# Uniqueness of addition in Lie algebras revisited

Ivan Arzhantsev  Faculty of Computer Science, HSE University, Pokrovsky
Boulevard 11, Moscow, 109028 Russia<EMAIL_ADDRESS>
###### Abstract.

We obtain new and improve old results on uniqueness of addition in Lie rings
and Lie algebras. A Lie ring $\mathfrak{R}$ is called a unique addition ring,
or a UA-Lie ring, if any commutator-preserving bijection from $\mathfrak{R}$
to an arbitrary Lie ring is additive. We describe wide classes of Lie rings
that are not UA-Lie ring. In the other direction, it is known that if a
finite-dimensional Lie algebra $\mathfrak{g}$ contains two elements whose
centralizers have trivial intersection, then $\mathfrak{g}$ is a UA-Lie ring.
We use this result to characterize UA-Lie rings among seaweed Lie algebras.
The paper includes many open problems and questions.

###### Key words and phrases:

Unique addition ring, Lie ring, centralizer, semisimple Lie algebra, parabolic
subalgebra, seaweed subalgebra, radical, Mal’cev splitting

###### 2020 Mathematics Subject Classification:

Primary 17B05, 17B30; Secondary 16B99, 17B20

##  1\. Introduction

Let us consider two algebraic structures $\mathcal{A}$ and $\mathcal{B}$ of
the same type. A homomorphism between $\mathcal{A}$ and $\mathcal{B}$ is a map
$\alpha\colon\mathcal{A}\to\mathcal{B}$ that preserves all operations of the
corresponding structure. It is useful to know that if $\alpha$ preserves some
of the operations, then all other operations are preserved automatically. And
it does happen sometimes.

We consider the case when the algebraic structure is a ring structure and the
map $\alpha$ is bijective. There is no chance that any bijection that
preserves addition will automatically preserve multiplication. But there is a
hope that for some rings $\mathfrak{R}$ all bijections from $\mathfrak{R}$ to
an arbitrary ring $\mathfrak{S}$ preserving multiplication are additive maps.
In this case one calls $\mathfrak{R}$ a unique addition ring, or a UA-ring for
short. The term is explained by the equivalent reformulation of the property
of unique addition: for the operation of multiplication $(\mathfrak{R},\cdot)$
there is a unique binary operation $(\mathfrak{R},+)$ such that
$(\mathfrak{R},+,\cdot)$ is a ring.

The study of associative rings with unique addition is a well-developed area
of ring theory with many meaningful results; see e.g. [9, 12, 14, 18, 19]. In
Section 2, we recall a proof of one of the main results in this field. Namely,
the theorem by Rickart and Stephenson states that for any associative ring $A$
with unit and any $n\geqslant 2$, the ring of matrices $\text{Mat}_{n\times
n}(A)$ is a UA-ring.

The main object of study in this paper are Lie rings and Lie algebras. In
Section 3 we give a definition of a Lie ring with a unique addition, or a UA-
Lie ring. Also we discuss a weaker form of this notion that sometimes appears
in the literature and causes certain misunderstandings.

First results on the uniqueness of addition in Lie rings appeared more than 20
years ago. They have been published in several papers, some of which are now
difficult to access. The results were not formulated in full generality, and
recently there has been a desire to review and systematize the results
obtained at that time, to find possible generalizations, as well as to give
brief and uniform proofs. This desire is realized in this work. In addition,
we obtain several new results. Besides the study of the uniqueness of addition
by itself, this topic is a good motivation to develop the structural theory of
Lie rings and Lie algebras over various fields.

In Section 4 we collect results showing that certain Lie rings are not UA-Lie
rings. Proposition 1 claims that if a Lie ring $\mathfrak{R}$ is not perfect,
has a non-trivial center and is big enough, then $\mathfrak{R}$ is not a UA-
Lie ring.

Starting from Section 5 we come to positive results. The crucial point is that
the class of UA-Lie rings is wide – many important Lie algebras turn out to be
Lie rings with unique addition, so this property is not exotic.

Let us say that a Lie ring $\mathfrak{R}$ satisfies the C-condition if there
are two elements in $\mathfrak{R}$ whose centralizers have trivial
intersection. In Theorem 2 following [4] we prove that any finite-dimensional
Lie algebra over an infinite field satisfying the C-condition is a UA-Lie
ring. To be honest, to check the C-condition is the only known way to prove
that a Lie algebra is a UA-Lie ring. At the same time, the class of Lie
algebras with the C-condition includes all semisimple Lie algebras and all
their parabolic subalgebras (Proposition 2). Moreover, the concept of Mal’tsev
splitting allows to impose certain admissibility conditions on the radical of
a finite dimensional Lie algebra under which the C-condition is met (Theorem
5).

In Section 7 we consider a popular class of regular subalgebras of semisimple
Lie algebras called seaweed subalgebras. In Proposition 4 we give an effective
necessary and sufficient condition for a seaweed Lie algebra to be a UA-Lie
ring. The results of this section did not apper in earlier publications.

The aim of Section 9 is to discuss a verion of the unique addition property
which takes into account the specifics of Lie algebras. We say that a Lie
algebra $\mathfrak{g}$ over a field $K$ is a UA-Lie algebra if any bijection
$\alpha\colon\mathfrak{g}\to\mathfrak{s}$ to a Lie algebra $\mathfrak{s}$ over
the same field $K$ is a semilinear map.

In the last short section we consider commutator-preserving injections between
Lie rings. Currently, we have no positive results in this direction, but we
expect they will appear soon. In particular, we conjecture that the rigidity
of the commutator structure of a semisimple Lie algebra will allow to prove
that every injective commutator-preserving map from such an algebra is
additive.

The text includes a number of open questions. They are designed to show what
kind of problems in this area can be worked on now. The main problem – find
necessary and sufficient conditions for a finite-dimensional Lie algebra
$\mathfrak{g}$ to be a UA-Lie ring – is still very far from being solved, and
this area contains many interesting and exciting tasks.

The author is grateful to the organizers of the European Non-Associative
Algebra Seminar for invitation to give a talk on May 15, 2023. The talk served
as an incentive to return to the subject and to write this paper.

##  2\. Associative rings with unique addition

We begin with the case of associative rings.

######  Definition 1.

An associative ring $(R,+,\cdot)$ is called a unique addition ring, or a UA-
ring for short, if for any ring $(S,\oplus,*)$ any isomorphism $\alpha\colon
R\to S$ of the multiplicative semigroups $(R,\cdot)$ and $(S,*)$ is an
isomorphism of rings, i.e. the condition $\alpha(a\cdot
b)=\alpha(a)*\alpha(b)$ for all $a,b\in R$ implies
$\alpha(a+b)=\alpha(a)\oplus\alpha(b)$ for all $a,b\in R$.

The semigroups $(R,\cdot)$ and $(S,*)$ contain unique zeroes, so
$\alpha(0_{R})=0_{S}$. If $R$ is a ring with unit, then the semigroups
$(R,\cdot)$ and $(S,*)$ contain unique units, and again $\alpha(1_{R})=1_{S}$.

Equivalently, an associative ring $(R,+,\cdot)$ is a UA-ring, if for the
semigroup $(R,\cdot)$ there exists a unique binary operation $+$ on $R$ making
$(R,+,\cdot)$ into a ring. Indeed, given two additions $+$ and $\oplus$ on
$R$, it follows from Definition 1 applied to the identity map from
$(R,+,\cdot)$ to $(R,\oplus,\cdot)$ that $a+b=a\oplus b$ for all $a,b\in R$.
Conversely, every isomorphism of semigroups $\alpha\colon R\to S$ defines a
new addition on $R$, namely $a\boxplus
b:=\alpha^{-1}(\alpha(a)\oplus\alpha(b)).$

Let us test this definition in the case of finite fields. It is
streightforward to check that for $q\leqslant 4$ the field ${\mathbb{F}}_{q}$
is a UA-ring. We show that for $q>4$ it is not the case.

Let $q=p^{n}$. We have
$\operatorname{Aut}(\mathbb{F}_{q})=\langle\psi\rangle$, where $\psi$ is the
Frobenius automorphism, and $|\operatorname{Aut}(\mathbb{F}_{q})|=n$. At the
same time, the number of automorphisms of the semigroup
$(\mathbb{F}_{q},\cdot)$ is the Euler function $\varphi(q-1)$. For $q>4$ we
have $n+1$ numbers $1,p,p^{2},\ldots,p^{n-1},p^{n}-2$ that are smaller than
$q-1$ and coprime with $q-1$. So $\varphi(q-1)>n$ and we can find a non-
additive automorphism of the semigroup $(\mathbb{F}_{q},\cdot)$.

We do not try to overview the theory of unique addition associative rings, but
just give one important result that may be considered as the starting point of
the story we are going to tell.

######  Theorem 1 (Rickart [18], Stephenson [19]).

Let $A$ be an associative ring with unit and ${n\geqslant 2}$. Then the matrix
ring $\text{Mat}_{n\times n}(A)$ is a UA-ring.

For convenience of the reader, we provide a short proof of this result
following [19].

Step 1. Let $\alpha\colon R\to S$ be an isomorphism of multiplicative
semigroups and $R=B\times C$, where $B$ and $C$ are left ideals in $R$. Then
$\alpha(b+c)=\alpha(b)\oplus\alpha(c)$ for all $b\in B,c\in C$.

###### Proof.

Let $1_{R}=b_{1}+c_{1}$ with $b_{1}\in B$, $c_{1}\in C$. We have
$b_{1}=b_{1}(b_{1}+c_{1})$, so $b_{1}^{2}=b_{1}$ and $b_{1}c_{1}=0$. Let
$\alpha(b_{1})\oplus\alpha(c_{1})=\alpha(d)$. We have

$\alpha(b_{1}d)=\alpha(b_{1})\alpha(d)=\alpha(b_{1})(\alpha(b_{1})\oplus\alpha(c_{1}))=\alpha(b_{1}^{2})\oplus\alpha(b_{1}c_{1})=\alpha(b_{1}),$

so $b_{1}d=b_{1}$, $c_{1}d=c_{1}$, and $d=(b_{1}+c_{1})d=b_{1}+c_{1}=1_{R}$.
For $x=b+c$ we have $b=xb_{1}$, $c=xc_{1}$, and

$\alpha(b)\oplus\alpha(c)=\alpha(xb_{1})\oplus\alpha(xc_{1})=\alpha(x)(\alpha(b_{1})\oplus\alpha(c_{1}))=\alpha(x)=\alpha(b+c).$

∎

Step 2. Let $s\colon B\to C$ be an $R$-homomorphism. Then
$\alpha(s(b)+c)=\alpha(s(b))\oplus\alpha(c)$ for all $b\in B$, $c\in C$.

###### Proof.

We have $B\times C=(\text{id}+s)(B)\times C=R$ and applying Step 1 three times
we obtain

$\alpha(b)\oplus\alpha(s(b)+c)=\alpha(b+s(b)+c)=\alpha((\text{id}+s)(b)+c)=$
$=\alpha((\text{id}+s)(b))\oplus\alpha(c)=\alpha(b)\oplus\alpha(s(b))\oplus\alpha(c),$

so the claim. ∎

Step 3. Denote by $E_{ij}$ the matrix units in $\text{Mat}_{n\times n}(A)$.
With $R=\text{Mat}_{n\times n}(A)$ we have $R=RE_{11}\times\ldots\times
RE_{nn}$, and the left ideals $RE_{ii}$ are pairwise isomorphic. Applying the
results of Step 1 and Step 2 to this decomposition, we complete the proof of
Theorem 1.

Let us consider one more variant of definition of a unique addition ring.

######  Definition 2.

An associative ring $(R,+,\cdot)$ is called a ring with weak unique addition,
or a wUA-ring, if any bijection $\alpha\colon R\to R$ preserving
multiplication preserves the addition as well.

Clearly, any UA-ring is a wUA-ring. But Definition 2 is weaker than Definition
1. Indeed, take two rings with zero multiplication (or commutative Lie rings):

$\mathfrak{R}_{1}=(\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z},+)\ \
\text{and}\ \ \mathfrak{R}_{2}=(\mathbb{Z}/4\mathbb{Z},+).$

Then any bijection on $\mathfrak{R_{1}}$ that preserves $0_{1}$ is additive,
but any bijection from $\mathfrak{R_{1}}$ to $\mathfrak{R_{2}}$ that sends
$0_{1}$ to $0_{2}$ is not additive. So $\mathfrak{R_{1}}$ is a wUA-ring, but
not a UA-ring.

It seems that Definition 1 is more natural and common in the literature.

