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We have the two norms $\|\cdot\|_a$ and $\|\cdot\|_b$ on the vectorspace V. They're equivalent if there exists a $k>0$ and $K>0$ so that $k\|\cdot\|_a\le\|\cdot\|_b\le$ K$\|\cdot\|_a$ for all $v\in V$.

I've proved this is an equivalence relation on the set of norms. I now have to prove, that $d_{\|\cdot\|_a}$ is equivalent to $d_{\|\cdot\|_b}$ if and only if $\|\cdot\|_a$ is equivalent to $\|\cdot\|_b$.

I've already proven the "if"-statement, I just have to prove the "only if"-statement now, and I don't really have any ideas how to do it. I would appreciate a little help very much.

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Previous comment removed due to updated wording. Thanks! –  hardmath Feb 19 '13 at 18:04
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I don't see how this question makes sense. The only way I understand this is by assuming $d_{\|\cdots\|_a}(x,y)=\|x-y\|_a$. So this is a tautology. What did I miss? –  1015 Feb 19 '13 at 18:05
    
@julien: I believe it is a simple argument, though not quite a tautology. The metric $d_{||\cdot ||_a}$ is a (real) function of two arguments while the norm $||\cdot ||_a$ takes only a single argument, just as you understand. –  hardmath Feb 20 '13 at 6:26

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up vote 4 down vote accepted

The definition of equivalent metrics is not given in the Question, and so let's work from a pretty basic notion of that, meaning the identity map on $V$ is a homeomorphism when metric $d_{||\cdot ||_a}$ is supplied for $V$ as the domain and metric $d_{||\cdot ||_b}$ for $V$ as the range.

For arbitrary metrics this notion of equivalence is too weak to conclude a "mutual coercivity" as desired here, so our argument will hinge on the fact that these two particular metrics are norm-derived. To illustrate the point, consider the metric $d_e$ derived from the Euclidean norm on $\mathbb{R}^n$ and as alternative "truncated" metric:

$$ d_f(x,y) = \min \{ d_e(x,y), 1 \} $$

Even though the two metrics are equivalent in the sense of the identity map being a homeomorphism (or equivalently, in giving the same topology) on $\mathbb{R}^n$, there does not exist a constant $K$ not depending on $x,y$ such that $\forall x,y \in \mathbb{R}^n$:

$$ d_e(x,y) \le K\; d_f(x,y) $$

Let's cite this previous MSE question for the proposition that two metric topologies are equivalent if and only if they have the same convergent sequences. Also let's simplify notation slightly, calling our norm-derived metrics:

$$ d_a(x,y) = ||x - y||_a $$ $$ d_b(x,y) = ||x - y||_b $$

Easy direction: Assume norms $||\cdot ||_a$ and $||\cdot ||_b$ are equivalent on vector space $V$, i.e. there exist positive constants $k,K$ as described in the Question. Then:

$$ k\; d_a(x,y) \le d_b(x,y) \le K\; d_a(x,y) \;\; \forall x,y \in V $$

From this pair of inequalities it is readily seen that a sequence $\{x_i\}$ converges with respect to one metric if and only if does so with respect to the other metric. Given $\epsilon \gt 0$, if $d_b(x_i,y) \lt k\epsilon$ for all $i \gt M$, then $d_a(x_i,y) \lt \epsilon$ for the same $i \gt M$. Similarly one shows the implication of convergence after exchanging metrics by replacing $k$ with $1/K$.

Less easy direction: Assume the metric topologies induced by $d_a,d_b$ are equivalent. We will show (by contradiction) that there exists $K \gt 0$ such that:

$$ d_b(x,y) \le K\; d_a(x,y) $$

and leave the existence of the constant $k$ in the other inequality as an exercise. If no such constant $K$ exists, then for each integer $i \gt 0$ we can find $x_i$ such that $d_a(x_i,0) < 2^{-i}$ and $d_b(x_i,0) = 2^i$, using the translation invariance and scaling properties of any norm-derived metric. Clearly ${x_i}$ converges to zero in the metric $d_a$, but the distances between points grow too rapidly for the same sequence to converge in the metric $d_b$, contradicting the equivalence of topologies. That is, by the "reverse triangle inequality":

$$ d_b(x_i,x_{i+1}) \ge | d_b(x_{i+1},0) - d_b(x_i,0) | = 2^{i+1} - 2^i = 2^i $$

whereas in a convergent sequence the distances between consecutive terms must tend to zero.

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