# On traceable iterated line graph and hamiltonian path index

Zhaohong Niu , Liming Xiong ,Weihua Yang School of Mathematical Sciences,
Shanxi University, Taiyuan, Shanxi, 030006, P.R. China. E-mail address:
<EMAIL_ADDRESS>Research is supported by the Natural Science Funds for Young
Scientists of China (No. 11501341 and 11701349).School of Mathematics and
Statistics, Beijing Institute of Technology, Beijing, 100081, P.R. China.
Research is supported by the Natural Science Funds of China (No.
11471037).Department of Mathematics, Taiyuan University of Technology,
Taiyuan, Shanxi, 030024, P.R. China. Research is supported by the Natural
Science Funds of China (No. 11671296).

###### Abstract

Xiong and Liu [L. Xiong and Z. Liu, Hamiltonian iterated line graphs, Discrete
Math. 256 (2002) 407-422] gave a characterization of the graphs $G$ for which
the $n$-th iterated line graph $L^{n}(G)$ is hamiltonian, for $n\geq 2$. In
this paper, we study the existence of a hamiltonian path in $L^{n}(G)$, and
give a characterization of $G$ for which $L^{n}(G)$ has a hamiltonian path. As
applications, we use this characterization to give several upper bounds on the
hamiltonian path index of a graph.

Keywords: iterated line graph; traceable; hamiltonian index; hamiltonian path
index

##  1 Introduction

The graphs considered in this paper are finite, undirected and loopless.
Undefined notation and terminology will follow [2].

The line graph $L(G)$ of a graph $G$ has $E(G)$ as its vertex set and two
vertices are adjacent in $L(G)$ if and only if they are adjacent as edges in
$G$. A trail $T$ of $G$ is dominated if each edge of $G$ is incident with at
least one vertex of $T$.

###### Theorem 1

(Harary and Nash-Williams, [7]) Let $G$ be a connected graph with at least
three edges. Then $L(G)$ is hamiltonian if and only if $G$ has a dominating
closed trail.

Theorem 1 characterized those graphs $G$ (expecially, non-hamiltonian graphs)
for which $L(G)$ is hamiltonian. It is natural to believe that for most
graphs, after applying the line graph operation iteratively a finite number of
times, the resulting will be hamiltonian. Chartrand [4] showed this is indeed
the case. He considered the $n$-iterated line graph $L^{n}(G)$ of $G$, and
introduced the hamiltonian index of a graph, denoted by $h(G)$, i.e., the
minimum number $n$ such that $L^{n}(G)$ is hamiltonian. Here the $n$-iterated
line graph of a graph $G$ is defined to be $L(L^{n-1}(G))$, where $L^{1}(G)$
denotes the line graph $L(G)$ of $G$, and $L^{n-1}(G)$ is assumed to have a
nonempty edge set. Chartrand showed that for any graph $G$ other than a path,
the hamiltonian index of $G$ exists. Since then, many results of determing the
exact values or upper bounds of $h(G)$ for special graphs are proved, such as
Chartrand and Wall [5] for trees (other than paths) and connected graphs $G$
with $\delta(G)\geq 3$, Kapoor and Stewart [10] for $K_{2,n}\,(n\geq 3)$,
Xiong and Liu [16] for connected graphs using the split blocks and Catlin’s
reduction method, Han et al. [6] for 2-connected graph with
$\kappa(G)\geq\alpha(G)-t$ (where $t$ is a nonnegative integer). Ryjáček et
al. [13] showed that the problem to decide whether the hamiltonian index of a
given graph is less than or equal to a given constant is NP-complete.

Let $G$ be a graph and $H$ a subgraph of $G$, then $V(H)$ and $E(H)$ denote
the sets of vertices and edges of $H$, respectively. Define $V_{i}(H)=\\{v\in
V(H):d_{H}(v)=i\\}$ and $W(H)=V(H)\backslash V_{2}(H)$. A branch in $G$ is a
nontrivial path with ends in $W(G)$ and with internal vertices, if any, of
degree 2. We denote by $B(G)$ the set of branches of $G$. Define
$B_{1}(G)=\\{b\in B(G):V(b)\cap V_{1}(G)\neq\emptyset\\}$. The distance
between two subgraphs $H_{1}$ and $H_{2}$ of $G$, denoted by
$d_{G}(H_{1},H_{2})$, is $\min\\{d_{G}(v_{1},v_{2}):v_{1}\in V(H_{1})\text{
and }v_{2}\in V(H_{2})\\}$.

For a positive integer $k$, Xiong and Liu [16] used $EU_{k}(G)$ to denote the
set of subgraphs $H$ of a graph $G$ that satisfy the following conditions, and
then investigated the existence of a characterization of those graphs $G$ for
which $L^{n}(G)$ is hamiltonian, which was mentioned in [3], and proved
Theorem 2.

  * (I)

$d_{H}(x)\equiv 0\,(\\!\\!\\!\\!\mod 2)$ for every $x\in V(H)$;

  * (II)

$V_{0}(H)\subseteq\bigcup_{i=3}^{\Delta(G)}V_{i}(G)\subseteq V(H)$;

  * (III)

$d_{G}(H_{1},H-H_{1})\leq k-1$ for every subgraph $H_{1}$ of $H$;

  * (IV)

$|E(b)|\leq k+1$ for every branch $b\in B(G)$ with $E(b)\cap E(H)=\emptyset$;

  * (V)

$|E(b)|\leq k$ for every branch $b\in B_{1}(G)$.

###### Theorem 2

(Xiong and Liu, [16]) Let $G$ be a connected graph with at least three edges
and $n\geq 2$. Then $L^{n}(G)$ is hamiltonian if and only if
$EU_{n}(G)\neq\emptyset$.

With the help of Theorem 2, one can deduce $h(G)\leq k$ if it is convenient to
check that $EU_{k}(G)$ is nonempty. See [9, 15, 17] for more upper bounds of
$h(G)$.

The further research of hamiltonianity has been producing several related
branches. As a form of weakening, the existence of hamiltonian paths of a
graph $G$ or a line graph $L(G)$ is also got a lot of attention. A graph $G$
is traceable if it has a hamiltonian path. Momège [12] showed that a connceted
graph $G$ with $\sigma_{2}(G)\geq 2n/3$ and $K_{1,4}$-free is traceable. He
and Yang [8] proved that there exist at least
$\max\\{1,\lfloor\frac{\,1\,}{\,8\,}\delta(G)\rfloor-1\\}$ edge-disjoint
hamiltonian paths between any two vertices in a hamiltonian-connected line
graph $L(G)$. Xiong and Zong [18] discussed the traceability of line graphs,
and obtained a similar result as Theorem 1. Note that the following result is
also an immediately corollary of Proposition 2.2 in [11], in which Lai et al.
discussed the hamiltonian connectedness of $L(G)$.

###### Theorem 3

(Xiong and Zong, [18], Lai, Shao and Zhan, [11]) Let $G$ be a connected graph
with at least three edges. Then the line graph $L(G)$ is traceable if and only
if $G$ has a dominating trail.

In this paper, inspired by Theorems 1, 2 and 3, and the closed relationship
between hamiltonian graphs and traceable graphs, we study the existence of a
hamiltonian path in the $n$-iterated line graph $L^{n}(G)$, i.e., the
traceability of $L^{n}(G)$.

For a graph $G$, we use $O(G)$ to denote the set of odd degree vertices of
$G$. The condition (I) guarantees that each vertex in $V(H)$ has even degree,
i.e., $O(H)=\emptyset$. When concerning about the traceability of $L^{n}(G)$,
this is too strong. That is, we can allow that $H$ has at most two odd degree
vertices. Moreover, by the fact that $H$ may contain a branch $b\in B_{1}(G)$
or a part of $b$, we just need the condition (V) holds for every branch $b\in
B_{1}(G)$ with $E(b)\cap E(H)=\emptyset$.

Let $EUP_{k}(G)$ denote the set of subgraphs $H$ of a graph $G$ that satisfy
the following conditions (I)′ and (V)′, and the conditions (II)-(IV) in the
definition of $EU_{k}(G)$.

  * (I)′

$|O(H)|\leq 2$;

  * (V)′

$|E(b)|\leq k$ for every branch $b\in B_{1}(G)$ with $E(b)\cap
E(H)=\emptyset$.

Obviously, $EU_{k}(G)\subseteq EUP_{k}(G)$. Now we are ready to present our
main result, the proof of which will be postponed to Section 3.

###### Theorem 4

Let $G$ be a connected graph with at least three edges and $n\geq 2$. Then
$L^{n}(G)$ is traceable if and only if $EUP_{n}(G)\neq\emptyset$.

Theorem 4 doesn’t hold for $n=1$. For example, Fig. 1 shows a graph $G$ with
$EUP_{1}(G)=\emptyset$ while $L(G)$ is traceable. By the definition of
$EUP_{1}(G)$, any subgraph $H$ in $EUP_{1}(G)$ should be connected,
$V_{3}(G)=\\{v_{3},v_{6},v_{9},v_{12}\\}\subseteq V(H)$, and the 4 branches
$v_{1}v_{2}v_{3}$, $v_{3}v_{4}v_{5}v_{6}$, $v_{6}v_{7}v_{8}v_{9}$ and
$v_{9}v_{10}v_{11}$ belong to $H$. Then $|O(H)|\geq 4$, a contradiction to
(I)′. Thus, $EUP_{1}(G)=\emptyset$, but $L(G)$ is traceable by the fact that
$v_{1}v_{2}\cdots\,\\!v_{11}$ is a dominating trail of $G$ and by Theorem 3.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$$v_{7}$$v_{8}$$v_{9}$$v_{10}$$v_{11}$$v_{12}$

Fig. 1. A graph $G$ with $EUP_{1}(G)=\emptyset$ while $L(G)$ is traceable.

Moreover, we also examine the hamiltonian path index of a graph, denoted by
$h_{p}(G)$, i.e., the minimum number $n$ such that $L^{n}(G)$ is traceable.
Regard $h_{p}(G)=0$ if $G$ is traceable.

In Section 2, we will present more terminology and notation, and some
auxiliary results. Theorem 4 will be proved in Section 3. As applications, we
will give some upper bounds of the hamiltonian path index $h_{p}(G)$ in the
last section.

##  2 Auxiliary results

In this section, we present several auxiliary results, which will be used in
the proof of Theorem 4.

The multi-graph of order 2 with two edges will be called 2-cycle. Let $G$ be a
graph and $H$ a subgraph of $G$, then $\bar{E}(H)$ denotes the set of all
edges of $G$ that are incident with vertices of $H$. If $u\in V(H)$, then
$E_{H}(u)$ denotes the set of all edges of $H$ that are incident with $u$, and
$d_{H}(u)=|E_{H}(u)|$. A graph is called a circuit if it is connected and
every vertex has an even degree. Regard $K_{1}$ as a circuit.

For any subgraph $C$ of $L(G)$, by $S(G,C)$ we denote the collection of
circuits $H$ of $G$, such that $L(G[\bar{E}(H)])$ contains $C$, and $C$
contains all elements of $E(H)$. Here and throughout, $G[S]$ denotes the
subgraph of $G$ induced by $S$, where $S\subseteq V(G)$ or $S\subseteq E(G)$.

###### Lemma 5

(Xiong and Liu, [16]) Each of the following holds.

  * (1)

If $C$ is a cycle of $L(G)$ with $|E(C)|\geq 3$, then $S(G,C)$ is nonempty.

  * (2)

If $G$ has a circuit $H$ such that $\bar{E}(H)$ has at least three edges, then
$L(G)$ has a cycle $C$ with $V(C)=\bar{E}(H)$.

###### Lemma 6

(Beineke, [1]) $K_{1,3}$ is not an induced subgraph of the line graph of any
graph.

The following lemma indicates the relationship between a branch of $G$ and the
corresponding branch of $L(G)$.

###### Lemma 7

(Xiong and Liu, [16]) Let $b=u_{1}u_{2}\cdots u_{s}\,(s\geq 3)$ be a path of
$G$ and $e_{i}=u_{i}u_{i+1}$. Then $b\in B(G)$ if and only if
$b^{\prime}=e_{1}e_{2}\cdots e_{s-1}\in B(L(G))$.

###### Lemma 8

Let $H$ be a subgraph of $G$ in $EUP_{k}(G)$ with a minimum number of
components. Then there exists no multiple edges in
$\bar{E}(H_{1})\cap\bar{E}(H_{2})$ for any two components $H_{1}$ and $H_{2}$
of $H$.

A similar result as Lemma 8 was proved by Xiong and Liu [16] for $H\in
EU_{k}(G)$. Then arguing similarly, one can obtain Lemma 8. Hence, we omit the
details here. An eulerian subgraph of $G$ is a circuit which contains at least
one cycle of length at least 3.

###### Lemma 9

(Xiong and Liu, [16]) Let $G$ be a connected graph and $C$ be an eulerian
subgraph of the line graph $L(G)$. Then there exists a subgraph $H$ of $G$
with

  * (1)

$d_{H}(x)\equiv 0\,(\\!\\!\\!\mod 2)$ for every $x\in V(H)$;

  * (2)

$d_{G}(x)\geq 3$ for every $x\in V(H)$ with $d_{H}(x)=0$;

  * (3)

for any two components $H^{0},H^{00}$ of $H$, there exists a sequence of
components $H^{0}=H_{1},H_{2},\dots,H_{s}=H^{00}$ of $H$ such that
$d_{G}(H_{i},H_{i+1})\leq 1$ for $i\in\\{1,2,\dots,s-1\\}$;

  * (4)

$L(G[\bar{E}(H)])$ contains $C$, and $C$ contains all elements of $E(H)$.

