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Let $A\subset \mathbb{R}^n$ be any bounded, open, convex, and the centre symmetry set having centre at $0$, that is if $x\in A$, then $-x\in A$.

Show that $$\|x\| = \inf \{k>0 : x/k \in A \}$$ is a norm on $\mathbb{R^n}$ and the induced open ball $B$ centered at $0$ and radius $1$ covers $A$.

I don't know how to show triangle inequality and $\|tx\|=|t| \cdot \|x\|$ and also how to show that the induced unit ball covers $A$.

Surb
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Derrick
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2 Answers2

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Let $A\subset \Bbb R^n$ be a bounded, symmetric, convex, open set.

Let $\|x\|=\inf\{k>0 : x/k\in A\}$.

$\bullet$ $\|0\|=0$
$0\in A$ because $A$ is symmetric and convex (because $0=x/2-x/2\in A$). Thus $0/k \in A$ for every $k>0$ implying that $\|0\|=0$.

$\bullet$ $\|t x\|=|t|\|x\|\quad \forall t\in \Bbb R$
Let $t>0$, then it holds $$ \|tx\| = \inf\Big\{k>0 : \frac{tx}{k}\in A\Big\}= t\inf\Big\{\frac{l}{t}>0 : \frac{x}{\frac{l}{t}}\in A\Big\}=t\|x\|.$$ Now, as $A$ is symmetric, we have $x/k\in A$ if and only if $-x/k\in A$ which implies that $\|x\|=\|-x\|$. Hence, if $t<0$, then $$\|tx\|=\||t|(-x)\|=|t|\|-x\|=|t|\|x\|.$$

$\bullet$ $\|x+y\|\leq \|x\| + \|y\|$
Let $x,y\in \Bbb R^n$, then for every $k,l>0$ such that $x/k\in A$ and $y/l\in A$ it holds $$ \frac{k}{k+l}\frac{x}{k}+\frac{l}{k+l}\frac{y}{l}=\frac{x+y}{k+l}\in A$$ by convexity of $A$. It follows that \begin{align*} \|x+y\| &= \inf\Big\{m>0 : \frac{x+y}{m}\in A\Big\}\\  &\leq \inf\Big\{k>0 : \frac{x}{k}\in A\Big\}+\inf\Big\{l>0 : \frac{y}{l}\in A\Big\}\\ &=\|x\|+\|y\|.\end{align*}

$\bullet$ $\|x\|=0 \ \implies \ x=0$
If $\|x\|=0$, then $x/k\in A$ for every $k>0$. But $A$ is bounded, and so, if $x\neq 0$, we get a contradiction by letting $k\to 0$. Hence, we must have $x=0$.

Combining the first and last point we get $\|x\|=0$ if and only if $x=0$ which shows that $\|\cdot \|$ is a norm on $\Bbb R^n$.

Remark: The reverse direction of the exercise can be shown. Indeed, if $\|\cdot\|'$ is a norm on $\Bbb R^n$, then $A'=\{x\mid \|x\|'< 1\}$ is a bounded, symmetric, convex, open set.

For the last part, let $B=\{x\in \Bbb R^n\mid \|x\|<1\}$ and $z\in A$. Then, we have $z=\frac{z}{1}\in A$ implying that $\|z\| <1$, i.e. $z\in B$. It follows that $A\subset B$.

ViktorStein
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Surb
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    I couldn't think of the right linear combination of x/k and y/l, thanks! I think this result is really cool as it gives the idea of some more than the usual $||x||_p$ norms :-) Makes one wonder if the classic norms are somewhat "common" in the space of all norms of $\mathbb{R}^n$... – Ayutac Jan 03 '17 at 20:47
  • @Ayutac :) this "linear combination" is a kind of classical trick in convex analysis. I should say that I also felt in love with this result when I first read it. Well, the $p$-norms are definitely very particular cases (there are MANY MANY other norms). They basically represent a set of measure $0$. However, it is often difficult to find an interesting application for these other (more crazy) norms. – Surb Jan 03 '17 at 20:57
  • Where have you used the openness? – ViktorStein Jun 05 '20 at 19:40
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    @ViktorGlombik Good question! Openness is needed to ensure that $|x|$ is well defined, i.e. that for every $x$, there exists $k>0$ such that $x/k \in A$. Actually, the assumption "$A$ is open" could be replaced with "$A$ has nonempty interior". However, having $A$ open, resp. closed with non-empty interior, makes the equivalence more clear with open, resp. closed, unit balls induced by norms (see the remark in my answer). – Surb Jun 06 '20 at 06:36
  • @Surb can you please explain the inequality at triangle inequallity? – convxy Nov 02 '20 at 18:07
  • @ronkurman By definition of $\inf$, there exist $k\geq|x|, l \geq |y|$ such that $x/k,y/l \in A$ and $k-|x| < \epsilon/2$, $l-|y| < \epsilon/2$. Now, as $A$ is convex, for any $t\in (0,1)$ we have $tx/k+(1-t)y/l \in A$. In particular, this is true for $t^= k/(k+l)$ and we have $t^x/k+(1-t^*)y/l=(x+y)/(k+l) \in A$. It follows that $|x+y| \leq k+l \leq |x|+|y| + \epsilon$. By letting $\epsilon\to 0$ you get the desired result. – Surb Nov 02 '20 at 19:03
  • @Surb Thank you very much :) – convxy Nov 03 '20 at 08:12
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The triangle inequality is a hard one, was just about to open a nearly duplicate of this question :-) My exercise reads:

Let $K\subseteq\mathbb{R}^n$ be a compact set such that $K$ is convex, symmetric and its open interior isn't empty. Set $||0||=0$ and $||x||=(\max\{t\in\mathbb{R}\;|\; tx\in K\})^{-1}$ for $x\neq 0$. Then $||.||$ is a norm on $\mathbb{R}^n$ and $\overline{K}_1(0) = K$.

Translated from Königsberger Analysis, 5th edition, p.44.

I would love contribution to this.

I can help however with the second and third part of the question.

$$||tx|| = \inf\{k>0:(tx)/k\in A\} = \inf\{k>0:x/(k/t)\in A\} \stackrel{(1)}{=} \inf\{k>0:x/(k/|t|)\in A\} = \inf\{|t|\ell>0:x/\ell\in A\} = |t|\inf\{\ell>0:x/\ell\in A\} = |t| \cdot||x||$$

(1) works if $A$ is symmetric, so we don't change the set the infimum is taken over. I'm not really sure what "the centre symmetry set having centre at 0" means but I guess my argument can be adjusted.

Now assume $x\in A$. Then $||x|| \leq 1$ because with $k=1$ we have $x/k=x\in A$. On the other hand, if $x\in\mathbb{R}^n$ and $||x||\leq 1$ and without loss of generality $x\neq 0$ then there exists a $1\leq k>0$ such that $x/k\in \overline{A}$ (because $\overline{A}$ is compact). We have $0\in A$ and since with $A$ convex $\overline{A}$ is also convex

$$k\cdot \frac{x}{k} + (1-k)\cdot 0 = x\in \overline{A}$$

holds. Therefore the closure of the open ball in point 0 with radius 1 equals $\overline{A}$. Because of the convexity and openess of $A$, the interior of $\overline{A}$ equals $A$ and therefore it is covered by the open ball.

The triangle question remains and I have no idea how to tackle it, although I'm pretty sure it relies on the convexity of $A$, but I couldn't quite capture it. Hints would be appreciated!

Ayutac
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