It is a well known fact that for $j\neq k$, we have $dB_t^k dB_t^j = 0. $
The argument for this is given by the fact that for independent Gaussians $G$ and $G'$ we have $\frac{1}{\sqrt{2}}(G \pm G')$ are independent Gaussians.
In Rene Schilling's Brownian Motion, it is given in the following sense.
We have a convergence in probability of $$ J_1 = \sum_{i,j}\sum_l\partial_i\partial_jf(\xi_l)\sum_{k=1}^d \sigma_{jk}(t_{l-1}) \sigma_{ik}(t_{l-1})(\Delta_l B^k)^2 \to \Sigma $$ where $d$ is the dimension and $$ \Sigma:=\sum_{i,j}\int_0^t \partial_i \partial_j f(X_s) \sum_{k=1}^d \sigma_{jk}(s)\sigma_{ik}(s)ds,$$ $\Delta_lB^k=B_{t_l}^k-B_{t_{l-1}}^k$. Also $\sigma=(\sigma_{jk})\in \mathbb{R}^{m\times d}$ is in $S_T$ which is the family of all simple processes on $[0,T]$ of the form $$f(t,\omega)=\sum_{j=1}^n \phi_{j-1}(\omega)1_{[s_{j-1},s_j)}(t)$$ where $n\ge 1$, $0=s_0\le s_1 \le \dots \le s_n\le T$ and $\phi_j \in L^\infty({\mathscr{F}_{s_j}})$.
Now in the theorem, we have $f\in C^2_b$and take $\xi_l$ to be intermediate values such that $f(\xi_l$) is measureable.
Here $t_l$ are the partition points of $[0,T]$ and we take the maximum of the partitions to $0$.
The text states that we get the above $J_1$ convergence to $\Sigma$ due to a previous result that gives $$ \sum_l g(\xi_l)(Y_{t_l}-Y_{t_{l-1}})^2 \to \int_0^T g(X_s)\tau^2(s)ds $$where $Y_t = Y_0 + \int_0^t \tau(s)dB_s + \int_0^t c(s)ds.$ Now the Ito Process $Y_t$ in this result is a $1$-dimensional Ito process, and the theorem we are trying to prove is for an $m$-dimensional Ito Process and in this result we have $(Y_{t_l}-Y_{t_{l-1}})^2=(\tau(t_{l-1})(B_{t_l}-B_{t_{l-1}})+c(t_{l-1})(t_l-t_{l-1}))^2$. And decompose $\sum_{l=1}^n (g(\xi_l))(Y_{t_l}-Y_{t_{l-1}})^2$ into three terms where the other two tends to $0$ and only $\sum_{l=1}^N g(\xi_l)\tau^2(t_{l-1})(B_{t_l}-B_{t_{l-1}})^2$ converges to $\int_0^T g(X_s)\tau^2(s)ds$. So how do we extend this result to the $m$-dimensional case to get the convergence in $J_1$?
Secondly, the text then says finally, we get for a suitable value $\Sigma'$ $$J_2 =\sum_{i,j}\sum_l\partial_i \partial_jf(\xi_l) \sum_{k \neq k'}\sigma_{jk}(t_{l-1})\sigma_{ik'}(t_{l-1})\Delta_l B^k \Delta_l B^{k'}\to \Sigma' - \Sigma'=0. $$ The text says this result follows from the convergence of $J_1$ (and we know from other parts of the proof that $J_2$ converges) and the independence of $B^k$ and $B^{k'}$ since we have $2(\Delta_lB^k)(\Delta_lB^{k'})= [\Delta_l(\frac{B^k + B^{k'}}{\sqrt(2)})]^2 - [\Delta_l(\frac{B^k - B^{k'}}{\sqrt(2)})]^2$ where $\frac{1}{\sqrt{2}} (B^k \pm B^{k'})$ are independent one-dimensional Brownian motions.
I can't wrap around my head for this. Why does the convergence of $J_1$ ensure that $J_2$ converges to some value minus itself? Isn't it possible that $\sum_{i,j} \sum_l \partial_i \partial_j f(\xi_l) \sum_{k<k^{'}} \sigma_{jk}(t_{l-1}) \sigma_{ik'}(t_{l-1})\Delta_lB^k \Delta_l B^{k'}$ diverges instead of converging to $\Sigma'$? And how can we interpret $(\Delta_lB^k)(\Delta_lB^{k'})= [\Delta_l(\frac{B^k + B^{k'}}{\sqrt(2)})]^2 - [\Delta_l(\frac{B^k - B^{k'}}{\sqrt(2)})]^2$ in terms of $\Delta_t B^k = B_{t_l}^k - B^k_{t_{l-1}}$ with both $k$ and $k'$ involved? $\Delta_t B^k = B_{t_l}^k - B^k_{t_{l-1}}$ is the kind of form of Brownian motion differences that allowed the convergence of $J_1$ but I don't know how it still works here.