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Consider finite-dimensional vector spaces, on which two norms $\|\|_1$ and $\|\|_2$ are always equivalent. Then a sequence of vectors ${\bf v}_k \to {\bf v}$ w.r.t. norm 1 iff ${\bf v}_k \to {\bf v}$ w.r.t. norm 2.

The question is about sequence $\frac{{{{\mathbf{v}}_k}}}{{{{\left\| {{{\mathbf{v}}_k}} \right\|}_1}}} \to \frac{{\mathbf{v}}}{{{{\left\| {\mathbf{v}} \right\|}_1}}}$ w.r.t. norm 1, will this limit imply $\frac{{{{\mathbf{v}}_k}}}{{{{\left\| {{{\mathbf{v}}_k}} \right\|}_2}}} \to \frac{{\mathbf{v}}}{{{{\left\| {\mathbf{v}} \right\|}_2}}}$ w.r.t. norm 2 (or maybe a weaker claim that one convergence implies the other but the limit might be different)? I have trouble showing this is true. If this is false, anyone can help provide a counterexample? Thanks!

For example, $(k,\sqrt k),k=1,2,...$ satisfies $\frac{{(k,\sqrt k )}}{{{{\left\| {(k,\sqrt k )} \right\|}_p}}} = \frac{{(k,\sqrt k )}}{{{{({k^p} + {k^{\frac{p}{2}}})}^{\frac{1}{p}}}}} \to (1,0)$ for any $p$-norm.

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Supposedly $v\ne0$. Let $u_k=v_k/\|v_k\|_1$ and $u=v/\|v\|_1\ne0$. By assumption, $u_k\to u$ (with respect to both norms, because all norms are equivalent) and hence $\|u_k\|_2\to \|u\|_2$ (because the norm function is continuous). Therefore $$ \left\|\frac{u_k}{\|u_k\|_2}-\frac{u}{\|u\|_2}\right\|_2 \le \frac{1}{\|u_k\|_2}\|u_k-u\|_2 +\left|\frac{1}{\|u_k\|_2}-\frac{1}{\|u\|_2}\right|\|u\|_2 \to0. $$ Consequently, $v_k/\|v_k\|_2=u_k/\|u_k\|_2\to u/\|u\|_2=v/\|v\|_2$.

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The statement is indeed true.

Let $\mathbf{v}_k$ be a sequence of vectors in a finite dimensional vector space $V$. Suppose that the sequence converges with respect to norm $\|\cdot \|_1$ to $\mathbf{v}$, i.e., $$\frac{\mathbf{v}_k}{\|\mathbf{v}_k\|_1}\rightarrow \mathbf{v}.$$

Since norms are continuous, it follows that we also have $$\frac{\|\mathbf{v}_k\|_2}{\|\mathbf{v}_k\|_1}\rightarrow \|\mathbf{v}\|_2.$$ In particular, the ratio of the two norms is a convergent sequence. From this it follows that $$\lim_{k\rightarrow \infty}\frac{\mathbf{v}_k}{\|\mathbf{v}_k\|_2} = \lim_{k\rightarrow \infty}\frac{\mathbf{v}_k}{\|\mathbf{v}_k\|_1}\frac{\|\mathbf{v}_k\|_1}{\|\mathbf{v}_k\|_2} = \frac{\mathbf{v}}{\|\mathbf{v}\|_2}.$$

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Define a function $f:V \to V$ by sending $v \mapsto \frac{\|v\|_1}{\|v\|_2}v$. This is a continuous function on $V \backslash \{0\}$, being the quotient of two continuous functions multiplied by the identity map (we are using the fact that both norms are continuous in the unique topology they define). Notice geometrically that $f$ sends the unit sphere of $\|\cdot \|_1$ to the unit sphere of $\|\cdot \|_2$.

By continuity, if $\frac{v_k}{\|v_k\|_1} \to \frac{v}{\|v\|_1}$, then $\frac{v_k}{\|v_k\|_2}=f\big(\frac{v_k}{\|v_k\|_1}\big) \to f\big(\frac{v}{\|v\|_1}\big) = \frac{v}{\|v\|_2}$.

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