# Showing that $p(x):=\inf\left\{a>0|\frac{1}{a}x\in K\right\}$ is a norm under certain circumstances

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

• 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? Nov 9 '17 at 14:53
• @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 $B(0,r) \subseteq K$. Nov 9 '17 at 19:44
• 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 $p(x_0) \ge \frac{\| x_0 \|}{M}$? Jun 19 '20 at 11:30
• Also with regards to what set do you take the complement? Is it $(0, \infty)$? Jun 19 '20 at 11:35
• @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 $\langle 0,\infty\rangle$. Jun 19 '20 at 21:55