# Norm induced by convex, open, symmetric, bounded set in $\Bbb R^n$.

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 $A$.

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 $$A\subset B$$.

• 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 $\mathbb{R}^n$... Jan 3 '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 3 '17 at 20:57
• Where have you used the openness? Jun 5 '20 at 19:40
• @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 6 '20 at 6:36
• @Surb can you please explain the inequality at triangle inequallity? Nov 2 '20 at 18:07

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!