# The exact bound for the reverse isodiametric problem in 3-space

Arkadiy Aliev

###### Abstract

Let $K$ be a convex body in $\mathbb{R}^{3}$. We denote the volume of $K$ by
$Vol(K)$ and the diameter of $K$ by $Diam(K).$ In this paper we prove that
there exists a linear bijection $T:\mathbb{R}^{3}\to\mathbb{R}^{3}$ such that
$Vol(TK)\geq\frac{\sqrt{2}}{12}Diam(TK)^{3}$ with equality if $K$ is a
simplex, which was conjectured by Ende Makai Jr. As a corollary, we prove that
$Vol(K)\geq\frac{1}{12}\omega(K)^{3}$, where $K\subset\mathbb{R}^{3}$ is a
convex body and $\omega(K)$ is the lattice width of $K$. In addition, there
exists a three-dimensional simplex $\Delta\subset\mathbb{R}^{3}$ such that
$Vol(\Delta)=\frac{1}{12}\omega(\Delta)^{3}.$

## Organisation of the paper

The paper is organised in the following way. In the first section, we give
basic definitions and recall well-known results, which we will use later in
the proof. In the second section, we review the literature and the best-known
results on the reverse isodiametric problem and non-separable lattices. In the
sections 3-8 we give the proof of the reverse isodiamteric problem in 3-space.

###### Contents

  1. 1 Preliminaries
  2. 2 Introduction
  3. 3 The scheme of the proof
  4. 4 Preliminary calculations
  5. 5 The case $\lambda_{1}=0$
  6. 6 Maximal sets
  7. 7 The case of peculiar sets
  8. 8 The case $a_{12}=0$

##  1 Preliminaries

We call a convex compact set $\mathbb{K}\subset\mathbb{R}^{n}$ with non-empty
interior a convex body. Denote the volume of $K$ by $Vol(K)$ and the diameter
of $K$ by $Diam(K).$ The difference body of $K$ is defined as
$K-K:=\\{x-y:x,y\in K\\}$. We denote the identity matrix $3\times 3$ by
$Id_{3}$.

######  Notation 1.1.

Let $u,v$ be two vectors in $\mathbb{R}^{3}$. We denote their cross product as
$[u;v].$

We will also need some definitions from the geometry of numbers.
$\Lambda\subset\mathbb{R}^{n}$ is a lattice if $\Lambda=A\mathbb{Z}^{n}$ for
some $A\in GL(n,\mathbb{R})$. We also define $d(\Lambda):=|detA|.$ A lattice
of translates of $K$ is defined as $\Lambda+K=\\{x+y|x\in\Lambda,y\in K\\}$
and its density is defined as

$D(\Lambda,K):=\frac{Vol(K)}{d(\Lambda)}.$

We say that a lattice of translates of $K$ is non-separable if each affine
$(n-1)$-subspace in $\mathbb{R}^{n}$ meets $x+K$ for some $x\in\Lambda$. Non-
separable lattices were introduced by L. Fejes Tóth in [1].

######  Definition 1.1.

For a convex body $K\subset\mathbb{R}^{n}$ we define

$d_{n,n-1}(K):=\min\\{d(\Lambda,K)|\Lambda+K\text{ is non-separable}\\}.$

######  Definition 1.2.

Lattice width $\omega_{K}:\mathbb{Z}^{n}\setminus{0}\to\mathbb{R}$ is defined
as

$\omega_{K}(u)=\max_{x,y\in K}\langle u,x-y\rangle.$
$\omega(K)=\min_{u\in\mathbb{Z}^{n}\setminus 0}\omega_{K}(u).$

The following propositions are well-known.

######  Proposition 1.1 ([2]).

Let the minimal volume ellipsoid containing $K-K\subset\mathbb{R}^{3}$ is the
unit ball $B$. Then there exist six points $u_{1},..,u_{6}\in(K-K)\cap\partial
B$ such that for some non-negative $\lambda_{1},..,\lambda_{6}$

$\sum_{i=1}^{6}\lambda_{i}u_{i}u_{i}^{T}=Id_{3}.$

We say that the set of vectors $u_{1},..,u_{6}$ decompose the identity matrix.

######  Proposition 1.2 ([1]).

$K+\mathbb{Z}^{3}$ is non-separable if and only if $\omega(K)\geq 1$. And
therefore $K+\mathbb{Z}^{3}$ is non-separable if and only if
$\frac{1}{2}(K-K)+\mathbb{Z}^{3}$ is non-separable.

######  Proposition 1.3 ([3]).

Let $B$ be the unit ball in $\mathbb{R}^{3}.$ Then

$d_{3,2}(B)=\frac{\pi}{6\sqrt{2}}.$

######  Proposition 1.4.

Let $K\subset\mathbb{R}^{d}$ be a convex body and
$\underline{z}^{i},\overline{z}^{i}\in K$ for $i=1,..,d.$ And let
$y_{i}=\overline{z}^{i}-\underline{z}^{i}.$ Then it holds that

$Vol(K)\geq\frac{|det(y_{1},..,y_{d})|}{d!}.$

###### Proof of Proposition 1.4.

Let us prove this statement by induction. If $d=1$ then there is nothing to
prove. Let $\Pi:=y_{d}^{\perp}$ and $P$ be the orthogonal projection onto
$\Pi$. Then

$|det(y_{1},..,y_{d})|=|y_{d}|\cdot|det(Py_{1},..,Py_{d-1})|$

and by (1.3) in [4] we have

$Vol(K)\geq\frac{1}{d}|y_{d}|Vol(PK)\geq\frac{1}{d}|y_{d}|\frac{|det(Py_{1},..,Py_{d-1})|}{(d-1)!}=\frac{|\det(y_{1},..,y_{d})|}{d!}.$

∎

##  2 Introduction

The isodiamteric inequality is a classic result in convex geometry, which
states that the Euclidean ball in $\mathbb{R}^{n}$ has the maximum volume
among all convex bodies of the same diameter, but it is impossible to give a
non-trivial lower bound for the volume of a convex body in terms of its
diameter. However, if one first minimizes the diameter using a volume-
preserving linear mapping, then a non-trivial bound exists.

######  Definition 2.1.

Let $K\subset\mathbb{R}^{n}$ be a convex body. Its isodiametric quotient is
defined as

$idq(K)=\frac{Vol(K)}{Diam(K)^{n}}.$

Endre Makai Jr. [3] posed the conjecture, which states that there exists a
linear image of a convex body $K$ such that its isodiametric quotient is at
least as large as that of a regular simplex.

######  Conjecture 2.1 (Makai Jr. [3]).

For every convex body $K\subset\mathbb{R}^{n}$ there exist a linear bijection
$T$ such that

$idq(TK)\geq\frac{\sqrt{n+1}}{n!2^{\frac{n}{2}}}.$

The two dimensional case of the conjecture was found by Behrend [5]. Cases of
dimension three and more still remain open and the best results in them were
obtained by Merino B.G. and Schymura M. in [6].

######  Proposition 2.1 (Merino-Schymura, [6]).

For a convex body $K$ there exists a linear bijection $T$ such that

$idq(TK)\geq\frac{\left(\frac{n+1}{2}\right)^{\frac{n}{2}}}{\binom{n(n+1)/2}{n}^{1/2}n!}\sim\sqrt{e}\left(\frac{2\pi}{n}\right)^{\frac{1}{4}}\frac{\sqrt{n+1}}{n!e^{\frac{n}{2}}}.$

In the three dimensional case they proved that for convex body
$K\subset\mathbb{R}^{3}$ there exists a linear bijection $T$ such that
$idq(TK)\geq\frac{\sqrt{10}}{30}$, which is very close to
$\frac{\sqrt{2}}{12}$. In this paper, we prove the Makai’s conjecture in
dimension three.

######  Theorem 2.1.

Let $K$ be a convex body in $\mathbb{R}^{3}$. Then there exists a linear
bijection $T:\mathbb{R}^{3}\to\mathbb{R}^{3}$ such that

$Vol(TK)\geq\frac{\sqrt{2}}{12}Diam(TK)^{3}.$

Note that if a convex body $K\subset\mathbb{R}^{n}$ is centrally symmetric,
then it is known that there exists a linear bijection $T$ such that
$idq(TK)\geq\frac{1}{n!}$ with equality if and only if $TK$ is a regular
crosspolytope (see [7]).

The estimates of $d_{n,n-1}$ are related to the Mahler’s conjecture and have
been well studied in the two-dimensional case (see [8, 9]). Namely, for a
convex body $K\subset\mathbb{R}^{2}$ we have

$\frac{3}{8}\leq d_{2,1}(K)\leq\frac{\pi\sqrt{3}}{8},$

and if $K$ is centrally symmetric, then

$d_{2,1}(K)\geq\frac{1}{2}.$

The proof of Mahler’s conjecture in dimension three gives us an estimate of
$d_{3,2}$ in the centrally symmetric case, but the general case remains open.
As a consequence of the reverse isodiametric inequality, we obtain the
following result.

######  Corollary 2.1.

Let $K$ be a convex body in $\mathbb{R}^{3}$. Then

$d_{3,2}(K)\geq\frac{1}{12}.$

In addition, if $K$ is a simplex, then $d_{3,2}(K)=\frac{1}{12}.$

The next corollary is an estimate of the volume of a convex body in terms of
its lattice width, which is of independent interest due to its connection to
the Newton polyhedrons of surfaces with $m$-singularity. There are estimates
of the area of the Newton polygons in two dimensions (see [10, 11]), but in
three dimensions the exact estimates remained unknown. We prove the following
result.

######  Corollary 2.2.

Let $K$ be a convex body in $\mathbb{R}^{3}$. Then

$Vol(K)\geq\frac{1}{12}\omega(K)^{3}.$

In addition, if
$K=conv\left\\{(0,0,0),\left(1,\frac{1}{2},\frac{1}{2}\right),\left(\frac{1}{2},1,\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{2},1\right)\right\\}$,
then $Vol(K)=\frac{1}{12}\omega(K)^{3}.$

##  3 The scheme of the proof

In this section we outline the proof of Theorem 2.1. First, we define the
required linear mapping $T$. Let $\mathcal{E}$ be the minimal volume ellipsoid
containing $K-K$ and $B$ be the unit ball in $\mathbb{R}^{3}.$ $T$ is a linear
bijection such that $T\mathcal{E}=B.$ Then by the famous John’s theorem we get
the existence of contact points $u_{1},..,u_{6}\in(TK-TK)\cap\partial B$ and
non-negative numbers $\lambda_{1},..,\lambda_{6},$ such that

$\sum_{i=1}^{6}\lambda_{i}u_{i}u_{i}^{T}=Id_{3}.$

Then $TK$ contains shifted copies of $u_{1},..,u_{6}$ and $Diam(TK)=1$.
Therefore by Proposition 1.4 it suffices to prove that
$\max_{i,j,k}|det(u_{i},u_{j},u_{k})|\geq\frac{1}{\sqrt{2}}.$

Let $\lambda_{6}=\max_{i}\lambda_{i}$ and $a_{ij}:=det(u_{i},u_{j},u_{6}).$
Then by Lemma 4.1 and Lemma 4.2 we have the following equalities.

