 Solution

Put ${\omega }_n=(n-\frac{1}{2})\pi$. The deterministic derivative is $g_n^{\prime }=-{\omega }_n{h}_n$, and $g_n(1)=0$, while $W_0=0$. The Itô product rule with a deterministic smooth function givesAlmost every Brownian path is continuous, hence belongs to $L^2[0,1]$. Its Fourier coefficient in the given orthonormal basis $(h_n)$ is therefore ${\xi }_n/{\omega }_n$. The Parseval identity for a Hilbertian basis gives, pathwise on a probability-one event,This is the squared-norm consequence of the Brownian half-integer sine expansion; completeness, rather than pointwise convergence of a Fourier series, is all that is needed.