#  $k-$smoothness on polyhedral Banach spaces

Subhrajit Dey, Arpita Mal and Kallol Paul  Department of Mathematics
Muralidhar Girls’ College
Kolkata 700029
West Bengal
INDIA<EMAIL_ADDRESS>Department of Mathematics
Jadavpur University
Kolkata 700032
West Bengal
INDIA<EMAIL_ADDRESS>Department of Mathematics
Jadavpur University
Kolkata 700032
West Bengal
INDIA<EMAIL_ADDRESS>
###### Abstract.

We characterize $k-$smoothness of an element on the unit sphere of a finite-
dimensional polyhedral Banach space. Then we study $k-$smoothness of an
operator $T\in\mathbb{L}(\ell_{\infty}^{n},\mathbb{Y}),$ where $\mathbb{Y}$ is
a two-dimensional Banach space with the additional condition that $T$ attains
norm at each extreme point of $B_{\ell_{\infty}^{n}}.$ We also characterize
$k-$smoothness of an operator defined between $\ell_{\infty}^{3}$ and
$\ell_{1}^{3}.$

###### Key words and phrases:

$k-$smoothness; linear operator; Banach space; polyhedral Banach space

###### 2010 Mathematics Subject Classification:

Primary 46B20, Secondary 47L05

The research of Arpita Mal is supported by UGC, Govt. of India. The research
of Prof. Kallol Paul is supported by project MATRICS (MTR/2017/000059) of DST,
Govt. of India.

##  1\. Introduction

The study of $k-$smoothness plays an important role to identify the structure
of the unit ball of a Banach space. The papers [1, 2, 3, 4] contain the study
of $k-$smooth points of many of the Banach spaces. There are several papers
including [1, 3, 4, 5, 6, 7, 9] that contain the study of $k-$smoothness of
operators on different spaces. In [7], authors have obtained a relation
between $k-$smoothness and extreme points of the unit ball of a polyhedral
Banach space. The purpose of this paper is to characterize the order of
smoothness of an element on the unit sphere of a finite-dimensional polyhedral
Banach space, we also study $k$-smoothness of an operator defined between
polyhedral Banach spaces. Let us first fix the notation and terminology.

Letters $\mathbb{X},$ $\mathbb{Y}$ denote Banach spaces. Throughout the paper
we assume the Banach spaces to be real. We denote the unit ball and the unit
sphere of $\mathbb{X}$ respectively by $B_{\mathbb{X}}$ and $S_{\mathbb{X}}$,
i.e., $B_{\mathbb{X}}=\\{x\in\mathbb{X}:\|x\|\leq
1\\},S_{\mathbb{X}}=\\{x\in\mathbb{X}:\|x\|=1\\}.$ Let
$\mathbb{L}(\mathbb{X},\mathbb{Y})$ denote the space of all bounded linear
operators between $\mathbb{X}$ and $\mathbb{Y}.$ For
$T\in\mathbb{L}(\mathbb{X},\mathbb{Y}),~{}M_{T}$ denotes the collection of all
unit vectors of $\mathbb{X}$ at which $T$ attains its norm, i.e.,
$M_{T}=\\{x\in S_{\mathbb{X}}:\|Tx\|=\|T\|\\}$. For a set $A,$ the cardinality
of $A$ is denoted by $|A|.$ The dual space of $\mathbb{X}$ is denoted by
$\mathbb{X}^{*}.$ An element $x\in S_{\mathbb{X}}$ is said to be an extreme
point of the convex set $B_{\mathbb{X}}$ if and only if $x=(1-t)y+tz$ for some
$y,z\in B_{\mathbb{X}}$ and $t\in(0,1)$ implies that $y=z=x.$ For
$x,y\in\mathbb{X},$ let $L[x,y]=\\{tx+(1-t)y:0\leq t\leq 1\\}$ and
$L(x,y)=\\{tx+(1-t)y:0<t<1\\}.$ The set of all extreme points of
$B_{\mathbb{X}}$ is denoted by $Ext(B_{\mathbb{X}}).$ An element $x^{*}\in
S_{\mathbb{X}^{*}}$ is said to be a supporting linear functional of $x\in
S_{\mathbb{X}},$ if $x^{*}(x)=1.$ For a unit vector $x,$ let $J(x)$ denote the
set of all supporting linear functionals of $x,$ i.e., $J(x)=\\{x^{*}\in
S_{\mathbb{X}^{*}}:x^{*}(x)=1\\}.$ The set $J(x)$ for $x\in S_{\mathbb{X}}$
plays a significant role to study the $k-$smoothness. By the Hahn-Banach
Theorem, it is easy to verify that $J(x)\neq\emptyset,$ for all $x\in
S_{\mathbb{X}}.$ We would like to mention that $J(x)$ is a weak*-compact
convex subset of $S_{\mathbb{X}^{*}}.$ A unit vector $x$ is said to be a
smooth point if $J(x)$ is singleton. $\mathbb{X}$ is said to be a smooth
Banach space if every unit vector of $\mathbb{X}$ is smooth. The set of all
extreme points of $J(x)$ is denoted by $Ext~{}J(x),$ where $x\in
S_{\mathbb{X}}.$ In 2005, Khalil and Saleh [3] defined $k-$smooth points as
follows: An element $x\in S_{\mathbb{X}}$ is said to be $k-$smooth or the
order of smoothness of $x$ is $k,$ if $J(x)$ contains exactly $k$ linearly
independent supporting linear functionals of $x$. In other words, $x$ is
$k-$smooth, if $\dim~{}span~{}J(x)=k.$ Moreover, from [4, Prop. 2.1], we get
that $x$ is $k-$smooth, if $k=\dim~{}span~{}Ext~{}J(x).$ Similarly,
$T\in\mathbb{L}(\mathbb{X},\mathbb{Y})$ is said to be $k-$smooth operator, if
$k=\dim~{}span~{}J(T)=\dim~{}span~{}Ext~{}J(T).$ Observe that, $1-$smooth
points of $S_{\mathbb{X}}$ are the smooth points of $S_{\mathbb{X}}.$ The
spaces that we are dealing with in this paper are mostly finite-dimensional
polyhedral Banach spaces, so it is appropriate to state the following
definitions.

######  Definition 1.1.

A polyhedron $P$ is a non-empty compact subset of $\mathbb{X}$ which is the
intersection of finitely many closed half-spaces of $\mathbb{X},$ i.e.,
$P=\cap_{i=1}^{r}M_{i},$ where $M_{i}$ are closed half-spaces in $\mathbb{X}$
and $r\in\mathbb{N}.$ The dimension $\dim(P)$ of the polyhedron $P$ is defined
as the dimension of the subspace generated by the differences $v-w$ of vectors
$v,w\in P.$

######  Definition 1.2.

A polyhedron $Q$ is said to be a face of the polyhedron $P$ if either $Q=P$ or
if we can write $Q=P\cap\delta M,$ where $M$ is a closed half-space in
$\mathbb{X}$ containing $P$ and $\delta M$ denotes the boundary of $M.$ If
$\dim(Q)=i,$ then $Q$ is called an $i-$face of $P$. If $\dim(P)=n,$ then
$(n-1)-$faces of $P$ are called facets of $P$ and $1-$faces of $P$ are called
edges of $P.$

######  Definition 1.3.

A finite-dimensional Banach space $\mathbb{X}$ is said to be polyhedral if the
unit ball $B_{\mathbb{X}}$ of $\mathbb{X}$ contains only finitely many extreme
points. Equivalently, a finite-dimensional Banach space $\mathbb{X}$ is a
polyhedral Banach space, if $B_{\mathbb{X}}$ is a polyhedron. In particular, a
two-dimensional polyhedral Banach space is said to be a polygonal Banach
space.

For a convex set $C,~{}\text{int}_{r}(C)$ denotes the relative interior of the
set $C,$ i.e., $x\in\text{int}_{r}(C)$ if there exists $\epsilon>0$ such that
$B(x,\epsilon)\cap\text{affine}(C)\subseteq C,$ where $\text{affine}(C)$ is
the intersection of all affine sets containing $C$ and an affine set is
defined as the translation of a vector subspace. A non-empty convex subset $F$
of $C$ is said to be a face of $C,$ if for $x,y\in C$ and $t\in(0,1),$
$(1-t)x+ty\in F\Rightarrow x,y\in F.$

In this paper, we first prove that a point on the relative interior of an
$i$-face of the unit ball of an $n-$dimensional polyhedral Banach space is
$(n-i)-$smooth. In [6], the authors completely characterized the
$k-$smoothness of an operator defined between two Banach spaces $\mathbb{X}$
and $\mathbb{Y},$ where $\dim(\mathbb{X})=\dim(\mathbb{Y})=2$ and in [5], the
authors characterized the $k-$smoothness of a bounded linear operator defined
between $\ell_{\infty}^{3}$ and a two-dimensional Banach space. We continue
our study in this direction and characterize the $k-$smoothness of a bounded
linear operator defined between $\ell_{\infty}^{n}$ and a two-dimensional
Banach space with the assumption that the linear operator attains its norm at
all the extreme points of the unit ball of $\ell_{\infty}^{n}.$ Then we
characterize $k-$smoothness of a bounded linear operator defined between
$\ell_{\infty}^{3}$ and $\ell_{1}^{3}.$

We state the following lemma [10, Lemma 3.1], characterizing $Ext~{}J(T),$
which will be used often. For simplicity we state the lemma for finite-
dimensional Banach spaces.

######  Lemma 1.4.

[10, Lemma 3.1] Suppose that $\mathbb{X},\mathbb{Y}$ are finite-dimensional
Banach spaces. Let $T\in\mathbb{L}(\mathbb{X},\mathbb{Y})$ and $\|T\|=1$ Then

$Ext~{}J(T)=\\{y^{*}\otimes x\in\mathbb{L}(\mathbb{X},\mathbb{Y})^{*}:x\in
M_{T}\cap Ext(B_{\mathbb{X}}),y^{*}\in Ext~{}J(Tx)\\},$

where $y^{*}\otimes x:\mathbb{L}(\mathbb{X},\mathbb{Y})\to\mathbb{R}$ is
defined by $y^{*}\otimes x(S)=y^{*}(Sx)$ for every
$S\in\mathbb{L}(\mathbb{X},\mathbb{Y}).$

##  2\. Main results

We begin this section with a relation between the order of smoothness of a
unit vector $x$ in a polyhedral Banach space and the dimension of the face $F$
such that $x$ is in the relative interior of $F.$

######  Theorem 2.1.

Let $\mathbb{X}$ be an $n$-dimensional polyhedral Banach space. Let $F$ be an
$i$-face of $B_{\mathbb{X}}.$ Let $x\in\text{int}_{r}(F).$ Then $x$ is
$(n-i)-$smooth.

