generalized derivative of Wiener process

Question

Defined a standard Wiener process $W = (W_t , \mathcal F_t)_{t≥0}$ and a deterministic, continuously differentiable function $f : [0, ∞) → \mathbb R$. Prove that $$f(t)W_t=\int_0^tW_sf'(s)ds+\int_0^tf(s)dW_s$$

Answer 1

If you know Ito Lemma, then you just consider $F(t,x) = f_t\cdot x$ and the differential of $F$ is $$ \mathrm dF(t,W_t) = f'_tW_t\mathrm dt + f_t\mathrm dW_t + \mathrm d[f,W]_t $$ and since $f$ is deterministic and continuous, the latter term is zero. Hence $$ \mathrm dF(t,W_t) = f'_tW_t\mathrm dt + f_t\mathrm dW_t \quad\Leftrightarrow \quad F(T,W_T) = F(0,W_0) + \int_0^T f'_tW_t\mathrm dt + \int_0^Tf_t\mathrm dW_t. $$