$a_{13}a_{24}-a_{14}a_{23}=a_{12}a_{34};$
$a_{13}a_{25}-a_{15}a_{23}=a_{12}a_{35};$
$a_{14}a_{25}-a_{15}a_{24}=a_{12}a_{45};$
$a_{14}a_{35}-a_{15}a_{34}=a_{13}a_{45};$
$a_{24}a_{35}-a_{25}a_{34}=a_{23}a_{45};$

$\sum_{1\leq i<j\leq 5}\lambda_{i}\lambda_{j}a_{ij}^{2}=1.$

Under this restrictions it suffices to prove that
$\max_{i,j}|a_{ij}|\geq\frac{1}{\sqrt{2}}.$ It turns out to be more convenient
to solve the following problem:

The optimization problem. Let $\\{a_{ij}\\}_{1\leq i<j\leq 5}$ be a set of
$10$ numbers, such that the list of equalities from Lemma 4.1 holds and
$\max_{1\leq i<j\leq 5}|a_{ij}|\leq 1$. Find the maximum value of

$\sum_{1\leq i<j\leq 5}\lambda_{i}\lambda_{j}a_{ij}^{2}.$

If the maximum value equals $2$, then we win. Our goal is to prove this.

######  Definition 3.1.

We say that the set $\\{a_{ij}\\}_{1\leq i<j\leq 5}$ is admissible, if
$\max_{1\leq i<j\leq 5}|a_{ij}|\leq 1$ and the following holds.

$a_{13}a_{24}-a_{14}a_{23}=a_{12}a_{34};$
$a_{13}a_{25}-a_{15}a_{23}=a_{12}a_{35};$
$a_{14}a_{25}-a_{15}a_{24}=a_{12}a_{45};$
$a_{14}a_{35}-a_{15}a_{34}=a_{13}a_{45};$
$a_{24}a_{35}-a_{25}a_{34}=a_{23}a_{45}.$

######  Definition 3.2.

We say that the admissible set $\\{a_{ij}\\}_{1\leq i<j\leq 5}$ is maximal if
it maximizes the value of $\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}a_{ij}^{2}$ among all admissible sets.

The proof of the estimate is divided into four parts. The first part, which is
Section 5, is related to the case where there is at least one zero among
$\lambda_{1},..,\lambda_{6}.$ The second part, which is Section 6, is related
to finding special conditions for maximal sets. If all $\lambda_{i}$ are
positive, then there are only two types of maximal sets: "peculiar sets" 6.1
and those with at least one zero. The third part, which is Section 7, is
related to the case where the set $\\{a_{ij}\\}$ is "peculiar". And the last
fourth part, which is Section 8 is related to the case where the set
$\\{a_{ij}\\}$ contains zero. Each of the cases is based on estimates of sums
of lambda products, which are proved in the Lemmas 4.3, 4.4, 4.5, 4.6. After
solving all this cases, Theorem 2.1 will be proved.

Summing up, the remaining steps are the following:

  1. 1.

Solve the case $\lambda_{1}=0.$ It is done in Proposition 5.1.

  2. 2.

If $\lambda_{i}>0$ for all $i$, prove that there are only two types of maximal
sets: "peculiar sets" and those with at least one zero. It is done in
Proposition 6.1.

  3. 3.

Solve the case of peculiar sets. It is done in Proposition 7.1.

  4. 4.

Solve the case $a_{12}=0$. It is done in Proposition 8.1.

During proofs, the words we may ignore some expression will often be used.
What we mean by that words is the following. Assume, for example, that we have
a function

$(x,y,z)\mapsto ax^{2}+by^{2}+cz^{2}$

with the domain $[-1;1]^{3}\cap\\{x+y+z=0\\}$ and $a,b,c>0$. We want to
estimate it as $a+b+c-\min(a,b,c).$ We know the obvious estimate $a+b+c$, but
we want to be more precise. This function is convex, therefore it suffices to
consider only

$(x,y,z)\in\\{(1,-1,0),(-1,1,0),(1,0,-1),(-1,0,1),(0,1,-1),(0,-1,1)\\}.$

In each of this cases either $a$ or $b$ or $c$ comes with zero coefficient.
And this means that we may ignore either $a$, or $b$, or $c$.

###### Proof of Corollaries 2.1 and 2.2.

This reasoning is well known, we present it for completeness.

Let $K+\Lambda$ is non-separable and let $T:\mathbb{R}^{3}\to\mathbb{R}^{3}.$
be a linear bijection such that

$Vol(TK)\geq\frac{\sqrt{2}}{12}Diam(TK)^{3}.$

Then $TK+T\Lambda$ is non-separable and therefore $\frac{1}{2}(TK-
TK)+T\Lambda$ is non-separable. Since $\frac{1}{2}(TK-TK)\subset
B_{\frac{1}{2}Diam(TK)}$, $B_{\frac{1}{2}Diam(TK)}+T\Lambda$ is non-separable.
Therefore its density is at least $\frac{\pi}{6\sqrt{2}}$ by Proposition 1.3.
Then we have the following chain of inequalities.

$\frac{Vol(K)}{det\Lambda}=\frac{Vol(TK)}{detT\Lambda}\geq\frac{\sqrt{2}}{12}\frac{Diam(TK)^{3}}{det(T\Lambda)}=\frac{8\sqrt{2}}{12\frac{4}{3}\pi}\frac{Vol(B_{\frac{1}{2}Diam(TK)})}{det(T\Lambda)}\geq\frac{8\sqrt{2}}{12\frac{4}{3}\pi}\frac{\pi}{6\sqrt{2}}=\frac{1}{12}.$

Therefore $d_{32}(K)\geq\frac{1}{12}.$ And, since $K+\mathbb{Z}^{3}$ is non-
separable if and only if $\omega(K)\geq 1$, we have
$Vol(K)\geq\frac{1}{12}\omega(K)^{3}.$ If $K$ is a simplex, then
$d_{3,2}(K)=\frac{1}{12}$ by Proposition 3.1 in [8].

∎

##  4 Preliminary calculations

Let $u_{1},..,u_{6}$ be unit vectors in $\mathbb{R}^{3}$ and for some non-
negative $\lambda_{1},..,\lambda_{6}$ we have

$\sum_{i=1}^{6}\lambda_{i}u_{i}u_{i}^{T}=Id_{3}.$

Then from the trace comparison we get $\sum_{i=1}^{6}\lambda_{i}=3.$ Note that
$u_{i}u_{i}^{T}\in\mathbb{R}^{3\times 3}.$

######  Notation 4.1.

We denote $det(u_{i},u_{j},u_{6})$ by $a_{ij}$. Note that $a_{ij}=-a_{ji}.$

We will need some properties of $a_{ij}$. Firstly, we have determinant-like
equations:

######  Lemma 4.1.

The following list of equalities holds.

$a_{13}a_{24}-a_{14}a_{23}=a_{12}a_{34};$
$a_{13}a_{25}-a_{15}a_{23}=a_{12}a_{35};$
$a_{14}a_{25}-a_{15}a_{24}=a_{12}a_{45};$
$a_{14}a_{35}-a_{15}a_{34}=a_{13}a_{45};$
$a_{24}a_{35}-a_{25}a_{34}=a_{23}a_{45}.$

###### Proof.

We can assume that $u_{6}=(0,0,1)^{T}$. Let us rewrite $u_{i}$ as
$u_{i}=(x_{i},y_{i},*).$ Then we have

$a_{12}=x_{1}y_{2}-x_{2}y_{1};$ $a_{13}=x_{1}y_{3}-x_{3}y_{1};$
$a_{14}=x_{1}y_{4}-x_{4}y_{1};$ $a_{23}=x_{2}y_{3}-x_{3}y_{2};$
$a_{24}=x_{2}y_{4}-x_{4}y_{2};$ $a_{34}=x_{3}y_{4}-x_{4}y_{3}.$

Direct calculation completes the proof. ∎

Secondly, from the decomposition of the identity matrix, we obtain the
following:

######  Lemma 4.2.

$\sum_{1\leq i<j\leq 5}\lambda_{i}\lambda_{j}a_{ij}^{2}=1.$

###### Proof.

From [12] we have the following equality:

$\sum_{1\leq i<j\leq
6}\lambda_{i}\lambda_{j}[u_{i};u_{j}][u_{i};u_{j}]^{T}=Id_{3},$

where $[u_{i};u_{j}]$ is the cross product of $u_{i},u_{j}$. Then since
$|u_{6}|=1$ we get

$\sum_{1\leq i<j\leq
6}\lambda_{i}\lambda_{j}[u_{i};u_{j}]\langle[u_{i};u_{j}],u_{6}\rangle=u_{6}$

$\sum_{1\leq i<j\leq 5}\lambda_{i}\lambda_{j}a_{ij}^{2}=\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}\langle[u_{i};u_{j}],u_{6}\rangle^{2}=1.$

∎

Now let us move on to optimizing the sum of lambda products, which we will
often use in the main proof.

######  Lemma 4.3.

Let $\lambda_{6}=\max_{i}\lambda_{i}$ and $k,l,m,n$ be four different numbers
from $\\{1,2,3,4,5\\}$.Then we have

$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}-\lambda_{k}\lambda_{l}-\lambda_{m}\lambda_{n}\leq 2.$

###### Proof.

We can assume that $\lambda_{1}\leq\lambda_{2}\leq..\leq\lambda_{5}$ and since
under such assumptions $\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{3}$ is the
smallest sum, we can also put $n=1,m=4,k=2,l=3$.

$(\lambda_{1}+\lambda_{4})(\lambda_{2}+\lambda_{3}+\lambda_{5})+\lambda_{5}(\lambda_{2}+\lambda_{3})\leq(\lambda_{1}+\lambda_{4})\left(\lambda_{2}+\lambda_{3}+\frac{\lambda_{5}+\lambda_{6}}{2}\right)+\frac{\lambda_{5}+\lambda_{6}}{2}(\lambda_{2}+\lambda_{3})$

Let $a=\lambda_{1}+\lambda_{4}$, $b=\lambda_{2}+\lambda_{3}$. Then $a+2b\leq
3$ and $2a+b\leq 3$. Our goal is to estimate the following expression

$a\left(b+\frac{3}{2}-\frac{a+b}{2}\right)+\left(\frac{3}{2}-\frac{a+b}{2}\right)b=\frac{3}{2}(a+b)-\frac{1}{2}(a^{2}+b^{2}).$

It suffices to consider the case where $a\leq b$ and $a+2b=3$.