###### Lemma 10

Each of the following holds.

  * (1)

If $P$ is a non-trivial path of $L(G)$, then $G$ has a trail $T^{\prime}$,
such that $L(G[\bar{E}(T^{\prime})])$ contains $P$, and $P$ contains all
elements of $E(T^{\prime})$.

  * (2)

If $G$ has a connected subgraph $H$ such that $|O(H)|=2$, then $L(G)$ has a
path $P$ with $V(P)=\bar{E}(H)$.

Proof. (1) The proof just need a slight modification of the proof of Theorem 1
in [7]. So we omit the details here.

(2) Suppose $O(H)=\\{u,v\\}$. If $|\bar{E}(H)|=1$, then $G\cong H\cong K_{2}$,
(2) holds trivially. So we may assume that $\bar{E}(H)$ has at least two
edges. Let $e^{*}=uv$ be a new edge, which doesn’t belong to $E(G)$. Note that
if $uv\in E(G)$, then $e^{*}$ and $uv\,(\in E(G))$ are multiple edges in
$G+e^{*}$. Hence, $H+e^{*}$ is a circuit of $G+e^{*}$ such that
$\bar{E}(H+e^{*})$ has at least three edges. By Lemma 5 (2), $L(G+e^{*})$ has
a cycle $C$ with $V(C)=\bar{E}(H+e^{*})$. Let $P=C-v_{e^{*}}$, where
$v_{e^{*}}$ is the vertex in $L(G+e^{*})$ corresponding to the edge $e^{*}$ in
$G$. Note that $\bar{E}(H)\cup\\{e^{*}\\}=\bar{E}(H+e^{*})$. $P$ is a path of
$L(G)$ with $V(P)=\bar{E}(H)$.

So Lemma 10 holds.$\qquad\square$

In [16], Theorem 2 was derived from the following two results by induction.
The first one indicates a close relationship between $EU_{k}(L(G))$ and
$EU_{k+1}(G)$, and the second one characterizes graphs with 2-iterated line
graphs are hamiltonian.

###### Theorem 11

(Xiong and Liu, [16]) Let $G$ be a connected graph and $k\geq 1$ be an
integer. Then $EU_{k}(L(G))\neq\emptyset$ if and only if
$EU_{k+1}(G)\neq\emptyset$.

###### Theorem 12

(Xiong and Liu, [16]) Let $G$ be a connected graph with at least three edges.
Then $L^{2}(G)$ is hamiltonian if and only if $EU_{2}(G)\neq\emptyset$.

##  3 Proof of Theorem 4

In this section, we will prove Theorem 4, which is a direct consequence of the
following two theorems. The symmetric difference of two non-empty sets $A$ and
$B$, denoted by $A\Delta B$, is the set $(A\cup B)\backslash(A\cap B)$.

###### Theorem 13

Let $G$ be a connected graph and $k\geq 1$ be an integer. Then
$EUP_{k}(L(G))\neq\emptyset$ if and only if $EUP_{k+1}(G)\neq\emptyset$.

Proof. Sufficiency. Supposing that $EUP_{k+1}(G)\neq\emptyset$. Note that if
$EU_{k+1}(G)\neq\emptyset$, then by Theorem 11, $EU_{k}(L(G))\neq\emptyset$,
and hence, $EUP_{k}(L(G))\neq\emptyset$ by the fact that $EU_{k}(G)\subseteq
EUP_{k}(G)$. So we may assume that $EU_{k+1}(G)=\emptyset$, which implies that
each subgraph of $G$ in $EUP_{k+1}(G)$ contains exactly two odd vertices.

Now let $H\in EUP_{k+1}(G)$ with a minimum number of components denoted by
$C_{1},C_{2},\dots,C_{t}$. If $H$ is connected, then $H\cong C_{1}$. Without
loss of generality, we let $|O(C_{1})|=2$, and then $C_{i}$ is a circuit for
$2\leq i\leq t$.

Since $|O(C_{1})|=2$ and $G$ is connected, we have $|\bar{E}(C_{1})|\geq 2$:
for otherwise, $|V(G)|=|V(C_{1})|=2$, Theorem 13 holds obviously. Hence, by
Lemma 10 (2), $L(G)$ has a nontrivial path $P$ with $V(P)=\bar{E}(C_{1})$. Now
we claim that $|\bar{E}(C_{i})|\geq 3$ for $2\leq i\leq t$: if $C_{i}$ is
nontrivial, then we are done; if $C_{i}$ is an isolated vertex, then by the
definition of $EUP_{n}(G)$, $d_{G}(C_{i})\geq 3$, our claim holds. By Lemma 5
(2), we can find a cycle $C_{i}^{\prime}$ in $L(G)$ with
$V(C_{i}^{\prime})=\bar{E}(C_{i})\,(2\leq i\leq t)$. Let

$H^{\prime}=P\cup\big{(}\bigcup_{i=2}^{t}C_{i}^{\prime}\big{)}.$

We will prove that $H^{\prime}\in EUP_{k}(L(G))$.

By Lemma 8 and the minimality of $t$, $E(P)\cap E(C_{i}^{\prime})=\emptyset$
and $E(C_{i}^{\prime})\cap E(C_{j}^{\prime})=\emptyset$ for $2\leq i,j\leq t$
with $i\neq j$, which implies that $|O(H^{\prime})|=2$. (I)′ holds.

Since $P$ is nontrivial, and $V(C_{i}^{\prime})=\bar{E}(C_{i})\geq 3\,(2\leq
i\leq t)$, $H^{\prime}$ contains no isolated vertex. Note that
$\bigcup_{i=3}^{\Delta(G)}V_{i}(G)\subseteq V(H)$ and
$V(H^{\prime})=\bigcup_{i=1}^{t}\bar{E}(C_{i})$. We have

$\bigcup_{i=3}^{\Delta(L(G))}V_{i}(L(G))\subseteq V(H^{\prime}).$

Hence, $H^{\prime}$ satisfies (II).

The details of $H^{\prime}$ satisfying (III), (IV) and (V)′ are almost the
same as the proof of Theorem 11 in [16], so we omit them here.

It follows that $H^{\prime}\in EUP_{k}(L(G))$.

Necessity. Supposing that $EUP_{k}(L(G))\neq\emptyset$. Note that if
$EU_{k}(L(G))\neq\emptyset$, then by Theorem 11, $EU_{k+1}(G)\neq\emptyset$,
and hence, $EUP_{k+1}(G)\neq\emptyset$. So we may assume that
$EU_{k}(L(G))=\emptyset$, which implies that each subgraph of $L(G)$ in
$EUP_{k}(L(G))$ contains exactly two odd vertices.

Let $H$ be a subgraph of $L(G)$ in $EUP_{k}(L(G))$ with a minimum number of
isolated vertices. Then $H$ contains no isolated vertices. For otherwise,
suppose $C_{1}=\\{e_{0}\\}$ is an isolated vertex of $H$, then by (II),
$d_{L(G)}(e_{0})\geq 3$. By Lemma 6, there exist $e_{1},e_{2}\in
N_{L(G)}(e_{0})$ such that $e_{1}e_{2}\in E(L(G))$. Now we construct a
subgraph $H_{0}$ of $L(G)$ as follows.

$H_{0}=\left\\{\begin{array}[]{ll}H+\\{e_{0}e_{1},e_{1}e_{2},e_{2}e_{0}\\}&\text{
if }e_{1}e_{2}\notin E(H),\\\
H+\\{e_{0}e_{1},e_{2}e_{0}\\}-\\{e_{1}e_{2}\\}&\text{ if }e_{1}e_{2}\in
E(H).\end{array}\right.$

Obviously $H_{0}\in EUP_{k}(L(G))$ has fewer isolated vertices than $H$ has, a
contradiction.

Let $H_{1},H_{2},\dots,H_{m}$ be the components of $H$, and without loss of
generality, let $|O(H_{1})|=2$. Since $H$ has no isolated vertices, $H_{i}$ is
an eulerian subgraph of $L(G)$ for $2\leq i\leq m$. Then $H_{1}$ can be
decomposed into a nontrivial path $P$ and several eulerian subgraphs. Let
$P,H_{1}^{1},H_{1}^{2},\dots,H_{1}^{q}$ be such a decomposition with $q$
minimized. Then $V(H_{1}^{i})\cap V(H_{1}^{j})=\emptyset$ for
$\\{i,j\\}\subseteq\\{1,2,\dots,q\\}$ with $i\neq j$.

For the path $P$, by Lemma 10 (1), $G$ has a trail $T^{\prime}$, such that
$L(G[\bar{E}(T^{\prime})])$ contains $P$, and $P$ contains all elements of
$E(T^{\prime})$. For any eulerian subgraph $H_{1}^{j}$ ($1\leq j\leq q$) or
$H_{i}$ ($i\in\\{2,3,\dots,m\\}$), by Lemma 9, there exists a subgraph
$C_{1}^{j}$ or $C_{i}$ of $G$, respectively, satisfying (1) to (4) of Lemma 9.
Let

$C=\big{(}\bigcup_{i=3}^{\Delta(G)}V_{i}(G)\big{)}\cup\big{(}T\Delta\big{(}\bigcup_{j=1}^{q}C_{1}^{j}\big{)}\big{)}\cup\big{(}\bigcup_{i=2}^{m}C_{i}\big{)},$

where $T\Delta(\bigcup_{j=1}^{q}C_{1}^{j})$ is the subgraph of $G$ with vertex
set $V(T\cup(\bigcup_{j=1}^{q}C_{1}^{j}))$ and edge set $E(T)\Delta
E(\bigcup_{j=1}^{q}C_{1}^{j})$.

We will prove that $C\in EUP_{k+1}(G)$.

Since $V(H_{i})\cap V(H_{j})=\emptyset$ for
$\\{i,j\\}\subseteq\\{1,2,\dots,m\\}$ with $i\neq j$, $V(H_{1}^{j})\subseteq
V(H_{1})$ ($1\leq j\leq q$), and $V(H_{1}^{i})\cap V(H_{1}^{j})=\emptyset$ for
$\\{i,j\\}\subseteq\\{1,2,\dots,q\\}$ with $i\neq j$, we have $E(C_{i})\cap
E(C_{j})=\emptyset$, $E(C_{1}^{i})\cap E(C_{1}^{j})=\emptyset$, and
$E(C_{i})\cap E(C_{1}^{j})=\emptyset$. It follows that $d_{C}(x)\equiv
0\,(\\!\\!\\!\\!\mod 2)$ for every $x\in V(C)$ excepting the end-vertices of
the path $T$, which implies that $C$ satisfies (I)′. Since $C_{i}$ and
$C_{1}^{j}$ satisfy Lemma 9 (2), $d_{G}(x)\geq 3$ for every $x\in V(C)$ with
$d_{C}(x)=0$. Thus, (II) holds.

Arguing similarly as the proof of Theorem 11, $C$ satisfies (III), (IV) and
(V)′.

It follows that $C\in EUP_{k+1}(G)$.

This completes the proof of Theorem 13.$\qquad\square$

###### Theorem 14

Let $G$ be a connected graph with at least three edges. Then $L^{2}(G)$ is
traceable if and only if $EUP_{2}(G)\neq\emptyset$.

Proof. Sufficiency. Supposing that $EUP_{2}(G)\neq\emptyset$. Note that if
$EU_{2}(G)\neq\emptyset$, then by Theorem 11, $L^{2}(G)$ is hamiltonian, we
are done. So we may assume that $EU_{2}(G)=\emptyset$, which implies that each
subgraph of $G$ in $EUP_{2}(G)$ contains exactly two odd vertices.

We choose an $H\in EUP_{2}(G)$ with a minimum number of components that are
denoted by $H_{1},H_{2},\dots,H_{t}$, and assume that $|O(H_{1})|=2$. Since
$H\in EUP_{2}(G)$ and $|E(G)|\geq 3$, we have $|\bar{E}(H_{1})|\geq 2$, and
$|\bar{E}(H_{i})|\geq 3$ for $i\in\\{2,3,\dots,t\\}$. Then by Lemma 10 (2), we
can find a nontrivial path $P$ of $L(G)$ such that $V(P)=\bar{E}(H_{1})$. By
Lemma 5 (2), we can find a cycle $C_{i}$ of $L(G)$ such that
$V(C_{i})=\bar{E}(H_{i})$, $i\in\\{2,3,\dots,t\\}$. Let

$T=P\cup\big{(}\bigcup_{i=2}^{t}C_{i}\big{)}.$

By Lemma 8 and the minimality of $t$, $P,C_{2},C_{3},\dots,C_{t}$ are edge-
disjoint. Hence, $T$ is a subgraph of $L(G)$ with exactly 2 odd vertices.
Since $d_{G}(H^{\prime},H-H^{\prime})\leq 1$ for every subgraph $H^{\prime}$
of $H$, $T$ is connected.

