I'm reading Wikipedia's article on Minkowski's functional. They state that if the set K used in defining Minkowski's functional pK is convex then pK is sub-additive. They argue as follows:
Suppose for the moment that pK(x) = pK(y) = r. Then for all ε > 0, we have x, y ∈ (r + ε) K = K' . The assumption that K is convex means K' is also. Therefore ½ x + ½ y is in K' . By definition of the Minkowski functional pK, one has
$p_K\left( \frac{1}{2} x + \frac{1}{2} y\right) \le r + \epsilon = \frac{1}{2} p_K(x) + \frac{1}{2} p_K(y) + \epsilon . $
But the left hand side is ½ pK(x + y), i.e. the above becomes
$p_K(x + y) \le p_K(x) + p_K(y) + \epsilon, \quad \mbox{for all} \quad \epsilon > 0.$
This is the desired inequality. The general case pK(x) > pK(y) is obtained after the obvious modification.
0) Let us examine this argument. When they say that "for all ε > 0, we have x, y ∈ (r + ε) K = K' " is the only place they use convexity, isn't that right? Furthermore in saying that $p_K\left( \frac{1}{2} x + \frac{1}{2} y\right)=$½ pK(x + y), they use homogeneity of the functional, don't they?
1) So how does subadditivity follow from convexity alone?
2) Furthermore, if we define the functional to be infinity whenever we take an infimum over the empty set, how does our set being absorbent guarantee finiteness?
3) Finally if K is balanced how come ($\lambda x \in K) \Leftrightarrow (x \in \frac{1}{|\lambda|} K$)?
How come $\inf \left\{r > 0: x \in r K \right\}
= \inf \left\{ \frac{1}{ | \lambda | }r > 0: x \in \frac{1}{|\lambda|}r K \right\}$? Do we need K to be balanced for the last equality?
Best Answer
1) They need convexity for $x,y \in K' \Rightarrow \frac12 x + \frac12y \in K'$.
2) Absorbent means that for each $x$, there exists $\alpha$ such that $x \in \alpha \, K$. Hence, you get finiteness.
3) Balanced means, that $K = -K$. From this, you obtain the equivalence. For the last inequality, you simply replace $r$ by $r/|\lambda|$. No balancedness needed.