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Let $E$ be a finite dimensional normed vector space. If $E$ is $\ell^1$-embeddable, then the norm satisfies Hlawka inequality $${\bf(H)}\qquad\|x+y\|+\|y+z\|+\|z+x\|\le\|x\|+\|y\|+\|z\|+\|x+y+z\|,\qquad\forall x,y,z\in E.$$ This is how one can prove that every Euclidian space satisfies (H).

Now consider the Euclidian space ${\mathbb R}^n$.

Does $E={\mathcal L}({\mathbb R}^n)$, endowed with the operator norm, satisfy (H) ? If so, is it $\ell^1$-embeddable ?

Edit. After Mateusz' answer, the two-dimensional case remains open.

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  • $\begingroup$ So the question is whether this inequality holds for arbitrary $2\times 2$ matrices under the operator norm? $\endgroup$
    – Suvrit
    Oct 19, 2014 at 2:05
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    $\begingroup$ Some time ago I wrote a paper on functional Hlawka inequality: $$f(x+y) + f(x+z) + f(y+z) \leq f(x+y+z) + f(x) + f(y) + f(z)$$ see link.springer.com/article/10.1007%2Fs00010-012-0178-2 I collected there the some basic facts I found on the inequality (H). The problem was studied by some people in 70's and 80's. And it seems that a characterization of the spaces in wchich (H) is true is not known. $\endgroup$ Nov 3, 2014 at 12:29

2 Answers 2

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It does not hold, at least for $n \geq 3$. The reason is that $M_n$ contains then a subspace isometric to $\ell_{\infty}^3$ (diagonal matrices with three non-zero entries) and for this space the inequality does not hold -- take $x=(1,1,0)$, $y=(0,1,1)$ and $z=(1,0,1)$.

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  • $\begingroup$ Thanks ! This must be well-known from specialist, I presume. $\endgroup$ Oct 17, 2014 at 13:35
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The inequality is true for $2 \times 2$ self-adjoint matrices, because of the following formula, which is reminiscent of $\max(|a|,|b|)=\frac{1}{2}|a+b|+\frac{1}{2}|a-b|$ $$ \|A\| = \frac{1}{2} | \mathrm{tr} A | + \frac{1}{\sqrt{2}} \| P(A) \|_2 $$ where $\|\cdot\|_2$ is the Hilbert-Schmidt (or Frobenius) norm and $P(A)= A - \frac{\mathrm{tr} A}{2} \mathrm{Id}$ is the orthogonal projection onto the subspace of trace $0$ matrices. Both terms in this sum satisfy Hlawka's inequality (the Hilbert--Schmidt norm is a Euclidean norm).

Geometrically the unit ball of $2 \times 2$ self-adjoint matrices looks like a tridimensional double-cone over a disk (extreme points are reflections and $\pm \mathrm{Id}$), which is a geometric way to interpret the formula above. In the non-self-adjoint case I don't have such a clear picture.

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