Let $(v,v_0,v_1,\dots)$ and $(w,w_0,w_1,\dots)$ be two infinite sequences of vectors of the closed unit ball of $\mathbb{R}^d$ (for the euclidean norm $\left\|.\right\|$, $d > 1$).
When $\left\|v_n-v\right\| \to 0$ and $\left\|w_n-w\right\| \to 0$, we can write $v_n^Tw_n-v^Tw=v_n^T(w_n-w)+(v_n-v)^Tw \leq \left\|v_n\right\|\left\|w_n-w\right\|+\left\|v_n-v\right\|\left\|w\right\| \leq \left\|w_n-w\right\| + \left\|v_n-v\right\| $ and conclude that $v_n^Tw_n$ converges towards $v^Tw$.
I am in a situation where I want to prove the convergence of $v_n^Tw_n$ towards $v^Tw$, without having the convergence of $\left\|v_n-v\right\|$ and $\left\|w_n-w\right\|$ towards $0$. In fact, I don't even have the convergence of $v_n^T(w_n-w)$ and $(v_n-v)^Tw$ towards $0$ (in numerical experiments, I have seen that these two terms appear to cancel out).
I would like to know what I could attempt in order to prove that $v_n^Tw_n \to v^Tw$ in such a case (e.g., is there a more refined decomposition of the scalar product that would make sense here?)
Edit : more details on the definition of the vectors. In my setting, $v$ can be written as a weighted sum of vectors of the unit ball, and w can be written as the product between the inverse of a sum of rank-1 matrices and v. Specifically, for some specific vectors $\mathbf{u_1},\dots,\mathbf{u_m}$ of the unit ball of $\mathbb{R}^d$ that are such that $\sum_{i=1}^{m}\mathbf{u_i}\mathbf{u_i}^T=I_d$, I can write $v=\sum_{i=1}^{m}\alpha_i\mathbf{u_i}$ for some weights $\alpha_i$ such that $\sum_{i=1}^{m}\alpha_i=1$. Furthermore, define some other weights $\beta_i$ of sum 1 and let $B=\sum_{i=1}^{m} \beta_i \mathbf{u_i}\mathbf{u_i}^T$. In my setting, $B$ is invertible, and we define $w=B^{-1}v.$ Consequently, we have $$v^Tw= \left(\sum_{i=1}^{m}\alpha_i\mathbf{u_i}\right)^T\left(\sum_{i=1}^{m} \beta_i \mathbf{u_i}\mathbf{u_i}^T\right)^{-1}\left(\sum_{i=1}^{m}\alpha_i\mathbf{u_i}\right).$$ Now, for each $i$, I have a sequence $\left(\mathbf{u_{n,i}}\right)_{n \geq 0}$ such that $\sum_{i=1}^{m}\mathbf{u_{n,i}}\mathbf{u_{n,i}}^T=I_d$ for every $n$. This sequence approaches $\mathbf{u_{i}}$ in a weak sense ; I can prove that $\left\|\mathbf{u_{n,i}}\right\| \to \left\|\mathbf{u_i}\right\|$, but I do not have $\left\|\mathbf{u_{n,i}}-\mathbf{u_i}\right\| \to 0$. In the end, what I'm interested in is the convergence of $$v_n^Tw_n:=\left(\sum_{i=1}^{m}\alpha_i\mathbf{u_{n,i}}\right)^T\left(\sum_{i=1}^{m} \beta_i \mathbf{u_{n,i}}\mathbf{u_{n,i}}^T\right)^{-1}\left(\sum_{i=1}^{m}\alpha_i\mathbf{u_{n,i}}\right)$$ towards $v^Tw$, for any choice of weights $\alpha$ and $\beta$, which appears to hold in my numerical experiments. On a side note, I also have that $\left\|v_n\right\| \to \left\|v\right\|$, but I do not think it sufficient to obtain what I need.