# Prove inequality $C |x| \leq \|x\|$

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 $$x$$.

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 $$\lVert x\rVert\geqslant C\lvert x\rvert$$.
• Note that this implies that all norms on $\mathbb R^n$ are equivalent. – Math1000 Mar 1 '19 at 23:53