##  3\. Lie rings with unique addition

Now we come to uniqueness of addition in Lie rings.

######  Definition 3.

A Lie ring $(\mathfrak{R},+,[,])$ is called a unique addition Lie ring, or a
UA-Lie ring for short, if any Lie ring $(\mathfrak{S},\oplus,\lceil,\rceil)$
any bijection $\alpha\colon\mathfrak{R}\to\mathfrak{S}$ preserving commutators
is an isomorphism of Lie rings, i.e. the condition
$\alpha([a,b])=\lceil\alpha(a),\alpha(b)\rceil$ for all $a,b\in\mathfrak{R}$
implies $\alpha(a+b)=\alpha(a)\oplus\alpha(b)$ for all $a,b\in\mathfrak{R}$.

We have $\alpha(0)=\alpha([0,0])=\lceil\alpha(0),\alpha(0)\rceil=0.$ Clearly,
bijections as in Definition 3 preserve any property of a Lie ring that can be
formulated in terms of commutators only. For example, commutative, nilpotent,
or solvable Lie rings go to Lie rings of the same type.

######  Remark 1.

If $\mathfrak{R}=\mathfrak{R}_{1}\oplus\mathfrak{R}_{2}$ is a direct sum of
Lie rings and $\mathfrak{R}$ is a UA-Lie ring, it is easy to check that
$\mathfrak{R}_{1}$ and $\mathfrak{R}_{2}$ are UA-Lie rings as well.

The following problem was communicated to the author by Konstantin Beidar in
1992.

######  Problem 1.

Prove that the Lie algebra $\mathfrak{sl_{n}}(K)$ is a UA-Lie ring.

Some results proved below grew out of trying to solve this problem.

Here we also have a weaker version of the unique addition property.

######  Definition 4.

A Lie ring $\mathfrak{R}$ is a Lie ring with weak unique addition, or a wUA-
Lie ring, if any bijection $\alpha\colon\mathfrak{R}\to\mathfrak{R}$
preserving multiplication preserves the addition as well.

Some results in [1]-[4], [6] and [17] are obtained for this weaker property,
and one of the aim of this paper is to return to more natural Definition 3.

##  4\. Negative results

In this section we collect examples of Lie rings that are not UA-Lie rings. In
order to show that a Lie ring $\mathfrak{R}$ is not a UA-Lie ring it suffices
to construct a non-additive bijection
$\alpha\colon\mathfrak{R}\to\mathfrak{R}$ that preserves commutators.

######  Lemma 1.

Let $(A,+)$ be a commutative group with $|A|>4$ and $a\neq b\in
A\setminus\\{0\\}$. Then the bijection $\alpha\colon A\to A$, $\alpha(a)=b$,
$\alpha(b)=a$ and $\alpha(x)=x$ for all $x\neq a,b$ is not additive.

###### Proof.

Take $c\in A$, $c\neq 0,a,b,b-a$. Then
$a+c=\alpha(a+c)\neq\alpha(a)+\alpha(c)=b+c$. ∎

######  Proposition 1.

Let $\mathfrak{R}$ be a Lie ring with $|\mathfrak{R}|>4$. Assume that
$[\mathfrak{R},\mathfrak{R}]\neq\mathfrak{R}$, $Z(\mathfrak{R})\neq\\{0\\}$
and the condition $Z(\mathfrak{R})\cap[\mathfrak{R},\mathfrak{R}]=\\{0\\}$
implies $|Z(\mathfrak{R})|>2$. Then $\mathfrak{R}$ is not a UA-Lie ring.

###### Proof.

Case 1. Let $\mathfrak{R}$ be commutative. Then we take a bijection from Lemma
1.

Case 2. Let $\mathfrak{R}$ be not commutative and
$Z(\mathfrak{R})\cap[\mathfrak{R},\mathfrak{R}]=\\{0\\}$. Take $0\neq
z_{1},z_{2}\in Z(\mathfrak{R})$, $z_{1}\neq z_{2}$, and
$a\in[\mathfrak{R},\mathfrak{R}]\setminus\\{0\\}$. We let

$\alpha(a+z_{1})=a+z_{2},\ \ \alpha(a+z_{2})=a+z_{1},\ \ \text{and}\ \
\alpha(x)=x\quad\forall x\neq a+z_{1},a+z-2.$

Since $a+z_{1},a+z_{2}\notin[\mathfrak{R},\mathfrak{R}]$,
$[x,a+z_{1}]=[x,a+z_{2}]$ for all $x\in\mathfrak{R}$ and
$[a+z_{1},a+z_{2}]=0$, the bijection $\alpha$ preserves commutators. On the
other hand, it is not additive by Lemma 1.

Case 3. Let $Z(\mathfrak{R})\cap[\mathfrak{R},\mathfrak{R}]\neq\\{0\\}$. Then
we take $0\neq z\in Z(\mathfrak{R})\cap[\mathfrak{R},\mathfrak{R}]$,
$a\notin[\mathfrak{R},\mathfrak{R}]$, and let

$\alpha(a)=a+z,\ \ \alpha(a+z)=a,\ \ \text{and}\ \ \alpha(x)=x\quad\forall
x\neq a,a+z.$

Again we have $a,a+z\notin[\mathfrak{R},\mathfrak{R}]$, $[x,a]=[x,a+z]$ for
all $x\in\mathfrak{R}$ and $[a,a+z]=0$, so the bijection $\alpha$ preserves
commutators. It is not additive by Lemma 1. ∎

######  Corollary 1.

Let $\mathfrak{R}$ be a nilpotent Lie ring with $|\mathfrak{R}|>4$. Then
$\mathfrak{R}$ is not a UA-Lie ring.

######  Corollary 2.

Let $K$ be a field with at least three elements. Then the Lie algebra
$\mathfrak{gl}_{n}(K)$ is not a UA-Lie ring for any $n\geqslant 2$.

######  Remark 2.

Let $K$ be the field ${\mathbb{F}}_{2}$. For $n=2$ we have
$E=[E_{12},{E_{21}}]$. So for $n=2k$ Proposititon 1 implies that
$\mathfrak{gl}_{n}({\mathbb{F}}_{2})$ is not a UA-Lie ring.

######  Question 1.

Is the Lie algebra $\mathfrak{gl}_{2k+1}({\mathbb{F}}_{2})$ a UA-Lie ring?

In Section 8 we will see many examples of solvable Lie rings that are UA-Lie
rings. So the condition $[\mathfrak{R},\mathfrak{R}]=\mathfrak{R}$ is not
necessary for a Lie ring to be a UA-Lie ring.

######  Question 2.

Does there exist a Lie ring with a trivial center that is not a UA-Lie ring?

######  Example 1.

Let $K$ be a field of characteristic different from $2$ and $\mathfrak{g}$ be
the 6-dimensional Lie algebra over $K$ defined as the semidirect product
$\mathfrak{sl}_{2}(K)\rightthreetimes\langle x,y,z\rangle$, where

$[x,y]=z,\quad[x,z]=[y,z]=0,$

and the homomorphism $\mathfrak{sl}_{2}(K)\to\operatorname{Der}(\langle
x,y,z\rangle)$ is given by the tautologial $\mathfrak{sl}_{2}$-representation
on $\langle x,y\rangle$ and the trivial $\mathfrak{sl}_{2}$-representation on
$\langle z\rangle$. We have $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ and
$Z(\mathfrak{g})=\langle z\rangle$.

######  Question 3.

Is the Lie algebra $\mathfrak{g}$ from Example 1 a UA-Lie ring?

##  5\. The C-condition

In this section we introduce the main technical tool to prove that a Lie
algebra $\mathfrak{g}$ is a UA-Lie ring. Let us consider the centralizer of an
element in a Lie ring:

$C(a)=\\{b\in\mathfrak{R}\,|\,[a,b]=0\\}.$

The following key lemma was found by Alexander Titov in his school years, see
[6, Lemma 2].

######  Lemma 2.

Let $\alpha\colon\mathfrak{R}\to\mathfrak{S}$ be a commutator-preserving
bijection of Lie rings. If for some $a,b\in\mathfrak{R}$ we have $C(a)\cap
C(b)=\\{0\\}$, then $\alpha(a+b)=\alpha(a)\oplus\alpha(b)$.

###### Proof.

Let $\alpha(a+b)=\alpha(a)\oplus\alpha(b)\oplus\alpha(c)$ for some
$c\in\mathfrak{R}$. We have

$\lceil\alpha(a),\alpha(b)\rceil=\alpha([a,b])=\alpha([a+b,b])=\lceil\alpha(a+b),\alpha(b)\rceil=$
$=\lceil\alpha(a)\oplus\alpha(b)\oplus\alpha(c),\alpha(b)\rceil=\lceil\alpha(a),\alpha(b)\rceil\oplus\lceil\alpha(c),\alpha(b)\rceil.$

Hence $0=\lceil\alpha(c),\alpha(b)\rceil=\alpha([c,b])$, and $c\in C(b)$. By
the same arguments, we have $c\in C(a)$, and so $c=0$. ∎

######  Definition 5.

A Lie ring $\mathfrak{R}$ satisfies the C-condition if there are elements
$a,b\in\mathfrak{R}$ with $C(a)\cap C(b)=\\{0\\}$.

######  Remark 3.

Any Lie ring with C-condition has trivial center. Conversely, any 2-generated
Lie ring (Lie algebra) with trivial center satisfies the C-condition. In
particular, every semisimple Lie algebra $\mathfrak{g}$ over a field of
characteristic zero has trivial center and is generated by two elements [10,
Theorem 6], so $\mathfrak{g}$ satisfies the C-property.

The following theorem is obtained in [4, Theorem 3.2]. Below we give a proof
of this results with some improvements and corrections.

######  Theorem 2.

Any finite-dimensional Lie algebra over an infinite field satisfying the
C-condition is a UA-Lie ring.

###### Proof.

For a Lie algebra $\mathfrak{g}$ with C-condition we let

$C(\mathfrak{g})=\\{a\in\mathfrak{g}\ |\ \exists\ b\in\mathfrak{g}\
\text{with}\ C(a)\cap C(b)=\\{0\\}\\}.$

######  Lemma 3.

For any $a\in C(\mathfrak{g})$ the set $U(a)=\\{b\in\mathfrak{g}\mid C(a)\cap
C(b)=\\{0\\}\\}$ is open and dense in Zariski topology on $\mathfrak{g}$.

###### Proof.

Consider the linear map

$\mathfrak{g}\to\mathfrak{g}\oplus\mathfrak{g},\ \ \ \ x\mapsto([x,a],[x,y]).$

At the point $y=b$ the rank of this map is maximal. So the rank is maximal on
a dense open subset $U(a)$ in $\mathfrak{g}$. ∎

Since the set $C(\mathfrak{g})$ is the union of subsets $U(a),a\in
C(\mathfrak{g})$, the set $C(\mathfrak{g})$ is open and dense in
$\mathfrak{g}$ as well.

Now let $\alpha\colon\mathfrak{g}\to\mathfrak{S}$ be a commutator-preserving
bijection of Lie rings.

Case 1. Suppose $a,b\in C(\mathfrak{g})$. The intersection $U(b)\cap(U(a)-b)$
of Zariski open subsets is nonempty and we take an element $c$ from this
intersection. By Lemma 2, we obtain

$\alpha(a)\oplus\alpha(b)\oplus\alpha(c)=\alpha(a)\oplus\alpha(b+c)=\alpha(a+b+c)=\alpha(a+b)\oplus\alpha(c)\oplus\alpha(d),$

with some $d\in\mathfrak{g}$, and
$\alpha(a)\oplus\alpha(b)=\alpha(a+b)\oplus\alpha(d)$. As we have seen in the
proof of Lemma 2, the equality
$\alpha(a+b+c)=\alpha(a+b)\oplus\alpha(c)\oplus\alpha(d)$ implies that the
elements $c$ and $d$ commute. Since $c$ runs through a dense open subset, we
conclude that $d\in Z(\mathfrak{g})$. But the algebra $\mathfrak{g}$ satisfies
the C-condition, so its center is trivial and $d=0$.

