of the same sequences). The special case $k\equiv1$ follows since $V_t=B_t$ almost surely by claim 1 of Wiener Integrals of Continuous Functions are Jointly Gaussian for $t>0$ and by property (i) of Standard Brownian Motion ($B_0=0$ almost surely) for $t=0$; covariances and mean-square limits are unaffected by almost-sure replacement.

Part 2. $(Y_t)$ is mean-square continuous by claim 6 of Basic Properties of the Mean-Square Riemann Integral, and the product families in the statement are mean-square continuous by claim 2 there. By the Abel identity with $U=Y$ ($Y_0=0$):

First sum. By claim 5 of Basic Properties of the Mean-Square Riemann Integral (applied on $[0,x_i]$ with intermediate point $x_{i-1}$), $Y_{x_i}-Y_{x_{i-1}}=\int_{x_{i-1}}^{x_i}H_u\,du$ almost surely, so by claims 1 and 2 there the first sum equals $\sum_i\int_{x_{i-1}}^{x_i}f(x_{i-1})H_u\,du$ almost surely, and

using claim 4 and the inductive application of claim 5 there. Second sum. Exactly as step (b) of part 1, with $Y$ in place of $V$ and, in place of the explicit modulus, the uniform mean-square continuity of $(Y_u)$ from Uniform Mean-Square Continuity on a Compact Interval: the second sum converges in mean square to $\int_0^tg(u)Y_u\,du$. Passing to the mean-square limit in the displayed identity and using uniqueness of mean-square limits yields $\int_0^tf(u)H_u\,du+\int_0^tg(u)Y_u\,du=f(t)Y_t$ almost surely, for every choice of versions. $\blacksquare$