Note that $H$ satisfies (II), $V(P)=\bar{E}(H_{1})$ and
$V(C_{i})=\bar{E}(H_{i})$ for $i\in\\{2,3,\dots,t\\}$. By the fact that any
edge in $G$, which corresponds to a vertex of degree at least 3 in $L(G)$,
must incident to a vertex of degree at least 3 in $G$,

$\bigcup_{i=3}^{\Delta(L(G))}V_{i}(L(G))\subseteq V(T).$

Since $H\in EUP_{2}(G)$, any branch $b\in B(L(G))$ with $E(b)\cap
E(C)=\emptyset$ has length at most 2, and any branch in $B_{1}(L(G))$ has
length at most 1. Then by Lemma 7, $\bar{E}(T)=E(L(G))$, which implies that
$T$ is a dominating trail of $L(G)$. Hence, $L^{2}(G)$ is traceable by Theorem
3.

Necessity. Supposing that $L^{2}(G)$ is traceable. By Theorem 3, $L(G)$ has a
dominating trail. Select a dominating trail $T$ of $L(G)$ with a maximum
number of vertices of degree at least 3.

Claim 1. $\bigcup_{i=3}^{\Delta(L(G))}V_{i}(L(G))\subseteq V(T)$.

The proof of Claim 14 is the same as the proof that $H$ has no isolated
vertices in Theorem 13, so we omit it here.

Then $T$ can be decomposed into a nontrivial path $P$ and several eulerian
subgraphs. Let $P,H_{1},H_{2},\dots,H_{q}$ be such a decomposition with $q$
minimized. Note that if $T$ is closed, then $T=\cup_{i=1}^{q}H_{i}$.

For the path $P$, by Lemma 10 (1), $G$ has a trail $T^{\prime}$, such that
$L(G[\bar{E}(T^{\prime})])$ contains $P$, and $P$ contains all elements of
$E(T^{\prime})$. For any eulerian subgraph $H_{i}$ ($1\leq i\leq q$), by Lemma
9, there exists a subgraph $C_{i}$ of $G$, satisfying (1) to (4).

Set

$H=\big{(}\bigcup_{i=3}^{\Delta(G)}V_{i}(G)\big{)}\cup\big{(}T^{\prime}\Delta\big{(}\bigcup_{i=1}^{q}C_{i}\big{)}\big{)},$

where $T^{\prime}\Delta(\bigcup_{i=1}^{q}C_{i})$ is the subgraph of $G$ with
vertex set $V(T^{\prime}\cup(\bigcup_{i=1}^{q}C_{i}))$ and edge set
$E(T^{\prime})\Delta E((\bigcup_{i=1}^{q}C_{i}))$. We will prove that $H\in
EUP_{2}(G)$. Before this, we present the following claim.

Claim 2. $d_{G}(x,H)\leq 1$ for any $x\in\bigcup_{i=3}^{\Delta(G)}V_{i}(G)$.

Proof of Claim 14. If $G$ is either a star or a cycle, then the conclusion
holds. For otherwise, then
$E_{G}(x)\cap(\cup_{i=3}^{\Delta(L(G))}V_{i}(L(G)))\neq\emptyset$ for every
vertex $x$ in $\cup_{i=3}^{\Delta(G)}V_{i}(G)$. Hence, by Claim 14, there
exists an edge $e_{x}$, which is incident to $x$ in $G$, has an endvertex in
$H$. Claim 14 holds.

Now we prove $H\in EUP_{2}(G)$. Obviously, $H$ satisfies (I)′. Since
$C_{i}\,(1\leq i\leq q)$ satisfies (2) of Lemma 9, and by the definition of
$H$, (II) holds. By Claim 14 and (3) of Lemma 9,
$d_{G}(H^{\prime},H-H^{\prime})\leq 1$ for every subgraph $H^{\prime}$ of $H$,
thus $H$ satisfies (III). It follows from Lemma 7 and $E(L(G))=\bar{E}(T)$
that $|E(b)|\leq 3$ for $b\in B(G)$ with $E(b)\cap E(H)=\emptyset$, and
$|E(b)|\leq 2$ for $b\in B_{1}(G)$ with $E(b)\cap E(H)=\emptyset$. $H$
satisfies (IV) and (V)′. Hence, $H\in EUP_{2}(G)$.

This completes the proof of Theorem 14.$\qquad\square$

Now we prove Theorem 4.

Proof of Theorem 4. We proceed by induction on $n$. Theorem 14 shows that
Theorem 4 holds for $n=2$.

Assume, as an inductive hypothesis, that the theorem is true for $n=k>2$,
i.e., $L^{k}(G)$ is traceable if and only if $EUP_{k}(G)\neq\emptyset$. Now
let $n=k+1$. Then $L^{k+1}(G)=L^{k}(L(G))$ is traceable if and only if
$EUP_{k}(L(G))\neq\emptyset$. Hence, by Theorem 13,
$EUP_{k}(L(G))\neq\emptyset$ if and only if $EUP_{k+1}(G)\neq\emptyset$.
Theorem 4 holds for $n=k+1$. Thus, the induction succeeds.$\qquad\square$

Note that Theorem 4 doesn’t hold for $n=1$ (see Section 1). Hence, when we
prove it by induction, the basis step is $n=2$ (Theorem 14).

##  4 Applications of Theorem 4

In this section, inspired by the massive upper bounds of $h(G)$, as
applications of Theorem 4, we will present some upper bounds on the
hamiltonian path index $h_{p}(G)$ of a graph $G$. The main idea is to show
that $EUP_{k}(G)\neq\emptyset$, and then by Theorem 4, $h_{p}(G)\leq k$.

Note that the hamiltonian index $h(G)$ exists for any connected graph $G$
other than a path and $h_{p}(G)\leq h(G)$. The hamiltonian path index
$h_{p}(G)$ exists for any connected graph.

Comparing the definition of hamiltonian with traceable, we know that being
hamiltonian is stronger than being traceable. Then one may believe that the
former needs less iterated steps, and hence, $h_{p}(G)<h(G)$. Unfortunately,
this is not true by the fact that $h_{p}(K_{1,n-1})=h(K_{1,n-1})=1$ ($n\geq
3$). Moreover, Fig. 2 shows a graph $G$ with $h_{p}(G)=h(G)=k$: one can check
that the unique cycle of $G$ is an element in $EU_{k}(G)$, but
$EUP_{k-1}(G)=\emptyset$ by the fact that any element in $EUP_{k-1}(G)$ cann’t
contain all the three pendent paths with length $k$. Hence, our trivial bound
$h_{p}(G)\leq h(G)$ is best possible.

a path oflength $k$a path of length $k$a path of length $k$

Fig. 2. A graph $G$ with $h_{p}(G)=h(G)=k$.

For a graph $G$, let $MT^{*}(G)$ be a trail of $G$ with the most number of
vertices, and in this sence, with the least number of vertices in
$\bigcup_{i=3}^{\Delta(G)}V_{i}(G)$. Denote $mt^{*}(G)=|V(MT^{*}(G))|$ and
$d_{\geq 3}^{*}(G)=|\bigcup_{i=3}^{\Delta(G)}V_{i}(G)\backslash
V(MT^{*}(G))|$.

###### Theorem 15

Let $G$ be a connected graph of order $n$. Then $h_{p}(G)\leq
n-mt^{*}(G)-d_{\geq 3}^{*}(G)+2$.

Proof. Since $G$ is connected, Theorem 15 holds for $|E(G)|<3$ trivially. So
we may assume that $|E(G)|\geq 3$.

Let $MT^{*}(G)$ be a trail of $G$ satisfying the hypotheses above. Denote
$k=n-mt^{*}(G)-d_{\geq 3}^{*}(G)+2$. Note that $k\geq 2$. By Theorem 4, it
suffices to prove that $EUP_{k}(G)\neq\emptyset$.

Let

$H=MT^{*}(G)\cup\big{(}\bigcup_{i=3}^{\Delta(G)}V_{i}(G)\big{)}.$

We will prove that $H\in EUP_{k}(G)$.

By the definition of $H$, $H$ satisfies (I)′ and (II). Note that
$|V(G)|-|V(H)|=n-(mt^{*}(G)+d_{\geq 3}^{*}(G))=k-2$. Then
$d_{G}(H_{1},H-H_{1})\leq k-1$ for every subgraph $H_{1}$ of $H$, $|E(b)|\leq
k-1\,(<k+1)$ for every branch $b\in B(G)$ with $E(b)\cap E(H)=\emptyset$, and
$|E(b)|\leq k-2\,(<k)$ for every branch $b\in B_{1}(G)$ with $E(b)\cap
E(H)=\emptyset$. Hence, $H$ satisfies (III), (IV) and (V)′. Theorem 15 holds.
$\qquad\square$

The bound of $h_{p}(G)$ in Theorem 15 is sharp. Fig. 3 shows a graph $G$ with
$h_{p}(G)=n-mt^{*}(G)-d_{\geq 3}^{*}(G)+2$, where $s$, $t$ are positive
integers and $t\geq s+5$ in the figure. Since $t\geq s+5$, $MT^{*}(G)$ is the
path $x_{1}x_{2}\cdots x_{t}w_{1}y_{t}\cdots y_{2}y_{1}$, and then
$mt^{*}(G)=2t+1$ and $d_{\geq 3}^{*}(G)=4$. Note that $n=|V(G)|=2t+s+5$. On
the one hand, by Theorem 15, we have $h_{p}(G)\leq n-mt^{*}(G)-d_{\geq
3}^{*}(G)+2=s+2$. On the other hand, we will explain $h_{p}(G)\geq s+2$ in the
following. For $k\leq s+1$, the $k$-th iterated line graph $L^{k}(G)$ is also
illustrated in Fig. 3, where the gray ellipse and triangle are the nontrivial
hamiltonian subgraph $S_{1}$ and $S_{2}$ of $L^{k}(G)$, respectively. Note
that $k\leq s+1$. We have $s-k+2\geq 1$, which means that either
$|V(S_{1})\cap V(S_{2})|=1$ (when $s-k+2=1$), or
$d_{L^{k}(G)}(S_{1},S_{2})\geq 1$ (when $s-k+2\geq 2$). In both cases,
$L^{k}(G)$ is not traceable. Hence, $h_{p}(G)\geq s+2$.

a path of $t$ verticesa path of $t$ verticesa path of$s$
vertices$x_{1}$$x_{2}$$x_{t}$$y_{1}$$y_{2}$$y_{t}$$z_{1}$$z_{s}$$w_{1}$$w_{2}$$w_{3}$$w_{4}$$w_{5}$$G$a
path of $t-k+1$ verticesa path of $t-k+1$ verticesa path of$s-k+2$
vertices$S_{1}$$S_{2}$$L^{k}(G)$ with $k\leq s+1$

Fig. 3. A graph $G$ with $h_{p}(G)=n-mt^{*}(G)-d_{\geq 3}^{*}(G)+2$

and its iterated line graph $L^{k}(G)$ with $k\leq s+1$.

By Theorem 15, we can obtain the following corollary.

###### Corollary 16

Let $G$ be a connected graph of order $n$. Then
$h_{p}(G)\leq\max\\{1,n-mt^{*}(G)\\}$.

Proof. Let $MT^{*}(G)$ be a maximum trail of $G$, and in this sence, with the
least number of vertices in $\bigcup_{i=3}^{\Delta(G)}V_{i}(G)$, and let
$k=n-mt^{*}(G)$. If $k\leq 1$, then $MT^{*}(G)$ is a dominating trail of $G$.
By Theorem 3, $L(G)$ is traceable. Hence, $h_{p}(G)\leq 1$. So we may assume
that $k\geq 2$. Now the proof is divided into three cases.

Case 1. $d_{\geq 3}^{*}(G)=0$.

Let $H=MT^{*}(G)$. We will prove that $H\in EUP_{k}(G)$, and then
$h_{p}(G)\leq k$. Obviously, $H$ satisfies (I)′, (II) and (III). Note that
$d_{\geq 3}^{*}(G)=0$. $G-H$ has exactly $k$ vertices. Then $|E(b)|\leq k+1$
for every branch $b\in B(G)$ with $E(b)\cap E(H)=\emptyset$, and $|E(b)|\leq
k$ for every branch $b\in B_{1}(G)$ with $E(b)\cap E(H)=\emptyset$. Hence, $H$
satisfies (IV) and (V)′.

Case 2. $d_{\geq 3}^{*}(G)=1$.

Let $v$ be the vertex of degree at least 3 in
$\bigcup_{i=3}^{\Delta(G)}V_{i}(G)\backslash V(MT^{*}(G))$. If $|N_{G}(v)|=1$,
then by $d_{\geq 3}^{*}(G)=1$, the neighbour of $v$ belongs to $MT^{*}(G)$.
Hence, the union of $MT^{*}(G)+v$ and two multiple edges incident to $v$ is a
longer trail than $MT^{*}(G)$, contrary to the maximality of $MT^{*}(G)$. Then
we may assume that $|N_{G}(v)|\geq 2$.

Let $H=MT^{*}(G)+v$. We will prove that $H\in EUP_{k}(G)$. Obviously, $H$
satisfies (I)′ and (II). Arguing similarly as the proof of Case 1, $H$
satisfies (IV) and (V)′. It remains to prove that $d_{G}(MT^{*}(G),v)\leq
k-1$. This holds by the fact that $G-H$ has exactly $k-1$ vertices,
$|N_{G}(v)|\geq 2$, and the shortest path between $v$ and $MT^{*}(G)$ contains
only one neighbour of $v$.

Case 3. $d_{\geq 3}^{*}(G)\geq 2$.

By Theorem 15, $h_{p}(G)\leq n-mt^{*}(G)-d_{\geq 3}^{*}(G)+2$. Then by
$d_{\geq 3}^{*}(G)\geq 2$, we have $h_{p}(G)\leq n-mt^{*}(G)$.

