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I am reading a proof of the claim that for a norm $\|\cdot\|$, we have $ C|x| \leq \|x\|$, where $|\cdot|$ is the Euclidean norm, $C$ is a constant greater than $0$ and $x \in \mathbb{R}^n$.

First of all, it is stated that the map $x \mapsto \|x\|$ is continuous. I haven't managed to show that and I'm also not sure why it is important/relevant.

Then it goes on to say that with $C:= \min_{|x|=1} \|x\|,$ we have $\biggr\|\frac{x}{|x|}\biggr\| \geq C, $ which then implies that $\|x\| \geq C|x|$. I understand the last implication but I'm not sure about the first part: Why we choose to define $C$ in such a way, and why this implies that $\biggr\|\frac{x}{|x|}\biggr\| \geq C. $ I suppose that if we only deal with $x$ such that $|x|=1$, then it makes some sense, but this should hold for all $x$.

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Since $x\mapsto\lVert x\rVert$ is continuous and since $\{x\in\mathbb{R}^n\,|\,\lvert x\rvert=1\}$ is compact, the function $x\mapsto\lVert x\rVert$ has a minimum there. Let $C$ be that minimum. Then, if $x\neq0$, since $\left\lvert\frac x{\lvert x\rvert}\right\rvert=1$, $\left\lVert\frac x{\lvert x\rvert}\right\rVert\geqslant C$, which is equivalent to the assertion $\lVert x\rVert\geqslant C\lvert x\rvert$.

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    $\begingroup$ Note that this implies that all norms on $\mathbb R^n$ are equivalent. $\endgroup$ – Math1000 Mar 1 '19 at 23:53

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