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Let's say we have \begin{align} Y_t=h(t,X_t) \end{align} and for simplicity \begin{align} dX_t=e\,dt+f\,dW_t \end{align} then by Ito's formula we have \begin{align} dY_t=\left(\frac{\partial h}{\partial t}+e\frac{\partial h}{\partial x}+\frac{1}{2}f^2\frac{\partial^2h}{\partial x^2}\right)dt+f\frac{\partial h}{\partial x}dW_t \end{align} if $h$ is a $C^2$ function. But what happens if $h$ isn't $C^2$? What if $h$ only has some Sobolev regularity $W^{1,2}$? or maybe just a $L^p$ function, do we still have similar result?

Many Thanks!!

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It's a matter of definition. If $h$ isn't $C^2$, $Y_t$ is not an Ito process. –  Sasha Sep 12 '12 at 13:42
    
dynamic89 : Even if Sasha's remark is correct, you can check Itô's lemma for (continuous) semimartingales, a good reference is Protter's book "Stochastic Integration and Differential Equations", or maybe George Lowther's blog (untitled "Almost Sure"). Best regards –  TheBridge Sep 14 '12 at 9:50

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