###### Proof.

Let $f\in Ext~{}J(x).$ Since $x\in\text{int}_{r}(F),$ we have for all $y\in
F,$ $f(y)=1.$ Therefore,

$F\subseteq\cap_{f\in Ext~{}J(x)}\\{y\in
S_{\mathbb{X}}:f(y)=1\\}=A~{}\text{(say)}.$

Clearly, $A$ is a face of $B_{\mathbb{X}}.$ If possible, suppose that
$F\subsetneqq A.$ Then there exists $z\in A\setminus F.$ Now, $x\in F\subseteq
A$ and $z\in A\Rightarrow tx+(1-t)z\in A$ for all $t\in[0,1],$ since $A$ is a
face. Using convexity argument of norm, it is easy to observe that
$\|x+\lambda(z-x)\|\geq\|x\|$ for all scalars $\lambda.$ Moreover, we have
$\|x+(z-x)\|=\|z\|=1.$ If possible, let $\|x+\lambda_{0}(z-x)\|=1$ for some
$\lambda_{0}<0.$ Then

$x=tz+(1-t)\\{x+\lambda_{0}(z-x)\\},~{}\text{
where~{}}t=\frac{-\lambda_{0}}{1-\lambda_{0}}\in(0,1).$

Since $F$ is a face and $x\in F,$ we get $z\in F,$ a contradiction. Thus
$\|x+\lambda(z-x)\|>1$ for all $\lambda<0.$ Let $Y=span\\{x,z\\}.$ Then
$\dim(Y)=2$ and $Y$ is a polygonal Banach space. Using $\|x+(z-x)\|=1$ and
$\|x+\lambda(z-x)\|>1$ for all $\lambda<0,$ it is easy to observe that $x\in
Ext(B_{Y}).$ Thus, by [6, Th. 3.5], we get $x$ is $2-$smooth in $Y.$ Let $h,g$
be two linearly independent elements of $Ext(B_{Y^{*}})$ such that
$h(x)=g(x)=1.$ If $h(z)=g(z)=1,$ then
$h(x-z)=g(x-z)=0\Rightarrow\ker(h)=\ker(g)\Rightarrow h=\lambda g,$ for some
scalar $\lambda,$ a contradiction. Without loss of generality, suppose that
$g(z)\neq 1.$ By [8, Lemma 1.2, Page 168], there exists $g_{1}\in
Ext(B_{\mathbb{X}^{*}})$ such that $g_{1}|_{Y}=g.$ Now, $g_{1}(x)=1$ and
$g_{1}\in Ext(B_{\mathbb{X}^{*}})\Rightarrow g_{1}\in Ext~{}J(x).$ Thus,
$g_{1}(z)\neq 1$ contradicts that $z\in A.$ Therefore, $A\setminus
F=\emptyset\Rightarrow A=F,$ i.e.,

$F=\cap_{f\in Ext~{}J(x)}\\{y\in S_{\mathbb{X}}:f(y)=1\\}=\cap_{f\in
Ext~{}J(x)}\left(x+\ker(f)\right)\cap S_{\mathbb{X}}.$

This implies that $i=\dim(F)=\dim(\cap_{f\in Ext~{}J(x)}ker(f)).$ Now, let $x$
be $k-$smooth. Let $\\{f_{1},f_{2},\dots,f_{k}\\}$ be the set of all linearly
independent vectors of $Ext~{}J(x).$ Then

$\displaystyle f\in Ext~{}J(x)\Rightarrow
f=\sum_{j=1}^{k}a_{j}f_{j},~{}(a_{j}\in\mathbb{R})$ $\displaystyle\Rightarrow$
$\displaystyle\cap_{j=1}^{k}\ker(f_{j})\subseteq\ker(f)$
$\displaystyle\Rightarrow$
$\displaystyle\cap_{j=1}^{k}\ker(f_{j})\subseteq\cap_{f\in
Ext~{}J(x)}ker(f)\subseteq\cap_{j=1}^{k}\ker(f_{j})$
$\displaystyle\Rightarrow$ $\displaystyle\cap_{f\in
Ext~{}J(x)}\ker(f)=\cap_{j=1}^{k}\ker(f_{j})$ $\displaystyle\Rightarrow$
$\displaystyle i=\dim(\cap_{j=1}^{k}\ker(f_{j}))=n-k$
$\displaystyle\Rightarrow$ $\displaystyle k=n-i.$

This completes the proof of the theorem. ∎

######  Remark 2.2.

Note that, if $\mathbb{X}$ is an $n$-dimensional polyhedral Banach space and
$F$ is a facet of $B_{\mathbb{X}},$ then from Theorem 2.1, we get for each
$x\in int_{r}(F),$ $x$ is smooth. On the other hand, if $F$ is a $0$-face,
i.e., $F=\\{x\\},$ then $x$ is $n-$smooth. In this case, clearly $x$ is an
extreme point of $B_{\mathbb{X}}.$ It is worth mentioning that Theorem 2.1
generalizes [6, Th. 3.5].

Now, we focus on the space of all operators defined between some particular
polyhedral Banach spaces. First we study $k-$smoothness of an operator defined
between $\ell_{\infty}^{n}$ and an arbitrary two-dimensional Banach space. To
do so we need the following two lemmas.

######  Lemma 2.3.

[6, Lemma 2.1] Suppose $\mathbb{X},\mathbb{Y}$ are finite-dimensional Banach
spaces. If $\\{x_{1},x_{2},\ldots,x_{m}\\}$ is a linearly independent subset
of $\mathbb{X}$ and $\\{y_{1}^{*},y_{2}^{*},\ldots,y_{n}^{*}\\}$ is a linearly
independent subset of $\mathbb{Y}^{*}$ then $\\{y_{i}^{*}\otimes x_{j}:1\leq
i\leq n,1\leq j\leq m\\}$ is a linearly independent subset of
$\mathbb{L}(\mathbb{X},\mathbb{Y})^{*}.$

######  Lemma 2.4.

Let $\mathbb{X}=\ell_{\infty}^{n}$ and $\mathbb{Y}$ be a two-dimensional
Banach space. Let $T\in S_{\mathbb{L}(\mathbb{X},\mathbb{Y})}$ be such that
$Rank(T)=2$ and $Ext(B_{\mathbb{X}})\subseteq M_{T}.$ Then the followings
hold:
(i) $T(B_{\mathbb{X}})$ is a convex set with $4$ extreme points.
(ii) If $T(B_{\mathbb{X}})$ is the convex hull of $\\{\pm z_{1},\pm z_{2}\\},$
then either for each $x\in Ext(B_{\mathbb{X}}),$ $Tx\in\pm L[z_{1},z_{2}]$ or
for each $x\in Ext(B_{\mathbb{X}}),$ $Tx\in\pm L[z_{1},-z_{2}].$

###### Proof.

$(i)$ Follows from [7, Remark 2.13].

$(ii)$ Suppose $e_{i}=(0,0,\ldots,1,0,\ldots,0)$ with $1$ at the $i-$th
coordinate and $0$ at the remaining places. Since, Rank$(T)=2,$ $Te_{i}\neq 0$
for some $i\in\\{1,2,\dots,n\\}.$ Without loss of generality, we assume $i=1,$
i.e., $Te_{1}\neq 0.$ It is well-known that $B_{\mathbb{X}}=conv(K\cup-K),$
where $K=\\{(1,u_{2},\dots,u_{n}):|u_{i}|\leq 1,2\leq i\leq n\\}.$ Now, $K$
can be expressed as $K=e_{1}+F,$ where $e_{1}=(1,0,\dots,0)$ and
$F=\\{(0,u_{2},\dots,u_{n}):|u_{i}|\leq 1,2\leq i\leq n\\}.$ Observe that
$x\in Ext(B_{\mathbb{X}})$ if and only if there exists $u\in Ext(F)$ such that
either $x=e_{1}+u$ or $x=-e_{1}+u.$ Clearly,
$T(B_{\mathbb{X}})=conv\Big{(}T(K)\cup T(-K)\Big{)},$ where
$T(K)=T(e_{1}+F)=Te_{1}+T(F)$ and $T(-K)=T(-e_{1}+F)=-Te_{1}+T(F).$ Now,
$T(F)$ must be a symmetric convex set about origin, since $F$ is so. If $T(F)$
has more than four extreme points then proceeding similarly as in [7, Lemma
2.11], it can be shown that there exists $v\in Ext(B_{\mathbb{X}})$ such that
$v\notin M_{T},$ a contradiction. Thus, $T(F)$ has at most four extreme
points.
First suppose $T(F)$ has only two extreme points say $\pm y,$ i.e., $T(F)$ is
the convex hull of $\\{y,-y\\}.$ Clearly, $T(B_{\mathbb{X}})$ is the convex
hull of $\pm Te_{1}\pm y.$ Now, for each $x\in Ext(F),Tx\in L[y,-y].$
Therefore, for each $z\in Ext(B_{\mathbb{X}}),$ $Tz\in\pm
L[Te_{1}+y,Te_{1}-y]$ and hence we are done.
Next, suppose $T(F)$ has four distinct extreme points say $\pm y_{1},\pm
y_{2}.$ We prove the rest of the lemma in the following two steps.

Step 1. We claim that either
$T(B_{\mathbb{X}})=conv\\{\pm(Te_{1}-y_{2}),\pm(Te_{1}+y_{1})\\}$ or
$T(B_{\mathbb{X}})=conv\\{\pm(Te_{1}+y_{2}),\pm(Te_{1}+y_{1})\\}.$

Since $y_{1},y_{2}\in Ext(T(F)),$ there exist $x_{1},x_{2}\in Ext(F)$ such
that $Tx_{1}=y_{1},Tx_{2}=y_{2}.$ Now, $T(F)=conv\\{\pm y_{1},\pm y_{2}\\}$
gives that $T(B_{\mathbb{X}})=conv\\{\pm Te_{1}\pm y_{1},\pm Te_{1}\pm
y_{2}\\}.$ Note that $\pm Te_{1}\pm y_{i}=\pm Te_{1}\pm Tx_{i},$ for $i=1,2$
and $\pm e_{1}\pm x_{i}\in Ext(B_{\mathbb{X}}).$ Therefore, $\|\pm Te_{1}\pm
y_{i}\|=1,$ for $i=1,2.$ Since $\mathbb{Y}$ is two-dimensional and
$\\{y_{1}+y_{2},y_{1}-y_{2}\\}$ is linearly independent,
$Te_{1}=a(y_{1}+y_{2})+b(y_{1}-y_{2}),$ where $a,b\in\mathbb{R}.$ We assert
that either $a=0,b\neq 0$ or $a\neq 0,b=0.$ Clearly, $a$ and $b$ can not be
simultaneously zero as $Te_{1}\neq 0.$ If possible, suppose that $a\neq
0,b\neq 0.$ If $a>0,b>0,$ then

$Te_{1}-y_{1}=\frac{2a}{2a+2b+1}(Te_{1}+y_{2})+\frac{2b}{2a+2b+1}(Te_{1}-y_{2})+\frac{1}{2a+2b+1}(-Te_{1}-y_{1})$

and $\frac{2a}{2a+2b+1},\frac{2b}{2a+2b+1},\frac{1}{2a+2b+1}\in(0,1).$
Moreover, we have, $\|Te_{1}+y_{2}\|=\|Te_{1}-y_{2}\|=\|Te_{1}+y_{1}\|=1.$
Using this, it can be easily observed that
$\|Te_{1}-Tx_{1}\|=\|Te_{1}-y_{1}\|<1,$ which contradicts that $e_{1}-x_{1}\in
Ext(B_{\mathbb{X}})\subseteq M_{T}.$ Similarly, considering the other possible
cases $a<0,b<0$ or $a<0,b>0$ or $a>0,b<0,$ we get a contradiction. This
establishes our assertion. Assume that $a=0,b\neq 0.$
Then we have

(2.1) $\displaystyle
Te_{1}-y_{1}=\frac{2b}{2b+1}(Te_{1}-y_{2})+\frac{1}{2b+1}(-Te_{1}-y_{1})~{}\text{and}$
(2.2) $\displaystyle-
Te_{1}-y_{2}=\frac{1}{2b+1}(Te_{1}-y_{2})+\frac{2b}{2b+1}(-Te_{1}-y_{1}).$

Thus, the only extreme points of $T(B_{\mathbb{X}})$ are $\pm(Te_{1}-y_{2})$
and $\pm(Te_{1}+y_{1}),$ i.e,
$T(B_{\mathbb{X}})=conv\\{\pm(Te_{1}-y_{2}),\pm(Te_{1}+y_{1})\\}.$
Assuming $a\neq 0,b=0$ and proceeding similarly, we can show that
$T(B_{\mathbb{X}})=conv\\{\pm(Te_{1}+y_{2}),\pm(Te_{1}+y_{1})\\}.$

Step 2. Claim that either $Tz\in\pm L[Te_{1}+y_{1},-Te_{1}+y_{2}]$ for each
$z\in Ext(B_{\mathbb{X}})$ or $Tz\in\pm L[Te_{1}+y_{1},-Te_{1}-y_{2}]$ for
each $z\in Ext(B_{\mathbb{X}}).$

Suppose that
$T(B_{\mathbb{X}})=conv\\{\pm(Te_{1}-y_{2}),\pm(Te_{1}+y_{1})\\}.$ Let $z\in
Ext(B_{\mathbb{X}})$ be arbitrary. We show that $Tz\in\pm
L[Te_{1}+y_{1},-Te_{1}+y_{2}].$ Now, there exists $x\in Ext(F)$ such that
either $z=x+e_{1}$ or $z=x-e_{1}.$ First let $z=x+e_{1}.$ Now, if $Tx$ is an
interior point of $T(F),$ then $Tz=Tx+Te_{1}$ will be an interior point of
$T(B_{\mathbb{X}})$ and hence $||Tz||<||T||,$ a contradiction as $z\in
Ext(B_{\mathbb{X}}).$ Thus, $Tx$ is on the boundary of $T(F),$ i.e., $Tx\in\pm
L[y_{1},y_{2}]\cup\pm L[y_{1},-y_{2}].$ If $Tx\in\pm L[y_{1},y_{2}],$ then
clearly $Tz=Tx+Te_{1}\in\pm L[Te_{1}+y_{1},-Te_{1}+y_{2}]$ and we are done.