Case 2. Suppose $b\in C(\mathfrak{g})$. Taking an element $a_{1}$ from
$C(\mathfrak{g})\cap(a-C(\mathfrak{g}))\cap(a+b-C(\mathfrak{g}))$ and letting
$a_{2}=a-a_{1}$, we obtain $a_{1},a_{2}\in C(\mathfrak{g})$ with
$a_{1}+a_{2}=a$ and $a_{2}+b\in C(\mathfrak{g})$. Using Case 1, we get

$\alpha(a)\oplus\alpha(b)=\alpha(a_{1})\oplus\alpha(a_{2})\oplus\alpha(b)=\alpha(a_{1})\oplus\alpha(a_{2}+b)=\alpha(a_{1}+a_{2}+b)=\alpha(a+b).$

Case 3. Let $a,b$ be arbitrary elements in $\mathfrak{g}$. Taking $b_{1}$ from
$C(\mathfrak{g})\cap(b-C(\mathfrak{g}))$ and letting $b_{2}=b-b_{1}$, we
obtain $b=b_{1}+b_{2}$ with $b_{1},b_{2}\in C(\mathfrak{g})$. Using Case 2, we
get

$\alpha(a+b)=\alpha(a+b_{1}+b_{2})=\alpha(a+b_{1})\oplus\alpha(b_{2})=$
$=\alpha(a)\oplus\alpha(b_{1})\oplus\alpha(b_{2})=\alpha(a)\oplus\alpha(b_{1}+b_{2})=\alpha(a)\oplus\alpha(b).$

This completes the proof of Theorem 2. ∎

If the center $Z(\mathfrak{R})\neq\\{0\\}$, then $\mathfrak{R}$ does not
satisfy the C-condition. The following example proposed by Mikhail Zaitsev [4,
Remark 3.5] demonstrates that the converse statement does not hold.

######  Example 2.

Consider a matrix Lie algebra

$\mathfrak{g}=\left\\{\left(\begin{array}[]{ccccc}a&0&a_{13}&a_{14}&a_{15}\\\
0&a&a_{23}&a_{24}&a_{25}\\\ 0&0&a&0&a_{35}\\\ 0&0&0&a&a_{45}\\\ 0&0&0&0&0\\\
\end{array}\right);\ a,a_{ij}\in K\right\\}.$

We claim that the center of this 9-dimensional Lie algebra is trivial, but the
C-condition does not hold. In order to check this, we consider the commutator
of two elements from $\mathfrak{g}$:

$[A,B]=\left(\begin{array}[]{ccccc}0&0&0&0&c_{15}\\\ 0&0&0&0&c_{25}\\\
0&0&0&0&c_{35}\\\ 0&0&0&0&c_{45}\\\ 0&0&0&0&0\\\ \end{array}\right),$

where

$c_{15}=ab_{15}+a_{13}b_{35}+a_{14}b_{45}-a_{15}b-a_{35}b_{13}-a_{45}b_{14},\quad
c_{35}=ab_{35}-a_{35}b$

and

$c_{25}=ab_{25}+a_{23}b_{35}+a_{24}b_{45}-a_{25}b-a_{35}b_{23}-a_{45}b_{24},\quad
c_{45}=ab_{45}-a_{45}b.$

If $A$ is in the center of $\mathfrak{g}$, then the condition $c_{15}=0$ for
all $B$ implies

$a=a_{13}=a_{14}=a_{15}=a_{35}=a_{45}=0,$

and the condition $c_{25}=0$ for all $B$ gives $a_{23}=a_{24}=a_{25}=0$, so
$A=0$.

Finally, the conditions $[A,D]=[B,D]=0$ are satisfied if we take
$d=d_{35}=d_{45}=0$ and

$(d_{13},d_{14},d_{15})=(d_{23},d_{24},d_{25})=(u,v,w),$

where $(u,v,w)$ is a non-zero solution of the system

$\left\\{\begin{aligned} a_{35}x+a_{45}y-az&=0\\\ b_{35}x+b_{45}y-bz&=0.\\\
\end{aligned}\right.$

######  Question 4.

Is the Lie algebra $\mathfrak{g}$ from Example 2 a UA-Lie ring?

##  6\. Semisimple Lie algebras and parabolic subalgebras

In this section we show that many important Lie algebras satisfy the
C-condition. We assume that the group field $K$ is an algebraically closed
field of characteristic zero.

######  Proposition 2.

Let $\mathfrak{g}$ be a semisimple Lie algebra over $K$ and $\mathfrak{p}$ be
a parabolic subalgebra in $\mathfrak{g}$. Then the Lie algebra $\mathfrak{p}$
satisfies the C-condition.

###### Proof.

Let $\mathfrak{b}$ be a Borel subalgebra in $\mathfrak{g}$ containing a Cartan
subalgebra $\mathfrak{h}$ and associated with a system of positive roots with
respect to $\mathfrak{h}$. We may assume that $\mathfrak{b}$ is contained in
$\mathfrak{p}$.

Take an element $x\in\mathfrak{h}$ that is not in the kernel of any root of
$\mathfrak{g}$ and let $y$ be the sum of root vectors over all positive roots
of $\mathfrak{g}$. Then $x,y\in\mathfrak{p}$ and we have $C(x)=\mathfrak{h}$
with $\mathfrak{h}\cap C(y)=\\{0\\}$. This proves that the Lie algebra
$\mathfrak{p}$ satisfies the C-condition. ∎

Theorem 2 and Proposition 2 immediately imply the following result, first
proved in [3, Theorem 1] and [4, Corollary 4.6] for a weaker definition of
unique addition.

######  Theorem 3.

Let $\mathfrak{g}$ be a semisimple Lie algebra over an algebraically closed
field $K$ of characteristic zero and $\mathfrak{p}$ be a parabolic subalgebra
in $\mathfrak{g}$. Then both $\mathfrak{g}$ and $\mathfrak{p}$ are UA-Lie
rings.

Clearly, the arguments given above imply that if $\mathfrak{g}$ is a direct
sum of some classical Lie algebras $\mathfrak{sl}_{n}(K)$,
$\mathfrak{so}_{n}(K)$ and $\mathfrak{sp}_{2n}(K)$ over any field $K$ of
characteristic zero, then $\mathfrak{g}$ is a UA-Lie ring.

Later a much more general result was proved by Mayorova.

######  Theorem 4.

[13, Theorem 6] Lie algebras of Chevalley type ($A_{n}$, $B_{n}$, $C_{n}$,
$D_{n}$, $E_{6}$, $E_{7}$, $E_{8}$, $F_{4}$, and $G_{2}$) over associative
commutative rings with $\frac{1}{2}$ (with $\frac{1}{2}$ and $\frac{1}{3}$ in
the case of $G_{2}$) are UA-Lie rings.

The proof of this result is based on explicit computations with root systems.

##  7\. Seaweed Lie algebras

The aim of this section is to generalize the idea of the proof of Proposition
2 to a wider class of Lie algebras. Let $\mathfrak{a}$ be a finite-dimensional
Lie algebra over an infinite field $K$ and $\mathfrak{h}\subseteq\mathfrak{a}$
be a commutative subalgebra. We say that the pair
$(\mathfrak{a},\mathfrak{h})$ is _suitable_ if
$\mathfrak{a}=\oplus_{i}\mathfrak{a}_{\lambda_{i}}$, where
$\lambda_{i}\in\mathfrak{h}^{*}$ are linear functions on $\mathfrak{h}$,
$\mathfrak{a}_{\lambda_{i}}:=\\{x\in\mathfrak{a}\,|\,[h,x]=\lambda_{i}(h)x\
\forall\,h\in\mathfrak{h}\\}$, we have
$\cap_{i}\operatorname{Ker}\lambda_{i}=\\{0\\}$, and
$\mathfrak{a}_{0}=\mathfrak{h}$.

######  Lemma 4.

Let $(\mathfrak{a},\mathfrak{h})$ be a suitable pair. Then the Lie algebra
$\mathfrak{a}$ is a UA-Lie ring.

###### Proof.

Take an element $h\in\mathfrak{h}$ such that $\lambda_{i}(h)\neq 0$ for any
non-zero $\lambda_{i}$. Then $C(h)=\mathfrak{h}$. If we let $y=\sum_{i}y_{i}$
with $0\neq y_{i}\in\mathfrak{a}_{\lambda_{i}}$, then
$C(y)\cap\mathfrak{h}=\\{0\\}$ and the algebra $\mathfrak{a}$ satisfies the
C-condition. So the assertion follows from Theorem 2. ∎

Lemma 4 can be applied in the following situation. Assume that the field $K$
is an algebraically closed field of characteristic zero, $\mathfrak{g}$ is a
semisimple Lie algebra over $K$ and $\mathfrak{h}$ is a Cartan subalgebra in
$\mathfrak{g}$. A subalgebra $\mathfrak{a}\in\mathfrak{g}$ is called _regular_
if $\mathfrak{a}$ containes $\mathfrak{h}$. In this case $\mathfrak{a}$ as a
vector space is a direct sum of $\mathfrak{h}$ and some root subspaces
$\mathfrak{g}_{\alpha_{i}}$ in $\mathfrak{g}$.

We call a regular subalgebra $\mathfrak{a}$ _ample_ if the roots $\alpha_{i}$
such that $\mathfrak{g}_{\alpha_{i}}$ is contained in $\mathfrak{a}$ span the
vector space $\mathfrak{h}^{*}$. Equivalently, this condition means that
$\cap_{i}\operatorname{Ker}\alpha_{i}=\\{0\\}$. Since there are no zero roots,
this condition is equivalent to the condition that the pair
$(\mathfrak{a},\mathfrak{h})$ is suitable.

######  Example 3.

Any parabolic subalgebra $\mathfrak{p}$ in a semisimple Lie algebra
$\mathfrak{g}$ is an ample regular subalgebra.

######  Proposition 3.

A regular subalgebra $\mathfrak{a}$ of a semisimple Lie algebra $\mathfrak{g}$
is a UA-Lie ring if and only if $\mathfrak{a}$ is ample.

###### Proof.

The ‘‘if’’ part follows from Lemma 4. As for the inverse implication, let
$\mathfrak{h}^{\prime}=\cap_{i}\operatorname{Ker}\alpha_{i}$. Clearly, the
subalgebra $\mathfrak{h}^{\prime}$ is contained in the center of
$\mathfrak{a}$.

The derived subalgebra $[\mathfrak{a},\mathfrak{a}]$ is spanned by all root
subspaces $\mathfrak{g}_{\alpha_{i}}$ and subspaces in $\mathfrak{h}$ that are
commutators of root subspaces
$[\mathfrak{g}_{\alpha_{i}},\mathfrak{g}_{-\alpha_{i}}]$ corresponding to
opposite roots. The latter subspaces lie in Lie subalgebras isomorphic to
$\mathfrak{sl}_{2}(K)$, and $\mathfrak{h}\cap[\mathfrak{a},\mathfrak{a}]$ is
the Cartan subalgebra of the semisimple part of the algebra $\mathfrak{a}$. It
proves that $[\mathfrak{a},\mathfrak{a}]\cap\mathfrak{h}^{\prime}=\\{0\\}$ and
if $\mathfrak{h}^{\prime}\neq 0$, then $\mathfrak{a}$ is not a UA-Lie ring by
Proposition 1. ∎

A remarkable class of regular subalgebras of a semisimple Lie algebra
$\mathfrak{g}$ form so-called seaweed subalgebras introduced by Dergachev and
Kirillov [8] in the case $\mathfrak{g}=\mathfrak{sl}_{n}(K)$ and by Panyushev
[15] in the general case; see also [7] for a nice recent survey.

Recall that a subalgebra $\mathfrak{a}$ in a semisimple Lie algebra
$\mathfrak{g}$ is called _seaweed_ if $\mathfrak{a}$ is the intersection of
two parabolic subalgebras $\mathfrak{p}$ and $\mathfrak{p}^{\prime}$ such that
$\mathfrak{g}=\mathfrak{p}+\mathfrak{p}^{\prime}$. Take a Borel subalgebra
$\mathfrak{b}\subseteq\mathfrak{g}$ and a Cartan subalgebra $\mathfrak{h}$ in
$\mathfrak{b}$. It is observed in [15] that up to conjugation one may assume
that $\mathfrak{p}$ contains $\mathfrak{b}$ and $\mathfrak{p}^{\prime}$
contains $\mathfrak{b}_{-}$, where $\mathfrak{b}_{-}$ is the opposite Borel
subalgebra. In particular, $\mathfrak{a}$ is a regular subalgebra in
$\mathfrak{g}$.

Proposition 3 immediatly implies the following result.

######  Proposition 4.

A seaweed subalgebra $\mathfrak{a}$ of a semisimple Lie algebra $\mathfrak{g}$
is a UA-Lie ring if and only if $\mathfrak{a}$ is ample.

##  8\. Solvable Lie algebras and Mal’cev splittings

In this section we show that the C-condition is fulfilled for a wide class of
solvable Lie algebras. Further we consider finite-dimensional Lie algebras
over an algebraically closed field $K$ of characteristic zero.