$\frac{3}{2}(a+b)-\frac{1}{2}(a^{2}+b^{2})=\frac{9}{2}b-\frac{5}{2}b^{2}\leq
2\text{ for $b\geq 1$.}$

∎

######  Lemma 4.4.

Let $\lambda_{6}=\max_{i}\lambda_{i}$ and $k,l,n$ be three different numbers
from $\\{1,2,3,4,5\\}$. Then we have

$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}-\lambda_{k}\lambda_{l}-\lambda_{l}\lambda_{n}-\lambda_{k}\lambda_{n}\leq\frac{9}{5}<2.$

###### Proof.

We can assume that $\lambda_{1}\leq\lambda_{2}\leq..\leq\lambda_{5}$ and that
$k=1,l=2,n=3$. Our goal is to find the upper bound for

$(\lambda_{1}+\lambda_{2}+\lambda_{3})(\lambda_{4}+\lambda_{5})+\lambda_{4}\lambda_{5}.$

Let $a:=\frac{1}{3}(\lambda_{1}+\lambda_{2}+\lambda_{3})$ and
$b=\frac{1}{3}(\lambda_{4}+\lambda_{5}+\lambda_{6})$. Then we have $a+b=1,$
$a\leq b$ and

$(\lambda_{1}+\lambda_{2}+\lambda_{3})(\lambda_{4}+\lambda_{5})+\lambda_{4}\lambda_{5}\leq
6ab+b^{2}=6b-5b^{2}\leq\frac{9}{5}.$

∎

As a consequence, we obtain the following lemma.

######  Lemma 4.5.

Let $\lambda_{6}=\max_{i}\lambda_{i}$ and $\lambda_{1}=0$. $k\neq
l\in\\{2,3,4,5\\}$ Then we have

$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}-\lambda_{k}\lambda_{l}\leq\frac{9}{5}<2.$

###### Proof.

$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}-\lambda_{k}\lambda_{l}=\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}-\lambda_{k}\lambda_{l}-\lambda_{k}\lambda_{1}-\lambda_{l}\lambda_{1}\leq\frac{9}{5}$

∎

######  Lemma 4.6.

Let $\lambda_{6}=\max_{i}\lambda_{i}$. Then we have

$\frac{3}{5}(\lambda_{1}\lambda_{5}+\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{3}+\lambda_{2}\lambda_{5}+\lambda_{3}\lambda_{4})+\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{4}+\lambda_{3}\lambda_{5}+\lambda_{4}\lambda_{5}\leq
2$

###### Proof.

Applying Lemma 4.3 for

$(k,l,m,n)=(1,4,2,3),(1,4,2,5),(1,5,2,3),(1,5,3,4),(2,5,3,4)$

gives us the following:

$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}-\lambda_{1}\lambda_{4}-\lambda_{2}\lambda_{3}\leq 2;$
$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}-\lambda_{1}\lambda_{4}-\lambda_{2}\lambda_{5}\leq 2;$
$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}-\lambda_{1}\lambda_{5}-\lambda_{2}\lambda_{3}\leq 2;$
$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}-\lambda_{1}\lambda_{5}-\lambda_{3}\lambda_{4}\leq 2;$
$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}-\lambda_{2}\lambda_{5}-\lambda_{3}\lambda_{4}\leq 2.$

The desired estimate is obtained by summation.

∎

##  5 The case $\lambda_{1}=0$

######  Proposition 5.1.

Let $\lambda_{1}=0$ and $\lambda_{6}=\max_{i}\lambda_{i}$. Then for an
admissible set $\\{a_{ij}\\}_{1\leq i<j\leq 5}$ we have

$\sum_{1\leq i<j\leq 5}\lambda_{i}\lambda_{j}a_{ij}^{2}\leq\frac{9}{5}<2.$

###### Proof.

We can assume that $\lambda_{2}\leq\lambda_{3}\leq\lambda_{4}\leq\lambda_{5}.$
Then we can evaluate pairs of terms as follows:

$\lambda_{2}\lambda_{4}a_{24}^{2}+\lambda_{3}\lambda_{5}a_{35}^{2}\leq\lambda_{2}\lambda_{4}(a_{24}a_{35})^{2}+\lambda_{3}\lambda_{5};$

$\lambda_{2}\lambda_{3}a_{23}^{2}+\lambda_{4}\lambda_{5}a_{34}^{2}\leq\lambda_{2}\lambda_{3}(a_{23}a_{45})^{2}+\lambda_{4}\lambda_{5};$

$\lambda_{2}\lambda_{5}a_{25}^{2}+\lambda_{3}\lambda_{4}a_{34}^{2}\leq\min(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4})(a_{25}a_{34})^{2}+\max(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4}).$

Let $u=a_{24}a_{35},v=a_{23}a_{45};w=a_{25}a_{34}.$ $u-v=w.$ Consider the
following cases:

Case 1: $uv\leq 0$.

$\lambda_{2}\lambda_{4}u^{2}+\lambda_{2}\lambda_{3}v^{2}\leq\lambda_{2}\lambda_{4}(u^{2}+v^{2})\leq\lambda_{2}\lambda_{4}(u-v)^{2}\leq\lambda_{2}\lambda_{4}.$

$\min(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4})(a_{25}a_{34})^{2}+\max(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4})\leq\lambda_{2}\lambda_{5}+\lambda_{3}\lambda_{4}$

Therefore we get

$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}a_{ij}^{2}\leq\lambda_{3}\lambda_{5}+\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{4}+\lambda_{2}\lambda_{5}+\lambda_{3}\lambda_{4}\leq\frac{9}{5},$

where the last inequality was obtained using Lemma 4.5.

Case 2: $uv>0$. Without loss of generality, we can assume that $u,v>0$.

Case 2.1 $w=u-v\leq 0$.

Then $uw\leq 0$ and

$\min(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4})w^{2}+\lambda_{2}\lambda_{4}u^{2}$

$\leq\min(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4})(w^{2}+u^{2})\leq\min(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4})(w-u)^{2}\leq\min(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4});$

Therefore we get

$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}a_{ij}^{2}\leq\lambda_{3}\lambda_{5}+\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}+\lambda_{2}\lambda_{5}+\lambda_{3}\lambda_{4}\leq\frac{9}{5},$

where the last inequality was obtained using Lemma 4.5.

Case 2.2 $w=u-v\geq 0$. then $vw\geq 0.$

$\min(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4})w^{2}+\lambda_{2}\lambda_{3}v^{2}$
$\leq\min(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4})(w^{2}+v^{2})\leq\min(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4})(w+v)^{2}\leq\min(\lambda_{2}\lambda_{5},\lambda_{3}\lambda_{4})$

Therefore we get

$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}a_{ij}^{2}\leq\lambda_{3}\lambda_{5}+\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{4}+\lambda_{2}\lambda_{5}+\lambda_{3}\lambda_{4}\leq\frac{9}{5},$

where the last inequality was obtained using Lemma 4.5.

∎

##  6 Maximal sets

Our main goal in this section is to find the special properties of maximal
sets.

######  Definition 6.1.

We call the admissible set $\\{a_{ij}\\}_{1\leq i<j\leq 5}$ peculiar if after
some permutation of indices it holds that

$|a_{12}|=|a_{13}|=|a_{24}|=|a_{35}|=|a_{45}|=1;$

$|a_{15}|+|a_{14}|\geq 1;$

$|a_{23}|=\frac{|a_{14}|+|a_{15}|-1}{|a_{14}a_{15}|};$
$|a_{25}|=\frac{1-|a_{15}|}{|a_{14}|};$
$|a_{34}|=\frac{1-|a_{14}|}{|a_{15}|}.$

######  Proposition 6.1.

Assume that $\lambda_{i}>0$ for all $i\in\\{1,2,3,4,5\\}$ and that the set
$A=\\{a_{ij}\\}_{1\leq i<j\leq 5}$ is maximal. Then $A$ is peculiar or $A$
contains zeros.

###### Proof.

Consider the maximal set $\\{a_{ij}\\}_{1\leq i<j\leq 5}$ and assume that it
does not contain zero. Then we fix $a_{12}$ and define
$u=(a_{13},a_{14},a_{15})^{T},v=(a_{23},a_{24},a_{25})^{T},w=(-a_{45},a_{35},-a_{34})^{T}$.
Note that the following relation holds:

$w=\frac{1}{a_{12}}[u;v].$

Moreover, if we have a set without zeros with this relation and all its
elements lie in $[-1;1]$, then the set is admissible.

We define positive matriсes $A,B,C\in\mathbb{R}^{3\times 3}$ as follows:

$A=diag(\lambda_{1}\lambda_{3},\lambda_{1}\lambda_{4},\lambda_{1}\lambda_{5});$

$B=diag(\lambda_{2}\lambda_{3},\lambda_{2}\lambda_{4},\lambda_{2}\lambda_{5});$

$C=diag(\lambda_{4}\lambda_{5},\lambda_{3}\lambda_{5},\lambda_{3}\lambda_{4}).$

Let us rewrite $\sum_{1\leq i<j\leq 5}\lambda_{i}\lambda_{j}a_{ij}^{2}$ in
terms of new variables $u,v,[u;v],a_{12}$.

$\sum_{1\leq i<j\leq 5}\lambda_{i}\lambda_{j}a_{ij}^{2}=\langle
Au,u\rangle+\langle Bv,v\rangle+\frac{1}{a_{12}^{2}}\langle
C([u;v]),[u;v]\rangle+\lambda_{1}\lambda_{2}a_{12}^{2}.$

Since the set $\\{a_{ij}\\}$ is admissible, we know that $u,v,w\in[-1;1]^{3}.$
Our goal now is to show that $u,v\in\partial[-1;1]^{3}.$

Consider $tu,\frac{1}{t}v$. Then our expression becomes

$\sum_{1\leq i<j\leq 5}\lambda_{i}\lambda_{j}a_{ij}^{2}(t)=\alpha
t^{2}+\beta\frac{1}{t^{2}}+\gamma,$

for some $\alpha,\beta>0.$ Then since $\alpha
t^{2}+\beta\frac{1}{t^{2}}+\gamma$ is strictly convex, its maximum is reached
only at the boundary. Therefore $u\in\partial[-1;1]^{3}$ or
$v\in\partial[-1;1]^{3}.$ Let us assume that $u\in\partial[-1;1]^{3}.$
Further, let us fix $u$ and $[u;v]$. Then all $v\in[-1;1]^{3}$ such that
$[u;v]$ is fixed form a segment $\\{(1-s)v_{0}+sv_{1}|t\in[0;1]\\}$ inside the
cube with $v_{0},v_{1}\in\partial[-1;1]^{3}.$ So let
$v_{s}=(1-s)v_{0}+sv_{1}.$ Then our expression becomes

$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}a_{ij}^{2}(s)=\tilde{\alpha}s^{2}+\tilde{\beta}s+\tilde{\gamma},$

where $\tilde{\alpha}=\langle B(v_{1}-v_{0}),v_{1}-v_{0}\rangle\geq 0.$ If
$\tilde{\alpha}=0$, then $v_{t}=v_{0}=v_{1}\in\partial[-1;1]^{3}$. If
$\tilde{\alpha}>0$, then, since
$\tilde{\alpha}s^{2}+\tilde{\beta}s+\tilde{\gamma}$ is strictly convex, its
maximum is reached only at the boundary. Therefore $v\in\partial[-1;1]^{3}$.

