Suppose $u(t)\in H^{1}(0,T)$ is a complex value function that $\forall t\in[0,T]$, satisfies: $du(x,t)=f(u(x,t))dt+g(u(x,t),t)dW$, where $W$ is a Brownian motion.
Calculate $d\Vert u(t)\Vert^p$ where $\Vert u(t)\Vert^2=\langle u,\bar{u}\rangle$.
\begin{align} d\Vert u\Vert^p &= d(\Vert u\Vert^2)^{\frac{p}{2}} \\ &= d(\langle u,\bar{u}\rangle)^{\frac{p}{2}} \\ &= \frac{p}{2}\langle u,\bar{u}\rangle^{\frac{p}{2}-1}d\langle u,\bar{u}\rangle+\frac{1}{2}\frac{p}{2}\frac{p-2}{2}\langle u,\bar{u}\rangle^{\frac{p}{2}-2}g(u(x,t),t)^2dt \\ &= p\Vert u\Vert^{p-2}\operatorname{Re}(\langle \bar{u},du\rangle)+\frac{p(p-2)}{8}\Vert u\Vert^{p-4}g(u(x,t),t)^2dt \end{align} then substitute $du$ to get the result. I get confused when considering the $u_x$, $u_{xx}$ terms in the Ito formula, the result from an article I read is: $$d\Vert u\Vert^p=p\Vert u\Vert^{p-2}\operatorname{Re}\langle du,u\rangle dt+p\Vert u\Vert^{p-2}\Vert g(u(t),t)\Vert^2dt+\frac{p(p-2)}{2}|\operatorname{Re}\langle g(u(t),t),u\rangle|^2\Vert u\Vert^{p-4}dt$$ I seemly get it, when I denote $V(x)=\Vert x\Vert^p,dV(u)=p\langle \bar{u},du\rangle\Vert u \Vert^{p-2},$ $d^2 V(u)=p(p-2)\langle u,du\rangle^2\Vert u\Vert^{p-4}+p\langle d\bar{u},du\rangle\Vert u \Vert^{p-2}$,
Where the first term corresponds to $V_{xx}$, I suspect this is just a formal derivation, Because $V$ is defined on the function space, the differential of it may not be well defined to the using of Ito's formula.