Let $f,h,g$ be continuous functions and $B$ a real Brownian motion. We suppose that a.s. $$\forall u \in \mathbb{R}_+,f(B_u)=f(B_0)+\int_0^ug(B_r)dB_r+\frac{1}{2}\int_0^uh(B_r)dr.$$

Prove that $f$ is a $C^2$-function, $f'=g$ and $f''=h.$

It seems this is a converse of Ito's formula.

Do you know how to prove it? What is the equivalent of this and the proof when we consider $q$-dimensional Brownian motion?

Let $f,h,g$ be continuous functions and $B$ a real Brownian motion. We suppose that a.s. $$\forall u \in \mathbb{R}_+,f(B_u)=f(B_0)+\int_0^ug(B_r)dB_r+\frac{1}{2}\int_0^uh(B_r)dr.$$

Prove that $f$ is a $C^2$-function, $f'=g$ and $f''=h.$

It seems this is a converse of Itô's formula.

Do you know how to prove it? What is the equivalent of this and the proof when we consider $q$-dimensional Brownian motion?