We showed that if the set $\\{a_{ij}\\}_{1\leq i<j\leq 1}$ is maximal and
$a_{12}\neq 0$, then

$(a_{13},a_{14},a_{15})^{T},(a_{23},a_{24},a_{25})^{T}\in\partial[-1;1]^{3}.$

Similarly, if $i,k,l,m$ are four distinct indices from $\\{1,2,3,4,5\\}$ and
the set $\\{a_{ij}\\}_{1\leq i<j\leq 1}$ is maximal, then
$(a_{ik},a_{il},a_{im})^{T}\in\partial[-1;1]^{3}$. Note that if $i>j$, then
$a_{ij}:=-a_{ji}$.

The remaining step is to use the relations from Lemma 4.1 and the condition

$(a_{ik},a_{il},a_{im})^{T}\in\partial[-1;1]^{3}$

to understand how does a maximal set look like.

There are at least two ones among $|a_{12}|,|a_{13}|,|a_{14}|,|a_{15}|$.
Assume without loss of generality that $|a_{12}|=|a_{13}|=1.$ Then there is at
least one $1$ among $|a_{23}|,|a_{24}|,|a_{25}|$. Since the case where
$|a_{24}|=1$ and the case where $|a_{25}|=1$ are the same, there are only two
cases: $|a_{23}|=1$ and $|a_{24}|=1.$

Case 23. ($|a_{12}|=|a_{13}|=|a_{23}|=1$). In this case there is at least one
$1$ among $|a_{42}|,|a_{43}|,|a_{45}|$. From symmetry, it suffices to consider
only two cases: $|a_{42}|=1$ and $|a_{45}|=1.$

Case 23.42 ($|a_{12}|=|a_{13}|=|a_{23}|=1$, $|a_{42}|=1$) The relation
$a_{13}a_{24}-a_{14}a_{23}=a_{12}a_{34}$ gives us $|a_{14}|+|a_{34}|=1.$
Further, there is at least one $1$ among $|a_{41}|,|a_{43}|,|a_{45}|$. If
$|a_{41}|=1$ or $|a_{43}|=1$, then from $|a_{14}|+|a_{34}|=1$ we can conclude
that the remaining equals $0$. Therefore the remaining case is $|a_{45}|=1.$
Then there is at least one $1$ among $|a_{15}|$, $|a_{25}|$, $|a_{35}|$.

Case 23.42.15($|a_{12}|=|a_{13}|=|a_{23}|=1$, $|a_{42}|=1$,
$|a_{45}|=|a_{15}|=1$) The relations

$a_{13}a_{24}-a_{14}a_{23}=a_{12}a_{34},$
$a_{13}a_{25}-a_{15}a_{23}=a_{12}a_{35},$
$a_{14}a_{35}-a_{15}a_{34}=a_{13}a_{45}$

give us $|a_{14}|+|a_{34}|=1;$ $|a_{25}|+|a_{35}|=1;$
$|a_{14}a_{35}|+|a_{15}a_{34}|=1$. Assume that $a_{14}\neq 0$, then
$|a_{35}|=1$ and therefore $a_{25}=0.$

Case 23.42.25($|a_{12}|=|a_{13}|=|a_{23}|=1$, $|a_{42}|=1$,
$|a_{45}|=|a_{25}|=1$) The relations

$a_{13}a_{24}-a_{14}a_{23}=a_{12}a_{34},$
$a_{13}a_{25}-a_{15}a_{23}=a_{12}a_{35},$
$a_{14}a_{35}-a_{15}a_{34}=a_{13}a_{45}$

give us $|a_{14}|+|a_{34}|=1;$ $|a_{15}|+|a_{35}|=1;$
$|a_{14}a_{35}|+|a_{15}a_{34}|=1$. Then $|a_{14}a_{15}|+|a_{34}a_{35}|=0$ and
therefore there is at least one zero in the set.

Case 23.42.35($|a_{12}|=|a_{13}|=|a_{23}|=1$, $|a_{42}|=1$,
$|a_{45}|=|a_{35}|=1$) The relations

$a_{13}a_{24}-a_{14}a_{23}=a_{12}a_{34},$
$a_{13}a_{25}-a_{15}a_{23}=a_{12}a_{35},$
$a_{14}a_{25}-a_{15}a_{24}=a_{12}a_{45};$

give us $|a_{14}|+|a_{34}|=1;|a_{25}|+|a_{15}|=1;|a_{14}a_{25}|+|a_{15}|=1.$
Assume that $a_{25}\neq 0$, then $|a_{14}|=1$ and therefore $a_{34}=0.$

Case 23.45 ($|a_{12}|=|a_{13}|=|a_{23}|=1$, $|a_{45}|=1$) Then there is at
least one $1$ among $|a_{14}|,|a_{24}|,|a_{34}|$ and all these cases are
similar to the Case $23.42$.

Case 24. ($|a_{12}|=|a_{13}|=|a_{24}|=1$). Then there is at least one $1$
among $|a_{32}|,|a_{34}|,|a_{35}|$. The case where $|a_{32}|=1$ is done in the
Case $23.42$, so there are only two left: $|a_{34}|=1$ and $|a_{35}|=1$.

Case 24.34 ($|a_{12}|=|a_{13}|=|a_{24}|=1$, $|a_{34}|=1$). Then the relation

$a_{13}a_{24}-a_{14}a_{23}=a_{12}a_{34}$

gives us $|a_{14}a_{23}|=0$.

Case 24.35 ($|a_{12}|=|a_{13}|=|a_{24}|=1$, $|a_{35}|=1$). Then there is at
least one $1$ among $|a_{41}|,|a_{43}|,|a_{45}|$. The case where
$|a_{12}|=|a_{24}|=|a_{14}|=1$ is similar to the case where
$|a_{12}|=|a_{13}|=|a_{23}|=1$, which is Case $23$. Therefore there are two
cases left: $|a_{43}|=1$ and $|a_{45}|=1$.

Case 24.35.43 ($|a_{12}|=|a_{13}|=|a_{24}|=1$, $|a_{35}|=1$, $|a_{43}|=1$).
Then the relation

$a_{13}a_{24}-a_{14}a_{23}=a_{12}a_{34}$

gives us $|a_{14}a_{23}|=0$.

Case 24.35.45 ($|a_{12}|=|a_{13}|=|a_{24}|=1$, $|a_{35}|=1$, $|a_{45}|=1$).
This is the case of peculiar sets. The relations give us the following system
of equations:

$1=|a_{34}|+|a_{14}a_{23}|;$

$1=|a_{25}|+|a_{15}a_{23}|;$

$1=|a_{15}|+|a_{14}a_{25}|;$

$1=|a_{14}|+|a_{15}a_{34}|;$

$1=|a_{23}|+|a_{25}a_{34}|.$

If we solve it with respect to $|a_{14}|,|a_{15}|$, we get

$|a_{15}|+|a_{14}|\geq 1;$

$|a_{23}|=\frac{|a_{14}|+|a_{15}|-1}{|a_{14}a_{15}|};$
$|a_{25}|=\frac{1-|a_{15}|}{|a_{14}|};$
$|a_{34}|=\frac{1-|a_{14}|}{|a_{15}|}.$

∎

##  7 The case of peculiar sets

In this case, we need two more technical lemmas.

######  Lemma 7.1.

Let

$\Omega=\\{(x,y)\in[0;1]^{2}|x\geq\frac{1}{2};y\geq\frac{1}{2};xy\leq\frac{1}{2};2y-xy\leq
1;2x-xy\leq 1;\\};$ $g:(x,y)\mapsto\left(\frac{1-x}{1-xy},1-xy\right).$

Then $g(\Omega)\subset\Omega.$

###### Proof.

We need to check five inequalities.

$2x-xy\leq 1\Leftrightarrow 1-xy\leq
2-2x\Leftrightarrow\frac{1}{2}\leq\frac{1-x}{1-xy};$

$xy\leq\frac{1}{2}\Leftrightarrow\frac{1}{2}\leq 1-xy;$

$x\geq\frac{1}{2}\Leftrightarrow
1-x\leq\frac{1}{2}\Leftrightarrow\frac{1-x}{1-xy}(1-xy)\leq\frac{1}{2};$

$y\geq\frac{1}{2}\Rightarrow x-2xy\leq 0\Rightarrow 2(1-xy)-(1-x)\leq 1.$

The remaining one requires additional calculations:

$2\frac{1-x}{1-xy}-(1-x)\stackrel{{\scriptstyle?}}{{\leq}}1\Leftrightarrow$
$2(1-x)-(1-x)(1-xy)\stackrel{{\scriptstyle?}}{{\leq}}1-xy\Leftrightarrow$
$2-2x-1+x+xy-x^{2}y\stackrel{{\scriptstyle?}}{{\leq}}1-xy\Leftrightarrow$

$-x+2xy-x^{2}y\stackrel{{\scriptstyle?}}{{\leq}}0\Leftrightarrow
x(-1+2y-xy)\leq 0$

∎

######  Lemma 7.2.

Assume that $(x,y)\in\Omega$. Then

$\max\left(x^{2},y^{2},(1-xy)^{2},\left(\frac{1-x}{1-xy}\right)^{2},\left(\frac{1-y}{1-xy}\right)^{2}\right)\leq\frac{9}{16}<\frac{3}{5}.$

###### Proof.

$1\leq 2x\leq 1+xy\leq\frac{3}{2}$ and $1\leq 2y\leq 1+xy\leq\frac{3}{2}$.
From Lemma 7.1 we have $\left(\frac{1-x}{1-xy},1-xy\right)\in\Omega,$ so the
same holds for them.

∎

######  Lemma 7.3.

Let $\lambda_{6}=\max_{i}\lambda_{i}$ and for $x,y\in[0;1]^{2}\setminus(1,1)$
we define $f(x,y)$ as follows:

$f(x,y):=\lambda_{1}\lambda_{5}\left(\frac{y-1}{xy-1}\right)^{2}+\lambda_{1}\lambda_{4}\left(\frac{x-1}{xy-1}\right)^{2}+\lambda_{2}\lambda_{3}(1-xy)^{2}+\lambda_{2}\lambda_{5}y^{2}+\lambda_{3}\lambda_{4}x^{2}.$

Then the following inequality is satisfied:

$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{4}+\lambda_{3}\lambda_{5}+\lambda_{4}\lambda_{5}+f(x,y)\leq
2.$

###### Proof.

Let us calculate the second derivative $\frac{\partial^{2}}{\partial
x^{2}}\left(\frac{1-x}{1-xy}\right)^{2}$.