If possible, let $Tx\in L(y_{1},-y_{2}),$ i.e.,
$Tx=(1-\lambda)y_{1}+\lambda(-y_{2}),0<\lambda<1.$ Then by Equation (2.2), we
have

$\displaystyle-Tx+Te_{1}$ $\displaystyle=$
$\displaystyle(1-\lambda)(Te_{1}-y_{1})+\lambda(Te_{1}+y_{2})$
$\displaystyle=$
$\displaystyle(1-\lambda)(Te_{1}-y_{1})+\lambda\Big{(}\frac{1}{2b+1}(-Te_{1}+y_{2})+\frac{2b}{2b+1}(Te_{1}+y_{1})\Big{)}$
$\displaystyle=$
$\displaystyle(1-\lambda)(Te_{1}-y_{1})+\frac{\lambda}{2b+1}(-Te_{1}+y_{2})+\frac{2b\lambda}{2b+1}(Te_{1}+y_{1}).$

Since $0<\lambda<1,$ we have,
$1-\lambda,\frac{\lambda}{2b+1},\frac{2b\lambda}{2b+1}\in(0,1).$ Moreover, we
have, $\|Te_{1}-y_{1}\|=\|-Te_{1}+y_{2}\|=\|Te_{1}+y_{1}\|=1.$ Using this, it
can be easily observed that $\|-Tx+Te_{1}\|<\|T\|,$ where $-x+e_{1}\in
Ext(B_{\mathbb{X}}),$ a contradiction.

Similarly, if $Tx\in-L(y_{1},-y_{2}),$ then we can show that
$\|Tz\|=\|Tx+Te_{1}\|<\|T\|,$ where $z=x+e_{1}\in Ext(B_{\mathbb{X}}),$ a
contradiction. Therefore, we must have $Tx\in\pm L[y_{1},y_{2}],$ i.e.,
$Tz=Tx+Te_{1}\in\pm L[Te_{1}+y_{1},-Te_{1}+y_{2}].$
Now if $z=x-e_{1},$ following the same line of arguments, we can show that
$Tz\in\pm L[Te_{1}+y_{1},-Te_{1}+y_{2}].$
On the other hand, if we assume that
$T(B_{\mathbb{X}})=conv\\{\pm(Te_{1}+y_{2}),\pm(Te_{1}+y_{1})\\},$ then
proceeding similarly, we can show that $Tz\in\pm
L[Te_{1}+y_{1},-Te_{1}-y_{2}]$ for each $z\in Ext(B_{\mathbb{X}}).$ This
completes the proof of the lemma. ∎

The following lemma is needed to prove the desired theorem. The proof of which
is trivial and hence we omit the proof.

######  Lemma 2.5.

Any face of $\ell_{\infty}^{n}$ having exactly $2^{k}$ extreme points contains
exactly $k+1$ linearly independent extreme points.

Now, we are ready to prove our desired result. We completely characterize
$k-$smoothness of an operator defined between $\ell_{\infty}^{n}$ and any two-
dimensional Banach space with the condition that the operator attains its norm
at each element of $Ext(B_{\ell_{\infty}^{n}}).$ We solve the problem in the
following two theorems. In the following theorem, we consider the case in
which image of each extreme point of $B_{\ell_{\infty}^{n}}$ is smooth.

######  Theorem 2.6.

Let $\mathbb{X}=\ell_{\infty}^{n}$ and $\mathbb{Y}$ be a two-dimensional
Banach space. Let $T\in S_{{\mathbb{L}}(\mathbb{X},\mathbb{Y})}$ be such that
$Ext(B_{\mathbb{X}})\subseteq M_{T}$ and $Tx$ is smooth for all $x\in
Ext(B_{\mathbb{X}}).$ Then the followings hold:
(i) If $Rank(T)=1,$ then $T$ is $n-$smooth.
(ii) Let $Rank(T)=2.$ If $Tx$ is an interior point of some line segment of
$T(B_{\mathbb{X}})$ for some $x\in Ext(B_{\mathbb{X}}),$ then $T$ is
$n-$smooth. Otherwise, $T$ is $(2n-2)-$smooth.

###### Proof.

Let us write $Ext(B_{\mathbb{X}})=\\{\pm x_{1},\pm x_{2},\dots,\pm
x_{2^{n-1}}\\},$ where $\\{x_{1},x_{2},\dots,x_{n}\\}$ is linearly
independent.
(i) Suppose $Rank(T)=1.$ Then $Tx_{i}=\pm Tx_{1}$ for all $2\leq i\leq
2^{n-1}.$ Let $J(Tx_{1})=\\{y^{*}\\}.$ Then for any
$i\in\\{1,2,\dots,2^{n-1}\\},$ either $J(Tx_{i})=\\{y^{*}\\}$ or
$J(Tx_{i})=\\{-y^{*}\\}.$ Now, if $T$ is $k-$smooth, then

$\displaystyle k$ $\displaystyle=$ $\displaystyle dim~{}span~{}J(T)$
$\displaystyle=$ $\displaystyle dim~{}span~{}Ext~{}J(T)$ $\displaystyle=$
$\displaystyle dim~{}span~{}\\{y^{*}\otimes x_{i}:1\leq i\leq 2^{n-1}\\}$
$\displaystyle=$ $\displaystyle dim~{}span~{}\\{y^{*}\otimes x_{i}:1\leq i\leq
n\\}$ $\displaystyle=$ $\displaystyle n,$

as $\\{y^{*}\otimes x_{i}:1\leq i\leq n\\}$ is linearly independent by Lemma
2.3. Hence, $T$ is $n-$smooth.

(ii) Suppose $Rank(T)=2.$ Then by Lemma 2.4, $T(B_{\mathbb{X}})$ is a convex
set with four extreme points. Let $\pm y_{1},\pm y_{2}$ be four distinct
extreme points of $T(B_{\mathbb{X}}).$

First suppose $Tx$ is an interior point of some line segment of
$T(B_{\mathbb{X}})$ for some $x\in Ext(B_{\mathbb{X}}),$ i.e., $Tx_{j}\in
L(y_{1},y_{2})$ for some $1\leq j\leq 2^{n-1}.$ Again by Lemma 2.4, we get
$Tx_{i}\in\pm L[y_{1},y_{2}]$ for all $1\leq i\leq 2^{n-1}.$ Let
$J(Tx_{j})=\\{y^{*}\\}.$ Then it is clear that for any
$i\in\\{1,2,\dots,2^{n-1}\\},$ either $J(Tx_{i})=\\{y^{*}\\}$ or
$J(Tx_{i})=\\{-y^{*}\\}.$ Then as in case (i) it is easy to show that $T$ is
$n-$smooth.

Next, suppose that $Tx$ is not an interior point of any line segment of
$T(B_{\mathbb{X}})$ for any $x\in Ext(B_{\mathbb{X}}).$ Then $Tx_{i}\notin
L(\pm y_{1},\pm y_{2})$ for any $1\leq i\leq 2^{n-1}.$ Thus, $Tx_{i}\in\\{\pm
y_{1},\pm y_{2}\\}$ for all $1\leq i\leq 2^{n-1}.$ Since, $Rank(T)=2,$ without
loss of generality, we may assume $Tx_{1}=y_{1},Tx_{2}=y_{2}.$ Let
$J(Tx_{1})=\\{z_{1}^{*}\\}$ and $J(Tx_{2})=\\{z_{2}^{*}\\}.$ Then for any
$i\in\\{1,2,\dots,2^{n-1}\\},$

$\displaystyle J(Tx_{i})=$
$\displaystyle\\{z_{1}^{*}\\}~{}or~{}\\{-z_{1}^{*}\\}~{}or~{}~{}\\{z_{2}^{*}\\}~{}or~{}\\{-z_{2}^{*}\\}.$

Let $S_{1}=\\{x_{i}\in Ext(B_{\mathbb{X}}):Tx_{i}=Tx_{1}\\}$ and
$S_{2}=\\{x_{i}\in Ext(B_{\mathbb{X}}):Tx_{i}=Tx_{2}\\}.$ Thus, we have
$S_{1}\cap S_{2}=\emptyset,$ $\pm S_{1}\cup\pm S_{2}=Ext(B_{\mathbb{X}})$ and
$|S_{1}\cup S_{2}|=|S_{1}|+|S_{2}|=2^{n-1}.$ Therefore, for any
$i\in\\{1,2,\dots,2^{n-1}\\},$

$\displaystyle J(Tx_{i})$ $\displaystyle=$
$\displaystyle\\{z_{1}^{*}\\},~{}\text{if~{}}x_{i}\in S_{1}$ $\displaystyle=$
$\displaystyle\\{z_{2}^{*}\\},~{}\text{if~{}}x_{i}\in S_{2}.$

Now, it is clear that $S_{1}$ as well as $S_{2}$ cannot contain $n$ linearly
independent vectors. For otherwise, we get $Rank(T)=1,$ a contradiction. Thus,
maximal linearly independent subsets of $S_{1}$ and $S_{2}$ contain at most
$n-1$ elements. Now, we show that $|S_{1}|=2^{n-2}$ and $|S_{2}|=2^{n-2}.$ If
possible, let $|S_{1}|<2^{n-2}.$ Then we must have $|S_{2}|>2^{n-2}.$ Observe
that $conv(S_{2})$ is a face of $B_{\mathbb{X}}.$ Clearly,
$Ext(conv(S_{2}))=S_{2},$ i.e., $|Ext(conv(S_{2}))|=|S_{2}|>2^{n-2}.$ Hence,
the face $conv(S_{2})$ of $B_{\mathbb{X}}$ contains at least $2^{n-1}$ extreme
points and hence by Lemma 2.5, $conv(S_{2})$ contains at least $n$ linearly
independent extreme points. Thus, $S_{2}$ contains at least $n$ linearly
independent vectors. This gives that $Rank(T)=1,$ a contradiction. Therefore,
$|S_{1}|\geq 2^{n-2}.$ Similarly, $|S_{2}|\geq 2^{n-2}.$ Thus,
$|S_{1}|=|S_{2}|=2^{n-2},$ i.e.,
$|Ext(conv(S_{1}))|=|Ext(conv(S_{2}))|=2^{n-2}.$ Now, by Lemma 2.5, $S_{1}$
and $S_{2}$ has exactly $n-1$ linearly independent vectors. Without loss of
generality, suppose that the set of all linearly independent vectors of
$S_{1}$ and $S_{2}$ are $\\{u_{1},u_{2},\ldots,u_{n-1}\\}$ and
$\\{u_{n},u_{n+1},\ldots,u_{2n-2}\\}$ respectively. Now, if $T$ is $k-$smooth,
then

$\displaystyle k$ $\displaystyle=$ $\displaystyle\dim~{}span~{}J(T)$
$\displaystyle=$ $\displaystyle\dim~{}span~{}Ext~{}J(T)$ $\displaystyle=$
$\displaystyle\dim~{}span~{}\\{z_{1}^{*}\otimes u_{i},z_{2}^{*}\otimes
u_{j}:u_{i}\in S_{1},u_{j}\in S_{2}\\}$ $\displaystyle=$
$\displaystyle\dim~{}span~{}\\{z_{1}^{*}\otimes u_{i},z_{2}^{*}\otimes
u_{j}:1\leq i\leq n-1,n\leq j\leq 2n-2\\}$ $\displaystyle=$ $\displaystyle
2n-2,\text{which can be easily verified}.$

Therefore, $T$ is $(2n-2)-$smooth. This completes the proof. ∎

In addition to Theorem 2.6, if we assume that the range space is strictly
convex and smooth, then we obtain the following corollary.

