Suppose that in a finite-dimensional normed space we have defined two (equivalent) norms $\|\cdot\|_1$ and $\|\cdot\|_2$. Is it true that if a sequence $(x_n)_{n \geq 1}$ converges to some point $x$ at a rate $n$ in the norm $\|\cdot\|_1$ (in the sense that $\|x_n-x\|_1 \sim \frac{1}{n}$), then we also have the same rate of convergence for the second norm, i.e. $\|x_n-x\|_2 \sim \frac{C}{n}$ for some constant $C$ ?
So far, I've been only able to deduce that $$C_1 \leq \lim_{n \to \infty} n\|x_n-x\|_2 \leq C_2$$ for some positive constants $C_1$ and $C_2$ by using the definition of equivalent norms. How can I prove that the limit actually exists and equals some constant $C$ ? Perhaps it is obvious, but I don't see it !
The limit doesn't need to exist. Consider on $\mathbb{R}^2$ the two norms
$$\lVert (x,y)\rVert_1 = \sqrt{x^2+y^2}\qquad\text{and}\qquad \lVert (x,y)\rVert_2 = \sqrt{x^2 + 2y^2},$$
and the sequence
$$x_n = \frac{1}{n}\biggl(\cos \frac{\pi n}{2}, \sin \frac{\pi n}{2}\biggr).$$
For all $n$ we have $\lVert x_n\rVert_1 = \frac{1}{n}$, but we have $\lVert x_{2n}\rVert_2 = \frac{1}{2n}$ and $\lVert x_{2n-1}\rVert_2 = \frac{\sqrt{2}}{2n-1}$.
The limit of $n \lVert x_n - x\rVert_a$ exists for all convergent sequences such that $\lim n\lVert x_n - x\rVert_1$ exists if and only if $\lVert\,\cdot\,\rVert_2$ is a scalar multiple of $\lVert\,\cdot\,\rVert_1$.