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I'm mainly dealing with $\mathbb{R}^n$.

$\nu(\cdot)$ and $\mu(\cdot)$ are equivalent iff

there exist constants $c_1,c_2>0$ such that for every $x \in \mathbb{R}^n$, $c_1\nu(x)\leq \mu(x)\leq c_2\nu(x)$.

I understand the definition. What I don't understand is why can we say that every element that satisfies a property where one norm is used, then it will satisfy the same property but with the other norm. Or this is not what is meant by equivalent norms?

By property I mean for example convergence, or continuity... What allows me to say that if two different normed spaces (different norms but same vector space), whenever on sequence converges in one of the normed spaces, the same sequence converges in the other normed space?

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    $\begingroup$ Note that two norms are equivalent if there exist constants c1, c2 > 0 such that for every x... not the other way around. $\endgroup$ Oct 24 '16 at 10:36
  • $\begingroup$ Also, what do you mean by "property"? Can you make an example of norm-dependent properties that fit your question? $\endgroup$ Oct 24 '16 at 10:38
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    $\begingroup$ Notice that when a vector space is equipped with two different norms, the outcomes are different as normed spaces. Now if both $(V, \| \cdot \|_1)$ and $(V, \| \cdot \|_2)$ are both normed spaced on the same vector space $V$, then the condition that $\| \cdot \|_1$ and $\| \cdot \|_2$ is precisely equal to the condition that the map $$\iota : (V, \| \cdot \|_1) \to (V, \| \cdot \|_2), \qquad \iota(x) = x $$ is homeomorphic. In this way, the norm equivalence captures the topological aspect of the norm. $\endgroup$ Oct 24 '16 at 10:38
  • $\begingroup$ "What I don't understand is why can we say that every element that satisfies a property where one norm is used, then it will satisfy the same property but with the other norm." I wouldn't put it this way, this is not the point. You are not interested in properties of elements, here, but in topological/metric properties of the space, as somebody else here has already pointed out. $\endgroup$
    – EM90
    Oct 24 '16 at 11:14
  • $\begingroup$ @LorenzoStella you're right. i've edited thanks $\endgroup$ Oct 24 '16 at 12:17
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First of all, this yours definition isn't right.

Definition: We say that $\nu(\cdot)$ and $\mu(\cdot)$ are equivalent norms iff there exist constants $c_1,c_2>0$ such that $c_1\nu(x)\leq \mu(x)\leq c_2\nu(x)$ for every $x \in \Bbb{R}^n.$

What does it mean?? It means that the topology induced by $\nu(\cdot)$ and the topology induced by $\mu(\cdot)$ are the same, that is, the families $\sigma_\nu$ and $\sigma_\mu$ of open sets in $(\Bbb{R}^n, \nu)$ and $(\Bbb{R}^n, \mu)$, respectively, will be equals.

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  • $\begingroup$ ... and since the topology in the present case determines which sequences are convergent one gets: a sequence is convergent with respect to $\nu$ if and only if it is convergent with respect to $\mu$. The same statement holds for the convergence of series or for continuity of functions. $\endgroup$
    – Hagen Knaf
    Oct 24 '16 at 11:07
  • $\begingroup$ @HagenKnaf That's my point. how does one go from what rdias wrote to what you wrote? That's the answer I'm looking for. $\endgroup$ Oct 24 '16 at 12:22
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Let's make a concrete example of a property that is invariant for equivalent norms: if $\|\cdot\|_1\sim\|\cdot\|_2$ on the space $V$, then you have the same convergent sequences in your space. This is actually a metric condition (i.e., invariant for equivalent of metrics), more than a norm condition. In fact, recall that any normed space $(V,\|\cdot\|)$ is naturally a metric space with distance $\mathrm{d}(x,y)=\|x-y\|$. I will then prove the statement only for sequences $\{x_n\}_n$ tending to zero. In particular, equivalence of norms tells you that, for any given radius and any given ball (say, centered in zero) w.r.t. the first norm, you can find another ball in the second norm containing it and a third ball in the second norm which is contained in it. For any given $r,s>0$, denote $B^1_r=\{x\in V\mid \|x\|_1< r\}$ and $B^2_r=\{x\in V\mid \|x\|_2< s\}$. Given $r>0$, you can find $s,t>0$ s.t. $B_s^2\subseteq B_r^1\subseteq B_t^2$.

Assume now that $x_n\to 0$ w.r.t. the topology induced by $\|\cdot\|_1$, i.e., by definition: for any $\epsilon >0$ there exists $N\in\mathbb{N}$ such that for all $n\geq N$ you have $x_n\in B_{\epsilon}^1$.

You want to prove a similar statement for $\|\cdot\|_2$.

Thus, given $\epsilon >0$, by the equivalence of norms, you can find $r>0$ s.t. $rB^1_{\epsilon}=\{r\cdot x\mid x\in B^1_{\epsilon}\}\subseteq B^2_{\epsilon}$ (we're just rescaling the ball). If $n$ is great enough, you have that $x_n\in rB_{\epsilon}^1$, thus also $x_n\in B_{\epsilon}^2$: now you are given $\epsilon'=r\epsilon$, hence you choose $M\in\mathbb{N}$ s.t. $x_n\in rB_{\epsilon}^1$ for $n\geq M$. Then you conclude.

Please, notice the fact that the topology induced by the norm is generated by the (open) balls. This gives the connection by the metric properties and the equivalence of norms.

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