TITLE: A sequence ${\bf x}_1,{\bf x}_2,{\bf x}_3,...$ converges geometrically in some norm, will it converge geometrically in any equivalent norm?
QUESTION [6 upvotes]: If a sequence ${\bf x}_1,{\bf x}_2,{\bf x}_3,...$ converges geometrically to a vector ${\bf u}$ in norm $\|\cdot\|_1$, i.e. $\|{\bf x}_i - {\bf u}\|_1 \le q\|{\bf x}_{i-1} - {\bf u}\|_1$ for some $q\in (0,1)$, will it converge geometrically in any norm $\|\cdot\|_2$ that is equivalent to $\|\cdot\|_1$?
Suppose $c_1\|\cdot\|_2 \le \|\cdot\|_1 \le c_2\|\cdot\|_2$ for some $0<c_1 \le c_2$. I think ${\bf x}_1,{\bf x}_2,{\bf x}_3,...$ converges to $\bf u$ in $\|\cdot\|_2$, through a simple $\epsilon$-arugment. However, I cannot see the convergence is also geometric in $\|\cdot\|_2$. If it is indeed not, can anyone help give a counterexample? Thanks!

REPLY [5 votes]: Take $\mathbf x_n=(\tfrac 1 {2^n},0)$ for odd $n$ and $\mathbf x_n=(\tfrac 1 {2^n},\tfrac 1 {2^n})$ for even $n$. Then $\mathbf x_n$ converges geometrically to zero in $\mathbb R^2$ with the $\|\cdot\|_\infty$ norm. But it does not converge geometrically with the $\|\cdot\|_1$ norm because $\mathbf x_{2n-1}$ and $\mathbf x_{2n}$ have the same $\|\cdot\|_1$ norm, $\tfrac{1}{2^{2n-1}}$.