######  Definition 6.

A solvable Lie algebra $\mathfrak{r}$ is Mal’cev split if
$\mathfrak{r}=\mathfrak{k}\oplus\mathfrak{n}$, where $\mathfrak{n}$ is a
nilpotent ideal, $\mathfrak{k}$ is a commutative subalgebra, and the adjoint
action of $\mathfrak{k}$ on $\mathfrak{n}$ is faithful and completele
reducible.

Here $\mathfrak{n}=\oplus_{i}\mathfrak{n}_{\lambda_{i}}$, where

$\mathfrak{n}_{\lambda_{i}}=\\{x\in\mathfrak{n}\,|\,[y,x]=\lambda_{i}(y)x\
\forall y\in\mathfrak{k}\\}$

and $\\{\lambda_{i}\\}$ are linear functions on $\mathfrak{k}$ with
$\cap_{i}\operatorname{Ker}\lambda_{i}=\\{0\\}$.

Any algebraic solvable Lie algebra is Mal’cev split. Moreover, every solvable
Lie algebra $\mathfrak{r}$ admits the Mal’cev splitting, i.e. an embedding
$\gamma\colon\mathfrak{r}\to M(\mathfrak{r})$, where
$M(\mathfrak{r})=\mathfrak{k}\oplus\mathfrak{n}$ is Mal’cev split and
$\gamma(\mathfrak{r})$ is an ideal in $M(\mathfrak{r})$ with
$M(\mathfrak{r})=\mathfrak{k}+\gamma(\mathfrak{r})=\gamma(\mathfrak{r})+\mathfrak{n}$.
The Mal’cev splitting of $\mathfrak{r}$ exists and is unique up to
isomorphism; see [11].

######  Definition 7.

A solvable Lie algebra $\mathfrak{r}$ is admissible if for
$M(\mathfrak{r})=\mathfrak{k}\oplus\mathfrak{n}$ there is $x\in\mathfrak{n}$
with $C(x)\cap\mathfrak{n}_{0}=\\{0\\}$.

######  Example 4.

Any parabolic subalgebra $\mathfrak{p}$ in a semisimple Lie algebra
$\mathfrak{g}$ is a Lie algebra with admissible radical.

The following result is obtained in [6] in a slightly weaker form.

######  Theorem 5.

Let $\mathfrak{g}$ be a finite-dimensional Lie algebra with admissible
radical. Then $\mathfrak{g}$ satisfies the C-condition. In particular,
$\mathfrak{g}$ is a UA-Lie ring.

###### Proof.

Let $\mathfrak{g}=\mathfrak{l}\rightthreetimes\mathfrak{r}$, where
$\mathfrak{l}$ is a maximal semisimple subalgebra in $\mathfrak{g}$ and
$\mathfrak{r}$ is the radical of $\mathfrak{g}$. We know from Proposition 2
that $\mathfrak{l}$ has the C-property.

######  Lemma 5.

The Lie algebra $M(\mathfrak{r})$ has the C-property.

###### Proof.

Take $a\in\mathfrak{k}$ with $\lambda_{i}(a)\neq 0$ for all non-zero
$\lambda_{i}$. Then $C(a)=\mathfrak{k}\oplus\mathfrak{n}_{0}$. By definition
of admissibility, there exists an element $x\in\mathfrak{n}$ such that
$C(x)\cap\mathfrak{n}_{0}=\\{0\\}$. This condition holds for all elements $x$
from a Zariski open dense subset $V$ in $\mathfrak{n}$, because if the rank of
the linear map $\mathfrak{n}_{0}\to\mathfrak{n}$, $y\mapsto[x,y]$ is maximal
at some point $x\in\mathfrak{n}$, it is maximal for all points $x$ from an
open dense subset. Take $b=\sum b_{\lambda_{i}}$ with $0\neq
b_{\lambda_{i}}\in\mathfrak{n}_{\lambda_{i}}$ and
$C(b)\cap\mathfrak{n}_{0}=\\{0\\}$.

We claim that $C(b)\cap(\mathfrak{k}\oplus\mathfrak{n}_{0})=\\{0\\}$. Indeed,
assume that $d+e\in C(b)$, $d\in\mathfrak{k}$, $e\in\mathfrak{n}_{0}$. Then
$[d+e,b]=0$. Since
$[\mathfrak{n}_{0},\mathfrak{n}_{\lambda_{i}}]\subseteq\mathfrak{n}_{\lambda_{i}}$,
we have $\lambda_{i}(d)b_{\lambda_{i}}=-[e,b_{\lambda_{i}}]$. But the operator
$\operatorname{ad}(e)$ is nilpotent, so all $\lambda_{i}(d)=0$. The condition
$\cap_{i}\operatorname{Ker}\lambda_{i}=\\{0\\}$ implies $d=0$, and the
condition $C(b)\cap\mathfrak{n}_{0}=\\{0\\}$ shows that $e=0$.

We conclude that $C(a)\cap C(b)=\\{0\\}$. ∎

######  Lemma 6.

The Lie algebra $\mathfrak{r}$ has the C-property.

###### Proof.

Take elements $a,b\in M(\mathfrak{r})$ as in Lemma 5. By definition of the
Mal’cev splitting, there are elements $p_{1},p_{2}\in\mathfrak{r}$ such that
$a=p_{1}+q$ and $b=r+p_{2}$ with $q\in\mathfrak{n}$ and $r\in\mathfrak{k}$.

We can assume that $q\in\mathfrak{n}_{0}$. Indeed, let
$q=\sum_{i}q_{\lambda_{i}}$. Then
$[a,q]=\sum_{i}\lambda_{i}(a)q_{\lambda_{i}}=[p_{1},q]\in\mathfrak{r}$ and
$\lambda_{i}(a)\neq 0$ for all non-zero $\lambda_{i}$. Adding to $q$ the
element $[a,q]\in\mathfrak{r}\cap\mathfrak{n}$ with a suitable coefficient, we
cancel one of the components $q_{\lambda_{i}}$ with non-zero $\lambda_{i}$.
Continuing this process, we come to $q\in\mathfrak{n}_{0}$.

We claim that $C(p_{1})\subseteq\mathfrak{k}\oplus\mathfrak{n}_{0}$. Indeed,
if $[a-q,l]=0$, then $\sum_{i}\lambda_{i}(a)l_{\lambda_{i}}=[q,l]$. But the
operator $\operatorname{ad}(q)$ is nilpotent and so all
$\lambda_{i}(a)l_{\lambda_{i}}=0$. For non-zero $\lambda_{i}$ we have
$\lambda_{i}(a)\neq 0$, so $l_{\lambda_{i}}=0$.

Finally, let us show that
$C(p_{2})\cap(\mathfrak{k}\oplus\mathfrak{n}_{0})=\\{0\\}$. If
$[b-r,t]=[b,t]=0$ with $t\in\mathfrak{k}\oplus\mathfrak{n}_{0}$, then $t\in
C(b)$. But it is shown in the proof of Lemma 5 that
$C(b)\cap(\mathfrak{k}\oplus\mathfrak{n}_{0})=\\{0\\}$.

Thus, we proved that $C(p_{1})\cap C(p_{2})=\\{0\\}$, ∎

######  Lemma 7.

If a finite-dimensional Lie algebra $\mathfrak{f}$ satifies the C-condition,
then there is a Zariski open dense subset
$U\subseteq\mathfrak{f}\oplus\mathfrak{f}$ such that $C(a)\cap C(b)=\\{0\\}$
for all $(a,b)\in U$.

###### Proof.

If for some pair $(a,b)$ the rank of the linear map
$\mathfrak{f}\to\mathfrak{f}\oplus\mathfrak{f}$, $c\mapsto([a,c],[b,c])$
equals $\dim\mathfrak{f}$ and, in particular, is maximal possible, then the
same holds for all pairs $(a,b)$ from a Zariski open dense subset
$U\subseteq\mathfrak{f}\oplus\mathfrak{f}$. ∎

Now we are going to prove the C-property for the Lie algebra $\mathfrak{g}$.
Take Zariski open dense subsets $U_{1}\subseteq\mathfrak{l}\oplus\mathfrak{l}$
and $U_{2}\subseteq\mathfrak{r}\oplus\mathfrak{r}$ as in Lemma 7. In
particular, if $(z_{1},z_{2})\in U_{2}$ then $C(z_{1})\cap
C(z_{2})\cap\mathfrak{r}=\\{0\\}$. It follows that we have $C(x_{1})\cap
C(x_{2})\cap\mathfrak{r}=\\{0\\}$ for all $(x_{1},x_{2})$ from a Zariski open
dense subset $W\subseteq\mathfrak{g}\oplus\mathfrak{g}$. Indeed, we can apply
the same arguments as in the proof of Lemma 7 to the linear map
$\mathfrak{r}\to\mathfrak{g}\oplus\mathfrak{g}$,
$z\mapsto([x_{1},z],[x_{2},z])$.

Choose elements $x_{1}=y_{1}+z_{1},x_{2}=y_{2}+z_{2}\in\mathfrak{g}$ with
$(y_{1},y_{2})\in U_{1}$, $z_{1},z_{2}\in\mathfrak{r}$, and $(x_{1},x_{2})\in
W$. This is possible because two Zariski open dense subsets
$U_{1}\oplus\mathfrak{r}\oplus\mathfrak{r}$ and $W$ in
$\mathfrak{g}\oplus\mathfrak{g}$ have non-empty intersection.

If $x=y+z\in C(x_{1})\cap C(x_{2})$, then $y\in C(y_{1})\cap C(y_{2})$, so
$y=0$. It implies $[z,x_{1}]=[z,x_{2}]=0$, and $z=0$ by assumption on the
subset $W$. It means that $C(x_{1})\cap C(x_{2})=\\{0\\}$, and Theorem 5 is
proved. ∎

##  9\. UA-Lie algebras

Now we try to take into account the specifics of Lie algebras in the
definition of the uniqueness of addition.

######  Definition 8.

Let $K$ be a field. Consider two $K$-vector spaces $V$ and $W$. A map
$\alpha\colon V\to W$ is semilinear if it is additive, i.e.
$\alpha(v_{1}+v_{2})=\alpha(v_{1})+\alpha(v_{2})$ for all $v_{1},v_{2}\in V$,
and there is an automorphism $\theta\in\text{Aut}(K)$ such that
$\alpha(\lambda v)=\theta(\lambda)\alpha(v)$ for all $\lambda\in K$ and $v\in
V$.

######  Definition 9.

A Lie algebra $(\mathfrak{g},+,[,])$ over a field $K$ is called a unique
addition Lie algebra, or a UA-Lie algebra, if for any Lie algebra
$(\mathfrak{s},\oplus,\lceil,\rceil)$ over the same field $K$ any bijection
$\alpha\colon\mathfrak{g}\to\mathfrak{s}$ preserving commutators is a
semilinear map.

######  Remark 4.

Some fields like $\mathbb{Q}$ or $\mathbb{R}$ have no nontrivial automorphism,
so in these cases any commutator-preserving bijection should be linear.

We also have a weaker version of the unique addition property.

######  Definition 10.

A Lie algebra $\mathfrak{g}$ is a Lie algebra with weak unique addition, or a
wUA-Lie algebra, if any bijection $\alpha\colon\mathfrak{g}\to\mathfrak{g}$
preserving commutators is a semilinear map.

######  Remark 5.

We have no example of two Lie algebras $\mathfrak{g}_{1}$ and
$\mathfrak{g}_{2}$ over a field $K$ such that there exists a semilinear
bijection $\mathfrak{g}_{1}\to\mathfrak{g}_{2}$, but the Lie algebras
$\mathfrak{g}_{1}$ and $\mathfrak{g}_{2}$ are not isomorphic.

Let us denote by $\mathfrak{s}_{2}(K)$ the 2-dimensional non-commutative Lie
algebra over $K$. The following result is proved in [1]-[2],

######  Proposition 5.

Let $K$ be a field of characteristic different from $2$. Then the Lie algebras
$\mathfrak{s}_{2}(K)$ and $\mathfrak{sl}_{2}(K)$ are wUA-Lie algebras.

Let us assume that $K$ has characteristic zero. Clearly, the Lie algebras
$\mathfrak{s}_{2}(K)$ and $\mathfrak{sl}_{2}(K)$ have the C-property, so they
are UA-Lie rings. Over the field ${\mathbb{Q}}$ of rational numbers any
additive map between vector spaces is linear, so we conclude that
$\mathfrak{s}_{2}({\mathbb{Q}})$ and $\mathfrak{sl}_{2}({\mathbb{Q}})$, as
well as any other Lie algebra over ${\mathbb{Q}}$ that is a UA-Lie ring, are
UA-Lie algebras.

