Let $\|\cdot\|_{\mathbb{R}^n}$ denote the Euclidean norm on $\mathbb{R}^n$ and let $g: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ be an inner product on $\mathbb{R}^n$ such that: $$\alpha \|x\|_{\mathbb{R}^n}^2 \le g(x, x) \le \beta \|x\|_{\mathbb{R}^n}^2$$ For some constants $\alpha, \beta \in \mathbb{R}$ and for all $x \in \mathbb{R}^n$. Now, for some compact interval $I \subset \mathbb{R}$, consider the sets
\begin{align*} L^2(I, \mathbb{R}^n) &= \left\{ x: I \to \mathbb{R}^n \middle| \int_{I} \|x|\|^2_{\mathbb{R}^n} < +\infty \right\} \\ L_g^2(I, \mathbb{R}^n) &= \left\{ x: I \to \mathbb{R}^n \middle| \int_{I} g(x, x) < +\infty \right\} \end{align*}
I believe that both of these spaces are Hilbert spaces with associated inner products
$$\left< x, y \right>_{L^2} = \int_I x \cdot y \quad \text{and} \quad \left< x, y \right>_{L^2_g} = \int_I g(x,y)$$
I am curious about how convergence in these spaces are related. Clearly, the (bi-Lipschitz) equivalence of the norms on $\mathbb{R}^n$ further implies the equivalence of the norms induced by the two inner products. It's easy to show $x \in L^2(I, \mathbb{R}^n)$ if and only if $x \in L^2_g(I, \mathbb{R}^n)$ and more generally that the boundedness of a sequence $(x_n)$ is equivalent in these spaces. I also see that strong convergence of a sequence is equivalent. However, I have yet to show that weak convergence is equivalent. Is it true that $x_n \rightharpoonup x$ in $L^2$ iff $x_n \rightharpoonup x$ in $L^2_g$?
Since the norms are equivalent, the identity map is a linear homeomorphism between the two spaces. Therefore, under a topological point of view, they are the same space. From this you get that they have the same dual space (the linear continuous functionals are clearly the same) and hence the same weak topology. So $x_n \rightharpoonup x$ in $L^2$ iff $x_n \rightharpoonup x$ in $L^2_g$.