Double stochastic integrals - relationship

Question

Suppose that $W$ is the standard Brownian motion and $f$ is a smooth function. Why is

\begin{equation} \int_u \int_s f(u) dW_u f(v) dW_v = \left(\int_u f(u) dW_u \right)^2 - \int_u f^2(u) du \end{equation}

Any pointers or name of this relationship?

Answer 1

Set $X_t := \int_0^t f(u) \, dW_u$, then by Itô's formula (with $f(x)=x^2$)

\begin{align*} X_t^2 &= 2 \int_0^t X_s \, dX_s + \frac{2}{2} \int_0^t \langle X \rangle_s \\ &= 2\int_0^t X_s f(s) \, dW_s + \int_0^t f(s)^2 \, ds \\ &= \int_0^t \left( \int_0^t f(r) \, dW_r \right) f(s) \, dW_s + \int_0^t f(s)^2 \, ds. \end{align*} (See @Zhoraster's comment below for the second "=".) Rearranging the terms, proves the assertion.