$\frac{\partial^{2}}{\partial
x^{2}}\left(\frac{1-x}{1-xy}\right)^{2}=\frac{2(1-y)((2x-3)y+1)}{(1-xy)^{4}}.$

Therefore if $(2x-3)y+1\geq 0$, then $\frac{\partial^{2}}{\partial
x^{2}}f(x,y)\geq 0$. Thus, for $(x,y)\in[0;1)^{2}\cap\\{(2x-3)y+1\geq 0\\}$ we
get the following estimate

$f(x,y)\leq\max\left(f\left(\max\left(0,\frac{3-\frac{1}{y}}{2}\right),y\right),f(1,y)\right).$

Our goal now is to find upper bounds for
$f(0,y),f\left(\frac{3-\frac{1}{y}}{2},y\right),f(1,y).$

$f(0,y)=\lambda_{1}\lambda_{5}+\lambda_{1}\lambda_{4}(1-x)^{2}+\lambda_{2}\lambda_{3}+\lambda_{2}\lambda_{5}\cdot
0+\lambda_{3}\lambda_{4}x^{2}$

$\leq\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{3}+\max(\lambda_{1}\lambda_{4},\lambda_{3}\lambda_{4}).$

Therefore we may ignore either $\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{5}$
or $\lambda_{3}\lambda_{4}+\lambda_{2}\lambda_{5}$.

$f(1,y)=\lambda_{1}\lambda_{5}+\lambda_{1}\lambda_{4}\cdot
0+\lambda_{2}\lambda_{3}(1-y)^{2}+\lambda_{2}\lambda_{5}y^{2}+\lambda_{3}\lambda_{4}$
$\leq\lambda_{1}\lambda_{5}+\lambda_{3}\lambda_{4}+\max(\lambda_{2}\lambda_{3},\lambda_{2}\lambda_{5}).$

Therefore we may ignore either $\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{3}$
or $\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{5}$.

For $y\geq\frac{1}{3}$ we also need to estimate
$f\left(\frac{3-\frac{1}{y}}{2},y\right)$.

$f\left(\frac{3-\frac{1}{y}}{2},y\right)=\frac{4}{9}\lambda_{1}\lambda_{5}+\frac{1}{9y^{2}}\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{3}\left(\frac{3-3y}{2}\right)^{2}+\lambda_{2}\lambda_{5}y^{2}+\lambda_{3}\lambda_{4}\left(\frac{3-\frac{1}{y}}{2}\right)^{2}.$

Assume that $y\in\left[\frac{1}{3};\frac{1}{2}\right]$. Then since

$\frac{\partial^{2}}{\partial
y^{2}}\left(3-\frac{1}{y}\right)^{2}=\frac{6-12y}{y^{4}}\geq 0,$

we have $\frac{\partial^{2}}{\partial
y^{2}}f\left(\frac{3-\frac{1}{y}}{2},y\right)\geq 0$ and

$f\left(\frac{3-\frac{1}{y}}{2},y\right)\leq\max\left(f\left(0,\frac{1}{3}\right),f\left(\frac{1}{2},\frac{1}{2}\right)\right).$

The estimation of $f(0,y)$ was already done so we only need to estimate
$f\left(\frac{1}{2},\frac{1}{2}\right).$

$f\left(\frac{1}{2},\frac{1}{2}\right)=\frac{4}{9}\lambda_{1}\lambda_{5}+\frac{4}{9}\lambda_{1}\lambda_{4}+\frac{9}{16}\lambda_{2}\lambda_{3}+\frac{1}{4}\lambda_{2}\lambda_{5}+\frac{1}{4}\lambda_{3}\lambda_{4}.$

Consider two cases: $\lambda_{1}\lambda_{5}\leq\lambda_{1}\lambda_{4}$ and
$\lambda_{1}\lambda_{4}\leq\lambda_{1}\lambda_{5}$. In the first case

$\frac{4}{9}\lambda_{1}\lambda_{5}+\frac{4}{9}\lambda_{1}\lambda_{4}\leq\lambda_{1}\lambda_{4};$

$\frac{1}{4}\lambda_{3}\lambda_{4}+\frac{9}{16}\lambda_{2}\lambda_{3}\leq\max(\lambda_{3}\lambda_{4},\lambda_{2}\lambda_{3}),$

so we may ignore either $\lambda_{1}\lambda_{5}+\lambda_{3}\lambda_{4}$ or
$\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{3}$. In the case where
$\lambda_{1}\lambda_{4}\leq\lambda_{1}\lambda_{5}$ the following estimates
hold.

$\frac{4}{9}\lambda_{1}\lambda_{5}+\frac{4}{9}\lambda_{1}\lambda_{4}\leq\lambda_{1}\lambda_{5};$

$\frac{1}{4}\lambda_{2}\lambda_{5}+\frac{9}{16}\lambda_{2}\lambda_{3}\leq\max(\lambda_{2}\lambda_{5},\lambda_{2}\lambda_{3}).$

So we may ignore either $\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{5}$ or
$\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{3}$.

As a result, we obtain that the desired estimate holds for all

$(x,y)\in[0;1]^{2}\cap\\{y\leq\frac{1}{2},(2x-3)y+1\geq 0\\}$

and similarly for all

$(x,y)\in[0;1]^{2}\cap\left(\\{y\leq\frac{1}{2},(2x-3)y+1\geq
0\\}\cup\\{x\leq\frac{1}{2},(2y-3)x+1\geq 0\\}\right)$

$=\left\\{(a,b)\in[0;1]^{2}|a\leq\frac{1}{2}\text{ or
}b\leq\frac{1}{2}\right\\}.$

Let us make a change of variables:

$c=\frac{1-x}{1-xy};$

$d=1-xy.$

Then

$y=\frac{1-d}{1-cd};$ $\frac{1-y}{1-xy}=\frac{1-c}{1-cd};$ $x=1-cd.$

The expression in terms of new variables has the form

$\lambda_{1}\lambda_{5}\left(\frac{1-c}{1-cd}\right)^{2}+\lambda_{1}\lambda_{4}c^{2}+\lambda_{2}\lambda_{5}\left(\frac{1-d}{1-cd}\right)^{2}+\lambda_{3}\lambda_{4}(1-cd)^{2}+\lambda_{2}\lambda_{3}d^{2}$

and similarly the desired inequality holds for $c\leq\frac{1}{2}$ or
$d\leq\frac{1}{2}$. That is $2x-xy\geq 1$ or $xy\geq\frac{1}{2}$. Then we also
get the estimate, if $2y-xy\geq 1$. The remaining case is when
$x\geq\frac{1}{2},y\geq\frac{1}{2},xy\leq\frac{1}{2},2x-xy\leq 1,2y-xy\leq 1$.
That is $(x,y)\in\Omega.$

If $(x,y)\in\Omega$ then by using Lemma 7.2 and Lemma 4.6 we get the following
chain of inequalities:

$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{4}+\lambda_{3}\lambda_{5}+\lambda_{4}\lambda_{5}+f(x,y)$
$\leq\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{4}+\lambda_{3}\lambda_{5}+\lambda_{4}\lambda_{5}+\frac{3}{5}(\lambda_{1}\lambda_{5}+\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{3}+\lambda_{2}\lambda_{5}+\lambda_{3}\lambda_{4})\leq
2.$

∎

######  Proposition 7.1.

Let $\\{a_{ij}\\}_{1\leq i<j\leq 5}$ is peculiar. Then we have

$\sum_{1\leq i<j\leq 5}\lambda_{i}\lambda_{j}a_{ij}^{2}\leq 2.$

###### Proof.

Since $|a_{12}|=|a_{13}|=|a_{24}|=|a_{35}|=|a_{45}|=1$ we need to estimate the
remaining part, which is

$\lambda_{1}\lambda_{5}a_{15}^{2}+\lambda_{1}\lambda_{4}a_{14}^{2}+\lambda_{2}\lambda_{5}\left(\frac{1-|a_{15}|}{|a_{14}|}\right)^{2}+\lambda_{3}\lambda_{4}\left(\frac{1-|a_{14}|}{|a_{15}|}\right)^{2}+\lambda_{2}\lambda_{3}\left(\frac{|a_{14}|+|a_{15}|-1}{|a_{14}a_{15}|}\right)^{2}.$

If $|a_{14}|+|a_{15}|=1$, then we instantly lose $\lambda_{2}\lambda_{3}$ and

$\lambda_{1}\lambda_{5}a_{15}^{2}+\lambda_{1}\lambda_{4}(1-|a_{15}|)^{2}\leq\max(\lambda_{1}\lambda_{5},\lambda_{1}\lambda_{4}).$

Therefore we may ignore either $\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{3}$
or $\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{3}$. Then the estimate follows
from Lemma 4.3.

If $|a_{14}|+|a_{15}|>1$, then let us define

$x:=\frac{1-|a_{14}|}{|a_{15}|};$ $y:=\frac{1-|a_{15}|}{|a_{14}|}.$

Then $x,y\in[0;1)$ and

$|a_{15}|=\frac{1-y}{1-xy};$ $|a_{14}|=\frac{1-x}{1-xy};$
$\frac{|a_{14}|+|a_{15}|-1}{|a_{14}a_{15}|}=1-xy.$

After changing the variables, our expression becomes the following

$\lambda_{1}\lambda_{5}\left(\frac{y-1}{xy-1}\right)^{2}+\lambda_{1}\lambda_{4}\left(\frac{x-1}{xy-1}\right)^{2}+\lambda_{2}\lambda_{3}(1-xy)^{2}+\lambda_{2}\lambda_{5}y^{2}+\lambda_{3}\lambda_{4}x^{2}.$

Applying Lemma 7.3 completes the proof.

∎

##  8 The case $a_{12}=0$

######  Lemma 8.1.

Let $x+y+z=0$ and $(x,y,z)\in[-1;1]^{3}$. Then for any $a,b,c\geq 0$ it holds
that

$ax^{2}+by^{2}+cz^{2}\leq a+b+c-\min(a,b,c).$

Therefore we may ignore either $a$ or $b$ or $c$.

###### Proof.

The function $(x,y,z)\mapsto ax^{2}+by^{2}+cz^{2}$ is convex and its domain is
just

$conv(\\{(1,-1,0),(-1,1,0),(1,0,-1),(-1,0,1),(0,1,-1),(0,-1,1)\\}).$

∎

######  Notation 8.1.

We write $\pm a\pm b$ if there is a fixed choice of signs, but we do not know
this choice. For example, assume that $|x|=|y|=1$ and we have the expression
$xa+yb$. Then we can rewrite it as $\pm a\pm b$.

######  Proposition 8.1.

Let $\\{a_{ij}\\}_{1\leq i<j\leq 5}$ is admissible and $a_{12}=0$. Then we
have

$\sum_{1\leq i<j\leq 5}\lambda_{i}\lambda_{j}a_{ij}^{2}\leq 2.$

###### Proof.

Assume that the set $\\{a_{ij}\\}$ maximizes the value of $\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}a_{ij}^{2}$ among all admissible sets with $a_{12}=0.$

Let $a_{13}=0$. Then either $a_{14}=0$ or $a_{23}=0$. If $a_{14}=0$ then
either $a_{15}=0$ or $a_{24}=0$.