This completes the proof of Corollary 16. $\qquad\square$

By the sharpness of $h_{p}(G)\leq n-mt^{*}(G)-d_{\geq 3}^{*}(G)+2$ and the
proof of Corollary 16, we know that the bound
$h_{p}(G)\leq\max\\{1,n-mt^{*}(G)\\}$ is sharp when $d_{\geq 3}^{*}(G)\leq 2$.

The diameter of a graph $G$, denoted by $\mathrm{diam}(G)$, is the greatest
distance between two vertices of G. Note that $\mathrm{diam}(G)+1\leq
mt^{*}(G)$. The following corollary is obvious.

###### Corollary 17

Let $G$ be a connected graph of order $n$. Then
$h_{p}(G)\leq\max\\{1,n-\mathrm{diam}(G)-1\\}$.

For a graph $G$, let $d^{\prime}_{G}(v)=|N_{G}(v)|$ and
$\Delta^{\prime}(G)=\max\\{d^{\prime}_{G}(v)\,:\,v\in V(G)\\}$. Note that if
$G$ is simple, then $d^{\prime}_{G}(v)=d_{G}(v)$ and
$\Delta^{\prime}(G)=\Delta(G)$. Let $d_{\geq
3}^{**}(G)=\max\\{|(\bigcup_{i=3}^{\Delta(G)}V_{i}(G))\backslash
N_{G}(v)|\,:\,v\in V(G)\text{ and }|N_{G}(v)|=\Delta^{\prime}(G)\\}$. A cycle
$C$ of $G$ is called pendent if
$|V(C)\cap(\bigcup_{i=3}^{\Delta(G)}V_{i}(G))|=1$. See Fig. 4 (a) for
illustrations of pendent cycles. Let $PC(G)$ be the set of pendent cycles of
$G$.

###### Theorem 18

Let $G$ be a connected graph of order $n$. Then

$h_{p}(G)\leq\bigg{\lfloor}\frac{\,n-\Delta^{\prime}(G)-d_{\geq
3}^{**}(G)\,}{\,3\,}\bigg{\rfloor}+3.$

Proof. Let $k=\lfloor(n-\Delta^{\prime}(G)-d_{\geq 3}^{**}(G))/3\rfloor+3$,
and $v$ a vertex of $G$ with $|N_{G}(v)|=\Delta^{\prime}(G)$ and
$|(\bigcup_{i=3}^{\Delta(G)}V_{i}(G))\backslash N_{G}(v)|$ maximized. Note
that $k\geq 3$. By Theorem 4, it suffices to prove that
$EUP_{k}(G)\neq\emptyset$.

If $\Delta^{\prime}(G)\leq 2$, then $G$ is traceable. Hence, $h_{p}(G)=0$, and
the bound holds trivially. Now we may assume that $\Delta^{\prime}(G)\geq 3$.
Then $B(G)$, the set of branches of $G$, has at least two elements.

Let $b_{1}\in B(G)$ be a branch with $|V(b_{1})\cap(V_{1}(G)\cup V_{2}(G))|$
maximized, and $b_{2}\in B(G)\backslash\\{b_{1}\\}$ a branch with
$|V(b_{2})\cap(V_{1}(G)\cup V_{2}(G))|$ maximized. Since $G$ is connected, we
can find a trail $T_{1}$ (may be trivial) which connects $b_{1}$ and $b_{2}$.
By the fact that the internal vertices (if any) of $b_{1}$ and $b_{2}$ have
degree 2 in $G$, $T=T_{1}\cup b_{1}\cup b_{2}$ is a trail of $G$.

Let

$H=\big{(}\bigcup_{i\geq 3}^{\Delta(G)}V_{i}(G)\big{)}\cup T\cup PC(G).$

We will prove that $H\in EUP_{k}(G)$.

Obviously, $H$ satisfies (I)′ and (II). By the choice of $b_{1}$ and $b_{2}$,
each of the other branches of $G$ has at most
$\lfloor(n-\Delta^{\prime}(G)-d_{\geq 3}^{**}(G))/3\rfloor+1$ vertices in
$V_{1}(G)\cup V_{2}(G)$, where the $+1$ is necessary since each branch may
have at most one neighbour of $v$. Then
$|E(b)|\leq\lfloor(n-\Delta^{\prime}(G)-d_{\geq 3}^{**}(G))/3\rfloor+2=k-1$
for every branch $b\in B(G)\backslash\\{b_{1},b_{2}\\}$, and hence, (III),
(IV) and (V)′ hold.

This completes the proof of Theorem 18.$\qquad\square$

The bound of $h_{p}(G)$ in Theorem 18 is sharp. Fig. 4 (b) shows a graph $G$
with $n=3s+16$, $\Delta^{\prime}(G)=6$ and $d_{\geq 3}^{**}(G)=13$, where $s$
is a positive integer and the three gray cycles are induced $K_{4}$. Then by
Theorem 18, $h_{p}(G)\leq\lfloor(n-\Delta^{\prime}(G)-d_{\geq
3}^{**}(G))/3\rfloor+3=s+2$. Note that $G$ has 3 branches of length $s+1$.
$L^{k}(G)$ is not traceable when $k<s+2$. Then $h_{p}(G)\geq s+2$, and hence,
$h_{p}(G)=s+2$, which implies the sharpness of the upper bound in Theorem 18.

If $G$ is a connected simple graph of order $n$, then Theorem 18 implies that
$h_{p}(G)\leq\lfloor(n-\Delta(G)-d_{\geq 3}^{**}(G))/3\rfloor+3$.

(a)$K_{4}$$K_{4}$$K_{4}$$s$ vertices$s$ vertices$s$ vertices(b)

Fig. 4. (a) A graph with 2 pendent cycles; and (b) A graph $G$ with

$h_{p}(G)=\lfloor(n-\Delta^{\prime}(G)-d_{\geq 3}^{**}(G))/3\rfloor+3$.

In [14], Saraz̆in proved the following upper bound of $h(G)$.

###### Theorem 19

(Saraz̆in, [14]) Let $G$ be a connected simple graph of order $n$. If
$\Delta(G)\geq 3$, then $h(G)\leq n-\Delta(G)$.

The following corollary, obtained by Theorem 18 immediately, implies that when
considering $h_{p}(G)$, the upper bound in Theorem 19 can be improved
evidently.

###### Corollary 20

Let $G$ be a connected simple graph of order $n$. Then

$h_{p}(G)\leq\bigg{\lfloor}\frac{\,n-\Delta(G)\,}{\,3\,}\bigg{\rfloor}+3.$

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# Broadening the convergence domain of Seventh-order method satisfying
Lipschitz and Hölder conditions

Akanksha Saxena, J. P. Jaiswal, K. R. Pardasani

Abstract. In this paper, the local convergence analysis of the multi-step
seventh order method is presented for solving nonlinear equations assuming
that the first-order Fréchet derivative belongs to the Lipschitz class. The
significance of our work is that it avoids the standard practice of Taylor
expansion thereby, extends the applicability of the scheme by applying the
technique based on the first-order derivative only. Also, this study provides
radii of balls of convergence, the error bounds in terms of distances in
addition to the uniqueness of the solution. Furthermore, generalization of
this analysis satisfying Hölder continuity condition is provided since it is
more relaxed than Lipschitz continuity condition. We have considered some
numerical examples and computed the radii of the convergence balls.

Mathematics Subject Classification (MSC2020). 65H10, 65J15, 65G99, 47J25.

Keywords and Phrases. Nonlinear equation, Banach space, local convergence,
Lipschitz continuity condition, Hölder continuity condition.

##  1\. Introduction

One of the most crucial problems of numerical analysis concerns with finding
efficiently and precisely the approximate locally unique solution $x^{*}$ of
the nonlinear equation having the form of

$T(x)=0,$ (1.1)

where $T$ is defined on a convex open subset $D$ of a Banach space $X$ with
values in a Banach space $Y$. Many problems in various fields of computational
science, engineering and other disciplines involves optimization, differential
equations, integral equations, radiative transfer theory, can be brought in a
form like $(\ref{eqn:11})$ using mathematical modeling. Analytical methods of
solving such type of problems are very scarce or almost non existent.
Therefore, many researchers only rely on iterative methods and they have
proposed a plethora of iterative methods.

The practice of numerical functional analysis for finding such solutions are
widely and substantially connected to Newton-like methods which is defined as
follows:

$x_{n+1}=x_{n}-[T^{\prime}(x_{n})]^{-1}T(x_{n})\ \ \ n\geq 0,$ (1.2)

is frequently used by various researchers as it has quadratic convergence (can
be see in the ref. [1]). The other properties of Newton’s method are
established in the article [2]. Moreover, in some applications involving stiff
systems, high-order methods are useful. Therefore, it is important to study
high-order methods.

The local convergence analysis of iterative procedures is based on the
information around a solution, to find estimates of the radii of the
convergence balls. There exists many studies which deals with the local and
the semilocal convergence analysis of Newton-like methods. In the last
decades, many iterative methods of increasing order of convergence have been
developed and have shown their efficiency, in numerical terms, like third
order [[3], [4]], fourth order [5] and fifth order [[6], [7]] etc.

Another issue of equal importance is obtaining the radius of convergence ball
as well as developing a theory to extend the convergence domain. Numerous
authors have studied the local convergence analysis using Taylor’s series but
don’t obtain the radii of convergence ball for the solution which can be seen
in the refs. [[8],[9]]. In terms of computing cost, it has a number of
drawbacks. These types of techniques are discussed by many authors, for
example, one can see the refs. [[1], [10]]. Using considered approach, it is
possible to compute the convergence radii and the error $\|x_{n}-x^{*}\|$
estimates.

However, lower-order classical methods have been usually re-examined for only
trying to increase its order of convergence. In particular, Sharma and Gupta
[11] constructed three steps method of order five, defined as follows:

$\displaystyle y_{n}$ $\displaystyle=$ $\displaystyle
x_{n}-\frac{1}{2}\Gamma_{n}T(x_{n}),$ $\displaystyle z_{n}$ $\displaystyle=$
$\displaystyle x_{n}-[T^{\prime}(y_{n})]^{-1}T(x_{n}),$ $\displaystyle
x_{n+1}$ $\displaystyle=$ $\displaystyle
z_{n}-[2[T^{\prime}(y_{n})]^{-1}-\Gamma_{n}]T(z_{n}),$ (1.3)

where, $\Gamma_{n}=[T^{\prime}(x_{n})]^{-1}$. The local convergence of above
multi-step Homeier’s-like method has been studied by Panday and Jaiswal [12]
with the help of Lipschitz and H$\ddot{o}$lder continuity conditions. In the
extension of three-step Homeier’s method, Xiao and Yin [13] developed a
fourth-step seventh order convergent method, which is given as:

$\displaystyle y_{n}$ $\displaystyle=$ $\displaystyle
x_{n}-\frac{1}{2}\Gamma_{n}T(x_{n}),$ $\displaystyle z_{n}^{(1)}$
$\displaystyle=$ $\displaystyle x_{n}-[T^{\prime}(y_{n})]^{-1}T(x_{n}),$
$\displaystyle z_{n}^{(2)}$ $\displaystyle=$ $\displaystyle
z_{n}^{(1)}-[2[T^{\prime}(y_{n})]^{-1}-\Gamma_{n}]T(z_{n}^{(1)}),$
$\displaystyle x_{n+1}$ $\displaystyle=$ $\displaystyle
z_{n}^{(2)}-[2[T^{\prime}(y_{n})]^{-1}-\Gamma_{n}]T(z_{n}^{(2)}).$ (1.4)

This method requires the evaluation of three function, two first order
derivatives and two matrix inversions per iteration. In this article, we will
present the answers with the range of initial guess $x^{*}$ that tell us how
close the initial guess should be required for granted convergence of the
method $(\ref{eqn:13})$.

In finding out the existence of the solution, we require the computation of
the first-order Fréchet derivative. However, the analysis of convergence is
established by applying the Taylor series approach based on higher-order
derivatives. These techniques, which need the higher-order derivatives, limit
the applicability of the algorithm. As an motivational illustration[15],
define a function $f$ on $D=[-\frac{1}{2},\frac{5}{2}]$ by

$\displaystyle f(x)=\begin{cases}x^{3}log(x^{2})+x^{5}-x^{4},&\text{if \
$x\neq 0$}\\\ 0,&\text{if \ $x=0$}.\end{cases}$ (1.5)

It is clearly notable that $f^{\prime\prime\prime}$ is unbounded on $D$.
Hence, the theory based on higher-order derivatives [14] fail to solve the
above problem. Also, one get no idea about the domain of convergence [15]. The
local convergence study gives valuable information regarding the radius of
convergence ball.

In this presented paper, motivated by the foregoing observations and ongoing
work in this area, we discuss the local convergence for the method
$(\ref{eqn:13})$, by following the approach based on $f^{\prime}$ to stay away
from the evaluation of higher-order Fréchet derivatives. Thereby, we enlarge
the utility of the method $(\ref{eqn:13})$ by using hypotheses only on the
first-order derivative of the function $T$ and Lipschitz and Hölder continuity
conditions for which earlier studies can not be used due to the computation of
higher-order Fréchet derivatives.

The arrangement of the whole text is as follows: the ”Local convergence
analysis of the method $(\ref{eqn:13})$ under Lipschitz condition” section
deals with the local convergence results of the method $(\ref{eqn:13})$.
Similarly, the ”Local convergence analysis of the method $(\ref{eqn:13})$
under Hölder condition” section deals with the local convergence results of
the method $(\ref{eqn:13})$ followed by the ”Applications with Numerical
Examples” section. The ”Conclusions” section is placed in the last section.

