Let $(X,\|\cdot\|_X)$ be a normed space and $K\subset X$ an open, convex set with $0\in K$. Let $p:X\rightarrow\mathbb{R}$ be the Minkowski-functional of $K$ defined by: $$p(x):=\inf\left\{a>0 | \frac{1}{a}x\in K\right\}$$i) Show that: If $K$ is symmetric$(-K=K)$ and bounded, then $p(x)$ is a norm in $X$, which is equivalent to $\|\cdot\|_X$

Do I have to show, that $p(x)$ is well defined? I thought about the following: If $x=0$: $$p(0)=\inf\{a>0|0\in K\}$$$$=\inf\{a>0\}$$$$=0$$ The positivity for $x\neq0$ is obvious. Then I found a theorem about the minkowski-functional:

Let $K$ be convex, open with $0\in K$ and $p(x)$ defined like above, then:

i) $p$ is sublinear,

ii) There is a $M>0 \forall x\in X: 0\le p(x)\le M\|x\|_X$

iii) $K=\{x\in X|p(x)<1\}$

With the sublinearity of $p$, the homogenity and the triangle inequality are clear. My problem here is: Where do I need the information, that $K$ is bounded?

Now I have to show that $p(x)$ is equivalent to $\|\cdot\|_X$. Which means:

There are $c,C>0$, so that $$c\|x\|_X\le p(x)\le C\|x\|_X$$ The theorem gives us a $C$ with $p(x)\le C\|x\|_X$ but I have no idea how to find the $c$ for the lower bound. Could someone help me with these problems?


To show that $p$ is well-defined, you have to show that the infimum exists: since $K$ is an open set and $0 \in K$, there exists $r > 0$ such that $\|x\|_X < r \implies x \in K$.

For $x \in X$ we have

$$\left\|\frac1a\cdot x\right\| = \frac1a \cdot \|x\| \xrightarrow{a \to \infty} 0$$

so for large enough $a$, it will be $\frac1a\cdot x \in K$ so the set $\left\{a > 0 : \frac1a\cdot x\in K \right\}$ is nonempty. It is certainly bounded from below by $0$, so the infimum exists. Furthermore, we have that $$\left\langle \frac{\|x\|}{r}, +\infty\right\rangle \subseteq \left\{a > 0 : \frac1a\cdot x\in K \right\}$$

This gives us $$p(x) \le \frac1r \|x\|, \quad\forall x\in X$$

The boundedness of $K$ is used precisely to find the constant $c$. Let $x_0 \in X \setminus \{0\}$. Since $K$ is bounded, there exists $M > r$ such that $\|x\| < M$ for all $x \in K$. For $a = \frac{\|x_0\|}{M}$ we have:

$$\left\|\frac1a\cdot x_0\right\| = \frac1a \cdot \|x_0\| = M$$

Therefore, $\frac1a\cdot x_0 \notin K$ for all $a \le \frac{\|x_0\|}{M}$ so $$\left\langle0, \frac{\|x_0\|}{M}\right]\subseteq \left\{a > 0 : \frac1a\cdot x_0\in K \right\}^c$$

Thus certainly $p(x_0) \ge \frac{\|x_0\|}{M}$.

Hence, since $x_0$ was arbitrary, we have $$\frac1{M}\|x\| \le p(x), \quad\forall x\in X$$

  • 1
    $\begingroup$ Thanks a lot. Just a few questions: What does $\langle\frac{||x||}{r},\infty\rangle$ mean? Is it an intervall? What is $r$ in this context? $\endgroup$
    – Tobi92sr
    Nov 9 '17 at 14:53
  • 1
    $\begingroup$ @Tobi92sr Yes, it is the interval $$\left\{t \in \mathbb{R} : t > \frac{\|x\|}r\right\}$$ $r$ was introduced earlier, it is a constant such that $\|x\| < r \implies x \in K$, or equivalently $B(0,r) \subseteq K$. $\endgroup$ Nov 9 '17 at 19:44
  • $\begingroup$ I see that for $a \le \frac{\| x_0 \|}{M}$ we have $$ \left\| \frac{1}{a} x_0 \right\| \ge \frac{M}{\| x_0 \|} \| x_0 \| = M.$$ This implies $\frac{1}{a} x_0 \not\in K$, i.e. $x_0 \not \in a K$. This implies $p(x_0) \ge a$, but as $a \le \frac{\| x_0 \|}{M}$, how do you conclude $p(x_0) \ge \frac{\| x_0 \|}{M}$? $\endgroup$
    – Ramanujan
    Jun 19 '20 at 11:30
  • $\begingroup$ Also with regards to what set do you take the complement? Is it $(0, \infty)$? $\endgroup$
    – Ramanujan
    Jun 19 '20 at 11:35
  • $\begingroup$ @ViktorGlombik $p(x_0) \ge a$ for all $a \le \frac{\|x_0\|}M$ so in particular for $a = \frac{\|x_0\|}{M}$ we have $p(x_0) \ge \frac{\|x_0\|}{M}$. Yes, the complement is taken within $\langle 0,\infty\rangle$. $\endgroup$ Jun 19 '20 at 21:55

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