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Show that all norm $\vert\vert .\vert\vert$ in $\mathbb{R}$ is of the form $\vert\vert x\vert\vert=a\cdot\vert x\vert$, where $a>0$ is a constant and $\vert x\vert$ is the absolute value of $x$.

Is obvious that all norm in $\mathbb{R}$ is of the form $a\vert x\vert$, but how I can formalized this?? I think I should define the function $f(x)=\vert\vert x\vert\vert-a\cdot\vert x\vert$, and maybe think in inner product to show that $f(x)\equiv 0$. Any hint pls!

Ahna Akbar
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    Note that you can assign any positive number to be the norm of $x=1$, say $|1| = a > 0$. Now, note that any norm must satisfy the property $|cx| = |c||x|$ for all real $x$ and $c$. What happens when you apply this condition to $x=1$? –  Oct 30 '15 at 01:30
  • It is a one-liner, but it is not completely obvious. Forget about inner products for a moment. Given a norm $\Vert \cdot \Vert$, how would you compute $a$? What could you then conclude about $\Vert x \Vert$ from the norm-axiom $\Vert \alpha \cdot x \Vert = |a| \cdot \Vert x \Vert$? – Andrey Tyukin Oct 30 '15 at 01:30
  • Suppose that $N$ is a norm in $\Bbb R$. Recall then that $N(ax)=|a|N(x)$ for every $x$ and thus $N(a1)=|a|N(1)$. – Pedro Oct 30 '15 at 01:30

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