##  2\. Local convergence analysis of the method $(\ref{eqn:13})$ under
Lipschitz condition

In this section, we describe the the local convergence analysis of the method
$(\ref{eqn:13})$ which is centered on some parameters and scalar functions.
Considering $\psi_{0}>0$ and $\psi>0$ be two parameters with
$\psi_{0}\leq\psi$, we define the functions
$\eta_{1},\eta_{2},\eta_{3},\eta_{4},p,H_{1},H_{2},H_{3}$ and $H_{4}$ on
interval $[0,\frac{1}{\psi_{0}})$ by

$\displaystyle\eta_{4}(a)$ $\displaystyle=$
$\displaystyle{\bigg{[}}\frac{\psi\eta_{3}(a)a}{2(1-\psi_{0}\eta_{3}(a)a)}$
(2.1)
$\displaystyle+\frac{1}{1-\psi_{0}\eta_{3}(a)a}\psi[\eta_{1}(a)+\eta_{3}(a)]\times\left(\frac{1}{1-p(a)}\bigg{[}1+\frac{\psi_{0}}{2}\eta_{3}(a)a\bigg{]}a\right)$
$\displaystyle+\left(\frac{1}{1-\psi_{0}a}.\frac{\psi[a+\eta_{1}(a)a]}{1-p(a)}\times\bigg{[}1+\frac{\psi_{0}}{2}\eta_{3}(a)a\bigg{]}\right){\bigg{]}}\eta_{3}(a),$

where

$\displaystyle\eta_{1}(a)$ $\displaystyle=$
$\displaystyle\frac{1}{1-\psi_{0}a}{\bigg{[}}\frac{\psi
a}{2}+\frac{1+(\psi_{0}/2)a}{2}{\bigg{]}},$ (2.2) $\displaystyle\eta_{2}(a)$
$\displaystyle=$ $\displaystyle\frac{1}{1-\psi_{0}a}\bigg{[}\frac{\psi
a}{2}+\frac{\psi[1+\eta_{1}(a)][\psi_{0}/2a+1]a}{1-p(a)}\bigg{]},$ (2.3)
$\displaystyle\eta_{3}(a)$ $\displaystyle=$
$\displaystyle{\bigg{[}}\frac{\psi\eta_{2}(a)a}{2(1-\psi_{0}\eta_{2}(a)a)}$
(2.4)
$\displaystyle+\frac{1}{1-\psi_{0}\eta_{2}(a)a}\psi[\eta_{1}(a)+\eta_{2}(a)]\times\left(\frac{1}{1-p(a)}\bigg{[}1+\frac{\psi_{0}}{2}\eta_{2}(a)a\bigg{]}a\right)$
$\displaystyle+\left(\frac{1}{1-\psi_{0}a}.\frac{\psi[a+\eta_{1}(a)a]}{1-p(a)}\times\bigg{[}1+\frac{\psi_{0}}{2}\eta_{2}(a)a\bigg{]}\right){\bigg{]}}\eta_{2}(a),$

and

$\displaystyle p(a)=\psi_{0}\eta_{1}(a)a.$ (2.5)

Let

$\displaystyle H_{1}(a)=\eta_{1}(a)-1,\ \ \ H_{2}(a)=\eta_{2}(a)-1,$ (2.6)
$\displaystyle H_{3}(a)=\eta_{3}(a)-1,\ \ \ H_{4}(a)=\eta_{4}(a)-1,$ (2.7)

and

$\rho_{1}=\frac{2}{2\psi+5\psi_{0}}<\frac{1}{\psi_{0}}.$ (2.8)

We observe that $\eta_{1}(\rho_{1})=1$ and can attain

$0\leq\eta_{1}(a)<1\ for\ a\in[0,\rho_{1}),\ \ \ 0<\rho_{1}<1/\psi_{0}.$ (2.9)

Now, we have that $H_{1}(0)=H_{2}(0)=H_{3}(0)=H_{4}(0)<0$ and
$H_{1}(a)\rightarrow+\infty$ or a positive constant as
$t\rightarrow\frac{1}{\psi_{0}}^{-}$.The intermediate value theorem confirms
the existence of the zeros of the function $H_{2}(a)$ in the interval
$(0,\frac{1}{\psi_{0}})$. We denote the smallest zero of $H_{2}(a)$ as
$\rho_{2}$. Also, $h_{2}(\rho_{1})>0$ and $\rho_{1}<\frac{1}{\psi_{0}}$, which
follows from that

$0<\rho_{2}<\rho_{1},\ \ \ 0<\eta_{2}(a)<1,for\ a\in[0,\rho_{2}).$ (2.10)

Following this procedure, there comes the existence of zeros of functions
$H_{i},\ i=1,2,3,4$ in the interval $(0,\rho_{0})$. Denote by $\rho_{i}$,
respectively, the smallest solution of functions $H_{i}^{\prime}s$. Define the
radius of convergence $\rho$ and for each $a\in[0,\rho)$ by

$\displaystyle\rho=min\\{\rho_{i}\\},\ i=1,2,3,4;\ \ \ 0\leq\eta_{i}(a)<1.$
(2.11)

Let $B(x^{*},\rho),\ \overline{B(x^{*},\rho)}$ stand, respectively for the
open and closed ball in $X$ such that $x^{*}\in X$ and of radius $\rho>0$.
Next, we present the local convergence analysis of method $(\ref{eqn:13})$
using the preceding notations and Lipschitz conditions.

######  Theorem 2.1.

Suppose that $T:D\subset X\rightarrow Y$ be a continuously first order Fréchet
differentiable operator. Presume $\psi_{0}>0$ and $\psi>0$ be given
parameters. Assume that there exists $x^{*}\in D$ for all $x,y\in D$ and
fulfill the below conditions:

$\displaystyle T(x^{*})=0,\ [T^{\prime}(x^{*})]^{-1}\in L(Y,X),$ (2.12)

where, $L(X,Y)$ is the set of bounded linear operators from $X$ to $Y$,

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}(T^{\prime}(x)-T^{\prime}(x^{*}))\|$
$\displaystyle\leq$ $\displaystyle\psi_{0}\|x-x^{*}\|.$ (2.13)
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}(T^{\prime}(x)-T^{\prime}(y))\|$
$\displaystyle\leq$ $\displaystyle\psi\|x-y\|,$ (2.14) $\displaystyle
B(x^{*},\rho)$ $\displaystyle\subseteq$ $\displaystyle D,$ (2.15)

where $\rho$ is defined by equation $(\ref{eqn:39})$. Then, the sequence
$\\{x_{n}\\}$ generated by the method $(\ref{eqn:13})$ for $x_{0}\in
B(x^{*},\rho)\backslash\\{x^{*}\\}$ is well defined in $B(x^{*},\rho)$ remains
in $B(x^{*},\rho)$ for each $n=0,1,2,\cdots$ and converges to $x^{*}$.
Consequently, the following relations holds for $n=0,1,2,\cdots$:

$\displaystyle\|y_{n}-x^{*}\|\leq\eta_{1}(\|x_{n}-x^{*}\|)\|x_{n}-x^{*}\|\leq\|x_{n}-x^{*}\|<\rho,$
(2.16)
$\displaystyle\|z_{n}^{(1)}-x^{*}\|\leq\eta_{2}(\|x_{n}-x^{*}\|)\|x_{n}-x^{*}\|\leq\|x_{n}-x^{*}\|<\rho,$
(2.17)
$\displaystyle\|z_{n}^{(2)}-x^{*}\|\leq\eta_{3}(\|x_{n}-x^{*}\|)\|x_{n}-x^{*}\|\leq\|x_{n}-x^{*}\|<\rho,$
(2.18)

and

$\displaystyle\|x_{n+1}-x^{*}\|\leq\eta_{4}(\|x_{n}-x^{*}\|)\|x_{n}-x^{*}\|\leq\|x_{n}-x^{*}\|<\rho,$
(2.19)

where the functions $\eta_{i},i=1,2,3,4$ are defined by the expressions
$(\ref{eqn:32})$ -$(\ref{eqn:32a})$. Moreover, if there exists
$\varrho\in[\rho,\frac{1}{\psi_{0}})$ such that
$\overline{B(x^{*},\varrho)}\subseteq D$. Then the limit point $x^{*}$ is the
only solution of equation $T(x)=0$ in
$D_{1}=D\cap\overline{B(x^{*},\varrho)}$.

###### Proof.

We shall show by mathematical induction that sequence $\\{x_{n}\\}$ is well
defined and converges to $x^{*}$. Using the hypotheses, $x_{0}\in
B(x^{*},\rho)\backslash\\{x^{*}\\}$, equation $(\ref{eqn:31})$ and inequality
$(\ref{eqn:312})$, we have

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}(T^{\prime}(x_{0})-T^{\prime}(x^{*}))\|\leq\psi_{0}\|x_{0}-x^{*}\|\leq\psi_{0}(\rho)<1.$
(2.20)

It follows from the above and the Banach lemma on invertible operator [17]
that $[T^{\prime}(x_{0})]^{-1}\in L(Y,X)$ or $T^{\prime}(x_{0})$ is invertible
and

$\displaystyle\|[T^{\prime}(x_{0})]^{-1}T^{\prime}(x^{*})\|\leq\frac{1}{1-\psi_{0}\|x_{0}-x^{*}\|}.$
(2.21)

Now, $y_{0}$ is well defined by the first sub-step of the scheme
$(\ref{eqn:13})$ and for $n=0,$

$\displaystyle y_{0}-x^{*}$ $\displaystyle=$ $\displaystyle
x_{0}-x^{*}-\frac{1}{2}[[T^{\prime}(x_{0})]^{-1}T(x_{0})]$ $\displaystyle=$
$\displaystyle\frac{1}{2}[[T^{\prime}(x_{0})]^{-1}T(x_{0})]+[T^{\prime}(x_{0})]^{-1}[T^{\prime}(x_{0})(x_{0}-x^{*})-T(x_{0})+T(x^{*})]||.$

Expanding $T(x_{0})$ along $x^{*}$ and taking the norm, we get

$\displaystyle\|y_{0}-x^{*}\|$ $\displaystyle\leq$
$\displaystyle\bigg{\|}\frac{1}{2}[T^{\prime}(x_{0})]^{-1}T^{\prime}(x^{*})\bigg{\|}\bigg{[}\int_{0}^{1}\|[T^{\prime}(x^{*}]^{-1}[T^{\prime}(x*+t(x_{0}-x^{*})]dt\|\|x_{0}-x^{*}\|\bigg{]}$
$\displaystyle+\|[T^{\prime}(x_{0})]^{-1}T^{\prime}(x^{*})\|\int_{0}^{1}\|[T^{\prime}(x^{*})]^{-1}[T^{\prime}(x_{0})-T^{\prime}(x^{*}+t(x_{0}-x^{*}))]dt\|\|x_{0}-x^{*}\|.$

Thus, we get

$\displaystyle\|y_{0}-x^{*}\|$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\psi_{0}\|x_{0}-x^{*}\|}\bigg{[}\frac{1}{2}{\bigg{[}}\frac{\psi_{0}}{2}\|x_{0}-x^{*}\|+1{\bigg{]}}$
(2.24) $\displaystyle+$
$\displaystyle\frac{\psi}{2}\|x_{0}-x^{*}\|\bigg{]}\|x_{0}-x^{*}\|\leq\eta_{1}(\|x_{0}-x^{*}\|)\|x_{0}-x^{*}\|<\rho.$

From the inequalities $(\ref{eqn:312})$ and $(\ref{eqn:323})$, we have

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}[T^{\prime}(y_{0})-T^{\prime}(x^{*})]\|$
$\displaystyle\leq$ $\displaystyle\psi_{0}\|y_{0}-x^{*}\|$ (2.25)
$\displaystyle\leq$
$\displaystyle\psi_{0}\eta_{1}(\|x_{0}-x^{*}\|)\|x_{0}-x^{*}\|$
$\displaystyle=$ $\displaystyle p(\|x_{0}-x^{*}\|)<1.$

Thus, by Banach lemma,

$\displaystyle\|[T^{\prime}(y_{0})]^{-1}T^{\prime}(x^{*})\|\leq\frac{1}{1-p(\|x_{0}-x^{*}\|)}.$
(2.26)

From the second sub-step of the method $(\ref{eqn:13})$, we have

$\displaystyle z_{0}^{(1)}-x^{*}$ $\displaystyle=$ $\displaystyle
x_{0}-x^{*}-[T^{\prime}(y_{0})^{-1}T(x_{0})]$ $\displaystyle=$ $\displaystyle
x_{0}-x^{*}-[T^{\prime}(x_{0})]^{-1}T(x_{0})+[T^{\prime}(x_{0})]^{-1}[T^{\prime}(y_{0})-T^{\prime}(x_{0})]T^{\prime}(y_{0})^{-1}T(x_{0}).$

