# Generalized Hlawka inequality

Let $E$ be a vector space over the real (the complex case is interesting too). We consider functions $f:E\rightarrow\mathbb R$ which satisfy the homogeneity property $$f(\lambda x)=|\lambda|\,f(x).$$ In particular, we have $f(0)=0$.

I am interested in the generalized Hlawka inequality (GH$n$) at order $n$. Given $n$ vectors $x_1,\ldots,x_n$, and $I\subset[\![1,n]\!]$, define $x_I=\sum_{i\in I}x_i$. Then (GH$n$) says that $$\sum_{I\subset[\![1,n]\!]}(-1)^{{\rm card}\, I}f(x_I)\le0,\qquad\forall x_1,\ldots,x_n\in E.$$ For instance, (GH1) and (GH2) are respectively $0\le f(x)$ and $f(x+y)\le f(x)+f(y)$; thus, a function $f$ satisfying (GH1) and (GH2) is a norm over $E$. (GH3) is Hlawka inequality, which is satisfied by Euclidian norms and some others, though not by all norms.

Let me point out that (GH$n$) is sharp, in the sense that the equality holds true when either one vector is $0$, or all the vectors are positively collinear. If $n$ is odd, it is also an equality when $\sum_1^nx_i=0$, while if $n$ is even, this choice in (GH$n$) yields (GH$(n-1)$). One may also derive (GH2) from (GH3) by choosing $x_3=x_1$.

First question :

Do Euclidian norms satisfy (GH$n$) for all $n$ ?

Second question :

Conversely, if a norm satisfies (GH$n$) for all $n$, is it Euclidian ?

Third question :

Can we derive (GH$(n-1)$) from (GH$n$) when $n$ is odd ? (True if $n=3$).

Edit. Guillaume, I liked your suggestion. It does give a (new ?) proof of the classical Hlawka inequality ($n=3$). However it fails as soon as $n=4$. A counter-example is $(1,1,1,-2)$, where the sum is $2$. Therefore the answer to the very first question above is No, even in one space-dimension.

• If these inequalities are true when E is 1-dimensional, then by integration they are valid for any $L^1$ space, and then also for the Euclidean space ($L^2$ embeds isometrically as a linear subspace in $L^1$) – Guillaume Aubrun Oct 13 '14 at 13:31
• See mathoverflow.net/questions/167685 , especially Bill Johnson's comments. – Emil Jeřábek supports Monica Oct 13 '14 at 14:31
• @Emil. Of course. – Denis Serre Oct 13 '14 at 14:37
• @Guillaume. This actually gives a proof of the classical Hlawka inequality ($n=3$). Notice that if this works, my second question has a negative answer. – Denis Serre Oct 13 '14 at 14:59

I studied the inequality in case $n=3$, see
You can see from this paper that not every subadditive function ($n=2$) satisfies inequality for $n=3$. So (GH3) is not equivalent to (GH1) and (GH2). I don't know what happens for greater $n$.