Is the following expression well defined $$ \frac{\partial}{\partial t} g(t,T) = \frac{\partial}{\partial t} \int_0^t \sigma(u,T) d W(u) = ~??$$
where $\sigma(\cdot, \cdot)$ is a deterministic function, $W(\cdot)$ is a $1$-dimentional Standard Brownian Motion, and $g(0,T) = 0$
If so, what is it equal to? We know that, in differential form $$d g(t,T) = \sigma(t,T) d W(t)$$ Now if we apply Ito's formula to $d g(t,T)$ we trivially get the same thing back i.e $d g(t,T) = d g(t, T)$ so if I were to guess, the above expression is actually $0$. Although I would like a formal argument/proof or explanation as I am not sure.
Any help would be greatly appreciated.
After consulting with some of my friends I have an answer to my own question.
The expression $g(t,T) = \int_0^t \sigma(u, T) d W(u)$ is a stochastic process and thus a function on $ [0, T] \times \Omega \to \mathbb{R}$. So when I was thinking about the "derivative" w.r.t $t$ of this stochastic process what I should have been thinking about is the derivative of the sample paths of the process i.e. for a fixed $\omega \in \Omega$, the sample path is a function on $ [0,T] \to \mathbb{R}$. So now the question is clearer at least: are the sample paths of a stochastic integral differentiable w.r.t $t$ ? The answer is I don't know in general and it probably depends on the $\sigma$ and $W_t$, but if $\sigma$ was constant and $W_t$ is Brownian motion, then the answer is obviously no since by definition the sample paths of Brownian motion are nowhere differentiable. For a different $\sigma$ or even different underlying process $W_t$ probably the answer could be different.