On taking norm of the equation $(\ref{eqn:326})$, we get

$\displaystyle\|z_{0}^{(1)}-x^{*}\|$ $\displaystyle\leq$
$\displaystyle\|x_{0}-x^{*}-[T^{\prime}(x_{0})]^{-1}T(x_{0})\|+\|[T^{\prime}(x_{0})]^{-1}T^{\prime}(x^{*})\|.$
(2.28)
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}[T^{\prime}(y_{0})-T^{\prime}(x_{0})]\|\|[T^{\prime}(y_{0})]^{-1}T^{\prime}(x^{*})\|\|[T^{\prime}(x^{*})]^{-1}T(x_{0})\|$
$\displaystyle\leq$
$\displaystyle\frac{1}{1-\psi_{0}\|x_{0}-x^{*}\|}\bigg{[}\frac{\psi}{2}\|x_{0}-x^{*}\|$
$\displaystyle+$
$\displaystyle\frac{[\psi\|y_{0}-x^{*}\|+\psi(\|x_{0}-x^{*}\|)][\frac{\psi_{0}}{2}\|x_{0}-x^{*}\|+1]}{1-p(\|x_{0}-x^{*}\|)}\bigg{]}\|x_{0}-x^{*}\|.$

Thus, we get

$\displaystyle\|z_{0}^{(1)}-x^{*}\|\leq\eta_{2}(\|x_{0}-x^{*}\|)\|x_{0}-x^{*}\|\leq\|x_{0}-x^{*}\|<\rho.$
(2.29)

From the next sub-step of the method $(\ref{eqn:13})$, we have

$\displaystyle
z_{0}^{(2)}-x^{*}=z_{0}^{(1)}-x^{*}-(2[T^{\prime}(y_{0})]^{-1}-[T^{\prime}(x_{0})]^{-1})T(z_{0}^{(1)})$
$\displaystyle=(z_{0}^{(1)}-x^{*}-[T^{\prime}(z_{0}^{(1)})]^{-1}T^{\prime}(z_{0}^{(1)}))+[T^{\prime}(z_{0}^{(1)})]^{(-1)}T^{\prime}(x^{*})T^{\prime}(x^{*})^{-1}[T^{\prime}(y_{0})-T^{\prime}(z_{0}^{(1)})]$
$\displaystyle.[T^{\prime}(y_{0})]^{(-1)}T^{\prime}(x^{*})[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(1)})+[T^{\prime}(x_{0})]^{(-1)}T^{\prime}(x^{*})[T^{\prime}(x^{*})]^{-1}[T^{\prime}(y_{0})-T^{\prime}(x_{0})]$
$\displaystyle.[T^{\prime}(y_{0})]^{(-1)}T^{\prime}(x^{*})[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(1)}).$
(2.30)

On expanding $T(z_{0}^{(1)})$ along $x^{*}$ and taking norm of the equation
$(\ref{eqn:327})$, we get

$\displaystyle\|z_{0}^{(2)}-x^{*}\|$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\psi_{o}\|z_{0}^{(1)}-x^{*}\|}\frac{\psi}{2}\|z_{0}^{(1)}-x^{*}\|.\|z_{0}^{(1)}-x^{*}\|+\frac{[\psi\|y_{0}-x^{*}\|+\psi\|z_{0}^{(1)}-x^{*}\|]}{1-\psi_{0}\|z_{0}^{(1)}-x^{*}\|}$
(2.31) $\displaystyle\times$
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(1)})\|+\frac{1}{1-\psi_{0}\|x_{0}-x^{*}\|}[\psi\|y_{0}-x^{*}\|+\psi\|x_{0}-x^{*}\|]$
$\displaystyle\times$
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(1)})\|$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\psi_{0}\|z_{0}^{(1)}-x^{*}\|}\frac{\psi}{2}\|z_{0}^{(1)}-x^{*}\|.\|z_{0}^{(1)}-x^{*}\|+\frac{[\psi\|y_{0}-x^{*}\|+\psi\|z_{0}^{(1)}-x^{*}\|]}{1-\psi_{0}\|z_{0}^{(1)}-x^{*}\|}$
$\displaystyle\times$
$\displaystyle\frac{1}{1-p(\|x_{0}-x^{*}\|)}\bigg{(}1+\frac{\psi_{0}}{2}\|z_{0}^{(1)}-x^{*}\|\bigg{)}\|z_{0}^{(1)}-x^{*}\|+\frac{[\psi\|y_{0}-x^{*}\|+\psi\|x_{0}-x^{*}\|]}{1-\psi_{0}\|x_{0}-x^{*}\|}$
$\displaystyle\times$
$\displaystyle\frac{1}{1-p(\|x_{0}-x^{*}\|)}\bigg{(}1+\frac{L_{0}}{2}\|z_{0}^{(1)}-x^{*}\|\bigg{)}\|z_{0}^{(1)}-x^{*}\|.$

Thus, we have

$\displaystyle\|z_{0}^{(2)}-x^{*}\|\leq\eta_{3}(\|x_{0}-x^{*}\|)\|x_{0}-x^{*}\|<\rho.$
(2.32)

Now, from the last sub-step of the method $(\ref{eqn:13})$, we have

$\displaystyle
x_{1}-x^{*}=z_{0}^{(2)}-x^{*}-(2[T^{\prime}(y_{0})]^{-1}-[T^{\prime}(x_{0})]^{-1})T(z_{0}^{(2)})$
$\displaystyle=(z_{0}^{(2)}-x^{*}-[T^{\prime}(z_{0}^{(2)})]^{-1}T^{\prime}(z_{0}^{(2)}))+[T^{\prime}(z_{0}^{(2)})]^{(-1)}T^{\prime}(x^{*})T^{\prime}(x^{*})^{-1}[T^{\prime}(y_{0})-T^{\prime}(z_{0}^{(1)})]$
$\displaystyle.[T^{\prime}(y_{0})]^{(-1)}T^{\prime}(x^{*})[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(2)})+[T^{\prime}(x_{0})]^{(-1)}T^{\prime}(x^{*})[T^{\prime}(x^{*})]^{-1}[T^{\prime}(y_{0})-T^{\prime}(x_{0})]$
$\displaystyle.[T^{\prime}(y_{0})]^{(-1)}T^{\prime}(x^{*})[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(2)}).$
(2.33)

On expanding $T(z_{0}^{(2)})$ along $x^{*}$ and taking norm of the equation
$(\ref{eqn:329})$, we get

$\displaystyle\|x_{1}-x^{*}\|$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\psi_{0}\|z_{0}^{(2)}-x^{*}\|}\frac{\psi}{2}\|z_{0}^{(2)}-x^{*}\|.\|z_{0}^{(2)}-x^{*}\|+\frac{[\psi\|y_{0}-x^{*}\|+\psi\|z_{0}^{(2)}-x^{*}\|]}{1-\psi_{0}\|z_{0}^{(2)}-x^{*}\|}$
(2.34) $\displaystyle\times$
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(2)})\|+\frac{1}{1-\psi_{0}\|x_{0}-x^{*}\|}[\psi\|y_{0}-x^{*}\|+\psi\|x_{0}-x^{*}\|]$
$\displaystyle\times$
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(2)})\|$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\psi_{0}\|z_{0}^{(2)}-x^{*}\|}\frac{\psi}{2}\|z_{0}^{(2)}-x^{*}\|.\|z_{0}^{(2)}-x^{*}\|+\frac{[\psi\|y_{0}-x^{*}\|+\psi\|z_{0}^{(2)}-x^{*}\|]}{1-\psi_{0}\|z_{0}^{(2)}-x^{*}\|}$
$\displaystyle\times$
$\displaystyle\frac{1}{1-p(\|x_{0}-x^{*}\|)}\bigg{(}1+\frac{\psi_{0}}{2}\|z_{0}^{(2)}-x^{*}\|\bigg{)}\|z_{0}^{(2)}-x^{*}\|+\frac{[\psi\|y_{0}-x^{*}\|+\psi\|x_{0}-x^{*}\|]}{1-\psi_{0}\|x_{0}-x^{*}\|}$
$\displaystyle\times$
$\displaystyle\frac{1}{1-p(\|x_{0}-x^{*}\|)}\bigg{(}1+\frac{\psi_{0}}{2}\|z_{0}^{(2)}-x^{*}\|\bigg{)}\|z_{0}^{(2)}-x^{*}\|.$

Thus, we have

$\displaystyle\|x_{1}-x^{*}\|\leq\eta_{4}(\|x_{0}-x^{*}\|)\|x_{0}-x^{*}\|<\rho,$
(2.35)

which shows that for $n=0,\ x_{1}\in B(x^{*},\rho).$ The function
$H_{4}(a)=\eta_{4}(a)-1$ gives $H_{4}(0)<0$ and $H_{4}(\rho_{3})>0$. Hence,
$H_{4}(t)$ has at least one root in $(0,\rho_{3})$. Let $\rho$ be the smallest
root of $H_{4}(t)$ in $(0,\rho_{3})$. Then, we have

$\displaystyle 0<\rho<\rho_{3}<\rho_{2}<\rho_{1}<1/\psi_{0}.$ (2.36)

and

$0<\eta_{4}(a)<1,for\ a\in[0,\rho).$ (2.37)

By simply replacing $x_{0},y_{0},z_{0}^{(1)},z_{0}^{(2)},x_{1}$ by
$x_{n},y_{n},z_{n}^{(1)},z_{n}^{(2)},x_{n+1}$ in the preceding estimates, we
arrive at inequalities $(\ref{eqn:316})-(\ref{eqn:318})$. By the estimate,

$\displaystyle\|x_{n+1}-x^{*}\|\leq\eta_{4}(\|x_{0}-x^{*}\|)\|x_{n}-x^{*}\|<\rho.$
(2.38)

We conclude that $\lim_{n\rightarrow\infty}x_{n}=x^{*}$ and $x_{n+1}\in
B(x^{*},\rho)$. Finally, to prove the uniqueness, let $y^{*}\in B(x^{*},\rho)$
where $y^{*}\neq x^{*}$ with $T(y^{*})=0$. Define
$F=\int_{0}^{1}T^{\prime}(x^{*}+t(y^{*}-x^{*}))dt$. On expanding $T(y^{*})$
along $x^{*}$ and using inequality $(\ref{eqn:312})$, we obtain

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}\int_{0}^{1}[T^{\prime}(x^{*}+t(y*-x^{*})-T^{\prime}(x^{*})]dt\|$
$\displaystyle\leq\frac{\psi_{0}}{2}\|y^{*}-x^{*}\|\leq\frac{\psi_{0}}{2}\varrho<1.$
(2.39)

So, by Banach lemma,
$\int_{0}^{1}[T^{\prime}(x^{*})]^{-1}[T^{\prime}(x^{*}+t(y^{*}-x^{*}))]dt$
exists and invertible leading to the conclusion $x^{*}=y^{*}$, which completes
the uniqueness part of the proof. ∎

##  3\. Local convergence analysis of the method $(\ref{eqn:13})$ under Hölder
condition

In this section, we move forward to present the local convergence analysis
using Hölder condition because there are numerous nonlinear equations for
which the assumptions based on Lipschitz condition fails to solve without
using higher-order derivatives. As an illustration, we review the following
problem [16].

$T(x)(s)=x(s)-\int_{0}^{1}G(s,t)\left(x(t)^{\frac{5}{2}}+\frac{x(t)^{2}}{2}\right)dt,$
(3.1)

where $T:C[0,1]\rightarrow C[0,1]$ and the kernel $G$ is the Green’s function
defined on the interval $[0,1]\times[0,1]$ by

$\displaystyle G(s,t)=\begin{cases}(1-s)t&t\leq s,\\\ s(1-t),&s\leq
t.\end{cases}$

Note that,

$\displaystyle\bigg{\|}\int_{0}^{1}G(s,t)\bigg{\|}\leq\frac{1}{8}.$

Then,

$\|G^{\prime}(x)-G^{\prime}(y))\|\leq\frac{1}{8}\left(\frac{5}{2}\|x-y\|^{\frac{3}{2}}+\|x-y\|\right).$
(3.2)

Clearly, it can be seen that $G^{\prime}$ does not satisfy the Lipschitz
continuity condition. However, $G^{\prime}$ is Hölder continuous. For such
kind of examples, we also derive the local convergence results. This analysis
also generalizes the local convergence analysis presented in the previous
section. Supposing $q\in(0,1]$, $\kappa_{0}>0$ and $\kappa>0$ be two
parameters with $\kappa_{0}\leq\kappa$, we define the functions
$\mu_{1},\mu_{2},\mu_{3},\mu_{4},p,M_{1},M_{2},M_{3}$ and $M_{4}$ on interval
$\left[0,\left(\frac{1}{\kappa_{0}}\right)^{\frac{1}{q}}\right)$ by