######  Corollary 2.7.

Let $\mathbb{X}=\ell_{\infty}^{n}$ and $\mathbb{Y}$ be a two-dimensional
strictly convex, smooth Banach space. Let $T\in
S_{{\mathbb{L}}(\mathbb{X},\mathbb{Y})}$ be such that
$Ext(B_{\mathbb{X}})\subseteq M_{T}.$ Then the followings hold:
(i) If $Rank(T)=1,$ then $T$ is $n-$smooth.
(ii) If $Rank(T)=2,$ then $T$ is $(2n-2)-$smooth.

###### Proof.

$(i)$ follows clearly from Theorem 2.6. We only prove $(ii)$. Let $Rank(T)=2.$
Then by Lemma 2.4, $T(B_{\mathbb{X}})$ is a convex set with four extreme
points. Without loss of generality, let $\pm Tx_{1},\pm Tx_{2}$ be four
distinct extreme points of $T(B_{\mathbb{X}}).$ If possible, suppose that
there exists $x\in Ext(B_{\mathbb{X}})$ such that $Tx$ is an interior point of
some line segment of $T(B_{\mathbb{X}}).$ Suppose that $Tx\in
L(Tx_{1},Tx_{2}).$ Since $x\in M_{T},\|Tx\|=1.$ Thus, it is clear that
$\|y\|=1$ for all $y\in L[Tx_{1},Tx_{2}],$ i.e., $L[Tx_{1},Tx_{2}]\subseteq
S_{\mathbb{Y}}.$ This contradicts that $\mathbb{Y}$ is strictly convex.
Therefore, there does not exist any $x\in Ext(B_{\mathbb{X}})$ such that $Tx$
is an interior point of some line segment of $T(B_{\mathbb{X}}).$ Hence, from
Theorem 2.6, we conclude that if $Rank(T)=2,$ then $T$ is $(2n-2)-$smooth. ∎

In the next theorem, we consider the remaining case in which the image of at
least one extreme point of $B_{\ell_{\infty}^{n}}$ is not a smooth point. Note
that in a two-dimensional Banach space, a non-zero vector is either smooth or
$2-$smooth.

######  Theorem 2.8.

Let $\mathbb{X}=\ell_{\infty}^{n}$ and $\mathbb{Y}$ be a two-dimensional
Banach space. Let $T\in S_{\mathbb{L}(\mathbb{X},\mathbb{Y})}$ be such that
$Ext(B_{\mathbb{X}})\subseteq M_{T}.$ Let $S=\\{x\in
Ext(B_{\mathbb{X}}):Tx~{}\text{is~{}not~{}smooth}\\}$ be non-empty and $S_{1}$
be the subset of $S$ containing all linearly independent vectors of $S.$ If
$|S_{1}|=k,$ then $T$ is $(n+k)-$smooth.

###### Proof.

Let $Ext(B_{\mathbb{X}})=\\{\pm x_{1},\pm x_{2},\ldots,\pm x_{2^{n-1}}\\}.$
First suppose that $Rank(T)=1.$ Since $S\neq\emptyset,$ assume $x_{1}\in S.$
Then for all $2\leq i\leq 2^{n-1},$ $Tx_{i}=\pm Tx_{1}.$ Let
$Ext~{}J(Tx_{1})=\\{z_{1}^{*},z_{2}^{*}\\}$ for some $z_{1}^{*},z_{2}^{*}\in
S_{\mathbb{Y}^{*}}.$ Then either $Ext~{}J(Tx_{i})=\\{z_{1}^{*},z_{2}^{*}\\}$
or $Ext~{}J(Tx_{i})=\\{-z_{1}^{*},-z_{2}^{*}\\}$ for all $2\leq i\leq
2^{n-1}.$ Clearly, $|S_{1}|=n.$ Now,

$\displaystyle\dim~{}span~{}Ext~{}J(T)$ $\displaystyle=$
$\displaystyle\dim~{}span~{}\\{z_{1}^{*}\otimes x_{i},z_{2}^{*}\otimes
x_{i}:1\leq i\leq 2^{n-1}\\}$ $\displaystyle=$
$\displaystyle\dim~{}span~{}\\{z_{1}^{*}\otimes x_{i},z_{2}^{*}\otimes
x_{i}:1\leq i\leq n\\}$ $\displaystyle=$ $\displaystyle 2n,~{}\text{using
~{}Lemma \ref{lemma-ind}}.$

Thus, in this case $T$ is $2n-$smooth and we are done.
Next, suppose that $Rank(T)=2.$ Then by Lemma 2.4, $T(B_{\mathbb{X}})$ is a
convex set with four extreme points. Without loss of generality, let $\pm
y_{1},\pm y_{2}$ be four distinct extreme points of $T(B_{\mathbb{X}}).$ We
consider the following two cases:
Case I : $S=Ext(B_{\mathbb{X}}).$
Clearly, in this case $|S_{1}|=n.$ Let $S_{1}=\\{x_{1},x_{2},\ldots,x_{n}\\}.$
Observe that if $Tx_{i}\in L(\pm y_{1},\pm y_{2}),$ for any $1\leq i\leq
2^{n-1},$ then $Tx_{i}$ will be smooth, which contradicts that
$S=Ext(B_{\mathbb{X}}).$ Thus, $Tx_{i}\in\\{\pm y_{1},\pm y_{2}\\}$ for all
$1\leq i\leq 2^{n-1}.$ Since $Rank(T)=2,$ $Tx_{i}=y_{1},Tx_{j}=y_{2}$ for some
$1\leq i\neq j\leq 2^{n-1}.$ Therefore, $y_{1},y_{2}$ are not smooth. Suppose
that $Ext~{}J(y_{1})=\\{z_{1}^{*},z_{2}^{*}\\}$ and
$Ext~{}J(y_{2})=\\{z_{3}^{*},z_{4}^{*}\\}.$ Then for each $1\leq i\leq
2^{n-1},$ $Ext~{}J(Tx_{i})$ is either $\\{z_{1}^{*},z_{2}^{*}\\}$ or
$\\{-z_{1}^{*},-z_{2}^{*}\\}$ or $\\{z_{3}^{*},z_{4}^{*}\\}$ or
$\\{-z_{3}^{*},-z_{4}^{*}\\}.$ Observe that $z_{3}^{*},z_{4}^{*}\in
span\\{z_{1}^{*},z_{2}^{*}\\},$ since $\dim(\mathbb{Y})=2.$ Hence, for any
$x\in\mathbb{X},z_{3}^{*}\otimes x,z_{4}^{*}\otimes x\in
span~{}\\{z_{1}^{*}\otimes x,z_{2}^{*}\otimes x\\}.$ Therefore,

$\displaystyle\dim~{}span~{}Ext~{}J(T)$ $\displaystyle=$
$\displaystyle\dim~{}span~{}\\{z_{1}^{*}\otimes x_{i},z_{2}^{*}\otimes
x_{i}:1\leq i\leq 2^{n-1}\\}$ $\displaystyle=$
$\displaystyle\dim~{}span~{}\\{z_{1}^{*}\otimes x_{i},z_{2}^{*}\otimes
x_{i}:1\leq i\leq n\\}$ $\displaystyle=$ $\displaystyle 2n.$

Thus, in this case $T$ is $2n-$smooth and we are done.
Case II : $S\subsetneqq Ext(B_{\mathbb{X}}).$
Without loss of generality, we may assume that
$S_{1}=\\{x_{1},x_{2},\ldots,x_{k}\\}$ and $S=\\{\pm x_{1},\pm
x_{2},\ldots,\pm x_{k},\pm x_{k+1},\ldots,\pm x_{m}\\},$ $(1\leq
m<2^{n-1},1\leq k\leq n).$ As in Case I, $Tx_{i}\in\\{\pm y_{1},\pm y_{2}\\}$
for all $1\leq i\leq m.$ Clearly, at least one of $y_{1},y_{2}$ is not smooth.
First we assume both of $y_{1},y_{2}$ are not smooth. Using Lemma 2.4, we get
either $Tx_{i}\in\pm L[y_{1},y_{2}]$ for each $m<i\leq 2^{n-1}$ or
$Tx_{i}\in\pm L[-y_{1},y_{2}]$ for each $m<i\leq 2^{n-1}.$ Without loss of
generality, assume that $Tx_{i}\in\pm L[y_{1},y_{2}]$ for each $m<i\leq
2^{n-1}$. Since, $y_{1},y_{2}$ are not smooth and $Tx_{i}$ are smooth,
$Tx_{i}\notin\\{\pm y_{1},\pm y_{2}\\}$ for $m<i\leq 2^{n-2}.$ Therefore,
$Tx_{i}\in\pm L(y_{1},y_{2})$ for each $m<i\leq 2^{n-1}$. Now, it is easy to
observe that either $J(Tx_{i})=\\{y^{*}\\}$ or $J(Tx_{i})=\\{-y^{*}\\}$ for
each $m<i\leq 2^{n-1}.$ Suppose $Ext~{}J(y_{1})=\\{y^{*},z_{1}^{*}\\}$ and
$Ext~{}J(y_{2})=\\{y^{*},z_{2}^{*}\\}.$ Then for each $1\leq i\leq m,$
$Ext~{}J(Tx_{i})$ is either $\\{y^{*},z_{1}^{*}\\}$ or
$\\{-y^{*},-z_{1}^{*}\\}$ or $\\{y^{*},z_{2}^{*}\\}$ or
$\\{-y^{*},-z_{2}^{*}\\}.$ Observe that $z_{2}^{*}\in
span\\{y^{*},z_{1}^{*}\\},$ since $\dim(\mathbb{Y})=2.$ Hence, for any
$x\in\mathbb{X},z_{2}^{*}\otimes x\in span~{}\\{y^{*}\otimes
x,z_{1}^{*}\otimes x\\}.$ Therefore,

$\displaystyle\dim~{}span~{}Ext~{}J(T)$ $\displaystyle=$
$\displaystyle\dim~{}span~{}\\{y^{*}\otimes x_{i},z_{1}^{*}\otimes x_{j}:1\leq
i\leq 2^{n-1},1\leq j\leq m\\}$ $\displaystyle=$
$\displaystyle\dim~{}span~{}\\{y^{*}\otimes x_{i},z_{1}^{*}\otimes x_{j}:1\leq
i\leq 2^{n-1},1\leq j\leq k\\}$ $\displaystyle=$ $\displaystyle
n+k,~{}(\text{by~{}Lemma~{}\ref{lemma-ind}}),$

since $\\{x_{i}:1\leq i\leq 2^{n-1}\\}$ contains only $n$ linearly independent
vectors. Therefore, $T$ is $(n+k)-$smooth.
Now, if we consider exactly one of $y_{1},y_{2}$ is not smooth, then following
same line of arguments, we can prove that $T$ is $(n+k)-$smooth. This
completes the proof of the theorem. ∎