The case $a_{12}=a_{13}=a_{14}=a_{15}=0$ is similar to the case
$\lambda_{1}=0$, which we did in Proposition 5.1.

The case where $a_{12}=a_{13}=a_{14}=a_{24}=0$ is the case with missing
$\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{4}$. Lemma 4.3 completes the proof
in this case.

The case where $a_{12}=a_{13}=a_{23}=0$ is the case with missing
$\lambda_{1}\lambda_{2}+\lambda_{2}\lambda_{3}+\lambda_{1}\lambda_{3}$. Lemma
4.4 completes the proof in this case.

Therefore we can assume that $a_{13},a_{14},a_{15},a_{23},a_{24},a_{25}$ are
not zeros. Then the relations become the following.

$a_{24}=\frac{1}{a_{13}}a_{14}a_{23};$ $a_{25}=\frac{1}{a_{13}}a_{15}a_{23};$

$a_{45}=\frac{1}{a_{13}}(a_{14}a_{35}-a_{15}a_{34}).$

Consider $u=(0,a_{14},a_{15}),v=(-a_{23},a_{34},a_{35})$. Again, we can move
$v$ inside $[-1;1]^{3}$ with $[u;v_{t}]\equiv[u;v]$ as in Proposition 6.1, and
hence, conclude that $v$ must lie on the boundary $\partial[-1;1]^{3}.$
Similarly
$(a_{42},a_{43},a_{45}),(a_{53},a_{52},a_{54}),(a_{13},a_{34},a_{35}),(a_{41},a_{43},a_{45}),(a_{53},a_{51},a_{54})\in\partial[-1;1]^{3}$.
Thus, in each of the following six lines, at least one element is equal to 1
or -1.

$a_{32},a_{34},a_{35};$ $a_{42},a_{43},a_{45};$ $a_{52},a_{53},a_{54};$

$a_{31},a_{34},a_{35};$ $a_{41},a_{43},a_{45};$ $a_{51},a_{53},a_{54}.$

Thus there are two possible cases: $|a_{ij}|\neq 1$ for $i,j\in\\{3,4,5\\}$
and $|a_{34}|=1$.

Case 1 ($|a_{ij}|\neq 1$ for $i,j\in\\{3,4,5\\}$ ). In this case

$|a_{13}|=|a_{14}|=|a_{15}|=|a_{23}|=|a_{24}|=|a_{25}|=1$

and $|a_{45}|=|\pm a_{35}\pm a_{34}|.$ Thus for some choice of signs, the
following is true

$\pm a_{45}\pm a_{35}\pm a_{34}=0.$

Therefore by Lemma 8.1 we may ignore either
$\lambda_{1}\lambda_{2}+\lambda_{3}\lambda_{4}$ or
$\lambda_{1}\lambda_{2}+\lambda_{3}\lambda_{5}$ or
$\lambda_{1}\lambda_{2}+\lambda_{4}\lambda_{5}$.

Case 2. ($|a_{34}|=1$). In this case there is at least one $1$ among
$|a_{52}|,|a_{53}|,|a_{54}|$ and among $|a_{51}|,|a_{53}|,|a_{54}|.$ And there
are only two cases: $|a_{53}|=1$ and $|a_{52}|=|a_{51}|=1.$

Case 2.1. ($|a_{34}|=1$, $|a_{53}|=1$). In this case we have the following
restrictions:

$|a_{24}|=\frac{1}{|a_{13}|}|a_{14}a_{23}|\leq 1;$
$|a_{25}|=\frac{1}{|a_{13}|}|a_{15}a_{23}|\leq 1;$

$|a_{45}|=\frac{1}{|a_{13}|}|\pm a_{14}\pm a_{15}|\leq 1.$

Then

$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}a_{ij}^{2}=(\lambda_{3}\lambda_{4}+\lambda_{3}\lambda_{5})+\lambda_{1}\lambda_{5}a_{15}^{2}$
$+\lambda_{2}\lambda_{5}\left(\frac{a_{15}a_{23}}{a_{13}}\right)^{2}+\lambda_{1}\lambda_{4}a_{14}^{2}+\lambda_{2}\lambda_{4}\left(\frac{a_{14}a_{23}}{a_{13}}\right)^{2}+\lambda_{1}\lambda_{3}a_{13}^{2}+\lambda_{2}\lambda_{3}a_{23}^{2}+\lambda_{4}\lambda_{5}\left(\frac{\pm
a_{14}\pm a_{15}}{a_{13}}\right)^{2}.$

If $|a_{24}|,|a_{25}|,|a_{23}|<1$, then we can increase $|a_{23}|.$ Therefore
there are two cases: $|a_{24}|=1$ and $|a_{23}|=1$.

Case 2.1.1. ($|a_{34}|=1$, $|a_{53}|=1$, $|a_{24}|=1$). The non-trivial part
of expression is

$\lambda_{1}\lambda_{5}a_{15}^{2}+\lambda_{2}\lambda_{5}\left(\frac{a_{15}}{a_{14}}\right)^{2}+\lambda_{1}\lambda_{4}a_{14}^{2}+\lambda_{1}\lambda_{3}a_{13}^{2}+\lambda_{2}\lambda_{3}\left(\frac{a_{13}}{a_{14}}\right)^{2}+\lambda_{4}\lambda_{5}\left(\frac{\pm
a_{14}\pm a_{15}}{a_{13}}\right)^{2}$

and the restrictions are

$|a_{15}|\leq|a_{14}|;$ $|a_{13}|\leq|a_{14}|;$ $|\pm a_{14}\pm
a_{15}|\leq|a_{13}|.$

If $|a_{14}|<1$ and $|\pm a_{14}\pm a_{15}|<|a_{13}|$, then it is possible to
increase both $|a_{14}|$ and $|a_{15}|$. Therefore there are two cases:
$|a_{14}|=1$ and $|\pm a_{14}\pm a_{15}|=|a_{13}|$.

Case 2.1.1.1. ($|a_{34}|=1$, $|a_{53}|=1$, $|a_{24}|=1$, $|a_{14}|=1$). The
non-trivial part of expression is

$(\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{5})a_{15}^{2}+(\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3})a_{13}^{2}+\lambda_{4}\lambda_{5}\left(\frac{\pm
1\pm a_{15}}{a_{13}}\right)^{2}$

and the restriction is

$|\pm 1\pm a_{15}|\leq|a_{13}|.$

Then $|a_{13}|\in\left[|\pm 1\pm a_{15}|;1\right]$. So it suffices to
substitute only $|a_{13}|=|\pm 1\pm a_{15}|$ and $|a_{13}|=1.$

Case 2.1.1.1.1. ($|a_{34}|=1$, $|a_{53}|=1$, $|a_{24}|=1$, $|a_{14}|=1$,
$|a_{13}|=|\pm 1\pm a_{15}|$). The non-trivial part of expression is

$(\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{5})a_{15}^{2}+(\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3})(1-|a_{15}|)^{2}\leq\max(\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{5},\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3}).$

Therefore we may ignore either
$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{5}$ or
$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3}$.

Case 2.1.1.1.2. ($|a_{34}|=1$, $|a_{53}|=1$, $|a_{24}|=1$, $|a_{14}|=1$,
$|a_{13}|=1$). The non-trivial part of expression is

$(\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{5})a_{15}^{2}+\lambda_{4}\lambda_{5}(1-|a_{15}|)^{2}\leq\max(\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{5},\lambda_{4}\lambda_{5}).$

Therefore we may ignore either
$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{5}$ or
$\lambda_{1}\lambda_{2}+\lambda_{4}\lambda_{5}$.

Case 2.1.1.2. ($|a_{34}|=1$, $|a_{53}|=1$, $|a_{24}|=1$, $|a_{13}|=|\pm
a_{14}\pm a_{15}|$). The non-trivial part of expression is

$\lambda_{1}\lambda_{5}a_{15}^{2}+\lambda_{2}\lambda_{5}\left(\frac{a_{15}}{a_{14}}\right)^{2}+\lambda_{1}\lambda_{4}a_{14}^{2}+\lambda_{1}\lambda_{3}(\pm
a_{14}\pm a_{15})^{2}+\lambda_{2}\lambda_{3}\left(\frac{\pm a_{14}\pm
a_{15}}{a_{14}}\right)^{2}$

$=(\lambda_{1}\lambda_{5}a_{14}^{2}+\lambda_{2}\lambda_{5})\left(\frac{a_{15}}{a_{14}}\right)^{2}+\lambda_{1}\lambda_{4}a_{14}^{2}+(\lambda_{1}\lambda_{3}a_{14}^{2}+\lambda_{2}\lambda_{3})\left(1-\left|\frac{a_{15}}{a_{14}}\right|\right)^{2}$
$\leq\lambda_{1}\lambda_{4}+\max(\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{5},\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3}).$

Therefore we may ignore either
$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{5}$ or
$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3}$.

Case 2.1.2. ($|a_{34}|=1$, $|a_{53}|=1$, $|a_{23}|=1$). The non-trivial part
of the expression is

$\lambda_{1}\lambda_{5}a_{15}^{2}+\lambda_{2}\lambda_{5}\left(\frac{a_{15}}{a_{13}}\right)^{2}+\lambda_{1}\lambda_{4}a_{14}^{2}+\lambda_{2}\lambda_{4}\left(\frac{a_{14}}{a_{13}}\right)^{2}+\lambda_{1}\lambda_{3}a_{13}^{2}+\lambda_{4}\lambda_{5}\left(\frac{\pm
a_{14}\pm a_{15}}{a_{13}}\right)^{2}$

$=(\lambda_{1}\lambda_{5}a_{13}^{2}+\lambda_{2}\lambda_{5})\left(\frac{a_{15}}{a_{13}}\right)^{2}+(\lambda_{1}\lambda_{4}a_{13}^{2}+\lambda_{2}\lambda_{4})\left(\frac{a_{14}}{a_{13}}\right)^{2}+\lambda_{4}\lambda_{5}\left(\frac{\pm
a_{14}\pm a_{15}}{a_{13}}\right)^{2}+\lambda_{1}\lambda_{3}a_{13}^{2}.$

Then by Lemma 8.1 we may ignore either
$\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{5}+\lambda_{1}\lambda_{2}$ or
$\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{4}+\lambda_{1}\lambda_{2}$ or
$\lambda_{1}\lambda_{2}+\lambda_{4}\lambda_{5}$.