$\displaystyle\mu_{4}(a)$ $\displaystyle=$
$\displaystyle{\bigg{[}}\frac{\kappa\mu_{3}(a)^{q}.a^{q}}{(q+1)(1-\kappa_{0}\mu_{3}(a)^{q}.a^{q})}$
(3.3)
$\displaystyle+\frac{1}{1-\kappa_{0}\mu_{3}(a^{q})a^{q}}\kappa[\mu_{1}(a)^{q}+\mu_{3}(a)^{q}]a^{q}\times\left(\frac{1}{1-p(a)}\bigg{[}1+\frac{\kappa_{0}}{q+1}\mu_{3}(a)^{q}.a^{q}\bigg{]}\right)$
$\displaystyle+\left(\frac{1}{1-\kappa_{0}a^{q}}.\frac{\kappa[1+\mu_{1}(a)^{q}]a^{q}}{1-p(a)}\times\bigg{[}1+\frac{\kappa_{0}}{q+1}\mu_{3}(a)^{q}.a^{q}\bigg{]}\right){\bigg{]}}\mu_{3}(a),$

where

$\displaystyle\mu_{1}(a)=\frac{1}{1-\kappa_{0}a^{q}}{\bigg{[}}\frac{\kappa
a^{q}}{q+1}+\frac{1+\frac{\kappa_{0}}{(q+1)}.a^{q}}{2}{\bigg{]}},$ (3.4)
$\displaystyle\mu_{2}(a)=\frac{1}{1-\kappa_{0}a^{q}}\bigg{[}\frac{\kappa
a^{q}}{q+1}+\frac{\kappa[1+\mu_{1}(a)^{q}][\frac{\kappa_{0}}{(q+1)}.a^{q}+1].a^{q}}{1-p(a)}\bigg{]},$
(3.5) $\displaystyle\mu_{3}(a)$ $\displaystyle=$
$\displaystyle{\bigg{[}}\frac{\kappa\mu_{2}(a)^{q}.a^{q}}{(q+1)(1-\kappa_{0}\mu_{2}(a)^{q}.a^{q})}$
(3.6)
$\displaystyle+\frac{1}{1-\kappa_{0}\mu_{2}(a^{q})a^{q}}\kappa[\mu_{1}(a)^{q}+\mu_{2}(a)^{q}]a^{q}\times\left(\frac{1}{1-p(a)}\bigg{[}1+\frac{\kappa_{0}}{q+1}\mu_{2}(a)^{q}.a^{q}\bigg{]}\right)$
$\displaystyle+\left(\frac{1}{1-\kappa_{0}a^{q}}.\frac{\kappa[1+\mu_{1}(a)^{q}]a^{q}}{1-p(a)}\times\bigg{[}1+\frac{\kappa_{0}}{q+1}\mu_{2}(a)^{q}.a^{q}\bigg{]}\right){\bigg{]}}\mu_{2},(a),$

and

$\displaystyle p(a)=\kappa_{0}\mu_{1}(a)^{q}a^{q}.$ (3.7)

Let

$\displaystyle M_{1}(a)=\mu_{1}(a)-1,\ \ \ M_{2}(a)=\mu_{2}(a)-1,$ (3.8)
$\displaystyle M_{3}(a)=\mu_{3}(a)-1,\ \ \ M_{4}(a)=\mu_{4}(a)-1.$ (3.9)

and

$\rho_{1}=\left(\frac{q+1}{2\kappa+\kappa_{0}(3+2q)}\right)^{\frac{1}{q}}<\left(\frac{1}{\kappa_{0}}\right)^{\frac{1}{q}}.$
(3.10)

We observe that $\mu_{1}(\rho_{1})=1$ and

$0\leq\mu_{1}(a)<1\ for\ a\in[0,\rho_{1}),\ \ \ \
0<\rho_{1}<(1/\kappa_{0})^{\frac{1}{q}}.$ (3.11)

We denote the smallest zero of $M_{2}(a)$ as $\rho_{2}$. Also,
$M_{2}(\rho_{1})>0$ and
$\rho_{1}<\left(\frac{1}{\kappa_{0}}\right)^{\frac{1}{q}}$, which follows from
that

$0<\rho_{2}<\rho_{1},\ \ \ 0<\mu_{2}(a)<1,for\ a\in[0,\rho_{2}).$ (3.12)

Following this procedure, there comes the existence of zeros of functions
$M_{i},\ i=1,2,3,4$ in the interval $(0,\rho_{0})$. Denote by $\rho_{i}$,
respectively, the smallest solution of functions $M_{i}^{\prime}s$. Define the
radius of convergence $\rho$ and for each $a\in[0,\rho)$ by

$\displaystyle\rho=min\\{\rho_{i}\\},\ i=1,2,3,4,\ \ \ 0\leq\mu_{i}(a)<1,$
(3.13)

Also, we assume that there exists $x^{*}\in D$ for all $x,y\in D$ and fulfill
the below conditions alongwith the assumption $(\ref{eqn:311})$:

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}(T^{\prime}(x)-T^{\prime}(x^{*}))\|$
$\displaystyle\leq$ $\displaystyle\kappa_{0}\|x-x^{*}\|^{q}.$ (3.14)
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}(T^{\prime}(x)-T^{\prime}(y))\|$
$\displaystyle\leq$ $\displaystyle\kappa\|x-y\|^{q},$ (3.15) $\displaystyle
B(x^{*},\rho)$ $\displaystyle\subseteq$ $\displaystyle D,$ (3.16)

######  Lemma 3.1.

If $T$ satisfies the assumptions $(\ref{eqn:413})$ and $(\ref{eqn:414})$,
consequently the inequalities given below hold for $x\in D$, $q=(0,1]$ and
$t\in[0,1]$:

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T^{\prime}(x)\|$ $\displaystyle\leq$
$\displaystyle 1+\kappa_{0}\|x-x^{*}\|^{q}.$ (3.17)
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}(T^{\prime}(x^{*}+t(x-x^{*}))\|$
$\displaystyle\leq$ $\displaystyle 1+\kappa_{0}t^{q}\|x-x^{*}\|^{q}.$ (3.18)
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T(x)\|$ $\displaystyle\leq$
$\displaystyle\left(1+\frac{\kappa_{0}}{q+1}\|x-x^{*}\|^{q}\right).\|x-x^{*}\|.$
(3.19)

###### Proof.

By considering the hypothesis $(\ref{eqn:413})$, we get

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T^{\prime}(x)\|$ $\displaystyle\leq$
$\displaystyle
1+\|[T^{\prime}(x^{*})]^{-1}T^{\prime}(x)-[T^{\prime}(x^{*})]^{-1}T^{\prime}(x^{*})\|$
(3.20) $\displaystyle\leq$ $\displaystyle 1+\kappa_{0}\|x-x^{*}\|^{q}.$

In the similar manner, we can derive

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}(T^{\prime}(x^{*}+t(x-x^{*}))\|$
$\displaystyle\leq$ $\displaystyle 1+\kappa_{0}\|x^{*}+t(x-x^{*})-x^{*}\|^{q}$
(3.21) $\displaystyle\leq$ $\displaystyle 1+\kappa_{0}t^{q}\|x-x^{*}\|^{q}.$

Next,

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T(x)$ $\displaystyle\leq$
$\displaystyle\int_{0}^{1}[\kappa_{0}\|x^{*}+t(x-x^{*})-x^{*}\|^{q}+1].\|x-x^{*}\|dt$
(3.22) $\displaystyle\leq$
$\displaystyle\left(1+\frac{\kappa_{0}}{q+1}\|x-x^{*}\|^{q}\right).\|x-x^{*}\|.$

∎

######  Theorem 3.1.

Suppose that there exists $x^{*}\in D$ and the Fréchet differentiable operator
$T:D\subset X\rightarrow Y$ satisfies the assumptions $(\ref{eqn:413})$ and
$(\ref{eqn:414})$. Then, the sequence $\\{x_{n}\\}$ generated by the method
$(\ref{eqn:13})$ for $x_{0}\in B(x^{*},\rho)\backslash\\{x^{*}\\}$ is well
defined in $B(x^{*},\rho)$ remains in $B(x^{*},\rho)$ for each
$n=0,1,2,\cdots$ and converges to $x^{*}$. Henceforward, the following
measures holds for $n=0,1,2,\cdots$:

$\displaystyle\|y_{n}-x^{*}\|\leq\mu_{1}(\|x_{n}-x^{*}\|)\|x_{n}-x^{*}\|\leq\|x_{n}-x^{*}\|<\rho,$
(3.23)
$\displaystyle\|z_{n}^{(1)}-x^{*}\|\leq\mu_{2}(\|x_{n}-x^{*}\|)\|x_{n}-x^{*}\|\leq\|x_{n}-x^{*}\|<\rho,$
(3.24)
$\displaystyle\|z_{n}^{(2)}-x^{*}\|\leq\mu_{3}(\|x_{n}-x^{*}\|)\|x_{n}-x^{*}\|\leq\|x_{n}-x^{*}\|<\rho,$
(3.25)

and

$\displaystyle\|x_{n+1}-x^{*}\|\leq\mu_{4}(\|x_{n}-x^{*}\|)\|x_{n}-x^{*}\|\leq\|x_{n}-x^{*}\|<\rho,$
(3.26)

where the functions $\mu_{i},i=1,2,3,4$ are defined by the expressions
$(\ref{eqn:42})$ \- $(\ref{eqn:45})$. Moreover, if there exists
$\varrho\in\left[\rho,\left(\frac{1+q}{\kappa_{0}}\right)^{\frac{1}{q}}\right]$
such that $\overline{B(x^{*},\varrho)}\subseteq D$. Then the limit point
$x^{*}$ is the only solution of equation $T(x)=0$ in
$D_{1}=D\cap\overline{B(x^{*},\varrho)}$.

###### Proof.

Assume that $\|x_{0}-x^{*}\|^{q}<\frac{1}{\kappa_{0}}$ and using the
hypotheses, $x_{0}\in B(x^{*},\rho)\backslash\\{x^{*}\\}$ and inequality
$(\ref{eqn:413})$, we have that

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}(T^{\prime}(x_{0})-T^{\prime}(x^{*}))\|\leq\kappa_{0}\|x_{0}-x^{*}\|^{q}<1.$
(3.27)

It follows from the above and the Banach lemma on invertible operator [17]
that $[T^{\prime}(x_{0})]^{-1}\in L(Y,X)$ or $T^{\prime}(x_{0})$ is invertible
and

$\displaystyle\|[T^{\prime}(x_{0})]^{-1}T^{\prime}(x^{*})\|\leq\frac{1}{1-\kappa_{0}\|x_{0}-x^{*}\|^{q}}.$
(3.28)

Now, $y_{0}$ is well defined by the first sub-step of the scheme
$(\ref{eqn:13})$ and for $n=0$ we get the approximation $(\ref{eqn:321})$.
Expanding $T(x_{0})$ along $x^{*}$ and taking the norm, we get the inequality
$(\ref{eqn:322a})$. Finally, using the assumption $(\ref{eqn:413})$ and the
inequality $(\ref{eqn:t41a})$, we get

$\displaystyle\|y_{0}-x^{*}\|$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\kappa_{0}\|x_{0}-x^{*}\|^{q}}\bigg{[}\frac{1}{2}{\bigg{[}}\frac{\kappa_{0}}{q+1}\|x_{0}-x^{*}\|^{q}+1{\bigg{]}}$
(3.29) $\displaystyle+$
$\displaystyle\frac{\kappa}{q+1}\|x_{0}-x^{*}\|^{q}\bigg{]}\|x_{0}-x^{*}\|$
$\displaystyle\leq$
$\displaystyle\mu_{1}(\|x_{0}-x^{*}\|)\|x_{0}-x^{*}\|<\rho.$

From the inequalities $(\ref{eqn:413})$ and $(\ref{eqn:t43})$, we have

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}[T^{\prime}(y_{0})-T^{\prime}(x^{*})]\|$
$\displaystyle\leq$ $\displaystyle\kappa_{0}\|y_{0}-x^{*}\|^{q}$ (3.30)
$\displaystyle\leq$
$\displaystyle\kappa_{0}\mu_{1}(\|x_{0}-x^{*}\|)^{q}\|x_{0}-x^{*}\|^{q}$
$\displaystyle=$ $\displaystyle p(\|x_{0}-x^{*}\|)<1.$

Again, from the second sub-step of the method $(\ref{eqn:13})$, we find the
approximation $(\ref{eqn:326})$. On taking the norm, we get

$\displaystyle\|z_{0}^{(1)}-x^{*}\|$ $\displaystyle\leq$
$\displaystyle\|x_{0}-x^{*}-[T^{\prime}(x_{0})]^{-1}T(x_{0})\|+\|[T^{\prime}(x_{0})]^{-1}T^{\prime}(x^{*})\|.$
(3.31)
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}[T^{\prime}(y_{0})-T^{\prime}(x_{0})]\|\|[T^{\prime}(y_{0})]^{-1}T^{\prime}(x^{*})\|\|[T^{\prime}(x^{*})]^{-1}T(x_{0})\|$
$\displaystyle\leq$
$\displaystyle\frac{1}{1-\kappa_{0}\|x_{0}-x^{*}\|^{q}}\bigg{[}\frac{\kappa}{q+1}\|x_{0}-x^{*}\|^{q}$
$\displaystyle+$
$\displaystyle\frac{[\kappa\|y_{0}-x^{*}\|^{q}+\kappa(\|x_{0}-x^{*}\|^{q})][\frac{\kappa_{0}}{q+1}\|x_{0}-x^{*}\|^{q}+1]}{1-p(\|x_{0}-x^{*}\|)}\bigg{]}\|x_{0}-x^{*}\|.$

Thus, we get

$\displaystyle\|z_{0}^{(1)}-x^{*}\|\leq\mu_{2}(\|x_{0}-x^{*}\|)\|x_{0}-x^{*}\|\leq\|x_{0}-x^{*}\|<\rho.$
(3.32)

From the next sub-step of the method $(\ref{eqn:13})$, we have the
approximation $(\ref{eqn:327})$. On expanding $T(z_{0}^{(1)})$ along $x^{*}$
and taking norm, we get