Let us give one more observation on this subject.

######  Lemma 8.

Let $K$ be an algebraically closed field of characteristic zero. Assume that
$\mathfrak{s}$ is a finite-dimensional Lie algebra over $K$. Then any
bijection $\alpha\colon\mathfrak{sl}_{2}(K)\to\mathfrak{s}$ preserving
commutators is a semilinear map.

###### Proof.

The Lie algebra $\mathfrak{s}$ inherits properties of $\mathfrak{sl}_{2}(K)$
that are expressible in terms of commutators. For instance, it is not solvable
and the centralizer of any non-zero element in $\mathfrak{s}$ is commutative.
By [5, Proposition 2], these two properties imply that $\mathfrak{s}$ is
isomorphic to $\mathfrak{sl}_{2}(K)$. So, the UA-property for
$\mathfrak{sl}_{2}(K)$ follows from the wUA-property. ∎

Summarizing this discussion, it is worth noting that very few examples of UA-
Lie algebra are known, and building such examples seems to be an important and
interesting task.

Finally, let us mention one more fact related to this subject. In [17],
Ponomarev proved the following result.

######  Theorem 6.

Let $K$ be an algebraically closed field of characteristic zero. Then a
parabolic subalgebra $\mathfrak{p}$ in a semisimple Lie algebra $\mathfrak{g}$
over $K$ is a wUA-Lie algebra if and only if the Lie algebra $\mathfrak{g}$ is
simple.

The proof uses Theorem 3 and is based on a description of automorphisms of
regular subalgebras of reductive Lie algebras as Lie rings obtained in [16].

##  10\. Commutator-preserving injections

One may weaken Definition 3 and assume that the map $\alpha$ is not a
bijection, but just an injection. This weakening significantly changes the
situation.

######  Lemma 9.

Assume that the field $K$ contains at least $3$ elements and $\mathfrak{g}$ is
a Lie algebra over $K$ with $\mathfrak{g}\neq[\mathfrak{g},\mathfrak{g}]$.
Then there is a Lie algebra $\mathfrak{s}$ over $K$ and an injective map
$\beta\colon\mathfrak{g}\to\mathfrak{s}$ that preserves commutators, but is
not additive.

###### Proof.

Let us fix a non-additive map $\gamma\colon K\to K$ with $\gamma(0)=0$. For
example, we may let $\gamma(a)=1$ for any non-zero $a\in K$.

Take a basis in $[\mathfrak{g},\mathfrak{g}]$ and complement it to a basis
$\\{x_{i}\\}$ in $\mathfrak{g}$. We assume that
$x_{1}\notin[\mathfrak{g},\mathfrak{g}]$. Let the Lie algebra $\mathfrak{s}$
be the direct sum of Lie algebras $\mathfrak{g}\oplus\langle z\rangle$, so $z$
is a central element. Define

$\beta\colon\mathfrak{g}\to\mathfrak{s},\quad\beta(x)=x+\gamma(a_{1})z,\quad\text{where}\quad
x=\sum_{i}a_{i}x_{i},\quad a_{i}\in K.$

Then we have

$\beta[x,y]=([x,y],0)=[\beta(x),\beta(y)]\quad\text{for all}\quad
x,y\in\mathfrak{g}.$

So the map $\beta$ preserves commutators, but it is not additive. ∎

######  Example 5.

As we have seen, the 2-dimensional non-commutative Lie algebra
$\mathfrak{s}_{2}(K)$ over an infinite field $K$ is a UA-Lie ring. At the same
time, Lemma 9 shows that a commutator-preserving injection need not be
additive in this case.

In view of the above, let us formulate the following question.

######  Question 5.

Let $\beta\colon\mathfrak{g}\to\mathfrak{S}$ be a commutator-preserving
injection from a semisimple Lie algebra $\mathfrak{g}$ to a Lie ring
$\mathfrak{S}$. Is $\beta$ additive?

At the moment we have no example of a Lie ring $\mathfrak{R}$ such that any
injective map $\beta\colon\mathfrak{R}\to\mathfrak{S}$ to a Lie ring
$\mathfrak{S}$ that preserves commutators is automatically additive.

## References

  * [1] Ivan Arzhantsev. Uniqueness of addition in Lie algebra $\mathfrak{sl}_{2}$. In "Lie Algebras, Rings and Related Topics" (Fong Yuen, A. Mikhalev, E. Zelmanov, Eds), Springer-Verlag Hong Kong Ltd. (2000), 1-4
  * [2] Ivan Arzhantsev. The uniqueness of addition in Lie algebra $\mathfrak{sl}_{2}$. Moscow Univ. Math. Bull. 55 (2000), no. 4, 29-31
  * [3] Ivan Arzhantsev. Uniqueness of addition in semisimple Lie algebras. Russian Math. Surveys 56 (2001), no. 3, 569-571
  * [4] Ivan Arzhantsev. Some results on uniqueness of addition in Lie algebras. Proceedings of the I Colloquium on Lie Theory and Applications (I.Bajo, E.Sanmartin, Eds), Universidad de Vigo (2002), 19-24
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  * [8] Vladimir Dergachev and Alexandre Kirillov. Index of Lie algebras of seaweed type. J. Lie Theory 10 (2000), no. 2, 331-343
  * [9] Richard Edward Johnson. Rings with unique addition. Proc. Amer. Math. Soc. 9 (1958), 57-61
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  * [11] Anatoly Mal’tsev. On solvable Lie algebras. Izv. Akad. Nauk SSSR, Ser. Mat. 9 (1945), 329-356; English transl.: Amer. Math. Soc. Transl. 27 (1950), 36 pages
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# An improved bound on Legendre approximation

Mehdi Hamzehnejad  Mohammad Mehdi Hosseini  Abbas Salemi<EMAIL_ADDRESS>Department of Mathematic, Kerman Graduate University of Advanced Technology,
Kerman, Iran.  Department of Applied Mathematics and Mahani Mathematical
Research Center, Shahid Bahonar University of Kerman, Kerman, Iran.

###### Abstract

In this paper, new relations between the derivatives of the Legendre
polynomials are obtained, and by these relations, new upper bounds for the
Legendre coefficients of differentiable functions are presented. These upper
bounds are sharp and cover more categories of differentiable functions.
Moreover, new and sharper bounds for the approximation error of the partial
sums of Legendre polynomials are provided. Numerical examples are given to
validate our theoretical results.

###### keywords:

Legendre polynomials, Legendre expansions, approximation error , error bound.

###### MSC:

[2010] 41A25, 41A10.

††journal: ……………………

##  1 Introduction

One of the most important properties of Legendre polynomials
$L_{n}(x),n=0,1,\ldots$ is their ability to approximate a function which has
fascinated a great attention in recent years. See for example: [1, 2, 3, 4, 5,
6, 7]. Let us suppose that a suitably smooth function $f$ has the following
expansion:

$f(x)=\sum_{n=0}^{\infty}a_{n}L_{n}(x),$ (1)

where $a_{n},$ Legendre coefficient, is defined by

$a_{n}=\left(n+\frac{1}{2}\right)\int_{-1}^{1}f(x)L_{n}(x)dx.$

Consider the approximation of differentiable functions using the partial sum

$f_{N}(x)=\sum_{n=0}^{N}a_{n}L_{n}(x).$

Several papers are available in literature dealing with this topic under
different smoothness assumptions of differentiable functions.

Suetin proved that if the function $f$ has $r$ continuous derivatives on
$[-1,1]$ and $f^{(r)}\in Lip\alpha$ with $r+\alpha\geq\frac{1}{2}$, then [1]

$\|f(x)-f_{N}(x)\|_{\infty}\leq\mathit{C}\frac{logN}{N^{r+\alpha-\frac{1}{2}}},\quad-1\leq
x\leq 1,$

where $\mathit{C}$ is positive constant independent of $n$.

Later on, in [2] the authors obtained error estimates which depend on the
Legendre coefficients of the function $f$. They showed that if
$f,f^{\prime},\cdots,f^{(r-1)}$ are absolutely continuous and the $r^{th}$
derivative $f^{(r)}$ is of bounded variation and

$V_{r}=\int_{-1}^{1}\frac{|f^{(r+1)}(x)|}{\sqrt{1-x^{2}}}dx<\infty,$ (2)

then

$\|f(x)-f_{N}(x)\|_{\infty}\leq\sqrt{\frac{\pi}{2}}\frac{V_{r}}{(r-1)\sqrt{N-r}\prod_{j=2}^{r}2N-2j+3}.$
(3)

Under the assumptions stated in [2], Wang [5] proved that if

$\widehat{V}_{r}=\int_{-1}^{1}\frac{|f^{(r+1)}(x)|}{\sqrt[4]{1-x^{2}}}dx<\infty,$
(4)

then

$\|f(x)-f_{N}(x)\|_{\infty}\leq\left\\{\begin{array}[]{lr}\frac{4\widehat{V}_{1}}{\sqrt{\pi(2N-5)}}&r=1\\\
\frac{2\widehat{V}_{r}}{\prod_{j=2}^{r}(N-j+\frac{1}{2})(r-1)\sqrt{\pi(2N-2r-1)}}&r\geq
2.\end{array}\right.$ (5)

In all the above estimates the rate of convergence is essentially
$N^{-r+\frac{1}{2}}$ that in each case the function $f$ depends on certain
smoothness properties.

Now we want to obtain a new upper bound for the Legendre coefficients without
the assumptions in (2) and (4) and provide new approximation error for more
categories of differentiable functions. For this purpose, we state new
relationship between derivatives of Legendre polynomials. Moreover, by using
the telescoping series and by induction, we obtain new error estimates in
uniform norm which are sharper than previous bounds presented in [2, 5, 7].

The rest of this paper is organized as follows. In Section 2, using some
properties of Legendre polynomials, we obtain a new expansion for derivatives
of Legendre polynomials and by using this expansion, we derive a new upper
bound for the Legendre coefficients of differentiable functions. Also, by
using the asymptotic of the Legendre coefficients, we provide new
approximation errors of a function $f(x)$ by partial sums of Legendre
polynomials. In Section 4, numerical results are presented and we compare the
proposed upper bound with the upper bounds were presented so far.

##  2 New approximation errors of Legendre polynomials

In this section some necessary properties of Legendre polynomials are
presented. Then we provide a new expansion for derivative of Legendre
polynomials. Using this expansion, new approximation errors of the partial
sums of the Legendre polynomials are provided.

Legendre polynomials are the eigenfunctions of the following Sturm-Liouville
problem [8]

$\left(\left(1-x^{2}\right)L_{n}^{\prime}(x)\right)^{\prime}=-n(n+1)L_{n}(x).$
(6)

Also, the following relation is hold [8]

$\left(1-x^{2}\right)L_{n}^{\prime}(x)=\frac{n(n+1)}{2n+1}\left(L_{n-1}(x)-L_{n+1}(x)\right).$
(7)

The following orthogonality is hold on $\left[-1,1\right]$:

$\int_{-1}^{1}L_{n}(x)L_{m}(x)dx=\frac{\delta_{mn}}{n+\frac{1}{2}},$

where $\delta_{mn}$ is the Kronecker delta.

By using Sturm-Liouville problem and relation (7), we provide a new expansion
for derivative of Legendre polynomials. For simplicity, we use the following
notation.

$\mathcal{L}_{n}(x):=\frac{\left(1-x^{2}\right)L_{n}^{\prime}(x)}{n(n+1)}.$
(8)

The following key lemma present a new relation between derivatives of Legendre
polynomials.

######  Lemma 1.

Let $\mathcal{L}_{n}(x)$ be as in (8). Then $\mathcal{L}_{n}(x)$ has the
following representation, for $1\leq r\leq n-1$.