We would like to mention that Theorem 2.6 and Theorem 2.8 improves on [5, Th.
3.10]. The study of $k-$smoothness of an operator defined between
$\ell_{\infty}^{n}$ and $\mathbb{Y}$ becomes more complicated when
$\dim\mathbb{Y}\geq 3.$ The rest of the paper is devoted to the study of
$k-$smoothness of an operator defined between two particular spaces
$\ell_{\infty}^{3}$ and $\ell_{1}^{3}.$ We denote the extreme points of
$B_{\ell_{\infty}^{3}}$ by $\pm x_{1}=\pm(1,1,1),\pm x_{2}=\pm(-1,1,1),\pm
x_{3}=\pm(-1,-1,1),\pm x_{4}=\pm(1,-1,1).$ $|M_{T}\cap
Ext(B_{\ell_{\infty}^{3}})|$ plays an important role in determining the order
of smoothness of $T.$ Observe that if $|M_{T}\cap
Ext(B_{\ell_{\infty}^{3}})|\leq 6,$ then the order of smoothness of $T$ can be
obtained using [6, Th. 2.2]. Therefore, we only consider the case for which
$|M_{T}\cap Ext(B_{\ell_{\infty}^{3}})|=8,$ i.e., $M_{T}\cap
Ext(B_{\ell_{\infty}^{3}})=\\{\pm x_{1},\pm x_{2},\pm x_{3},\pm x_{4}\\}.$
Note that, for $1\leq i\leq 4,$ $Tx_{i}$ is $k-$smooth, where
$k\in\\{1,2,3\\}.$ Suppose $S_{k}=\\{x\in M_{T}\cap
Ext(B_{\ell_{\infty}^{3})}:Tx~{}is~{}k-smooth\\},$ where $k\in\\{1,2,3\\}.$
Clearly, $|S_{1}|+|S_{2}|+|S_{3}|=8.$ In the following theorem, we consider
the case when $|S_{1}|=8.$

######  Theorem 2.9.

Let $\mathbb{X}=\ell_{\infty}^{3}$ and $\mathbb{Y}=\ell_{1}^{3}.$ Let $T\in
S_{\mathbb{L}(\mathbb{X},\mathbb{Y})}$ be such that $M_{T}\cap
Ext(B_{\mathbb{X}})=\\{\pm x_{1},\pm x_{2},\pm x_{3},\pm x_{4}\\}.$ Let
$|S_{1}|=8.$ Then the following hold:
$(i)$ If $\pm J(Tx_{i})=\pm J(Tx_{j})$ for all $x_{i},x_{j}\in S_{1}$, then
$T$ is $3-$smooth.
$(ii)$ Otherwise, $T$ is $4-$smooth.

###### Proof.

$(i)$ Suppose the given condition is satisfied. Let $\pm J(Tx_{i})=\\{\pm
y^{*}\\}$ for $1\leq i\leq 4.$ Now, if $T$ is $k-$smooth, then

$\displaystyle k$ $\displaystyle=$ $\displaystyle\dim~{}span~{}J(T)$
$\displaystyle=$ $\displaystyle\dim~{}span~{}Ext~{}J(T)$ $\displaystyle=$
$\displaystyle\dim~{}span~{}\\{y^{*}\otimes x_{1},y^{*}\otimes
x_{2},y^{*}\otimes x_{3},y^{*}\otimes x_{4}\\}$ $\displaystyle=$
$\displaystyle\dim~{}span~{}\\{y^{*}\otimes x_{1},y^{*}\otimes
x_{2},y^{*}\otimes x_{3}\\}$ $\displaystyle=$ $\displaystyle
3,~{}(\text{using~{} Lemma~{}\ref{lemma-ind}}).$

Hence, $T$ is $3-$smooth.
$(ii)$ Let $\pm J(Tx_{i})=\\{\pm y_{i}^{*}\\}$ for $1\leq i\leq 4.$ Since
$(i)$ is not satisfied, without loss of generality, we assume
$y_{1}^{*}\neq\pm y_{2}^{*},$ i.e., $\\{y_{1}^{*},y_{2}^{*}\\}$ is linearly
independent. Let $y_{3}^{*}=ay_{1}^{*}+by_{2}^{*}$ and
$y_{4}^{*}=cy_{1}^{*}+dy_{2}^{*},$ where $a,b,c,d\in\mathbb{R}.$ Since
$\|y_{3}^{*}\|=1,$ $a$ and $b$ cannot be zero simultaneously. Similarly, $c$
and $d$ cannot be zero simultaneously. Now, if $T$ is $k-$smooth, then

$\displaystyle k$ $\displaystyle=$ $\displaystyle\dim~{}span~{}J(T)$
$\displaystyle=$ $\displaystyle\dim~{}span~{}Ext~{}J(T)$ $\displaystyle=$
$\displaystyle\dim~{}span~{}\\{y_{1}^{*}\otimes x_{1},y_{2}^{*}\otimes
x_{2},y_{3}^{*}\otimes x_{3},y_{4}^{*}\otimes x_{4}\\}.$

Using the relations $x_{4}=x_{1}-x_{2}+x_{3},$
$y_{3}^{*}=ay_{1}^{*}+by_{2}^{*},$ $y_{4}^{*}=cy_{1}^{*}+dy_{2}^{*}$ and Lemma
2.3, it is easy to verify that $\\{y_{1}^{*}\otimes x_{1},y_{2}^{*}\otimes
x_{2},y_{3}^{*}\otimes x_{3},y_{4}^{*}\otimes x_{4}\\}$ is linearly
independent. Therefore, $k=4$ and $T$ is $4-$smooth. ∎

Proceeding similarly we can find the $k-$smoothness of operator $T$ for other
feasible cases. We skip the details of proof to avoid monotonicity. The
following two tables illustrates the various possible cases of $k-$smoothness
under different conditions on $S_{1},S_{2}$ and $S_{3}.$ The first table
contains the cases when $S_{3}=\emptyset,$ i.e., $Tx_{i}$ is either $1-$smooth
or $2-$smooth but not $3-$smooth.

$|S_{1}|$ | $|S_{2}|$ | $|S_{3}|$ | Further conditions on the operator $T$ |  $T$ is
---|---|---|---|---
|  |  |  |  $k$-smooth
|  |  |  | with $k=$
8 | 0 | 0 | $\pm J(Tx_{i})=\pm J(Tx_{j})~{}\forall~{}x_{i},x_{j}\in S_{1}$ | 3
Otherwise | 4
6 | 2 | 0 |  $\pm J(Tx_{i})=\pm J(Tx_{j})~{}\forall~{}x_{i},x_{j}\in S_{1}$ and |
$\pm J(Tx_{i})\subseteq\pm Ext~{}J(Tx_{k})$,$~{}\forall x_{i}\in S_{1},x_{k}\in S_{2}$ | 4
|  |  | Otherwise | 5
4 | 4 | 0 |  $\pm J(Tx_{i})=\pm J(Tx_{j})~{}\forall~{}x_{i},x_{j}\in S_{1}$ and |
$\pm J(Tx_{i})\subseteq\pm Ext~{}J(Tx_{k})$,$\forall x_{i}\in S_{1},\forall x_{k}\in S_{2}$ | 5
|  |  | Otherwise | 6
2 | 6 | 0 |  $|\cap_{x_{k}\in S_{2}}\pm J(Tx_{k})|\geq 2$ and |
$\pm J(Tx_{i})\subseteq\pm Ext~{}J(Tx_{k})$,$\forall x_{i}\in S_{1},\forall x_{k}\in S_{2}$ | 6
|  |  | Otherwise | 7
0 | 8 | 0 | $|\cap_{i=1}^{4}\pm J(Tx_{i})|=4~{}$ | 6
Either $|\cap_{i=1}^{4}\pm J(Tx_{i})|=2~{}$ or |
$|\pm Ext~{}J(Tx_{i})\cap\pm Ext~{}J(Tx_{j})|\neq 2,$ for $1\leq i\neq j\leq 4$ | 7
|  |  | Otherwise | 8

The next table exhibits the cases when $S_{3}\neq\emptyset.$

$|S_{1}|$ | $|S_{2}|$ | $|S_{3}|$ | Further conditions on the operator $T$ |  $T$ is
---|---|---|---|---
|  |  |  |  $k$-smooth
|  |  |  | with $k=$
6 | 0 | 2 | $\pm J(Tx_{i})=\pm J(Tx_{j}),\forall~{}x_{i},x_{j}\in S_{1}$ | 5
Otherwise | 6
4 | 2 | 2 |  $\pm J(Tx_{i})=\pm J(Tx_{j}),\forall~{}x_{i},x_{j}\in S_{1}$ and |
$\pm J(Tx_{i})\subseteq\pm Ext~{}J(Tx_{j})$, $\forall~{}x_{i}\in S_{1},x_{j}\in S_{2}$ | 6
|  |  | Otherwise | 7
$2$ | $4$ | 2 | - | 7
0 | 6 | 2 | $|\cap_{x_{i}\in S_{2}}\pm Ext~{}J(Tx_{i})|=4$ | 7
Otherwise | 8
$4$ | $0$ | $4$ | - | $7$
$0$ | $4$ | $4$ | - | $8$
$0$ | $0$ | $8$ | - | $9$

Finally we would like to note that the following possibilities are not
feasible: (i) $|S_{1}|=2,|S_{2}|=2,|S_{3}|=4,$ (ii) $|S_{1}|=2,|S_{3}|=6$ and
(ii) $|S_{2}|=2,|S_{3}|=6.$

## References

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  * [4] B. -L. Lin and T. S. S. R. K. Rao, Multismoothness in Banach spaces, Int. J. Math. Math. Sci., Vol. 2007, Article ID 52382, 12 pages.
  * [5] A. Mal, S. Dey and K. Paul Characterization of $k-$smoothness of operators defined between infinite-dimensional spaces, Linear Multilinear Algebra, (2020) DOI: 10.1080/03081087.2020.1844130.
  * [6] A. Mal and K. Paul, Characterization of $k-$smooth operators between Banach spaces, Linear Algebra Appl., 586 (2020) 296-307.
  * [7] A. Mal, K. Paul and S. Dey, Characterization of extreme contractions through $k-$smoothness of operators, arXiv:2006.15318v1 [math.FA] 27 Jun 2020.
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# Global displacement induced by rigid motion simulation during MRI
acquisition.

Ghiles Reguig<EMAIL_ADDRESS>Paris Brain Institute – ICM, Siemens
Healthineers.ParisFrance ,  Marc Lapert  Siemens HealthinnersParisFrance ,
Stéphane Lehéricy  Paris Brain Institute – ICM, INSERM U 1127, CNRS UMR 7225,
Sorbonne Université; Department of Neuroradiology, Pitié-Salpêtrière Hospital,
Public Assistance - Paris Hospitals (AP-HP), Paris, FranceParisFrance and
Romain Valabregue  Paris Brain Institute – ICM, INSERM U 1127, CNRS UMR 7225,
Sorbonne UniversitéParisFrance<EMAIL_ADDRESS>
###### Abstract.

