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Let $(V,\rVert \lVert_V)$ be a $k$-dimensional linear space over $\mathbb R$. Let $\Phi:\mathbb R^k\to V$ a bijection. Choose a norm $\lVert \rVert$ for $\mathbb R^k$. Prove using Bolzano-Weierstrass that $$ \inf_{\{\vec x\in\mathbb R^k:\lVert \vec x\rVert=1\}}\lVert\Phi(\vec x)\rVert_V>0. $$ Conclude that there is a constant $c>0$, such that $c\Vert \vec x \Vert\leq\Vert \Phi(\vec x)\Vert_V$ for each $\vec x\in\mathbb R^k$.

I've chosen the Euclidean norm on the $k$-dimensional Euclidean space $\mathbb R^k$: \begin{align} \lVert\vec x\rVert=\sqrt{x_1^2+\dots+x_k^2}. \end{align} I've already shown the first part of the question. Now I need to find this $c>0$. I don't know how I can use the result of the first part. We know that for vectors $\vec x$ with norm 1, it holds that $\inf \lVert\Phi(\vec x)\rVert_V=r$, for some $r>0$. We could have shown this for any $t\in\mathbb R_{>0}$, I think; so for any vector $\vec x\neq \vec 0$, it holds that $\inf \lVert\Phi(\vec x)\rVert_V=r_{\vec x}$, for some $r_{\vec x}>0$. I really don't know how to continue from here on...

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    First give us a complete question. With the hypothesis stated here, it's completly false – 2017-02-23
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    $\Phi$ can be $any$ continuous map? Then it's not true. – 2017-02-23
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    My apologies @Tryss! $\Phi$ is a bijective, linear mapping – 2017-02-23

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Let $\inf_{x, \|x\| = 1} \| \Phi(x) \| = c > 0$

That means that $\forall x \text{ with } \|x\| = 1, \| \Phi(x) \| \geq c$

Not take $y \in \Bbb R^k, y\neq 0$, then

$$\| \Phi(y) \| = \| \Phi(\|y\| \frac{y}{\|y\|}) \| = \|y\|\| \Phi( \frac{y}{\|y\|}) \|$$

But $\| \frac{y}{\|y\|} \| = 1$, so

$$\| \Phi(y) \| = \|y\|\| \Phi( \frac{y}{\|y\|}) \| \geq c \|y\|$$

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    Ah right, so we use the linearity of both $\Phi$ and the norm. I would have not guessed this myself! Thanks a lot for working it out. – 2017-02-23