$\displaystyle\mathcal{L}_{n}(x)$
$\displaystyle=\left(\frac{(-1)^{r}\mathcal{L}_{n-r}(x)}{\lambda^{1}_{n,r}}+\frac{(-1)^{r-1}\mathcal{L}_{n-r+2}(x)}{\lambda^{2}_{n,r}}+\cdots\right.$
$\displaystyle\left.-\frac{\mathcal{L}_{n+r-2}(x)}{\lambda^{2^{r-1}}_{n,r}}+\frac{\mathcal{L}_{n+r}(x)}{\lambda^{2^{r}}_{n,r}}\right)^{(r)},$
(9)

where

$\lambda^{j}_{n,r}=\left\\{\begin{array}[]{lr}\prod_{i=1}^{r}(2n-2i+3)&j=1,2\\\
\prod_{i=1}^{r}(2n+2i-1)&j=2^{r}-1,2^{r}\end{array}\right.$

Moreover, for $1\leq j\leq 2^{r-1}$,

$\lambda^{1}_{n,r}=\lambda^{2}_{n,r}\leq\lambda^{2j-1}_{n,r}=\lambda^{2j}_{n,r}\leq\lambda^{2^{r}-1}_{n,r}=\lambda^{2^{r}}_{n,r}.$

###### Proof.

By combining (6) and (8), we obtain

$\left(\mathcal{L}_{n}(x)\right)^{\prime}=-L_{n}(x).$ (10)

If we plug (10) in (7), we get

$\mathcal{L}_{n}(x)=\frac{1}{2n+1}\left(-\mathcal{L}_{n-1}(x)+\mathcal{L}_{n+1}(x)\right)^{\prime}.$
(11)

Then for $r=1$, $\mathcal{L}_{n}(x)$ has the following respresentation.

$\mathcal{L}_{n}(x)=\left(-\frac{\mathcal{L}_{n-1}(x)}{(2n+1)}+\frac{\mathcal{L}_{n+1}(x)}{(2n+1)}\right)^{(1)}=\left(-\frac{\mathcal{L}_{n-1}(x)}{\lambda^{1}_{n,1}}+\frac{\mathcal{L}_{n+1}(x)}{\lambda^{2}_{n,1}}\right)^{(1)},$
(12)

where $\lambda^{1}_{n,1}=\lambda^{2}_{n,1}.$ By using (11) in two numerators
of (12), we get the following representation for $\mathcal{L}_{n}(x)$

$\displaystyle\mathcal{L}_{n}(x)$
$\displaystyle=\left(\frac{\mathcal{L}_{n-2}(x)}{\lambda^{1}_{n,1}(2n-1)}-\frac{\mathcal{L}_{n}(x)}{\lambda^{1}_{n,1}(2n-1)}-\frac{\mathcal{L}_{n}(x)}{\lambda^{2}_{n,1}(2n+3)}+\frac{\mathcal{L}_{n+2}(x)}{\lambda^{2}_{n,1}(2n+3)}\right)^{(2)}$
$\displaystyle=\left(\frac{\mathcal{L}_{n-2}(x)}{\lambda^{1}_{n,2}}-\frac{\mathcal{L}_{n}(x)}{\lambda^{2}_{n,2}}-\frac{\mathcal{L}_{n}(x)}{\lambda^{3}_{n,2}}+\frac{\mathcal{L}_{n+2}(x)}{\lambda^{4}_{n,2}}\right)^{(2)},$
(13)

where
$\lambda^{1}_{n,2}=\lambda^{2}_{n,2}\leq\lambda^{3}_{n,2}=\lambda^{4}_{n,2}.$
By induction, we assume that (1) holds for $1\leq r\leq n-2$. We will show
that (1) holds for $r+1$. Using (11) in all numerators of (1), we get the
following representation for $\mathcal{L}_{n}(x)$ with $2^{r+1}$ terms as
follows:

$\displaystyle\mathcal{L}_{n}(x)$
$\displaystyle=\left(\frac{(-1)^{r+1}\mathcal{L}_{n-r-1}(x)}{\lambda^{1}_{n,r}(2n-2r+1)}+\frac{(-1)^{r}\mathcal{L}_{n-r+1}(x)}{\lambda^{1}_{n,r}(2n-2r+1)}+\cdots\right.$
$\displaystyle\left.-\frac{\mathcal{L}_{n+r-1}(x)}{\lambda^{2^{r}}_{n,r}(2n+2r+1)}+\frac{\mathcal{L}_{n+r+1}(x)}{\lambda^{2^{r}}_{n,r}(2n+2r+1)}\right)^{(r+1)},$

Easy computation shows that

$\lambda^{2j-1}_{n,r+1}=\lambda^{2j}_{n,r+1}=\left\\{\begin{array}[]{lr}\lambda^{1}_{n,r}(2n-2r+1)&j=1\\\
\lambda^{j}_{n,r}(\gamma_{j})&2\leq j\leq 2^{r}-1\\\
\lambda^{2^{r}}_{n,r}(2n+2r+1)&j=2^{r}\end{array}\right.$

where $2n-2r+3\leq\gamma_{j}\leq 2n+2r-1.$ Also, by assumption, we know that
$\lambda^{1}_{n,r}\leq\lambda^{j}_{n,r}\leq\lambda^{2^{r}}_{n,r}.$ Then

$\lambda^{1}_{n,r+1}=\lambda^{2}_{n,r+1}\leq\lambda^{2j-1}_{n,r+1}=\lambda^{2j}_{n,r+1}\leq\lambda^{2^{r+1}-1}_{n,r+1}=\lambda^{2^{r+1}}_{n,r+1},~{}~{}1\leq
j\leq 2^{r}.$

Therefore, the result holds. ∎

######  Lemma 2.

Suppose that $S_{n}^{0}=\frac{2}{2n+1}$ and for $1\leq r\leq n-1$

$S_{n}^{r}=\frac{2}{\lambda^{1}_{n,r}(2n-2r+1)}+\frac{2}{\lambda^{2}_{n,r}(2n-2r+5)}+\cdots+\frac{2}{\lambda^{2^{r}-1}_{n,r}(2n+2r-3)}+\frac{2}{\lambda^{2^{r}}_{n,r}(2n+2r+1)}.$
(14)

Then for $r\geq 1$

$S_{n}^{r}=\frac{1}{2n+1}\left(S_{n-1}^{r-1}+S_{n+1}^{r-1}\right).$ (15)

###### Proof.

To prove relation (15), We can see that

$S_{n}^{1}=\frac{2}{\lambda^{1}_{n,1}(2n-1)}+\frac{2}{\lambda^{2}_{n,1}(2n+3)}=\frac{1}{2n+1}\left(\frac{2}{2n-1}+\frac{2}{2n+3}\right)=\frac{1}{2n+1}\left(S_{n-1}^{0}+S_{n+1}^{0}\right).$
(16)

This complete the proof for $r=1$. Also, for $r=2$ we obtain

$\displaystyle S_{n}^{2}$
$\displaystyle=\frac{2}{\lambda^{1}_{n,2}(2n-3)}+\frac{2}{\lambda^{2}_{n,2}(2n+1)}+\frac{2}{\lambda^{3}_{n,2}(2n+3)}+\frac{2}{\lambda^{4}_{n,2}(2n+5)}$
$\displaystyle=\frac{1}{2n+1}\left(\frac{2}{(2n-1)(2n-3)}+\frac{2}{(2n-1)(2n+1)}+\frac{2}{(2n+3)(2n+1)}+\frac{2}{(2n+3)(2n+5)}\right)$
$\displaystyle=\frac{1}{2n+1}\left(\frac{2}{\lambda^{1}_{n-1,1}(2n-3)}+\frac{2}{\lambda^{2}_{n-1,1}(2n+1)}+\frac{2}{\lambda^{1}_{n+1,1}(2n+1)}+\frac{2}{\lambda^{2}_{n+1,1}(2n+5)}\right)$
$\displaystyle=\frac{1}{2n+1}\left(S_{n-1}^{1}+S_{n+1}^{1}\right).$ (17)

This complete the proof for $r=2$. By induction, suppose (15) holds for
nonnegative integer $r$. We will show that (15) holds for $r+1$. So we have

$\displaystyle\frac{1}{2n+1}\left(S_{n-1}^{r}+S_{n+1}^{r}\right)$
$\displaystyle=\frac{1}{2n+1}\left(\frac{2}{\lambda^{1}_{n-1,r}(2n-2r-1)}+\frac{2}{\lambda^{2}_{n-1,r}(2n-2r+3)}+\cdots+\frac{2}{\lambda^{2^{r}-1}_{n-1,r}(2n+2r-5)}+\frac{2}{\lambda^{2^{r}}_{n-1,r}(2n+2r-1)}\right.$
$\displaystyle\left.+\frac{2}{\lambda^{1}_{n+1,r}(2n-2r+3)}+\frac{2}{\lambda^{2}_{n+1,r}(2n-2r+7)}+\cdots+\frac{2}{\lambda^{2^{r}-1}_{n+1,r}(2n+2r-1)}+\frac{2}{\lambda^{2^{r}}_{n+1,r}(2n+2r+3)}\right).$
(18)

From Lemma 1, for $1\leq j\leq 2^{r-1}$ we have
$\lambda^{2j-1}_{n,r}=\lambda^{2j}_{n,r}$ . Also we know that

$\displaystyle\lambda^{1}_{n,r+1}=\lambda^{2}_{n,r+1}=(2n+1)\lambda^{1}_{n-1,r},\quad\lambda^{2^{r}-1}_{n,r+1}=\lambda^{2^{r}}_{n,r+1}=(2n+1)\lambda^{2^{r}-1}_{n+1,r}.$

So (2) becomes to

$\displaystyle\frac{1}{2n+1}\left(S_{n-1}^{r}+S_{n+1}^{r}\right)$
$\displaystyle=\left(\frac{2}{\lambda^{1}_{n,r+1}(2n-2r-1)}+\frac{2}{\lambda^{2}_{n,r+1}(2n-2r+3)}+\cdots+\frac{2}{\lambda^{2j-1}_{n,r+1}(2n+2r-5)}+\frac{2}{\lambda^{2j}_{n,r+1}(2n+2r-1)}\right.$
$\displaystyle\left.+\frac{2}{\lambda^{2i-1}_{n,r+1}(2n-2r+3)}+\frac{2}{\lambda^{2i}_{n,r+1}(2n-2r+7)}+\cdots+\frac{2}{\lambda^{2^{r}-1}_{n,r+1}(2n+2r-1)}+\frac{1}{\lambda^{2^{r}}_{n,r+1}(2n+2r+3)}\right)$
$\displaystyle=S_{n}^{r+1}.$ (19)

which complete the proof. ∎

######  Lemma 3.

Under the assumptions of Lemma 2, the following relation holds for $r\geq 0$

$S_{n}^{r}=\frac{2^{r+1}}{\prod_{j=1}^{r+1}(2n-2r+4j-3)}$ (20)

###### Proof.

We know that $S_{n}^{0}=\frac{2}{2n+1}$ and

$S_{n}^{1}=\frac{1}{2n+1}\left(S_{n-1}^{0}+S_{n+1}^{0}\right)=\frac{4}{(2n-1)(2n+3)}.$
(21)

By induction, suppose (20) holds integer $r$. We will show that (20) holds for
$r+1$. Applying the relation (15) we obtain

$\displaystyle\frac{1}{2n+1}\left(S_{n-1}^{r}+S_{n+1}^{r}\right)$
$\displaystyle=\frac{1}{2n+1}\left(\frac{2^{r+1}}{(2n-2r-1)\left[\prod_{j=1}^{r}(2n-2r+4j-1)\right]}+\frac{2^{r+1}}{\left[\prod_{j=1}^{r}(2n-2r+4j-1)\right](2n+2r+3)}\right)$
$\displaystyle=\frac{1}{2n+1}\left(\frac{2^{r+1}(2n+2r+3+2n-2r-1)}{(2n-2r-1)\left[\prod_{j=1}^{r}(2n-2r+4j-1)\right](2n+2r+3)}\right)$
$\displaystyle=\frac{2^{r+2}}{\prod_{j=1}^{r+2}\left(2n-2(r+1)+4j-3\right)}=S_{n}^{r+1}.$
(22)

Then the result holds. ∎

Let $n\geq 1$, from [9, eq. 27], we see that

$|L_{n-1}(x)-L_{n+1}(x)|<\frac{2\mathbf{A}(1-x^{2})^{\frac{1}{4}}}{\sqrt{n+\frac{1}{3}}},\quad\mathbf{A}=0.825031.$
(23)

Also, by using (7) and (8) we obtain the following inequality of derivative of
Legendre polynomials for $x\in[-1,1].$

$|\mathcal{L}_{n}(x)|\leq\frac{1}{2n+1}\left(|L_{n-1}(x)-L_{n+1}(x)|\right)$
(24)

Therefore, by using (23) and (24), we get

$|\mathcal{L}_{n}(x)|<\frac{\mathbf{A}(1-x^{2})^{\frac{1}{4}}}{(n+\frac{1}{2})\sqrt{n+\frac{1}{3}}},\quad-1\leq
x\leq 1.$ (25)

In the next theorem, new upper bounds for Legendre coefficients of
differentiable functions are derived.