## Abstract

Purpose: Simulation of motion in the frequency domain during the MRI
acquisition process is an important tool to study the effect of object motion.
It is natural in this context to rely on a voxel-to-voxel difference metric
between the original image and its motion-simulated version to quantify the
alteration. These metrics are very sensitive to any mis-coregistration between
the two images. As we observed global displacements of the corrupted image for
specific simulated motions, we focused this work on the critical choice of a
reference position of the object for the motion simulation that avoids any
global displacement.
Methods: We used the motion simulation framework and we studied the different
proposed solutions. Classically, the position around the k-space center is
considered as a reference with two analytical solutions to correct for global
image shifts: demeaning the image in the Fourier domain either using the exact
position at k-space center or using a weighted average around the center.
Results: We demonstrated with motion simulated examples that these analytical
solutions did not yield the same results as a direct co-registration method.
Indeed, short transitions around the k-space center did not induce any global
displacement, in contrast with what both analytical solutions assumed. We also
observed a dependence of the global shift to both the motion amplitude and the
translation direction.
Conclusion: In the absence of a theoretical solution, we concluded that an
extra co-registration step after motion simulation was needed in order to
compare the simulation with its reference image.

Motion simulation, Rigid motion, coregistration, Brain MRI, loss function

##  1\. Introduction

The rigid movement of the head is the predominant artifact of 3D cerebral MRI
acquisitions, due to its long acquisition time. As motion occurs while
sampling the frequency domain, the effects on the image are complex and hard
to predict. Quality control studies, which aimed to automatically detect
motion-artifacted images (among other artifacts) always relied on a ground
truth defined by an expert ((Alfaro-Almagro et al.,, 2018; Esteban et al.,,
2017)). This restricts to a binary and noisy definition of the artifact
severity with a poor inter-rater reproducibility (0̃.85 in (Esteban et al.,,
2017)). On the other hand, a realistic motion simulation framework provides a
tool to better define the motion artifact severity, by relying on a voxel-to-
voxel difference metric (L1, L2, NCC, MI, SSIM …) which quantifies the
difference between the motion simulated image and its original version. Those
difference metrics are also used as loss functions for deep learning
approaches and constitute a key element of the learning strategy of motion
detection/correction tasks (Shaw et al.,, 2020; Duffy et al.,, 2021; Johnson
and Drangova,, 2019; Liu et al.,, 2020; Pawar et al.,, 2019; Tamada et al.,,
2020).
The objective of this work is to point out an important bias of those metrics
as they are known to be very sensitive to co-registration errors. Indeed
simulating a motion during the acquisition process may induce a global
displacement of the image, due to the fact that subject position is defined
relatively to a chosen reference. Changing the reference, which is equivalent
to subtracting a constant position, would result in the same motion corrupted
volume but with a global shift. Given a particular motion time course
representing the different subject’s positions during data acquisition, the
global mean displacement (or a reference position 111Choosing the correct
reference position is equivalent to predict the global displacement and
choosing it as the reference. So both terms “reference position” and “global
displacement” can be used interchangeably.) has to be subtracted, so that the
simulated motion corrupted volume remains at the same global position as in
the original volume.
To the best of our knowledge, the global displacement problem has not been
explicitly explored in the literature and is often bypassed in studies relying
on motion simulation even though a voxel-to-voxel difference metric was used.
Few works mentioned specific solutions. In the retromoco toolbox ((Gallichan,,
2021)), the mean reference position is chosen at the k-space center. (Bazin et
al.,, 2020) proposed to take as a reference the average position over eleven
lines around the k-space center. (Shaw et al.,, 2020) took a weighted average
of the motion time course, with the weights defined as the absolute value of
the Fourier Transform coefficients. Although the formulation is different, the
latter is equivalent to the proposition of (Todd et al.,, 2015), where a
partition-weighted integrated motion was used. A more radical solution is to
avoid any motion around the k-space center as in (Duffy et al.,, 2021). But
the chosen 10% portion around the k-space center is quite arbitrary and
prevents the study of motion occurring near the k-space center. Finally, we
only found one work that proposed to add an extra co-registration step of the
motion-simulated image ((Johnson and Drangova,, 2019)).
Although the motion simulation is well known, different strategies were
proposed for its implementation. (Liu et al.,, 2020) simulated only
translation by adding a pseudo periodic phase shift perturbation to the FFT
signal as proposed by (Tamada et al.,, 2020) for breathing artifacts
simulation on liver imaging. It is however less suited for the rigid brain
motion as it is important to also consider the effect of head rotation. A
similar approach was used by (Duffy et al.,, 2021) by considering a Gaussian
random walk perturbation of the k-space lines (translations and rotations).
A second approach was used by (Johnson and Drangova,, 2019; Pawar et al.,,
2019; Shaw et al.,, 2020). They assumed a piecewise motion, where the object
takes n different positions during acquisition. The n motions are applied in
the image domain and the ffts are computed for each resliced image. Then a new
k-space is built by concatenating blocks corresponding to the duration of a
given position and a final fft provides the motion-corrupted volume. Although
this strategy is correct, the computation time needed to reslice each position
and to compute the corresponding Fourier Transform makes it impractical for
more than a few positions (maximal number of positions used is 5 in (Shaw et
al.,, 2020) and 32 in (Pawar et al.,, 2019)). This constrains the method to
study only over simplified motion.
The third approach that we followed in this work, which is equivalent in
results to the second approach described above, relies on the known effect of
motion on the k-space: a translation in the image domain corresponds to a
multiplication of k-space data with linear phase ramps proportional to the
translation amplitude, and rotation around the volume center in image domain
corresponds to the same rotation around the DC component in k-space
((Gallichan,, 2021; Loktyushin et al.,, 2013; Zahneisen and Ernst,, 2016)).
The main advantage of this approach is that we can consider as input, motion
time course at any resolution.
The first contribution of this work is to propose an open source
implementation of this third approach in the torchio library (Pérez-García et
al.,, 2021), to help its use by the machine learning community. The second
contribution is to demonstrate the importance of global displacement induced
by motion simulation: we showed that none of the proposed solutions to correct
it are in perfect agreement with the direct co-registration method (considered
as ground truth). The first experiment was done in one dimension in order to
show the problem in a simplified setting. In a second experiment, we explored
the case of 3D brain MRI acquisition for the simplest motion: one transient
position change. This allowed us to systematically explore the effects of four
factors: the direction, the amplitude, the duration and the position in
k-space where the motion occurs. Finally, we explored the more realistic case
of random motion pattern and further demonstrated the need of an external co-
registration step to avoid an over estimation of the artifact severity.

##  2\. Methods

###  2.1. Motion representation

####  2.1.1. Motion Time course

The motion time course is defined as the temporal positions of the object
during acquisition. To describe the motion time course, we used Euler
representation with 6 parameters (3 rotations and 3 translations) defined over
time. To model realistic motion, we used the motion model as described in
(Gallichan,, 2021) composed of three components: a Perlin noise to model
‘slow’ random motions, a step displacement to model sudden motions and a
transient motion to model swallowing movements. Two examples are given in
Figure 5. For Experiment 2, a two-position problem modeled by a transient step
motion with varying durations and k-space locations was considered (see Figure
2).

####  2.1.2. Motion amplitude, max displacement

For a given motion time course, the amplitude of the movement was computed as
the maximum difference between all positions. For the translation part we
computed the differences of the translation vector.

$AmplitudeTranslation(mm)=\max_{i,j>i}(|\overrightarrow{T_{i}}-\overrightarrow{T_{j}}|)$

where $i$ and $j$ are the time step indices and $\overrightarrow{T}$ the
translation vector.
To accurately set a chosen maximum amplitude, the time course was divided by
this maximal pairwise distance between all positions and multiplied by the
desired amplitude. The same was applied to constrain the rotation amplitude
(i.e. the norm of the vector $R_{x,y,z}$ differences). Note that this motion
amplitude is independent of the chosen reference.

####  2.1.3. Averaging affine transforms

As we will see in section 2.2.2, a weighted average of the motion time course
is needed for the reference position computation. The average is computed on
each Euler parameters separately. Even though this is correct for
translations, it is not accurate for rotations. Averaging rotations is indeed
a very difficult task, with no analytical form as reviewed previously by
(Hanson,, 2020). In machine vision ((Kavan et al.,, 2006)) or for entropy
estimation ((Huggins,, 2014)), the advantage of using dual quaternions
formalism was demonstrated to properly average rotations. Different averaging
propositions were compared but no large differences were found for small
rotations ($<10^{\circ}$). In this work, only the simple Euler parameters
average was used.

###  2.2. Motion artifact simulation

####  2.2.1. Motion during k-space acquisition

We followed the approach described by (Loktyushin et al.,, 2013). We first
applied the rotation on the kspace grid and performed a type 2 nufft (non
uniform fast Fourier transform) ((Barnett et al.,,
2019))222https://github.com/flatironinstitute/finufft/. Then the linear phase
shift proportional to the translation amplitude was added to the complex data
before the final inverse fft. The derivation calculation is detailed in the
appendix. The algorithm requires having the motion time course defined on each
kspace point. In this work, we limited ourselves to the study of a segmented
3D acquisition sequence of the MPRAGE type for which the frequency and the
first phase dimension are acquired within a short time ($<300ms$). Hence, we
supposed that no motion occurred while acquiring these two dimensions. It was
thus only necessary to define the motion time course (or interpolate it) at
the resolution of the third phase direction (y axis), and then to apply the
motion on k-space planes instead of k-space points. We propose an open source
implementation within the torchio library (Pérez-García et al.,, 2021), which
regroups different medical imaging augmentation tools in a pytorch compatible
environment333At the time of writing, this is not yet incorporated, but a
compatible version is available on our github repository :
https://github.com/romainVala/torchio.

####  2.2.2. Computation of the induced global displacement

Two different propositions have been made to define a reference position:
either the position at the k-space center (Gallichan,, 2021) or a weighted
average based on the Fast Fourier Transform (FFT) coefficients of the image
(Shaw et al.,, 2020; Todd et al.,, 2015) from which we derived three ways of
computing the reference: center, wFT_mean and wFT2_mean. An extra reference
position (coreg_shift), derived from a co-registration algorithm, was used as
the ground truth.

center:

the reference was the position at the k-space center.

wFT_mean (or wFT):

the reference was a weighted average of the motion time course, where the
coefficients wFT were the square root of the FFT signal power of the original
image at each phase step,

$wFT_{y}=\sqrt{\sum_{x,z}FT^{2}_{xyz}}$

where $FT_{xyz}$ were the discrete Fourier Transform coefficients. All weights
on the x and z dimensions were summed as the simulated motion only occurred in
the slowest sampling dimension (y axis). This could be easily extended to the
x, y and z dimensions if the motion was defined on all k-space points.
Although different, the formulation was equivalent to the proposition made by
(Shaw et al.,, 2020).

wFT2_mean (or wFT2):

the reference was computed similarly to wFT but using squared weights. Figure
2 shows that this was very similar to the center choice.

coreg_shift:

the reference was the co-registration between the motion simulated image and
the original one. This was used as ground truth. For the one-dimensional (1D)
experiment the shift was measured by minimizing the L1 difference for all
possible pixel shifts. In the 3D experiment, the rigid co-registration from
Elastix software ((Klein et al.,, 2009)) was used.

As stated in the introduction, this global displacement needs to be considered
only for a voxel-to-voxel difference metric. Without loss of generality we
will report the L1 metric as a measure of the artifact’s severity. L1 is
defined as the mean of the absolute voxel-to-voxel difference between the
motion simulation and its original version. $L1_{wFT}$, $L1_{wFT2}$ and
$L1_{coreg\\_shift}$ denoted respectively the artifact severity computed using
wFT, wFT2 and coreg_shift as the reference position.

