Consider an SDE:
$$dX_t = \mu(t,X_t)dt + \sigma(t,X_t)dW_t, \hspace{1cm} t \in [0:T]$$
where $\mu$ and $\sigma$ are Lipschitz continuous in $t$ and $x$. Define some Lipschitz continuous function $g$.
(Lets assume that $X_t$ is Markov process). By the Markov property of $X$: $\mathbb{E}(g(X_T)|\mathcal{F}_t) = f(t,X_t)$ for some measurable function $f(t,x)$
If $f$ was differentiable in $t$ and twice differentiable in $x$ by Ito's formula then:
$$f(T,X_{T}) = f(t,X_t) + \int_t^{T} \left(\frac{\partial f}{\partial t} + \mu(s,X_s)\frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2(s,X_s)\right) ds + \int_t^{T}\frac{\partial f}{\partial x}\sigma(s,X_s)dW_s$$
$$g(X_{T}) = f(t,X_t) + \int_t^{T}\left(\frac{\partial f}{\partial t} + \mu(s,X_s)\frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2(s,X_s)\right) ds + \int_t^{T}\frac{\partial f}{\partial x}\sigma(s,X_s)dW_s$$
But the process $f(t,X_t)$ is a martingale and thus
$$g(X_T) - f(t,X_t) = \int_t^{T}\frac{\partial f}{\partial x}\sigma(s,X_s)dW_s$$
My question is, suppose now that we don't have it as given that $f$ is differentiable in $x$ and differentiable in $t$, can we deduce that
$$g(X_T) - f(t,X_t) = \int_t^{T}\frac{\partial f}{\partial x}(s,X_s)\sigma(s,X_s)dW_s$$ ?
Motivation:
Intuitively conditional expectation should smoothen function $g$ knowing that $\mu, \sigma, g$ are Lipschitz continuous, and maybe it is possible to prove that function $f(t,x)$ is differentiable.
From martingale representation theorem there must be some adapted process $Z_s$ such that
$$g(X_T) - f(t,X_t) = \int_t^{T}Z_sdW_s$$
And if $f(t,x)$ is differentiable in $x$ intuitively it must be
$Z_s = \frac{\partial f}{\partial x}(s,X_s)\sigma(s,X_s) $