A function {\phi:[0,\infty[\rightarrow[0,\infty[} is called {p}-homogeneous, if for any {t>0} it holds that {\phi(tx) = t^{p}\phi(x)}. This implies that {\phi(x) = x^{p}\phi(1)} and thus, all {p}-homogeneous functions on the positive reals are just multiples of powers of {p}. If you have such a power of {p} you can form the {p}-norm

\displaystyle \|x\|_{p} = \left(\sum_{k}|x_{k}|^{p}\right)^{1/p}.

By Minkowski’s inequality, this in indeed a norm for {p\geq 1}.

If we have just some function {\phi:[0,\infty[\rightarrow[0,\infty[} that is not homogeneous, we could still try to do a similar thing and consider

\displaystyle \phi^{-1}\left(\sum_{k}\phi(|x_{k}|)\right).

It is easy to see that one needs {\phi(0)=0} and {\phi} increasing and invertible to have any chance that this expression can be a norm. However, one usually does not get positive homogeneity of the expression, i.e. in general

\displaystyle \phi^{-1}\left(\sum_{k}\phi(t|x_{k}|)\right)\neq t \phi^{-1}\left(\sum_{k}\phi(|x_{k}|)\right).

A construction that helps in this situation is the Luxemburg-norm. The definition is as follows:

Definition 1 (and lemma). Let {\phi:[0,\infty[\rightarrow[0,\infty[} fulfill {\phi(0)=0}, {\phi} be increasing and convex. Then we define the Luxemburg norm for {\phi} as

\displaystyle \|x\|_{\phi} := \inf\{\lambda>0\ :\ \sum_{k}\phi\left(\frac{|x_{k}|}{\lambda}\right)\leq 1\}.

Let’s check if this really is a norm. To do so we make the following observation:

Lemma 2 If {x\neq 0}, then {c = \|x\|_{\phi}} if and only if {\sum_{k}\phi\left(\frac{|x_{k}|}{c}\right) = 1}.

Proof: Basically follows by continuity of {\phi} from the fact that for {\lambda >c} we have {\sum_{k}\phi\left(\frac{|x_{k}|}{\lambda}\right) \leq 1} and for {\lambda<c} we have {\sum_{k}\phi\left(\frac{|x_{k}|}{\lambda}\right) > 1}. \Box

Lemma 3 {\|x\|_{\phi}} is a norm on {{\mathbb R}^{d}}.

Proof: For {x=0} we easily see that {\|x\|_{\phi}=0} (since {\phi(0)=0}). Conversely, if {\|x\|_{\phi}=0}, then {\limsup_{\lambda\rightarrow 0}\sum_{k}\phi\left(\tfrac{|x_{k}|}{\lambda}\right) \leq 1} but since {\lim_{t\rightarrow\infty}\phi(t) = \infty} this can only hold if {x_{1}=\cdots=x_{d}= 0}. For positive homogeneity observe

\displaystyle \begin{array}{rcl} \|tx\|_{\phi} & = & \inf\{\lambda>0\ :\ \sum_{k}\phi\left(\frac{|tx_{k}|}{\lambda}\right)\leq 1\}\\ & = & \inf\{|t|\mu>0\ :\ \sum_{k}\phi\left(\frac{|x_{k}|}{\mu}\right)\leq 1\}\\ & = & |t|\|x\|_{\phi}. \end{array}

For the triangle inequality let {c = \|x\|_{\phi}} and {d = \|y\|_{\phi}} (which implies that {\sum_{k}\phi\left(\tfrac{|x_{k}|}{c}\right)\leq 1} and {\sum_{k}\phi\left(\tfrac{|y_{k}|}{d}\right)\leq 1}). Then it follows

\displaystyle \begin{array}{rcl} \sum_{k}\phi\left(\tfrac{|x_{k}+y_{k}|}{c+d}\right) &\leq& \sum_{k}\phi\left(\tfrac{c}{c+d}\tfrac{|x_{k}|}{c} +\tfrac{d}{c+d}\tfrac{|y_{k}|}{d}\right)\\ &\leq& \tfrac{c}{c+d}\underbrace{\sum_{k} \phi\left(\tfrac{|x_{k}|}{c}\right)}_{\leq 1} + \tfrac{d}{c+d}\underbrace{\sum_{k}\phi\left(\tfrac{|y_{k}|}{d}\right)}_{\leq 1}\\ &\leq& 1 \end{array}

and this implies that {c+d \geq \|x+y\|_{\phi}} as desired. \Box

As a simple exercise you can convince yourself that {\phi(t) = t^{p}} lead to {\|x\|_{\phi} = \|x\|_{p}}.

Let us see how the Luxemburg norm looks for other functions.

Example 1 Let ‘s take {\phi(t) = \exp(t)-1}.

095_luxemburg_balls-figure0.png

The function {\phi} fulfills the conditions we need and here are the level lines of the functional {x\mapsto \phi^{-1}\left(\sum_{k}\phi(|x_{k}|)\right)} (which is not a norm):

095_luxemburg_balls-figure1

[Levels are {0.5, 1, 2, 3}]

The picture shows that this functional is not a norm ad the shape of the “norm-balls” changes with the size. In contrast to that, the level lines of the respective Luxemburg norm look like this:

095_luxemburg_balls-figure2

[Levels are {0.5, 1, 2, 3}]

 

Advertisements