##  3\. Data

###  3.1. Synthetic 3D brain data

We relied on synthetic data, since it offers a total control of key image
properties: signal to noise ratio (SNR) and contrast. 3D brain data was
synthesized with tissue masks extracted from Freesurfer surfaces of gray and
white matter ((Dahnke et al.,, 2013)). The FSL first tool was used for the
estimation of 7 subcortical structures (caudate, putamen, palidum, thalamus,
hyppocampy, amygdala, accubens) for each side (left or right). Surfaces were
then used by the toblorone software to estimate a Partial volume map for each
structure (Kirk et al.,, 2019). We concatenated the partial volume maps with
posterior estimation of cerebellar white and gray matter, skull skin and
background extracted by CAT12 segmentation pipeline ((Dahnke et al.,, 2013)).
Each partial volume map was multiplied by a signal value drawn from a Gaussian
distribution ($std=1e^{-3}$) and all maps were summed to obtain the synthetic
data. This approach was very similar to the one proposed by (Billot et al.,,
2021) to train a segmentation task. The main difference lied in the use of
more realistic partial volume maps instead of binary tissue masks. Therefore,
no extra smoothing step of the label was required. This provided ideal
synthetic acquisition, as if the different tissues were fully homogeneous.
Synthetic data allowed us to study independently different key aspects of the
volume: the SNR, the contrast and the specific shape of the brain. The latter
can be explored by adding an affine or elastic transformation to the label or
by choosing different Human Connectome Projet (HCP) subjects ((Glasser et
al.,, 2013)). This provided a wide variability of brain shapes. We generated
two types of synthetic images using either a T1 like contrast
($CSF=0.1;GM=0.6;WM=1$) or a random contrast where the intensities of each
tissue were drawn from a uniform distribution in the $]0,1[$ range.

##  4\. Results

###  4.1. 1D translation

Figure 1. The same motion simulated for two different 1D objects. The first line shows the transient motion step in orange at the k-space center (magnitude of 10 pixels, duration 20), and the FFT of the object in blue (arbitrary units). The middle line represents the original object in blue, and the motion corrupted version in orange. The third line represents the cost function L1 between the original object and its motion-corrupted version as a function of pixel-wise shift. The minimum of this function represents the global displacement induced by the transient motion (coreg_shift). The two objects (left and right columns) are similar except for the ramp which is of 10 pixels for the left panel and of 2 pixels on the right.  | coreg_shift | wFT_mean | wFT2_mean | center
---|---|---|---|---
Object with ramp=2 | $1$ | $7.23$ | $9.92$ | $10$
Object with ramp=10 | $8$ | $9.15$ | $9.98$ | $10$
Table 1. Displacement in pixels computed for the motion applied on the two
objects of Figure 1

To illustrate the global displacement issue, two different 1D objects were
used to simulate a single motion occurring at the center of the k-space.
The results are shown in Figure 1. The object with a sharp ramp (right column)
had a larger spatial frequency content, thus motion in the center of k-space
disturbed less of the spatial frequencies of the object (the frequencies
outside the motion step in orange are not perturbed, first row). The global
shift observed was small (one pixel). For the object with a larger ramp (left
column), most of the frequency content of the object was disturbed by motion
(orange step, first row), resulting in a global shift of 8 pixels.
The main result was that for a given motion, the induced global displacement
varied depending on the original object spatial frequency content. This was
counter intuitive because one would have expected that a given movement would
move any object by the same amount regardless of its shape. Furthermore, all
numerical solutions as defined in section 2.2.2 and reported in Table 1
predicted a large displacement ($>7pixels$) for motion occurring in the center
of k-space.
With two specific examples, we therefore showed that the theoretical
predictions did not correspond to the co-registration ground truth. The 1D
case was not further explored as the arbitrary choice of the shape of the
object may strongly alter the results. In the next section, we explored the
global displacement in the case of 3D motion during the k-space acquisition of
a 3D brain MRI.

###  4.2. Transient motion during 3D brain acquisition

Figure 2. First plot in first row shows the motion time course
parametrisation: the displacement amplitude, the duration sigma and the
position of k-space xend. The reference position was set as the position at
the beginning of the k-space. In the three last plots of first row, the
predicted displacements were represented relative to the motion amplitude (in
%) as defined in section 2.2.2. Note that all different amplitude overlap in
those cases. In the second and third row we plotted the coreg_shift value
(computed by co-registration relative to input motion amplitude). We compared
three different translation directions X, Y, Z. (X and Z on the second row and
Y on the third raw). For all simulations the chosen phase encoding direction
was the Y-axis.

In this experiment, we studied the effect of transient motion by considering a
two state problem: the subject instantaneously moved from a reference position
to a second position for some time and then returned to the original position.
The motion time course was defined as a boxcar function. It was thus
characterized by four parameters: the displacement direction, the amplitude,
the duration (sigma) and the position in the k-space where the subject came
back to the initial reference position (xend) as defined in Figure 2. We
explored 456 different motion configurations as a combination of 4 amplitudes
([1, 2, 4, 8] mm or degree), 6 durations $sigma=[4,8,16,32,64,128]$ and 19
different xend positions from 128 to 384. The duration sigma and the position
xend were defined as the number of k-space points for a phase dimension of
512. The motion time course was then interpolated to the real phase dimension
of the image (218 in our case). A sigma of 4 and 128 represented a
perturbation of 0.8 % and 25 % of the total k-space, respectively. We applied
each motion to four different synthetic data consisting of simulations of two
distinct HCP subjects using two different contrasts (a T1-like contrast and a
random contrast).

Figure 3. Relative changes of the L1:
$\frac{L1_{wFT}–L1_{coreg\\_shift}}{L1_{coreg\\_shift}}$ (top line) and
$\frac{L1_{wFT2}–L1_{coreg\\_shift}}{L1_{coreg\\_shift}}$ (bottom line).
$L1_{wFT}$, $L1_{wFT2}$, $L1_{coreg\\_shift}$ are the L1 metrics after
simulating motion with different reference positions: wFT, wFT2, coreg_shift.
We simulated the same set of motion as in Figure 2, but only for a translation
along the X and Z directions.

In this experiment we choose always the same reference as the first time
point. Prediction of the correct reference is then equivalent to predict the
induced global displacement of the motion simulated volume. Figure 2 (first
row) shows the theoretical displacement as predicted by the different methods
(Cf section 2.2.2) for all motion time courses of this experiment. As the
displacement was by definition proportional to the amplitude of movement, the
predicted shift was represented as a percentage of the input movement
amplitude. It then becomes independent of the given amplitude. There was also
no dependence on the direction on which the movement occurred, since the
averages were computed independently on each Euler parameter. A small
dependence on subject shape and contrast was observed, as shown by the small
variance over the different simulations (100% confidence interval), this was
due to the weights derived from the Fourier Transform which depended on the
subject shape and the image contrast. The predicted displacement by the wFT2
and center methods were very similar and will be considered as equivalent for
the rest of this work (we will consider only wFT2).
Figure 2 second row, shows the measured coreg_shift in % (relative to the
input motion amplitude) for the X and Z translations directions, which had
similar effects. An important difference when comparing the measured shift to
theoretical predictions was the dependence on the amplitude of motion: the
1-mm amplitude looked similar to the wFT predictions, whereas the 8 mm
amplitude looked similar to wFT2 predictions. In both cases, there were
important differences for short duration motions (sigma ¡ 16) around k-space
center: a small shift was measured by the co-registration whereas wFT and wFT2
predicted a large one.
Figure 2 third row, shows the coreg_shift for a translation in the y
direction. The Y translation was very different compared to the X or Z
translations, specifically at the k-space center, where almost no shift was
measured. This was not due to a spatial specificity of the brain in the y
direction, but to the fact that the translation was co-linear to the phase
encoding direction. Indeed changing the phase encoding direction for the X
axis, then measuring the global shift for the X translation yields similar
results to the ones in Figure 2 third row (results not shown). This
demonstrated a strong interaction between the motion translation direction and
the phase encoding direction. To better understand this interaction, we refer
the reader to Appendix which shows that applying motion in k-space is
equivalent to applying a convolutional filter (Figure 6).
The results shown in Figure 2 demonstrated the discrepancy between the
observed global shift (obtained by direct coregistration) and the two
theoretical solutions wFT and wFT2. Next, we showed how the choice of a
reference influenced the severity of artifacts measured by the L1 difference
metric. Motion shown in Figure 2 was simulated for three different reference
positions (wFT, wFT2 or coreg_shift) and the L1 metric was computed for each
case. Figure 3 shows that wFT and wFT2 references lead to higher L1 changes
compared to coreg_shift. We plotted the relative changes
$\frac{L1_{wFT}–L1{coreg\\_shift}}{L1_{coreg\\_shift}}$ for motion translation
in the X and Z directions (and the same metrics for wFT2). Positive values
were expected as the co-registration algorithm minimized the L1 metric (even
though the mutual information was used as cost function) as shown in Figure 3.
We observed larger L1 errors for wFT when applied to large translation
amplitudes and for wFT2 when applied to small translation amplitudes. Using
wFT or WFT2 as reference lead to an overestimation of the severity of the
artifact measured by an L1 metric due to global displacement.
To summarize, we have shown that wFT or wFT2 predicted a large global
displacement for a motion occurring around the k-space center, in contrast to
the coregistration that showed no or small displacements for a short
transition ($\leq 16$). These differences induced an overestimation of the L1
severity. It might have been amplified by the simplicity of the motions
simulated. We therefore conducted a last experiment using more realistic
random motions.

###  4.3. Random motion during 3D brain acquisition

Figure 4. Scatter plots of L1 when choosing wFT as reference on the left and
wFT2 on the right, compared with L1 measured by co-registration. 50 random
motions of four amplitude levels, on 12 different subjects using 2 types of
contrasts were simulated.

To further confirm the importance of choosing the correct method for computing
the reference position, we proposed to study more realistic motion. We
simulated motion as a combination of perlin noise, swallow motions and sudden
displacements as explained in section 2.1.2. We compared the L1 values given
by the two reference positions: either wFT or wFT2 versus coreg_shift. As in
the previous experiment (Figure 3), we observed larger differences for the
wFT2 reference. The wFT method correlated more with the co-registration, but
some differences remained: for instance a given motion could lead to
$L1_{coreg\\_shift}=4$ but $L1_{wFT}=6$ and $L1_{wFT2}=8$.
To illustrate how the $L1_{wFT2}$ ranking of motion severity can be
misleading, we plotted on Figure 5, two motion simulated volumes with the same
level of artifact severity ($L1_{wFT2}$ 10) but very different
$L1_{coreg\\_shift}$ scores. The $L1_{coreg\\_shift}$ score was more
representative of the visual severity. The discrepancy observed for the volume
on the right is due to a mis-coregistration induced by choosing the wFT2
method as a reference. It is worth to note the swallow peak occurring at the
k-space center on the randomly simulated time course. This short transient
motion around k-space center was the reason why choosing the wFT2 reference
(and to a lower extent the wFT reference) induced the global displacement
instead of correcting for it.

Figure 5. Example of two simulated motion artifacts with the corresponding
time courses (amplitude of 8 mm and 8°). Both images have similar L1 when
computed with the wFT2 reference. But after co-registration the L1 scores were
really different. $L1_{coreg\\_shift}$ was more coherent with the visual
inspection of both images: the left one looks much more corrupted.

