2
$\begingroup$

I'm having trouble understanding the definition of Minkowski functional.

Let $K$ be a symmetric (i.e. if it contains $x$ it also contains $-x$) convex body in a linear space $V$. We define a function $p$ on $V$ as $$p(x) = \inf \{ \lambda \in \mathbb{R}_{> 0} : x \in \lambda K \}$$ Where $p$ is called the Minkowski functional.

Lax's gives the defintion of gauge (with respect to origin) as follows:

If $K \subset V$ is a convex set in a vector space with an interior point, the gauge $p_K$ is given by: $$p_K(x) = \inf a \quad a>0,\frac{x}{a} \in K$$

I'm having trouble breaking down the definitions in "plain english". My attempt:

  1. Minkowski functional -- choose a point in $K$. Take all reals, $\lambda$. $\lambda K$ is then a "scaled version of $K$". $p(x)$ is the "smallest" $\lambda$ such that $x$ is still in $\lambda K$. If I think about a unit ball at the origin and $x$ being the origin, is $p(x) = 0$? The set of all $\lambda$ , according to my understanding, will be all positive reals - the inf of which is zero.
  2. Gauge -- I'm getting mixed up by the effect of $a$ in the denominator.
$\endgroup$
1
$\begingroup$

First of all, I think the key to the trouble of your understanding is the following relation: $$ \forall a>0\; \frac{x}{a} \in K \iff x \in aK $$ So the definition you wrote for the Minkowski functional and the gauge Lax defines are essentially the same, note that you missed some key esssential properties: In the definition for the Minkowski functional you have the assume that $0 \in K$ is an interior point (we call such sets absorbing), since only then every vector $v \in V$ has some scalar small enough such that $v \in \lambda K$, otherwise the infimum maybe empty. The same thing goes for gauge: Lax explicitly mentions that he assumes the interior point is $0$.

Note that your intuition in 1) is reasoable (yes $p(0)=0$), but we are not only talking about points in $K$, but in the whole of $V$, this is used in the Geometric Version of the Hahn Banach Theorem for example.

$\endgroup$

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.