I need the martingale part of the put payoff (not $C^2$..). Where $S_t=exp(\sigma W_t \frac{\sigma^2t}{2})$
$d[(S_t K)^+ ]$ ??
I guess I need to use local times but how?
I need the martingale part of the put payoff (not $C^2$..). Where $S_t=exp(\sigma W_t \frac{\sigma^2t}{2})$
I guess I need to use local times but how? 


Thanks you all! (proof, for $\phi(t,S_t)=(KS_t)^+$: Step 1 smoothing : $\phi_\epsilon(x)=1_{S_t\leq K+\epsilon}\cdot\phi(x)+1_{S_t\in]K\epsilon,K]} \cdot \psi(x)$, where $\psi(x)=\frac{1}{\epsilon^2}(Kx)^2(Kx2\epsilon)$ . This function is $C^1$, and also $C^2$ excepting in a countable set. Step 2 Itô on $\phi_\epsilon(S_t)= \phi_\epsilon(S_0)+\int^t_0\phi_\epsilon^{'}(S_t)dS_t+\frac{1}{2}\int^t_0 1_{S_t\in[K\epsilon,K]}\phi_\epsilon^{''}(S_t)d\langle S\rangle _t$ because $\phi _\epsilon=0$ out of $[K\epsilon,K]$ Let's denote by $L_t=lim_{\epsilon \to0}{\frac{1}{2\epsilon^2}*\int{_{K\epsilon}^K(3S_t+4\epsilon3K)dS_t}}$ it's a finite variation process since it is increasing Step 3 We have that $\phi_\epsilon(S_t)\phi_\epsilon(S_0)\int^t_0\phi_\epsilon^{'}(S_t)dS_t \space \xrightarrow {L^2} \space\phi(S_t) \phi(S_0) \int^t_0\phi^{'}(S_t)dS_t $ (because $\int^t_0\phi_\epsilon^{'}(S_t) 1_{S_u\in[0,K\epsilon]}dS_u \xrightarrow {L^2}\int_0^t\phi^{'}(S_u)du$ by using Itô isometry) Finally We get the formula 'à la bridge' namely
and the martingale part is



Hi, Simply use ItôTanaka's formula I guess this should give something like : $df(S_t)=D_f(S_t)dS_t+\frac{1}{2}dL^s_tf''(ds)$ with $f(S)=(SK)^+$ so $D_f(S)=1_{]K,+\infty}(S)$ and $f''(ds)=\delta_K(ds)$ This gives if I am not mistaken : $d(S_tK)^+=df(S_t)=1_{]K,+\infty}(S_t)dS_t+\frac{1}{2}dL^K_t$ With $L^K_t$ being the local time of your geometric Brownian Motion $S$ around level $K$ at time $t$. Regards Edit
NB: $f''(ds)$ stands for second derivatives in the distributionsense. The use of ItôTanaka's formula allows to avoid the derivation of the Mollifierstype argument for the direct proof of the result (which is quite cumbersome in my opinion). I should add that ItoTanaka's formula is applicable to every $f$ that is the differnce of two convex functions if I remember well, which is the case here with $f(x)=(XK)^+$. 

