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?
Attempt: we can suppose $B$ bounded by $c$, otherwise, we localize, $f$ is continuous, therefore there exists $(f_k)_k$ of class $C^2$ converging uniformly to $f$ on compacts, applying Ito's formula we find that $$\forall u \in \mathbb{R}_+,f_k(B_u)=f_k(B_0)+\int_0^uf'_k(B_r)dB_r+\frac{1}{2}\int_0^uf''_k(B_r)dr.$$ and $\int_0^uf'_k(B_r)dB_r+\frac{1}{2}\int_0^uf''_k(B_r)dr$ converges to $\int_0^ug(B_r)dB_r+\frac{1}{2}\int_0^uh(B_r)dr.$
How to continue, is a way to prove that $f'$ converges uniformly to $g$? De we also need to approximate it?
A start: Let $H$ be a primitive of $h$, and $K$ a primitive of $H$. That is, $H'(x)=h(x)$ for all $x$, and $K'(x) = H(x)$ for all $x$. Use Ito to check that $f(B_t) - K(B_t)$ is a (continuous) local martingale. From this it follows that $K-f$ must have the form $x\mapsto a+bx$ for some real constants $a$ and $b$.