Show that a sequence can converge linearly with one norm but not another. Show that superlinear convergence is independent of the norm.
I must find a sequence of elements $x^k$ such that
$$||x^{k+1}-a||_1\le r||x^k-a||_1$$
and $\lim x^k=a$
for $k>k_0$ for some norm $||.||_1$ but not for $||.||_2$
I tried thinking in $\mathbb{R}$ but this didn't work because all norms are equal there. I think that using the euclidean and taxi norm it may work.
Maybe something like $(\cos \frac{x}{k},\cos \frac{x}{k})\to (1,1)$. In the euclidean norm, we get $||x^{k+1}-a||=\sqrt{(\cos \frac{x}{k}-1)^2+(\cos \frac{x}{k}-1)^2}$ for the enclidean norm, but I don't think it works linearly.
I also have to prove $$\lim_{k\to\infty} \frac{e_{k+1}}{e_k}=0$$
where $e_k = ||x^k-a||$ and $\lim_{k\to\infty} x^k = a$, which is the definition of superlinear convergence.
Why this shouldn't depend on the norm?
Best Answer
Consider $x^k = 0.5^k e \in \mathbb{R}^n$ with $e$ the vector of all ones.
Try to prove it using $||x||_2 \leq ||x||_1 \leq \sqrt{n} ||x||_2$ (or this generalized form).