Case 2.2. ($|a_{34}|=1$, $|a_{51}|=|a_{52}|=1$). In this case we have the
following restrictions:

$|a_{14}|=|a_{24}|;$ $|a_{13}|=|a_{23}|;$

$|a_{45}|=\frac{1}{|a_{13}|}(1-|a_{14}a_{35}|).$

Then

$\sum_{1\leq i<j\leq
5}\lambda_{i}\lambda_{j}a_{ij}^{2}=(\lambda_{3}\lambda_{4}+\lambda_{1}\lambda_{5}+\lambda_{2}\lambda_{5})$
$+(\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{4})a_{14}^{2}+(\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3})a_{13}^{2}+\lambda_{3}\lambda_{5}a_{35}^{2}+\lambda_{4}\lambda_{5}\frac{1}{a_{13}^{2}}(1-|a_{14}a_{35}|)^{2}.$

We have $|a_{45}|\leq 1$, thus

$\frac{1-|a_{13}|}{|a_{14}|}\leq|a_{35}|\leq 1$

and since this function is convex with respect to $|a_{35}|$ it suffices to
substitute $|a_{35}|=\frac{1-|a_{13}|}{|a_{14}|}$ and $|a_{35}|=1$.

Case 2.2.1. ($|a_{34}|=1$, $|a_{51}|=|a_{52}|=1$,
$|a_{35}|=\frac{1-|a_{13}|}{|a_{14}|}$). The non-trivial part of the
expression is

$(\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{4})a_{14}^{2}+(\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3})a_{13}^{2}+\lambda_{3}\lambda_{5}\left(\frac{1-|a_{13}|}{|a_{14}|}\right)^{2}$

and the remaining restrictions are $|a_{13}|\leq 1,|a_{14}|\leq
1,|a_{13}|+|a_{14}|\geq 1.$ Therefore

$1-|a_{13}|\leq|a_{14}|\leq 1$

and since the function is convex with respect to $|a_{14}|$, it suffices to
substitute $|a_{14}|=1-|a_{13}|$ and $|a_{14}|=1$.

Case 2.2.1.1. ($|a_{34}|=1$, $|a_{51}|=|a_{52}|=1$,
$|a_{35}|=\frac{1-|a_{13}|}{|a_{14}|}$, $|a_{14}|=1-|a_{13}|$). The non-
trivial part of the expression is

$(\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{4})(1-|a_{13}|)^{2}+(\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3})a_{13}^{2}\leq\max(\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{4},\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3}).$

Therefore we may ignore either
$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3}$ or
$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{4}$ and
Lemma 4.4 completes the proof of this case.

Case 2.2.1.2. ($|a_{34}|=1$, $|a_{51}|=|a_{52}|=1$,
$|a_{35}|=\frac{1-|a_{13}|}{|a_{14}|}$, $|a_{14}|=1$). The non-trivial part of
the expression is

$(\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3})a_{13}^{2}+\lambda_{3}\lambda_{5}(1-|a_{13}|)^{2}\leq\max(\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3},\lambda_{3}\lambda_{5}).$

Therefore we may ignore either
$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3}$ or
$\lambda_{1}\lambda_{2}+\lambda_{3}\lambda_{5}$. Lemmas 4.3 and 4.4 complete
the proof of this case.

Case 2.2.2. ($|a_{34}|=1$, $|a_{51}|=|a_{52}|=1$, $|a_{35}|=1$). The non-
trivial part of the expression is

$(\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{4})a_{14}^{2}+(\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3})a_{13}^{2}+\lambda_{4}\lambda_{5}\frac{1}{a_{13}^{2}}(1-|a_{14}|)^{2}.$

This case is similar to the Case $2.2.1$ and we may ignore either
$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3}$ or
$\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{4}$ or
$\lambda_{1}\lambda_{2}+\lambda_{4}\lambda_{5}.$ ∎

## Acknowledgement

I would like to thank my research advisor Nikita Kalinin for the help with the
preparation of this article.

## References

  * [1] L. Fejes Tóth and E. Makai Jr., On the thinnest non-separable lattice of convex plates, Studia Sci. Math. Hungar. 9 (1974).
  * [2] Ball, K. (1997) An Elementary Introduction to Modern Convex Geometry. In: Flavors of Geometry, Math. Sci.Res. Inst. Publ., 31, Cambridge University Press, Cambridge. http://library.msri.org/books/Book31/files/ball.pdf
  * [3] E. Makai, Jr.:On the thinnest non-separable lattice of convex bodies. Studia Sci. Math. Hungar. 13 (1978),19-27. MR 38a:52016
  * [4] Gaoyong Zhang (1991). Restricted chord projection and affine inequalities. Geometriae Dedicata, 39(2), 213–222. doi:10.1007/bf00182294
  * [5] Felix Behrend, Über einige Affininvarianten konvexer Bereiche, Math. Ann. 113 (1937), no. 1, 713–747.
  * [6] Merino, B.G., Schymura, M. On the reverse isodiametric problem and Dvoretzky–Rogers-type volume bounds. RACSAM 114, 136 (2020). https://doi.org/10.1007/s13398-020-00867-7
  * [7] Franck Barthe, An extremal property of the mean width of the simplex, Math. Ann. 310 (1998), no. 4, 685–693.
  * [8] Endre Makai Jr. and Horst Martini, Density estimates for k-impassable lattices of balls and general convex bodies in $\mathbb{R}^{n}$,https://arxiv.org/abs/1612.01307, 2016.
  * [9] Arkadiy Aliev. New estimates for $d_{2,1}$ and $d_{3,2}$. https://arxiv.org/abs/2207.09552
  * [10] Nikita Kalinin, Tropical approach to Nagata’s conjecture in positive characteristic, Discrete and Computational Geometry, 58(1), 158-179
  * [11] Nikita Kalinin, The Newton polygon of a planar singular curve and its subdivision, Journal of Combinatorial Theory, Series A, 137, 226–256
  * [12] Grigory Ivanov, Tight frames and related geometric problems https://arxiv.org/abs/1804.10055.

E-mail address<EMAIL_ADDRESS>


# An adversarial example for Direct Logit Attribution:
memory management in gelu-4l

James Dao
<EMAIL_ADDRESS>
&Yeu-Tong Lau
<EMAIL_ADDRESS>
&Can Rager
<EMAIL_ADDRESS>&Jett Janiak
<EMAIL_ADDRESS>

###### Abstract

We provide concrete evidence for memory management in a 4-layer transformer.
Specifically, we identify clean-up behavior, in which model components
consistently remove the output of preceeding components during a forward pass.
Our findings suggest that the interpretability technique Direct Logit
Attribution provides misleading results. We show explicit examples where this
technique is inaccurate, as it does not account for clean-up behavior.

##  1 Introduction

Previously, Elhage et al. [1] suggested a mechanism for memory management and
speculated it occurs due to the high demand on residual stream bandwidth. In
the following section 1.1, we define clean-up behavior, in which attention
heads and MLPs (which we collectively call nodes) clear information from the
residual stream that is only used in early layers of the network. Then, we
identify specific nodes that write and remove information from the residual
stream of the gelu-4l model in section 2. Finally, we select prompts that
result in misleading DLA scores in section 3.

###  1.1 Clean-up behavior

We characterize clean-up behavior as four steps during a forward pass:

  1. 1.

A writer node or embedding writes a specific direction to the residual stream

  2. 2.

Subsequent nodes use this direction for further computation

  3. 3.

A clean-up node clears this direction by writing its negative value to the
residual stream

  4. 4.

The direction is used in the later part of the model, in one or two of the
following ways:

     * •

Subsequent nodes write to the direction, using this free subspace for
communicating information different to the information which has been cleaned
up

     * •

The unembedding matrix reads the direction, which directly affects the output
logits

Throughout this work, we analyze the removal of the writer node output (steps
1 - 3). In future work, we will address how the cleared space is used by later
nodes after clean-up (step 4).

###  1.2 Direct Logit Attribution

Background: The final state of the residual stream is the sum of all outputs
of nodes and embeddings in the model [1]. The final state of the residual
stream is mapped to the logit distribution by applying Layer Norm and
Unembedding. The logit difference of two tokens is equivalent to the
difference in log probabilities for both tokens and therefore directly
interpretable in terms of predicting the next token.

Direct Logit Attribution (DLA) [2] has been used to identify the direct
contribution of individual nodes to the prediction of the correct next token
[3, 4, 5, 6]. This is done by applying unembedding directly to the output of
any node, after accounting for Layer Norm. For example, a single node makes
the prediction of token A more likely than the prediction of token B, if it
increases the logit difference
$\text{logit}_{\text{A}}-\text{logit}_{\text{B}}$. However, DLA does not
account for the fact that later nodes depend on the output of earlier nodes.
Clean-up behavior is one possible reason why DLA may be misleading –
especially for early nodes – as a node output may be consistently cleaned up
by later nodes und thus not contribute to the final logit difference
$\text{logit}_{\text{A}}-\text{logit}_{\text{B}}$.

##  2 Evidence for Clean-Up Behavior

###  2.1 Metrics and Terminology

We introduce the Projection Ratio to compare the extent to which the direction
in the residual stream has been overwritten. This metric is a normalized
scalar projection. Let $\mathbf{a}$, $\mathbf{b}$ be elements of the residual
stream $\mathbb{R}^{d_{\text{model}}}$. The Scalar Projection
$\text{SP}(\mathbf{a},\mathbf{b})$ returns a scalar value representing the
resulting signed length after projecting $\mathbf{a}$ onto the direction of
$\mathbf{b}$. The Projection Ratio $\text{PR}(\mathbf{a},\mathbf{b})$ returns
a scalar value representing $a_{\parallel\mathbf{b}}$ as a ratio of the norm
of $\mathbf{b}$.

$\displaystyle\text{SP}(\mathbf{a},\mathbf{b})$
$\displaystyle=a_{\parallel\mathbf{b}}=\mathbf{a}\cdot\hat{\mathbf{b}}=\frac{\mathbf{a}\cdot\mathbf{b}}{||\mathbf{b}||}$
(1) $\displaystyle\text{PR}(\mathbf{a},\mathbf{b})$
$\displaystyle:=\frac{a_{\parallel\mathbf{b}}}{||\mathbf{b}||}=\frac{\mathbf{a}\cdot\mathbf{b}}{||\mathbf{b}||^{2}}$
(2)

One intuitive understanding of $\text{PR}(\mathbf{a},\mathbf{b})$ is that it
tells you what proportion of $\mathbf{b}$ is present in $\mathbf{a}$. For
instance, we can split up $\mathbf{a}$ into components parallel and orthogonal
to $\mathbf{b}$:
$\mathbf{a}=\mathbf{a}_{\parallel\mathbf{b}}+\mathbf{a}_{\perp\mathbf{b}}=\text{PR}(\mathbf{a},\mathbf{b})\
\mathbf{b}+\mathbf{a}_{\perp\mathbf{b}}$ where
$\mathbf{b}\cdot\mathbf{a}_{\perp\mathbf{b}}=0$ . Both metrics SP and PR are
negative if the cosine similarity between $\mathbf{a}$ and $\mathbf{b}$ is
negative.