##  5\. Discussion

All the experiments demonstrated that the global displacement could not always
be predicted by the theoretical solutions proposed in the literature (wFT or
center). Furthermore using those theoretical references may lead to large
overestimation of voxel-to-voxel metrics such as the L1. More specifically we
showed that short transient motions around k-space center did not induce a
global displacement (only a contrast change). This implied that the reference
position estimated with wFT and wFT2 may induce a global shift instead of
correcting for it, because their weights are maximum at k-space center. It
will then lead to a large overestimation of the L1 metric as demonstrated in
Figure 3 and 5. in Figure 3, short transitions are for a duration sigma from 4
to 16. For a typical MPRAGE acquisition time of 8 minutes this corresponds to
a motion duration of 3.7 to 15 seconds which is long enough to be expected in
practice.
An even stronger argument is to consider the interaction between the
translation direction and the phase encoding direction. When both are aligned,
the convolution motion filter that apply to the image does not show two
separate lines (Figure 6) compared to the other two translation directions.
This is why less spatial mixing and less global displacement are observed in
figure 2 third row. Since theoretical prediction of global displacement are,
by definition, independent of the translation direction, they must be
erroneous.
Whether or not including an extra co-registration step after the motion
simulation will change the deep learning performance is out the scope of this
paper, but it will certainly depend on the chosen motion parameters. As we
could not derive a proper theoretical or analytical solution we strongly
advise to add an extra co-registration step after the simulation when using
translation sensitive measures. Note that this is unfortunate, because it
increase the computational demand of the motion simulation, and add some noise
regard to the exact reference.
In this work we considered the Elastix co-registration algorithm as the ground
truth to quantify the global displacement. However, as the co-registration
results from an optimization method there may be some numerical errors or
local minima convergence issues. We visually performed a quality control of
different results and did not find any incoherence in our experiments. Even
for an amplitude of 1 mm (very small shifts), a visual control of all motions
from experience 2 showed an effective global motion which was no more visible
after the co-registration. When specific translation directions were studied,
the amplitude of the shift in the other directions (which should be zero) were
inspected as well and an error of less than 4 % of the initial amplitude was
found. We were then confident that co-registration did not stop in local
optimum.
In experiment 2, we focused on an over simplified problem: a two position
motion. From a theoretical point of view, we expected the reference position
to be either position 1 or position 2, depending on how long the object was in
each respective position during acquisition. This behavior was more in line
with the wFT2 (or center) predictions. But the results for small amplitudes
that showed a global displacement in between the two positions were more
puzzling. A possible interpretation was that for short amplitudes, the motion
artifact induced mainly a local blurring which lead to an in between position
in the image domain. However, for larger amplitudes, spatially distinct parts
of the object were mixed and the solution was then more binary. With smaller
amplitudes, the observed global displacement was more in line with the wFT
prediction, except again for short transitions.
The induced global displacement may not be an issue if no difference metric is
used. This is indeed the case in the autofocusing litterature, where a global
metric was derived from the images and the motion was inferred by minimizing
this metric (Atkinson et al.,, 1997). Since no comparison was done with the
original image, the global displacement did not matter in this case.

##  6\. Conclusion

When dealing with motion artifact simulation, one often needs a measure of the
artifact severity which is classically achieved by a voxel-to-voxel difference
of the images with and without artifact. In this case it is of particular
importance to properly choose the reference position of the motion time course
in hand, to avoid any mis-co-registration between the two images.
Theoretical solutions proposed in the literature used the object position
around the center of the k-space as a reference: either the position at kpace
center ( wFT2), or a weighted average (wFT). We showed here that this
assumption no longer holds for short transient motion at k-space center, where
no global displacement was observed.
We also showed a strong dependence of the problem to the amplitude and the
direction of the motion which makes a theoretical solution more challenging.
It is quite unfortunate not to find any theoretical or analytical solution
yet, as this constrains to add a co-registration step which implies an extra
computational cost and numerical imprecision.

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## Annex: technical notes : motion simulation in MRI

#### Numerical formulation of Motion Simulation

Demonstration of accurate motion simulation start from signal equation :

(1)
$S(\overrightarrow{k_{0}})=\int_{V}\mathrm{d}\overrightarrow{r_{0}}\rho(\overrightarrow{r_{0}}){\rm
e}^{\textrm{i}\overrightarrow{r_{0}}^{t}\overrightarrow{k_{0}}}$

Using the following homogeneous coordinate
$\displaystyle\mathbf{r}_{0}=(\overrightarrow{r_{0}},1)=(r_{x}^{0},r_{y}^{0},r_{z}^{0},1)$
and
$\displaystyle\mathbf{k}_{0}=(\overrightarrow{k_{0}},0)=(k_{x}^{0},k_{y}^{0},k_{z}^{0},0)$,
the signal equation can be written :

(2)
$S(\mathbf{k}_{0})=\int_{V}\mathrm{d}\overrightarrow{r_{0}}\rho(\mathbf{r}_{0}){\rm
e}^{\textrm{i}\mathbf{r}_{0}^{t}\mathbf{k}_{0}}$

With homogenous coordinate, motion can be express easily trough use of affine
transformation $\displaystyle A$, which express as :

(3) $A(t)=\left(\begin{array}[]{cc}R(t)&T(t)\\\ 0&1\end{array}\right),$

with $\displaystyle R(t)$ the rotational compound of motion and $\displaystyle
T(t)$ the translational compound and $\displaystyle t$ express the time
dependance of motion. The inverse of the affine express as :

(4) $A^{-1}(t)=\left(\begin{array}[]{cc}R^{t}(t)&-R^{t}(t)T(t)\\\
0&1\end{array}\right)$

Effect of motion on the object $\displaystyle\rho$ can be expressed trough
action of affine transform as
$\displaystyle\rho(A^{-1}(\overrightarrow{k_{0}})\mathbf{r_{0}})$. The
measured signal is thus :

(5)
$S(\mathbf{k}_{0})=\int_{V}\mathrm{d}\overrightarrow{r_{0}}\rho(A^{-1}(\overrightarrow{k_{0}})\mathbf{r}_{0}){\rm
e}^{\textrm{i}\mathbf{r}_{0}^{t}\mathbf{k}_{0}}$

The exponential part is not change as this corresponds to laboratory
coordinate, only the object $\displaystyle\rho$ position change : this is the
active point of view. Numerical implementation of motion simulation in the
active point of view is numerically tedious as the rotated/translated object
need to be computed for each k-space points. This amount of computation can be
reduced while using the passive point of view. Notes that passive point of
view can be expressed in two ways : a first on using NuFFT type 2 (type 2 in
the sense from uniform space to a dual non uniform space) from image k-space
followed by inverse FFT and a second on that use FFT and then a NuFFT type 1
(type 1 in the sense from non uniform dual space to uniform space) from
k-space to image. The first step to express this view point is to use the
change of coordiante $\displaystyle\mathbf{r_{1}}=A^{-1}\mathbf{r_{0}}$ :

(6) $\displaystyle\displaystyle S(\mathbf{k_{0}})$
$\displaystyle\displaystyle=\int_{V}\mathrm{d}\overrightarrow{r_{0}}\rho(\mathbf{r}_{1}){\rm
e}^{\textrm{i}(A(\overrightarrow{k_{0}})\mathbf{r}_{1})^{t}\mathbf{k}_{0}}$
(7)
$\displaystyle\displaystyle=\int_{V}\mathrm{d}\overrightarrow{r_{0}}\rho(\mathbf{r}_{1}){\rm
e}^{\textrm{i}\mathbf{r}_{1}^{t}A^{t}(\overrightarrow{k_{0}})\mathbf{k}_{0}}$
(8)
$\displaystyle\displaystyle=\int_{V}\mathrm{d}\overrightarrow{r_{1}}\left|J(\overrightarrow{r_{1}})\right|\rho(\mathbf{r}_{1}){\rm
e}^{\textrm{i}\mathbf{r}_{1}^{t}A^{t}(\overrightarrow{k_{0}})\mathbf{k}_{0}}$

The Jacobian $\displaystyle J(\overrightarrow{r_{1}})$ is the identity matrix
since $\displaystyle A$ depends only on $\displaystyle\overrightarrow{k_{0}}$
and not on $\displaystyle\overrightarrow{r_{0}}$. The determinant
$\displaystyle\left|J(\overrightarrow{r_{1}})\right|$ is one and the signal
equation simplify to :

(9)
$S(\mathbf{k_{0}})=\int_{V}\mathrm{d}\overrightarrow{r_{1}}\rho(\mathbf{r}_{1}){\rm
e}^{\textrm{i}\mathbf{r}_{1}^{t}A^{t}(\overrightarrow{k_{0}})\mathbf{k}_{0}}$

Expanding the term
$\displaystyle\mathbf{r}_{1}^{t}A^{t}(\overrightarrow{k_{0}})\mathbf{k}_{0}$
lead to :
$\displaystyle\overrightarrow{x_{1}}R^{t}(\overrightarrow{k_{0}})\overrightarrow{k_{0}}+T^{t}(\overrightarrow{k_{0}})\overrightarrow{k_{0}}$.
The signal equation can now be expressed as :

(10) $S(\overrightarrow{k_{0}})={\rm
e}^{\textrm{i}\overrightarrow{T}^{t}(\overrightarrow{k_{0}})\overrightarrow{k_{0}}}\int_{V}\mathrm{d}\overrightarrow{r_{1}}\rho(\overrightarrow{r_{1}}){\rm
e}^{\textrm{i}\overrightarrow{r_{1}}^{t}\left(R^{t}\overrightarrow{k_{0}}\right)}$

In the equation above, the term
$\displaystyle\left(R^{t}\overrightarrow{k_{0}}\right)$ corresponds to
inhomogeneous grid in discrete case, thus the integral correspond to a non
uniform Fourier transform of type 2 (e.g. from uniform to non uniform
coordinate). Numerically the implementation is a NuFFT algorithm followed by
application of a linear phases. The motion corrupted image is then obtained
via an inverse Fourier transform of the signal. Note that a second approach
could be derived trough the change of coordinate
$\displaystyle\mathbf{k}_{1}=A^{t}(\overrightarrow{k_{0}})\mathbf{k}_{0}$ This
will lead to a nufft of type 1 but in this case the Jacobian is no more the
identiy matrix and it is difficult to derive an analytic solution.

Figure 6. The first row shows the simulated motion corrupted volume for 3
similar input motion time courses: a transient motion at the kspace center
($\displaystyle amplitude=4$, $\displaystyle sigma=10$). Each columns
corresponds to a different translation direction, from left to right the X, Y
and Z direction. In the second row we plotted the corresponding motion filter
G as described in equation 13. The slowest phase encoding direction is the Y
axis.

##  7\. Effect of translation

In case of pure translation motion, the effect on the resulting image can be
expressed in image domain as a convolution with a motion filter. Starting from
equation 10, the assumption of pure translation lead to $\displaystyle
R(\overrightarrow{r_{1}})=\textrm{1}\\!\\!\\!\\!1$ :

(11) $\displaystyle\displaystyle S(\overrightarrow{k_{0}})$
$\displaystyle\displaystyle={\rm
e}^{\textrm{i}\overrightarrow{T}^{t}(\overrightarrow{k_{0}})\overrightarrow{k_{0}}}\int_{V}\mathrm{d}\overrightarrow{r_{1}}\rho(\overrightarrow{r_{1}}){\rm
e}^{\textrm{i}\overrightarrow{r_{1}}^{t}\overrightarrow{k_{0}}}$ (12)
$\displaystyle\displaystyle={\rm
e}^{\textrm{i}\overrightarrow{T}^{t}(\overrightarrow{k_{0}})\overrightarrow{k_{0}}}\operatorname{FT}\left[\rho(\overrightarrow{r_{1}})\right]$

Doing Fourier transform of this signal and using the convolution properties of
Fourier transform we obtain :

(13)
$\tilde{\rho}(\overrightarrow{r_{1}})=\rho(\overrightarrow{r_{1}})\otimes\operatorname{FT^{-1}}\left[{\rm
e}^{\textrm{i}\overrightarrow{T}^{t}(\overrightarrow{k_{0}})\overrightarrow{k_{0}}}\right]=\rho(\overrightarrow{r_{1}})\otimes
G.$

Thus in image domain, in case of pure translation, motion acts as a filter
$\displaystyle G$ convolving the original images. Plotting this convolutional
filter helps to understand why the global displacement is different when the
translation direction is the same as the slowest varying phase encoding
direction. In the experiment reported in Figure 6, we considered a transient
motion occurring around kspace center with an amplitude of 4 and a duration of
10 and we only change the translation direction (X Y and Z). The main
observation is that for X and Z direction there is a clear spatial mixing due
to the double ray of the filter, whereas only one ray is observed when
translation is the same as the phase encoding direction (Y axis)