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Let $X$ be a vector space. Two norms $\|\cdot\|,\|\cdot\|':X\to\mathbb{R}$ are equivalent, if there are constants $\alpha,\beta >0$, such that for all $x\in X$ holds:

$\alpha\|x\|\leq \|x\|'\leq \beta\|x\|$

Show that all norms on $\mathbb{R}$ are equivalent.

A detailed hint is given:

It is enough to show, that an arbitrary norm $\|\cdot\|'$ is equivalent to the Euclidean norm $\|\cdot\|$. [Why is that the case?]

Consider the set $M:=\{x\in\mathbb{R}:\|x\|'\leq 1\}$ and show, that it exists an $r\in (0,\infty)$, such that $M=[-r,r]$. It is possible to show, that $r\|x\|'=\|x\|$ for all $x\in\mathbb{R}$.

My first question is, why it is enough to show, that an arbitrary norm is equivalent to the Euclidean norm?

Then I want to follow the hint and find $r$ with $M=[-r,r]$, but I am stuck and do not know how to start here...

I would appreciate a hint to get me started.

Thanks in advance.

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    $\begingroup$ Hint: $||x|| = || (x)(1) || = |x| * ||1||$ (where $||.||$ is an arbitrary norm on $\mathbb{R}$) $\endgroup$ – Joe Jul 17 '18 at 14:04
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    $\begingroup$ As a first step, you may want to try proving that every norm on $\mathbb{R}$ has the form $\|x\| = c |x|$, where $c > 0$ is a fixed constant and $|\cdot|$ is the ordinary absolute value. As for your question "Why is it enough to show that an arbitrary norm is equivalent to the Euclidean norm?", the answer is that equivalence is transitive. $\endgroup$ – Connor Harris Jul 17 '18 at 14:05
  • $\begingroup$ This also holds for $\Bbb R^n$ for any $n\in \Bbb N.$ $\endgroup$ – DanielWainfleet Jul 17 '18 at 18:06
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1) If you are given any two norms $\Vert\cdot \Vert_1,\Vert\cdot \Vert_2$ on $\mathbb R$ which are both equivalent to the Euclidean norm, they are also equivalent to each other. Just take your conditions for equivalence: If $$\alpha_1\leq\frac{\Vert\cdot \Vert_1}{\Vert\cdot \Vert}\leq \beta_1, $$ and $$\alpha_2\leq\frac{\Vert\cdot \Vert_2}{\Vert\cdot \Vert}\leq \beta_2,$$ then $$ \frac{\Vert\cdot\Vert_1}{\Vert\cdot\Vert_2}=\frac{\Vert\cdot\Vert_1\Vert\cdot\Vert}{\Vert\cdot\Vert_2\Vert\cdot\Vert}=\frac{\frac{\Vert\cdot\Vert_1}{\Vert\cdot\Vert}}{\frac{\Vert\cdot\Vert_2}{\Vert\cdot\Vert}}$$ and here you can use the two equations above to obtain your desired estimates. I'm of course cheating a little here but that's the heuristic behind it.

2) For finding your $r$ you can argue in the same way. Normalize one norm by the other. This is then bounded by a constant that you can take as your $r$. Can you proceed from here?

Before voting down, comment and give me the chance to improve the answer ;)

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  • $\begingroup$ Do you mean it like this? $\frac{\|x\|'}{\|x\|}=\frac{|x|\|1\|'}{|x|}=\|1\|'\leq r$ $\endgroup$ – Cornman Jul 17 '18 at 16:49
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To your first question, one can see that the equivalence of norms is (as the name would imply) an equivalence relation, and thus if two norms are equivalent to the same norm, then by transitivity they are equivalent to each other.

To your other question, you can see that $M$ is symmetric by Homogeneity of a norm. You can show that the norm $\vert \cdot \vert'$ is a continuous map, and as such on a compact set, attains it's maximum which we'll denote by $r$. Again using homogeneity, you can see that the maximum is attained at $x=\pm 1$, and since the image of an interval under a continuous map is an interval, you'll get that $M=[-r,r]$ as you intended.

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The reason that it is sufficient to show that an arbitrary norm is equivalent to the Euclidean norm is that equivalence of norms is an equivalence relation, meaning that if $||\cdot||'$ and$ ||\cdot||''$ are both equivalent to the Euclidean norm, then they are equivalent to each other using transitivity. If you're unsure, you should verify for yourself that norm equivalence is an equivalence relation.

To show that there exists such an $r$, as has already been suggested you might consider the fact that $||x||' = |x| \cdot ||1||'$ for all $x$.

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