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Theorem:Let $(X,\|\|_1)$ and $(X,\|\|_2)$ are normed spaces. Let $\|\|_1$ and $\|\|_2$ are equivalent $<=>$ there exist $c_1,c_2>0$ such that $c_1\|x\|_1\leq\|x\|_2\leq c_2\|x\|_1$ , $\forall x\in X$.

In this theorem if $\|x\|_1$ and $\|x\|_2$ are real numbers and are finite then we can always get $c_1$ such that $c_1\|x\|_1\leq\|x\|_2$ and $c_2$ such that $\|x\|_2\leq c_2\|x\|_1$ by archimedian property. If one of $\|x\|_1$ and $\|x\|_2$ is infinite then the above relation will not hold.Thus, if $\|x\|_1$ and $\|x\|_2$ is finite $\forall x\in X$ then $\|\|_1$ and $\|\|_2$ are equivalent. Am I right or wrong?

Thanks!

uuuuuuuuuu
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  • How do you define equivalent here? The condition $c_1\lVert\cdot\rVert_1 \leq \lVert\cdot\rVert_2 \leq c_2 \lVert\cdot\rVert_1$ is sometimes taken to be the definition of equivalence. – Henricus V. Mar 24 '16 at 06:06
  • Def of equivalent: Let $X$ be a non-empty set and $d_1$ and $d_2$ are two metrics on $X$. $d_1$ is said to be equivalent to $d_2$ if the topology $\tau _1$ corresponding to $d_1$ is equal to the topology $\tau _2$ corresponding to $d_2$. i.e., U is $d_1-$open $<=>$ U is $d_2-$open. – uuuuuuuuuu Mar 24 '16 at 06:12
  • Then this result does not hold for general metric spaces. See http://math.stackexchange.com/questions/1123503/does-topological-equivalence-of-metrics-imply-strong-equivalence – Henricus V. Mar 24 '16 at 06:19
  • It does hold for complete spaces though, I believe. Is this correct or no? – User8128 Mar 24 '16 at 06:22
  • @HenryW. It seems it should hold in any space where the topology is uniquely determined by convergent sequences because that condition clearly implies that a sequence is convergent in $| \cdot |_1$ iff it is convergent in $| \cdot |_2$. Are norm topologies not uniquely determined by convergent sequences? – User8128 Mar 24 '16 at 06:27
  • Here I am considering the normed space only. I am asking that any two norms in a normed space are equal or not given that the two norms are finite for all $x\in X$. – uuuuuuuuuu Mar 24 '16 at 06:29

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