Moreover, we use the notation “L0H2” to denote the attention head that resides
in layer 0 with head index 2. All heads in layer 2 are collectively denoted by
“L2HX”. In this section, we will always have the output of the writer head
L0H2 as vector $\mathbf{b}$, while vector $\mathbf{a}$ will be replaced either
by residual stream or by the output of the clean-up heads.

###  2.2 Identifying a writer node: output of head L0H2 is being consistently
cleaned up

First, we scanned the full gelu-4l model for consistent clean-up behavior. For
each node, we checked whether the output of a node is present in later states
of the residual stream. We measure clean-up by projecting the residual stream
before and after each attention and MLP layer onto the output direction of
L0H2. An intuitive understanding: When projecting the state of the residual
stream after layer 2 onto L0H2, the projection ratio represents how much of
L0H2’s output is still present in the residual stream after layer 2. If the
projection ratio is zero, the residual stream is orthogonal to the output of
L0H2. In other words, the output direction of L0H2 is not present in the
residual stream. It may have been moved to another subspace or totally cleared
from the residual stream.

The projection of the residual stream onto L0H2 is shown in Figure 1 (top). We
identified the output of L0H2 is consistently being cleaned up across 300
forward passes. In Figure 1 (bottom), we can track the presence of L0H2’s
information in the residual stream as it moves through the transformer model:

  * •

Initially we see a projection ratio of ~0 at resid_pre0, as L0H2 (which
resides between resid_pre0 and resid_mid0) has not written to the residual
stream yet

  * •

After L0H2 writes to the residual stream (at resid_mid0), the projection ratio
goes to ~1. It is not exactly 1 since the output of other heads in layer 0
might not be fully orthogonal to the output of L0H2

  * •

After resid_mid0, a projection ratio of ~1 suggests that L0H2’s information is
present in the residual stream until resid_post1 (inclusive)

  * •

Attention heads in layer 2 (which resides between resid_post1 and resid_mid2)
appear to remove the information the L0H2 originally wrote, resulting in a
much smaller projection ratio of ~0. This happens consistently across 300
prompts (randomly sampled from the model’s training dataset), but there is
some variance across different sequence positions.

The function of L0H2 is not totally clear, but it resembles a positional
information head [7, 8].

Figure 1: Projection of residual stream at various locations onto the
direction of the outputs of various heads in attention layer 0. “resid_mid0”
refers to the residual stream location before the MLP in layer 0, and
“resid_post0” refers to the residual stream location after the MLP in layer0.
Projection ratios are aggregated using median across batch (n=300) and
position (n=1024). The bottom subplot’s shaded region represents the 25th and
75th quantiles.

###  2.3 Identifying cleaners: six layer 2 attention heads are cleaning up
L0H2

We find that six attention heads (L2H2, L2H3, L2H4, L2H5, L2H6, L2H7) are
cleaning up the output of L0H2. In Figure 2 below, we see the aforementioned
attention heads have a consistently negative projection ratio, implying they
are writing to the residual stream in the opposite direction of L0H2. We
believe that most of the variance as seen by the error bars is due to clean-up
behavior being sensitive with respect to position rather than sensitivity with
respect to prompt.

Figure 2: Projection of all attention heads and MLPs onto the direction of
L0H2 output. Projection ratios are aggregated using median across prompts
(n=300) and position (n=1024), with error bars at the 25th and 75th quantiles.
L0H2 is intentionally omitted. The median of the sum of projection ratios of
the 6 heads is -0.903.

###  2.4 Verifying causal relation: cleaning behavior depends on the writer
output

We verify the causal relation between clean-up heads and writer heads by
patching the residual stream as an input to the OV circuits of layer 2
attention heads. The OV circuit is responsible for what directions are written
to the residual stream by attention heads [1]. By patching the OV circuit, we
check what information heads L2HX write to the residual stream without the
presence of L0H2.

We repeat experiments from sections 2.2 and 2.3 and compare results between
the clean and a patched run. In the patched run, we alter the Value input
(hook_v_input in the transformerlens python library) of every head in layer 2
by subtracting the output of L0H2. This is equivalent to zero ablating L0H2
for only the OV circuits of the attention heads in layer 2. Every other
component in the transformer will still “see” L0H2’s original output.

Figure 3(a) shows that the projection ratio of residual stream onto L0H2
remains high after the attention block in layer 2 in the patched run,
indicating a significant fraction of the clean-up is indeed input-dependent.
Figure 3(b) compares patched runs to clean runs for each individual head. The
clean-up behavior disappears when the cleaner head does not “see” the output
of L0H2.

(a) L0H2 is only cleaned up if fed into OV circuits of attention heads in
layer 2. Patching refers to subtracting L0H2’s output from the Value input of
attention heads in layer 2. Projection ratios are aggregated using median
across batch (n=300) and position (n=1024), with a shaded region at the 25th
and 75th quantiles.

(b) Patching refers to subtracting L0H2’s output from the Value input of
attention heads in layer 2. Projection ratios are aggregated using median
across batch (n=300) and position (n=1024), with error bars at the 25th and
75th quantiles.

Figure 3: Verifying the causal relationship between writer and cleaner nodes
by patching the output of the writer node.

##  3 Implication for Direct Logit Attribution

As mentioned in the introduction (section 1), the DLA method applies final
layer norm and unembedding directly to the node output. The DLA values will be
high, if the node output happens to be aligned with the unembedding direction
of certain tokens.

In the gelu-4l model however, the output of L0H2 is largely removed from the
residual stream after the attention block in layer 2 as we’ve seen in Figure
1. In subsection 3.1 below, we present four examples where significant DLA
values for L0H2 could be easily misinterpreted as a significant direct
contribution to the final logits. Furthermore, since reading and writing to
residual stream are linear operations, applying DLA to clean-up heads will
yield significant values as well, but with a flipped sign. We show these
results in subsection 3.2.

###  3.1 Adversarial examples of high DLA values without direct contribution

We adversarially selected four tokens " bottom", " State", " __", and "
Church" and constructed four prompts such that the model predicts one of the
tokens with highest probability.

  1. 1.

prompt: "It’s in the cupboard, either on the top or on the"
top predictions: " bottom", " top" (logit difference 1.07)

  2. 2.

prompt: "I went to university at Michigan"
top predictions: " State", " University" (logit difference 1.89)

  3. 3.

prompt: "class MyClass:\n\tdef"
top predictions: " __", " get" (logit difference 3.02)

  4. 4.

prompt: "The church I go to is the Seventh-day Adventist"
top predictions: " Church", " church" (logit difference 0.94)

In our opinion, it is extremely unlikely that a node can directly111without
composition, or by writing directly to the final logits improve the model
predictions in prompts 1-4 without considering token information222or based
purely on positional information. However, patching the input to L0H2 with
unrelated text doesn’t affect the DLA values (see Figure 4(a)). We therefore
conclude that L0H2 is not contributing directly to the model predictions in
prompts 1-4, despite significant DLA values.

The invariance of L0H2’s DLA to input tokens is unusual. We reran the patching
experiment for four other attention heads that exhibit the highest DLA values
for the respective prompt in Figure 4(b). In contrast to L0H2, the DLA values
for these heads are severely affected by the patch.

(a) Direct Logit Attribution of L0H2 on four adversarial examples. Patched
refers to replacing the input to L0H2 with unrelated text with the same number
of tokens (300 examples). The orange bars show median with error bars at the
25th and 75th quantiles.

(b) Direct Logit Attribution of other heads on adversarial examples. Patched
refers to replacing the input to a head with one from a run on unrelated text
with the same number of tokens (300 examples). The orange bars show median
with error bars at the 25th and 75th quantiles.

Figure 4: Direct Logit Attribution of selected nodes before and after patching
the input.

###  3.2 Negative correlation between the DLA of writer and cleaners

Finally, we examine the DLA values of the clean-up heads and how they are
related to the DLA of the writer head. Instead of using prompts 1-4, we
collect 30 samples from the model’s training dataset and determine the logit
difference between top 2 predictions at every position, which gives roughly
30,000 data points. We consider the clean-up heads in aggregate, by summing
the DLA values of each individual head.

This experiment yields a low, but non-negligible negative correlation
coefficient of $r=-0.159$. One possible explanation for the small magnitude of
$r$ is that the cleaners perform many tasks unrelated to clean-up. The
unrelated tasks may have an effect on the cleaner’s DLA values which can’t be
explained by the writer’s DLA. To extract a part of the cleaners DLA that is
related to the clean-up, we consider only part of the cleaners output that
comes from V-composition [1] with the writer head. Specifically, for a given
clean-up head $c$, we compute

$\sum_{i}h^{w}_{i}A^{c}_{i}W^{c}_{\text{OV}}$ (3)

where $h^{w}_{i}$ is the writer head’s output at position $i$ (after layer 2
Layer Norm) and $A^{c}_{i}$ is the cleaner’s attention weight at position $i$.
Intuitively, this V-composition output tells us how the directions written by
the cleaners depend on the output of the writer. Its DLA values show how much
the clean-up heads overwrite the DLA of the writer head. As we can see in
Figure 5, the cleaner’s DLA shows strong negative correlation ($r=-0.716$)
with the writer’s DLA. In other words, writer’s contribution to final logits
is mostly333a perfect clean-up behavior would yield r = -1 and a slope of -1
canceled by the clean-up heads.

Figure 5: Correlation between DLA of writer head and DLA of [clean-up heads
output dependent on V-composition with the writer head]. The negative
correlation coefficient $r$ suggests that output of a writer node is
consistently removed from the residual stream by subsequent clean-up nodes.

##  4 Conclusion

In this post, we presented a concrete example of memory management in
transformers. Furthermore, we constructed adversarial examples for the DLA
method that relies on directions being preserved in residual stream through
the forward pass. We propose three directions for future work: First, we’re
interested in how the cleaned-up space is used by later nodes during a forward
pass. Secondly, we want to further examine the role of head L0H2 in the
gelu-4l model. Finally, we will look for clean-up behavior in other models.
Previous research suggests that especially early heads and late heads exhibit
high DLA values [8, 9]. We want to verify whether high DLA values of early
heads were misinterpreted in these works.

##  5 Acknowledgements and Contributions

Our research benefited from discussions, feedback, and support from many
people, including Chris Mathwin, Neel Nanda, Jacek Karwowski, Callum
McDougall, Joseph Bloom, Alan Cooney, Arthur Conmy, Matthias Dellago, Eric
Purdy and Stefan Heimersheim. This work was produced as part of ARENA 2.0 and
the SERI ML Alignment Theory Scholars Program - Spring 2023 Cohort.

All of the authors contributed equally to this post. Jett proposed and led the
project, while James, Can, and Yeu-Tong ran most of the experiments.

##  6 Resources

Please check out our notebook for figure recreation and to examine your own
model for clean-up behavior accessible via:
https://colab.research.google.com/drive/16Kp-4iH330a1dF6F0ntPK7kfNuxqkvfZ.

We examined the "gelu-4l" model accessible via: http://neelnanda.io/toy-
models.

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