My question is about a simplification made in the proof of Ito's formula.
In the proof of Ito's formula in my textbook it says that since $\int_0^T f dBs$ is defined as
$$\int_0^T f dB_s = \lim_{n\to \infty} \int_0^T f_n dB_s$$
(with limit in probability) where $f_n$ is a sequence of step-functions such that
$$\int_0^t (f-f_n)^2 ds \to 0$$
in probability, then if $X_t$ is an Ito process of the form
$$X_t = X_0 + \int_0^t u(s,\omega)ds + \int_0^t v(s,\omega)dB_s$$
then to prove Ito's formula, we can assume $u$ and $v$ are elementary functions. ie, they have the form
$$f(s,\omega) = \sum_j e_j (\omega) 1\{t \in [t_j, t_{j+1})\}$$
So I understand that if $X_t$ is defined by
$$X_t^n = \int_0^t f_n dB_s$$
where $f_n$ is a step function. Then clearly $f_n$ is an elementary function. So the fact we can assume $u,v$ are elementary is due the fact $X_t$ is the limit of integrals of step functions. Where I am confused is how this limit can be interchanged with the function $g \in C^2$. That is, why is it the case that
$$g(X_t) = g(\lim_n X_t^n) = \lim_n g(X_t^n)$$
I suspect it is because:
-
$u,v$ are assumed to be almost surely integrable (definition of Ito process)
-
$g$ is a continuous mapping
And so it follows by some form of the continuous mapping theorem? But I am still unsure of the exact justification.
I am guessing this is the reason we need $g(t,x)$ to be twice continuously differentiable? So that we can take the limit out of all:
$$g, \frac{\partial g}{\partial t}, \frac{\partial g}{\partial x}, \frac{\partial^2 g}{\partial x^2}$$
Using the continuous mapping theorem? Is this correct?
Best Answer
If $X_t $ is of the form $$ X_t = \int_0^t f(s,\omega)dB_s(\omega),\quad t\geq 0, $$ then by using localization technique, we may assume $X_t$ is a $L^2$-bounded martingale. In that case, there is a sequence of elementary processes $\{f_n\}$ such that $$ E[\int_0^T |f_n(s,\cdot)-f(s,\cdot)|^2ds]\to 0 $$ for all $T>0$. Let $X_{n,t} = \int_0^t f_n(s,\omega)dB_s(\omega)$. Then by martingale maximal inequality, we have $$ E[\sup_{t\in [0,T]}|X_t -X_{n,t}|^2]\leq C E[\int_0^T |f_n(s,\cdot)-f(s,\cdot)|^2ds]\to 0. $$ This implies that $X_{n.t} \to_p X_t$ uniformly on every compact interval $[0,T]$. We may further assume that $g\in C^2_0$ via localization method. Then, $$ g(X_{n,t}), g'(X_{n,t}), g''(X_{n,t}) $$ converge in probability to $$ g(X_t), g'(X_t), g''(X_t) $$ locally uniformly.