$\displaystyle\|z_{0}^{(2)}-x^{*}\|$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\kappa_{0}\|z_{0}^{(1)}-x^{*}\|^{q}}\frac{\kappa}{q+1}\|z_{0}^{(1)}-x^{*}\|^{q}.\|z_{0}^{(1)}-x^{*}\|+\frac{[\kappa\|y_{0}-x^{*}\|^{q}+\kappa\|z_{0}^{(1)}-x^{*}\|^{q}]}{1-\kappa_{0}\|z_{0}^{(1)}-x^{*}\|^{q}}$
(3.33) $\displaystyle\times$
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(1)})\|+\frac{1}{1-\kappa_{0}\|x_{0}-x^{*}\|^{q}}[\kappa\|y_{0}-x^{*}\|^{q}+\kappa\|x_{0}-x^{*}\|^{q}]$
$\displaystyle\times$
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(1)})\|$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\kappa_{0}\|z_{0}^{(1)}-x^{*}\|^{q}}\frac{\kappa}{q+1}\|z_{0}^{(1)}-x^{*}\|^{q}.\|z_{0}^{(1)}-x^{*}\|+\frac{[\kappa\|y_{0}-x^{*}\|^{q}+\kappa\|z_{0}^{(1)}-x^{*}\|^{q}]}{1-\kappa_{0}\|z_{0}^{(1)}-x^{*}\|^{q}}$
$\displaystyle\times$
$\displaystyle\frac{1}{1-p(\|x_{0}-x^{*}\|)}\bigg{(}1+\frac{\kappa_{0}}{q+1}\|z_{0}^{(1)}-x^{*}\|^{q}\bigg{)}\|z_{0}^{(1)}-x^{*}\|+\frac{[\kappa\|y_{0}-x^{*}\|^{q}+\kappa\|x_{0}-x^{*}\|^{q}]}{1-\kappa_{0}\|x_{0}-x^{*}\|^{q}}$
$\displaystyle\times$
$\displaystyle\frac{1}{1-p(\|x_{0}-x^{*}\|^{q})}\bigg{(}1+\frac{\kappa_{0}}{q+1}\|z_{0}^{(1)}-x^{*}\|^{q}\bigg{)}\|z_{0}^{(1)}-x^{*}\|.$

Thus, we have

$\displaystyle\|z_{0}^{(2)}-x^{*}\|\leq\kappa_{3}(\|x_{0}-x^{*}\|)\|x_{0}-x^{*}\|<\rho,$
(3.34)

Finally, from the last sub-step of the method $(\ref{eqn:13})$ we have the
approximation $(\ref{eqn:329})$. On expanding $T(z_{0}^{(2)})$ along $x^{*}$
and taking norm, we get

$\displaystyle\|x_{1}-x^{*}\|$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\kappa_{0}\|z_{0}^{(2)}-x^{*}\|^{q}}\frac{\kappa}{q+1}\|z_{0}^{(2)}-x^{*}\|^{q}.\|z_{0}^{(2)}-x^{*}\|+\frac{[\kappa\|y_{0}-x^{*}\|^{q}+\kappa\|z_{0}^{(2)}-x^{*}\|^{q}]}{1-\kappa_{0}\|z_{0}^{(2)}-x^{*}\|^{q}}$
(3.35) $\displaystyle\times$
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(2)})\|+\frac{1}{1-\kappa_{0}\|x_{0}-x^{*}\|^{q}}[\kappa\|y_{0}-x^{*}\|^{q}+\kappa\|x_{0}-x^{*}\|^{q}]$
$\displaystyle\times$
$\displaystyle\|[T^{\prime}(x^{*})]^{-1}T(z_{0}^{(2)})\|$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\kappa_{0}\|z_{0}^{(2)}-x^{*}\|^{q}}\frac{\kappa}{q+1}\|z_{0}^{(2)}-x^{*}\|^{q}.\|z_{0}^{(2)}-x^{*}\|+\frac{[\kappa\|y_{0}-x^{*}\|^{q}+\kappa\|z_{0}^{(2)}-x^{*}\|^{q}]}{1-\kappa_{0}\|z_{0}^{(2)}-x^{*}\|^{q}}$
$\displaystyle\times$
$\displaystyle\frac{1}{1-p(\|x_{0}-x^{*}\|)}\bigg{(}1+\frac{\kappa_{0}}{q+1}\|z_{0}^{(2)}-x^{*}\|^{q}\bigg{)}\|z_{0}^{(2)}-x^{*}\|+\frac{[\kappa\|y_{0}-x^{*}\|^{q}+\kappa\|x_{0}-x^{*}\|^{q}]}{1-\kappa_{0}\|x_{0}-x^{*}\|^{q}}$
$\displaystyle\times$
$\displaystyle\frac{1}{1-p(\|x_{0}-x^{*}\|)}\bigg{(}1+\frac{\kappa_{0}}{q+1}\|z_{0}^{(2)}-x^{*}\|^{q}\bigg{)}\|z_{0}^{(2)}-x^{*}\|.$

Thus, we have

$\displaystyle\|x_{1}-x^{*}\|\leq\mu_{4}(\|x_{0}-x^{*}\|)\|x_{0}-x^{*}\|<\rho,$
(3.36)

which shows that for $n=0,\ x_{1}\in B(x^{*},\rho).$ The function
$M_{4}(a)=\mu_{4}(a)-1$ gives $M_{4}(0)<0$ and $M_{4}(\rho_{3})>0$. Hence,
$M_{4}(t)$ has at least one root in $(0,\rho_{3})$. Let $\rho$ be the smallest
root of $M_{4}(t)$ in $(0,\rho_{3})$. Then, we have

$\displaystyle
0<\rho<\rho_{3}<\rho_{2}<\rho_{1}<\left(\frac{1}{\kappa_{0}}\right)^{\frac{1}{q}}.$
(3.37)

and

$0<\mu_{4}(a)<1,for\ a\in[0,\rho).$ (3.38)

By simply replacing $x_{0},y_{0},z_{0}^{(1)},z_{0}^{(2)},x_{1}$ by
$x_{n},y_{n},z_{n}^{(1)},z_{n}^{(2)},x_{n+1}$ in the preceding estimates, we
arrive at inequalities $(\ref{eqn:415})-(\ref{eqn:418})$. By the estimate

$\displaystyle\|x_{n+1}-x^{*}\|\leq\mu_{4}(\|x_{0}-x^{*}\|)\|x_{n}-x^{*}\|<\rho.$
(3.39)

We conclude that $\lim_{n\rightarrow\infty}x_{n}=x^{*}$ and $x_{n+1}\in
B(x^{*},\rho)$. Finally, to prove the uniqueness, let $y^{*}\in B(x^{*},\rho)$
where $y^{*}\neq x^{*}$ with $T(y^{*})=0$. Define
$F=\int_{0}^{1}T^{\prime}(x^{*}+t(y^{*}-x^{*}))dt$. On expanding $T(y^{*})$
along $x^{*}$ and using inequality $(\ref{eqn:413})$, we obtain

$\displaystyle\|[T^{\prime}(x^{*})]^{-1}\int_{0}^{1}[T^{\prime}(x^{*}+t(y*-x^{*})-T^{\prime}(x^{*})]dt\|$
$\displaystyle\leq\frac{\kappa_{0}}{q+1}\|y^{*}-x^{*}\|^{q}\leq\frac{\kappa_{0}}{q+1}\varrho^{q}<1.$
(3.40)

So, by Banach lemma,
$\int_{0}^{1}[T^{\prime}(x^{*})]^{-1}[T^{\prime}(x^{*}+t(y^{*}-x^{*}))]dt$
exists and invertible leading to the conclusion $x^{*}=y^{*}$, which completes
the uniqueness part of the proof. ∎

##  4\. Applications with numerical examples

In this section, two numerical examples are worked out to demonstrate the
efficiency of our local convergence analysis by giving the radii of
convergence for the scheme $(\ref{eqn:13})$. We obtain better results using
our technique.

Example 4.1[12] Returning back to the illustration example given in the
introduction of this study of Lipschitz, The unique solution is $x^{*}=1$. The
consecutive derivatives of $f$ are

$\displaystyle f^{\prime}(x)$ $\displaystyle=$ $\displaystyle
3x^{2}logx^{2}+5x^{4}-4x^{3}+2x^{2},$ $\displaystyle f^{\prime\prime}(x)$
$\displaystyle=$ $\displaystyle 6xlogx^{2}+20x^{3}-12x^{2}+10x,$
$\displaystyle f^{\prime\prime\prime}(x)$ $\displaystyle=$ $\displaystyle
6logx^{2}+60x^{2}-24x+22.$

It can be easily visible that $f^{\prime\prime\prime}$ is unbounded on $D$.
Nevertheless, all the assumptions of the Theorem $(\ref{thm:31})$ for the
iterative method $(\ref{eqn:13})$ are satisfied and hence applying the
convergence results with $x^{*}=1$, we obtain $\psi_{0}=\psi=96.6628$. Here,
we will use the iterative method $(\ref{eqn:13})$ and compare it with the
scheme given by Cordero et al. [21] and we denote it by CHMT. From the above
Table $(\ref{tab:1})$, we can see that the radius $\rho$ of convergence
computed for the method $(\ref{eqn:13})$ seems to be finer than the method
CHMT which local convergence was discussed in the article [22]. As a result,
the approach being evaluated is more powerful.

Table 1. Comparison of convergence radius (Example 4.1) $Radius$ | Method $(\ref{eqn:13})$ | CHMT$(\theta=-2)$
---|---|---
$\rho_{1}$ | 0.00295578 | 0.006689
$\rho_{2}$ | 0.00246894 | 0.005750
$\rho_{3}$ | 0.00217353 | 0.003001
$\rho_{4}$ | 0.00208131 | 0.001943
$\rho$ | 0.00208131 | 0.001943

Example 4.2[23] In order to show the applicability of the results presented in
this paper, we consider the following Planck’s radiation law problem [24]:

$\displaystyle\phi(\lambda)=\frac{8\pi cP\lambda^{-5}}{e^{\frac{cP}{\lambda
BT}}-1},$

which calculates the energy density within an isothermal blackbody, where

  * •

$\lambda$ is the wavelength of the radiation

  * •

$T$ is the absolute temperature of the blackbody

  * •

$B$ is Boltzmann’s constant

  * •

$P$ is the Planck’s constant

  * •

$c$ is the speed of light.

Suppose, we would like to determine wavelength $\lambda$ which corresponds to
maximum energy density $\phi(\lambda)$. Therefore, solving for maxima we
define

$\displaystyle f(x)=e^{-x}-1+\frac{x}{5}.$

We have $x^{*}=4.965114$ and $F^{\prime}(x^{*})=0.193023$. Then, on using
assumptions $(\ref{eqn:413})-(\ref{eqn:414})$ we have that, $q=1$,
$\kappa_{0}=0.0608658<\kappa=0.094888.$ Table $(\ref{tab:5})$ displays the
radius $\rho$ of convergence by the discussed method $(\ref{eqn:13})$ along
with the existing scheme KFS analyzed in the reference [23]. We discovered
that when compared the provided method enlarges the radius of the convergence
ball.

Table 2. Comparison of convergence radius (Example 4.2) $Radius$ | Method $(\ref{eqn:13})$ | KFS
---|---|---
$\rho_{1}$ | 4.04772 | 9.23282
$\rho_{2}$ | 2.99797 | 2.40532
$\rho_{3}$ | 2.58569 | 1.11454
$\rho_{4}$ | 2.45972 |
$\rho$ | 2.45972 | 1.11454

Example 4.3[16] Returning back to the illustration example given in the
introduction of this study of Hölder, in view of $(\ref{eqn:42a})$, the
previous results, which required $G^{(4)}$ to be bounded, are no longer valid.
However, our findings are applicable and therefore our study, on the other
hand, is valuable in ensuring that these techniques for this problem are
converging. Here, $q=1$ and
$\kappa_{0}=\kappa=\frac{1}{8}\left(\frac{5}{2}\sqrt{2}+1\right)$. Indeed,
using the above choices of functions $\kappa,\ \kappa_{0}$ and expression
$(\ref{eqn:41})$, the radius $\rho$ of convergence is computed in Table
$(\ref{tab:6})$.

Table 3. Convergence radius for (Example 4.3) $Radius$ | Method $(\ref{eqn:13})$
---|---
$\rho_{1}$ | 0.503957
$\rho_{2}$ | 0.420951
$\rho_{3}$ | 0.378541
$\rho_{4}$ | 0.363397
$\rho$ | 0.363397

Thus, we guarantee the convergence of the method $(\ref{eqn:13})$ with radius
$\rho=0.363397$.

##  5\. Conclusions

In the presented work, we have analyzed the local convergence analysis of the
efficient seventh order method for solving the nonlinear equation in Banach
spaces. A convergence theorem for existence and uniqueness of the solution has
been established followed by its error bounds giving the benefit that the
iterative method always converges to the solution. This local convergence
analysis is applicable in solving such problems for which higher-order
derivative based previous studies fail. Later, we relaxed these assumptions
and derive convergence results under Hölder condition for solving different
types of nonlinear integral equations, which are not solvable by the previous
approach.

## References

  * [1] Traub, J.F., (1977), Iterative Methods for the solution of equations, Chelsea Publishing Company, New York.
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Akanksha Saxena
Department of Mathematics
Maulana Azad National Institute of Technology
Bhopal, M.P. India-462003.
Email<EMAIL_ADDRESS>

J. P. Jaiswal
Department of Mathematics
Guru Ghasidas Vishwavidyalaya ( A Central University)
Bilaspur, C.G. India-495009.
Email<EMAIL_ADDRESS>

K. R. Pardasani
Department of Mathematics
Maulana Azad National Institute of Technology
Bhopal, M.P. India-462003.
Email<EMAIL_ADDRESS>