######  Theorem 1.

Let $f,f^{\prime},\cdots,f^{(r-1)}$ be absolutely continuous and the $r^{th}$
derivative $f^{(r)}$ is of bounded variation on $[-1,1]$. Then the upper bound
for the Legendre coefficients of the function $f$ for $n\geq r$ is as follows:

$\left|a_{n}\right|<\frac{\mathbf{A}(n+\frac{1}{2})U_{r}}{\sqrt{n-r+\frac{1}{3}}\prod_{j=1}^{r+1}(n-r+2j-\frac{3}{2})},$
(26)

where, $U_{r}=\int_{-1}^{1}|f^{(r+1)}(x)|dx$.

###### Proof.

By using (6), the Legendre coefficients for the function $f(x)$ are as follows

$a_{n}=\left(n+\frac{1}{2}\right)\int_{-1}^{1}f(x)L_{n}(x)dx=-(n+\frac{1}{2})\int_{-1}^{1}f(x)\left(\mathcal{L}_{n}(x)\right)^{\prime}dx.$

Using Integration by parts

$a_{n}=-(n+\frac{1}{2})\left[f(x)\mathcal{L}_{n}(x)\right]_{-1}^{1}+(n+\frac{1}{2})\int_{-1}^{1}f^{\prime}(x)\mathcal{L}_{n}(x)dx.$
(27)

Since $\mathcal{L}_{n}(1)=\mathcal{L}_{n}(-1)=0,$ we obtain that the first
part in (27) vanishes and we get

$a_{n}=(n+\frac{1}{2})\int_{-1}^{1}f^{\prime}(x)\mathcal{L}_{n}(x)dx.$ (28)

Using (25) and applying $U_{0}=\int_{-1}^{1}\left|f^{\prime}(x)\right|dx$, we
get

$|a_{n}|\leq\frac{\mathbf{A}U_{0}}{\sqrt{n+\frac{1}{3}}}.$ (29)

which complete the proof for $r=0$.

For $r\geq 1$, applying (1) in (28) we have

$\displaystyle a_{n}=(n+\frac{1}{2})\int_{-1}^{1}f^{\prime}(x)$
$\displaystyle\left(\frac{(-1)^{r}\mathcal{L}_{n-r}(x)}{\lambda^{1}_{n,r}}+\frac{(-1)^{r-1}\mathcal{L}_{n-r+2}(x)}{\lambda^{2}_{n,r}}+\cdots\right.$
$\displaystyle\left.-\frac{\mathcal{L}_{n+r-2}(x)}{\lambda^{2^{r-1}}_{n,r}}+\frac{\mathcal{L}_{n+r}(x)}{\lambda^{2^{r}}_{n,r}}\right)^{(r)}dx.$
(30)

Since $\mathcal{L}_{n+j}(1)=\mathcal{L}_{n+j}(-1)=0,-r\leq j\leq r,$ with $r$
integrations by parts and vanishing the first term at each steps, we get

$\displaystyle a_{n}=(n+\frac{1}{2})\int_{-1}^{1}f^{(r+1)}(x)$
$\displaystyle\left(\frac{(-1)^{r}\mathcal{L}_{n-r}(x)}{\lambda^{1}_{n,r}}+\frac{(-1)^{r-1}\mathcal{L}_{n-r+2}(x)}{\lambda^{2}_{n,r}}+\cdots\right.$
$\displaystyle\left.-\frac{\mathcal{L}_{n+r-2}(x)}{\lambda^{2^{r-1}}_{n,r}}+\frac{\mathcal{L}_{n+r}(x)}{\lambda^{2^{r}}_{n,r}}\right)dx.$
(31)

Using (25) and applying the given condition
$U_{r}=\int_{-1}^{1}|f^{(r+1)}(x)|dx$, we obtain the following relation with
$2^{r}$ terms.

$\displaystyle|a_{n}|$
$\displaystyle<(n+\frac{1}{2})\left(\frac{2\mathbf{A}}{\lambda^{1}_{n,r}(2n-2r+1)\sqrt{n-r+\frac{1}{3}}}+\frac{2\mathbf{A}}{\lambda^{2}_{n,r}(2n-2r+5)\sqrt{n-r+\frac{7}{3}}}+\cdots\right.$
$\displaystyle\left.+\frac{2\mathbf{A}}{\lambda^{2^{r}-1}_{n,r}(2n+2r-3)\sqrt{n+r-\frac{5}{3}}}+\frac{2\mathbf{A}}{\lambda^{2^{r}}_{n,r}(2n+2r+1)\sqrt{n+r+\frac{1}{3}}}\right)U_{r}$
$\displaystyle\leq\frac{\mathbf{A}(n+\frac{1}{2})U_{r}}{\sqrt{n-r+\frac{1}{3}}}\left(\frac{2}{\lambda^{1}_{n,r}(2n-2r+1)}+\cdots+\frac{2}{\lambda^{2^{r}}_{n,r}(2n+2r+1)}\right)$
$\displaystyle=\frac{\mathbf{A}(n+\frac{1}{2})U_{r}S_{n}^{r}}{\sqrt{n-r+\frac{1}{3}}}$

By Lemma 3, we obtain the following upper bound

$\displaystyle|a_{n}|$
$\displaystyle<\frac{2^{r+1}\mathbf{A}(n+\frac{1}{2})U_{r}}{\prod_{j=1}^{r+1}(2n-2r+4j-3)\sqrt{n-r+\frac{1}{3}}}$
$\displaystyle=\frac{\mathbf{A}(n+\frac{1}{2})U_{r}}{\prod_{j=1}^{r+1}(n-r+2j-\frac{3}{2})\sqrt{n-r+\frac{1}{3}}}.$

This complete the proof. ∎

In the next theorem, new approximation errors using the partial sums of the
Legendre polynomials is provided.

######  Theorem 2.

Let $f,f^{\prime},\cdots,f^{(r-1)}$ be absolutely continuous and the $r^{th}$
derivative $f^{(r)}$ is of bounded variation on $[-1,1]$. Then for $r\geq 1$
and $N\geq r+1$

$\|f(x)-f_{N}(x)\|_{\infty}\leq\left\\{\begin{array}[]{lr}\frac{2\mathbf{A}U_{1}}{\sqrt{N-\frac{2}{3}}}&r=1\\\
\frac{\mathbf{A}(N+\frac{3}{2})U_{r}}{\prod_{j=1}^{r}(N-r+2j-\frac{3}{2})(r-\frac{1}{2})\sqrt{N-r+\frac{1}{3}}}&r~{}odd\\\
\frac{\mathbf{A}(N+\frac{1}{2})U_{r}}{\prod_{j=1}^{r}(N-r+2j-\frac{3}{2})(r-\frac{1}{2})\sqrt{N-r+\frac{1}{3}}}&r~{}even.\end{array}\right.$
(32)

###### Proof.

Applying the inequality $|L_{n}(x)|\leq 1$ for all $x\in\left[-1,1\right]$, we
have

$\|f(x)-f_{N}(x)\|_{\infty}=|\sum_{n=N+1}^{\infty}a_{n}L_{n}(x)|\leq\sum_{n=N+1}^{\infty}|a_{n}||L_{n}(x)|\leq\sum_{n=N+1}^{\infty}|a_{n}|.$
(33)

By using (26) for $r=1$, we obtain

$\displaystyle\|f(x)-f_{N}(x)\|_{\infty}\leq\sum_{n=N+1}^{\infty}\frac{\mathbf{A}(n+\frac{1}{2})U_{1}}{(n-\frac{1}{2})(n+\frac{3}{2})\sqrt{n-\frac{2}{3}}}$
$\displaystyle\leq\sum_{n=N+1}^{\infty}\frac{\mathbf{A}U_{1}}{(n-\frac{1}{2})\sqrt{n-\frac{2}{3}}}\leq\sum_{n=N+1}^{\infty}\frac{\mathbf{A}U_{1}}{(n-\frac{1}{2})^{\frac{3}{2}}\sqrt{1-\frac{1}{6(n-\frac{1}{2})}}}$
$\displaystyle\leq\frac{\mathbf{A}U_{1}}{\sqrt{1-\frac{1}{6(N-\frac{1}{2})}}}\sum_{n=N+1}^{\infty}\frac{1}{(n-\frac{1}{2})^{\frac{3}{2}}}\leq\frac{\mathbf{A}U_{1}}{\sqrt{1-\frac{1}{6(N-\frac{1}{2})}}}\int_{n=N}^{\infty}\frac{1}{(x-\frac{1}{2})^{\frac{3}{2}}}dx$
$\displaystyle=\frac{2\mathbf{A}U_{1}}{\sqrt{N-\frac{2}{3}}}$ (34)

By using (26) for odd $r\geq 3$, we get

$\displaystyle\|f(x)-f_{N}(x)\|_{\infty}\leq\sum_{n=N+1}^{\infty}\frac{\mathbf{A}(n+\frac{1}{2})U_{r}}{(n-r+\frac{1}{2})\cdots(n-\frac{1}{2})(n+\frac{3}{2})\cdots(n+r+\frac{1}{2})\sqrt{n-r+\frac{1}{3}}}$
$\displaystyle\leq\sum_{n=N+1}^{\infty}\frac{\mathbf{A}U_{r}}{(n-r+\frac{1}{2})\cdots(n-\frac{1}{2})(n+\frac{7}{2})\cdots(n+r+\frac{1}{2})\sqrt{n-r+\frac{1}{3}}}$
$\displaystyle=\mathbf{A}U_{r}\sum_{n=N+1}^{\infty}\frac{1}{(n+r+\frac{1}{2})^{r+\frac{1}{2}}(1-\frac{2r}{(n+r+\frac{1}{2})})\cdots(1-\frac{r+1}{(n+r+\frac{1}{2})})(1-\frac{r-3}{(n+r+\frac{1}{2})})\cdots(1-\frac{2}{(n+r+\frac{1}{2})})\sqrt{1-\frac{2r+\frac{1}{6}}{(n+r+\frac{1}{2})}}}$
$\displaystyle\leq\frac{\mathbf{A}U_{r}}{(1-\frac{2r}{(N+r+\frac{1}{2})})\cdots(1-\frac{r+1}{(N+r+\frac{1}{2})})(1-\frac{r-3}{(N+r+\frac{1}{2})})\cdots(1-\frac{2}{(N+r+\frac{1}{2})})\sqrt{1-\frac{2r+\frac{1}{6}}{(N+r+\frac{1}{2})}}}\sum_{n=N+1}^{\infty}\frac{1}{(n+r+\frac{1}{2})^{r+\frac{1}{2}}}$
$\displaystyle\leq\frac{\mathbf{A}U_{r}}{(1-\frac{2r}{(N+r+\frac{1}{2})})\cdots(1-\frac{r+1}{(N+r+\frac{1}{2})})(1-\frac{r-3}{(N+r+\frac{1}{2})})\cdots(1-\frac{2}{(N+r+\frac{1}{2})})\sqrt{1-\frac{2r+\frac{1}{6}}{(N+r+\frac{1}{2})}}}\int_{N}^{\infty}\frac{1}{(x+r+\frac{1}{2})^{r+\frac{1}{2}}}dx$
$\displaystyle=\frac{\mathbf{A}U_{r}}{(N-r+\frac{1}{2})\cdots(N-\frac{1}{2})(N+\frac{7}{2})\cdots(N+r-\frac{3}{2})(r-\frac{1}{2})\sqrt{N-r+\frac{1}{3}}}$
$\displaystyle=\frac{\mathbf{A}(N+\frac{3}{2})U_{r}}{\prod_{j=1}^{r}(N-r+2j-\frac{3}{2})(r-\frac{1}{2})\sqrt{N-r+\frac{1}{3}}}.$

By a similar proof for even $r$, we obtain

$\displaystyle\|f(x)-f_{N}(x)\|_{\infty}\leq\frac{\mathbf{A}(N+\frac{1}{2})U_{r}}{\prod_{j=1}^{r}(N-r+2j-\frac{3}{2})(r-\frac{1}{2})\sqrt{N-r+\frac{1}{3}}}.$

This complete the proof. ∎

##  3 New results on the Legendre coefficients

In [4] the authors obtained an optimal upper bound for the Legendre
coefficients without the assumptions in (2) and (4) as follow

$|a_{n}|\leq\frac{U_{r}(n+\frac{1}{2})\Gamma(\frac{n-r}{2})}{2^{r}\sqrt{\pi}(n+r+1)\Gamma(\frac{n+r+1}{2})},$
(35)

where, $U_{r}=\int_{-1}^{1}|f^{(r+1)}(x)|dx$. This upper bound is really sharp
and the authors provided this bound using the Rodrigues’ formula.

