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How can I prove the following?

Every norm in $\mathbb{R}$ has the form $||x||=a|x|$ where $a>0$ and $|x|$ is the absolute value of $x$. Conclude that every norm in $\mathbb{R}$ comes from a inner product.

First I think in prove by contradiction, but I don't know how to express that $||x||$ doesn't has the form $a|x|$, so maybe that's not the way.

I'm just starting to study metric spaces, and I really don't know how to solve this. If there is some hint to guide me I'll appreciate it.

Mateus Rocha
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1 Answers1

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Let $a=||1||>0$. So $||x||=||1\cdot x||=||1|| \,|x|=a|x|$.

Mateus Rocha
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  • Hello Mateus! Quick question: Why does $|1 \cdot x|=|1||x|$ ? When I first wrote this equality, i thought that $|1 \cdot x|=|1||x|$ – Lucas Apr 18 '21 at 22:34
  • I thought that $1$ should be inside the absolute value, since it is the scalar quantity in question. However, since $x \in \mathbb{R}$, then $x$ is also a scalar quantity... is that why $x$ is inside an absolute value? Also, why that implies that every norm in $\mathbb{R}$ comes from an inner product? Cheers from another brazilian! – Lucas Apr 18 '21 at 22:37
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    Hi, Lucas! Basically, any norm is "multiplicative", I mean, if $\lambda$ is a scalar, so $||\lambda\cdot x||=| \lambda |\cdot ||x||$. Here, $| \cdot |$ denotes the absolute value of the scalar $\lambda$.

    The key is to "change" the roles. Consider $x$ as the scalar multiplying the number $1$ (since we are working in $\Bbb{R}$, every element is a scalar itself).

     – Mateus Rocha Apr 20 '21 at 01:06
  • I got it, thanks! The only part I still cannot understand is how the proposition implies that every norm in $\mathbb{R}$ comes from an inner product. Can you clarify that? Please – Lucas Apr 20 '21 at 01:14