Let $S=(S_t)_{t\geq 0}$ be a continuous local martingale.
Does it make sense to ask if in general we have : $\langle dS \rangle _t=d\langle S\rangle_t$ ?
My supposition :
I think if it makes sense to ask this question it is not hold true :
If we take a standard Brownian motion $B$ we have for all $t\geq 0$ : $$d\langle B\rangle_t = dt$$ and $$ \mathbb E ((dB_t)^2) = \mathbb E ((B_t-B_{t-\epsilon})^2) \text{ when } \epsilon\rightarrow 0^+ $$ then : \begin{align} \mathbb E ((B_t-B_{t-\epsilon})^2) &= \mathbb E (B_t^2) + \mathbb E (B_{t-\epsilon}^2) - 2\mathbb E (B_tB_{t-\epsilon}) \\ &= t+(t-\epsilon) \overset {\epsilon\rightarrow 0}{=} 2t \\ &= \langle dB\rangle_t\neq t = d\langle B\rangle_t \end{align}
But I am not really convinced, someone may tell me if I am wrong?
Edit: Calculus wrong, look at the comments.