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Definition 2.22. We say that two norms $\|\cdot\|$ and $\|\cdot\|'$ are equivalent if there exist $C_1,C_2>0$ such that $$C_1\|x_1\|' \le \|x\| \le C_2 \|x\|',$$ for all $x\in V$.

If two norms $\|\cdot\|$ and $\|\cdot\|'$ are not equivalent, then how can we construct a sequence such that $\|x_n\|/\|x_n\|' \to 0$? From the definition, we have for all $C_1, C_2 \gt 0$ there is a $x\in V$ such that $C_1\|x\|'\gt \|x\|$ or $C_2\|x\|'\lt \|x\|$. If we have the first strict inequality for all cases then the construction is obvious, but we could get the second strict inequality, in which case I don't see a way to construct the sequence. So how is this possible? This question stems from the following proof in showing that if two norms are not equivalent then they do not generate the same topology.

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Now suppose that $\|\cdot\|$ and $\|\cdot\|'$ are not equivalent. Then without loss of generality there exists a sequence of points $x_n\in V$, $n\ge 1$, such that $\|x_n\|/\|x_n\|' \to 0$ as $n\to+\infty$. Set $y_n=x_n/\|x_n\|'$. Then $y_n\to0$ in $(V,\|\cdot\|)$ but $I(y_n)$ does not converge to $0$ in $(V,\|\cdot\|'$ (since $\|y_n\|'=1$ for all $n$). Thus $I$ is not continuous, so there exists $U$ which is $\|\cdot\|'$-open such that $U=I^{-1}(U)$ is not $\|\cdot\|$-open, i.e., the topologies are different.

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  • $\begingroup$ It is a good practice if you quote some source to add also the reference. (And a link, if it is available online.) $\endgroup$ – Martin Sleziak Mar 12 '16 at 9:40
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It is easier if you "separate" $C_1$ and $C_2$, i.e., $\Vert\cdot\Vert$ and $\Vert\cdot\Vert'$ are equivalent if and only if there exists $C_2>0$ such that $\Vert x\Vert\leq C_2\Vert x\Vert'$ for all $x$ and there exists $C_1>0$ such that $\Vert x\Vert'\leq (1/C_1)\Vert x\Vert$ for all $x$ (equivalently, there exists $D$ with $\Vert x\Vert'\leq D\Vert x\Vert$ for all $x$: take $D=C_1^{-1}$).

Now for the contrapositive: if $\Vert\cdot\Vert$ and $\Vert\cdot\Vert'$ are not equivalent, then either for every $C>0$ there exists $x$ with $\Vert x\Vert>C\Vert x\Vert'$ or for every $C>0$ there exists $x$ with $\Vert x\Vert'>C\Vert x\Vert$. By swapping the roles of $\Vert\cdot\Vert$ and $\Vert\cdot\Vert'$, assume the second case. Let $C=C_n$ run through the positive integers and pick $x_n$ satisfying that condition.

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