4
$\begingroup$

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?

$\endgroup$
1
3
$\begingroup$

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 $$\frac1{M}\|x\| \le p(x), \quad\forall x\in X$$

$\endgroup$
5
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.