In Lemma (1), using the Sturm-Liouville problem, we presented a new relation
between derivatives of Legendre polynomials. Now, using this lemma, we want to
provide a new upper bound for the Legendre coefficients and we will show that
this new upper bound match the ones obtained in [4]. First, we state a lemma
which provides an upper bound for Gegenbauer polynomials.

######  Lemma 4.

[10] Let $\lambda\geq 1.$ Then

$(1-x^{2})^{\lambda-\frac{1}{2}}\left|C_{n}^{(\lambda)}(x)\right|\leq\frac{\Gamma(\frac{n}{2}+\lambda)}{\Gamma(\lambda)\Gamma(\frac{n}{2}+1)},\quad-1\leq
x\leq 1,$ (36)

where $C_{n}^{(\lambda)}(x)$ is the Gegenbauer polynomial of degree $n$.

It is well-known that $L_{n}(x)=C_{n}^{(\frac{1}{2})}(x)$. From [11, p. 992],
we know that

$\frac{d}{dx}C_{n}^{(\lambda)}(x)=2\lambda C_{n-1}^{(\lambda+1)}(x).$ (37)

So, we can see that

$\frac{d}{dx}L_{n}(x)=\frac{d}{dx}C_{n}^{(\frac{1}{2})}(x)=C_{n-1}^{(\frac{3}{2})}(x).$
(38)

From (8), Lemma (4) and (38) we obtain

$|\mathcal{L}_{n}(x)|\leq\frac{\Gamma(\frac{n+2}{2})}{n(n+1)\Gamma(\frac{3}{2})\Gamma(\frac{n+1}{2})}=\frac{\Gamma(\frac{n}{2})}{\sqrt{\pi}(n+1)\Gamma(\frac{n+1}{2})},\quad-1\leq
x\leq 1.$ (39)

This upper bound is sharp and interesting, because by using (39), we obtain
the same upper bound for Legendre coefficients as in [4]. Now, we present the
following lemma.

######  Lemma 5.

Under the assumptions of Lemma (1), and using the inequlity (39), the
following relation holds:

$\left(\frac{\Gamma(\frac{n-r}{2})}{\sqrt{\pi}\lambda^{1}_{n,r}(n-r+1)\Gamma(\frac{n-r+1}{2})}+\cdots+\frac{\Gamma(\frac{n+r}{2})}{\sqrt{\pi}\lambda^{2^{r}}_{n,r}(n+r+1)\Gamma(\frac{n+r+1}{2})}\right)=\frac{\Gamma(\frac{n-r}{2})}{2^{r}\sqrt{\pi}(n+r+1)\Gamma(\frac{n+r+1}{2})},$

where, $\lambda^{1}_{n,0}=1$ and for $r\geq 1$,
$\lambda^{j}_{n,r},j=1,\ldots,2^{r}$ are defined in Lemma (1).

###### Proof.

The equality is evident for $r=0$. In the case that $r=1$, we get

$\displaystyle\frac{\Gamma(\frac{n-1}{2})}{\sqrt{\pi}\lambda^{1}_{n,1}n\Gamma(\frac{n}{2})}+\frac{\Gamma(\frac{n+1}{2})}{\sqrt{\pi}\lambda^{2}_{n,1}(n+2)\Gamma(\frac{n+2}{2})}=\frac{\Gamma(\frac{n-1}{2})}{2\sqrt{\pi}\lambda^{1}_{n,1}\Gamma(\frac{n+2}{2})}+\frac{(n-1)\Gamma(\frac{n-1}{2})}{2\sqrt{\pi}\lambda^{2}_{n,1}(n+2)\Gamma(\frac{n+2}{2})}$
$\displaystyle=\frac{\Gamma(\frac{n-1}{2})}{2\sqrt{\pi}\Gamma(\frac{n+2}{2})}\left(\frac{1}{\lambda^{1}_{n,1}}+\frac{(n-1)}{\lambda^{2}_{n,1}(n+2)}\right)=\frac{\Gamma(\frac{n-1}{2})}{2\sqrt{\pi}(n+2)\Gamma(\frac{n+2}{2})},$

where we used $\Gamma(t+1)=t\Gamma(t)$. Also, in the case that $r=2$, the
following equality is obtained.

$\displaystyle\frac{\Gamma(\frac{n}{2}-1)}{\sqrt{\pi}\lambda^{1}_{n,2}(n-1)\Gamma(\frac{n}{2}-\frac{1}{2})}+\frac{\Gamma(\frac{n}{2})}{\sqrt{\pi}\lambda^{2}_{n,2}(n+1)\Gamma(\frac{n}{2}+\frac{1}{2})}+\frac{\Gamma(\frac{n}{2})}{\sqrt{\pi}\lambda^{3}_{n,2}(n+1)\Gamma(\frac{n}{2}+\frac{1}{2})}+\frac{\Gamma(\frac{n}{2}+\frac{1}{2})}{\sqrt{\pi}\lambda^{4}_{n,2}(n+3)\Gamma(\frac{n}{2}+\frac{3}{2})}$
$\displaystyle=\underbrace{\frac{\Gamma(\frac{n}{2}-1)}{2\sqrt{\pi}\lambda^{1}_{n,2}\Gamma(\frac{n}{2}+\frac{1}{2})}+\frac{(n-2)\Gamma(\frac{n}{2}-1)}{2\sqrt{\pi}\lambda^{2}_{n,2}(n+1)\Gamma(\frac{n}{2}+\frac{1}{2})}}+\underbrace{\frac{\Gamma(\frac{n}{2})}{2\sqrt{\pi}\lambda^{3}_{n,2}\Gamma(\frac{n}{2}+\frac{3}{2})}+\frac{(n-1)\Gamma(\frac{n}{2}-\frac{1}{2})}{2\sqrt{\pi}\lambda^{4}_{n,2}(n+3)\Gamma(\frac{n}{2}+\frac{3}{2})}}$
$\displaystyle=\frac{\Gamma(\frac{n}{2}-1)(2n-1)}{2\sqrt{\pi}\lambda^{1}_{n,2}(n+1)\Gamma(\frac{n}{2}+\frac{1}{2})}+\frac{\Gamma(\frac{n}{2})(2n+3)}{2\sqrt{\pi}\lambda^{3}_{n,2}(n+3)\Gamma(\frac{n}{2}+\frac{3}{2})}$
$\displaystyle=\frac{\Gamma(\frac{n}{2}-1)}{4\sqrt{\pi}(2n+1)\Gamma(\frac{n}{2}+\frac{3}{2})}+\frac{(n-2)\Gamma(\frac{n}{2}-1)}{4\sqrt{\pi}(2n+1)(n+3)\Gamma(\frac{n}{2}+\frac{3}{2})}$
$\displaystyle=\frac{\Gamma(\frac{n-2}{2})}{4\sqrt{\pi}(n+3)\Gamma(\frac{n+3}{2})}.$

Continuing the process for $r\geq 3$ show that the result holds. ∎

Applying (39) in the Lemma (1) and by similar proof of Theorem (1), we obtain
the following theorem.

######  Theorem 3.

Under the assumptions of Theorem (1), the upper bound for the Legendre
coefficients of the function $f$ for $n\geq r$ is as follows:

$\left|a_{n}\right|\leq\frac{U_{r}(n+\frac{1}{2})\Gamma(\frac{n-r}{2})}{2^{r}\sqrt{\pi}(n+r+1)\Gamma(\frac{n+r+1}{2})}.$

The upper bound obtained in Theorem (3) match the ones obtained in [4]. Note
that a new upper bounds for $\mathcal{L}_{n}(x)$ help us to obtain a new upper
bound for Legendre coefficients and improve the result in Theorem (3).

##  4 Comparison Results

This section aims to draw a comparison between our proposed upper bounds and
the upper bounds provided so far. In [7], the author presented the following
upper bound for the Legendre coefficients without the assumptions in (2) and
(4)

$|a_{n}|\leq\frac{2U_{r}\prod_{j=1}^{r}h_{n-j}}{\sqrt{2\pi(n-r)}},$ (40)

where, $h_{n-j}=\frac{1}{n-j+\frac{1}{2}}$. The author showed that the
approximation errors obtained in [7] is sharper than ones presented in [2, 5].

######  Remark 1.

Now, we compare our proposed upper bound (26) with the upper bound (40).

Assume that

$\gamma_{n,r}=\frac{\mathbf{A}(n+\frac{1}{2})}{\Pi_{j=1}^{r+1}(n-r+2j-\frac{3}{2})\sqrt{n-r+\frac{1}{3}}},\quad\quad\theta_{n,r}=\frac{2\prod_{j=1}^{r}h_{n-j}}{\sqrt{2\pi(n-r)}}.$

The numerical results for the values of $\gamma_{n,r}$ and $\theta_{n,r}$ are
listed in Table 1. The numerical results obtained from this table indicate
that in all cases, the value of $\gamma_{n,r}$ is smaller than thats of
$\theta_{n,r}$. In particular, for the case that $r$ is close to $n$, the
value of $\gamma_{n,r}$ is much smaller than $\theta_{n,r}$. Therefore, it
reveals from this remark that the proposed upper bound (26) is sharper than
the upper bound (40).

Table 1: Comparison between $\gamma_{n,r}$ and $\theta_{n,r}$.

n | r | $\gamma_{n,r}$ | $\theta_{n,r}$
---|---|---|---
5 | 4 | 0.0023 | 0.0135
10 | 2 | 0.0033 | 0.0035
10 | 7 | $9.96\times 10^{-8}$ | $1.35\times 10^{-6}$
20 | 5 | $7.68\times 10^{-8}$ | $1.27\times 10^{-7}$
20 | 14 | $6.81\times 10^{-19}$ | $1.73\times 10^{-16}$
30 | 15 | $3.09\times 10^{-23}$ | $1.43\times 10^{-21}$
30 | 24 | $3.69\times 10^{-35}$ | $1.94\times 10^{-30}$

In the following, we compare the approximation error of the partial sums of
Legendre polynomials evaluated by Theorem 2 and the bound presented in [7].
The author proved that

$\|f(x)-f_{N}(x)\|_{\infty}\leq\left\\{\begin{array}[]{lr}\frac{4U_{1}}{\sqrt{2\pi(N-1)}}&r=1,\\\
\frac{2U_{r}}{\prod_{j=2}^{r}(N-j+\frac{3}{2})(r-1)\sqrt{2\pi(N-r+1)}}&r\geq
2,\end{array}\right.$ (41)

which is sharper than ones presented in [2, 5].

######  Example 1.

Let $j\geq 2$ and $-1<t<1.$ Consider the function
$f_{j}(x)=\frac{1}{j!}(x-t)^{j-1}|x-t|$. This function and its derivatives are
absolutely continuous on $[-1,1]$ and $f^{(j-1)}_{j}(x)=|x-t|$. Also,
$f_{j}^{(j)}(x)=2H(x-t)-1,$ where $H(x)$ is the Heaviside step function which
is of bounded variation and $f_{j}^{(j+1)}(x)=2\delta(x-t),$ where
$\delta(x-t)$ is the Dirac delta function. Then

$U_{j}=\int_{-1}^{1}\left|f_{j}^{(j+1)}(x)\right|dx=\int_{-1}^{1}2\delta(x-t)dx=2,\\\
$

In Table 2 the comparison results between Theorem 2 and (41) are listed for
some values of $N,j$.

Table 2: Comparison results between two approximation errors for the function
$f_{j}(x)$.

$N$ | $j$ | Inequality (41) | Inequality (32) | $N$ | $j$ | Inequality (41) | Inequality (32)
---|---|---|---|---|---|---|---
15 | 3 | $1.394\times 10^{-3}$ | $9.868\times 10^{-4}$ | 20 | 3 | $6.069\times 10^{-4}$ | $4.421\times 10^{-4}$
15 | 10 | $1.895\times 10^{-10}$ | $6.824\times 10^{-12}$ | 20 | 10 | $2.352\times 10^{-12}$ | $2.126\times 10^{-13}$
15 | 12 | $1.321\times 10^{-11}$ | $7.989\times 10^{-14}$ | 20 | 12 | $2.682\times 10^{-14}$ | $7.544\times 10^{-16}$

##  5 Conclusion

In this paper a new relation between derivatives of Legendre polynomials is
presented in Lemma 1 and by using this key lemma and Lemmas 2, 3, new upper
bounds for the Legendre coefficients of differentiable functions are obtained
(see Theorem 1). We compare these upper bounds by previous upper bounds which
are presented in [7]. Moreover, we provide a new upper bound error on the
approximation of a function $f(x)$ by truncated Legendre polynomial series.

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