# On the Existence of Solutions of Nonlinear Fredholm Integral Equations from Kantorovich's Technique
## 2 [PERSON]'s Technique
### The Auxiliary Scalar Function  
We begin by analysing the operator \(\mathfrak{F}(x)\) given in (2). So, from (2), it follows that the Frechet derivatives of operator \(\mathfrak{F}\) are  
\[[\mathfrak{F}^{\prime}(x)y](s)=y(s)-\varpi p\int_{a}^{b}\mathfrak{F}(s,t)x(t )^{p-1}y(t)\,dt,\]  
\[[\mathfrak{F}^{(k)}(x)(y_{1}y_{2}\cdots y_{k})](s)=-\varpi p(p-1)\cdots(p-k+1 )\int_{a}^{b}\mathfrak{F}(s,t)x(t)^{p-k}y_{1}(t)y_{2}(t)\cdots y_{k}(t)\,dt,\]  
for \(k=2,3,\ldots,[p]\), where \([p]\) denotes the integer part of the real number \(p\geq 2\). In addition,  
\[\|\mathfrak{F}^{(k)}(x)\|\leq|\varpi|\binom{p}{k}k!\,S\|x\|^{p-k},\]  
where \(S=\max_{s\in[a,b]}\int_{a}^{b}|\mathfrak{F}(s,t)|\,dt\) with the infinity-norm. Next, taking into account that \(\|x\|\leq\|x_{0}\|+\|x-x_{0}\|\), it follows  
\[\|\mathfrak{F}^{(k)}(x)\|\leq|\varpi|\binom{p}{k}k!\,S\|x\|^{p-k}\leq|\varpi |\binom{p}{k}k!\,S\left(\|x_{0}\|+\|x-x_{0}\|\right)^{p-k}\]  
\[\leq|\varpi|\binom{p}{k}k!\,S\left(\|x_{0}\|+t-t_{0}\right)^{p-k}, \tag{5}\]  
provided that \(\|x-x_{0}\|\leq t-t_{0}\). Moreover, for \(p\geq 3\), we denote  
\[\|\mathfrak{F}^{(i)}(x_{0})\|\leq|\varpi|\binom{p}{i}i!\,S\|x_{0}\|^{p-i}=M_{ i},\]  
for \(i=2,\ldots,k-1\). On the other hand, we observe that the existence of the operator \([\mathfrak{F}^{\prime}(x_{0})]^{-1}\) must be guaranteed in the first step of [PERSON]'s method, since \(x_{1}=x_{0}-[\mathfrak{F}^{\prime}(x_{0})]^{-1}\mathfrak{F}(x_{0})\). The existence of \([\mathfrak{F}^{\prime}(x_{0})]^{-1}\) follows from the Banach lemma on invertible operators, so the operator \([\mathfrak{F}^{\prime}(x_{0})]^{-1}\) exists and is such that  
\[\|[\mathfrak{F}^{\prime}(x_{0})]^{-1}\|\leq\frac{1}{1-\|I-\mathfrak{F}^{\prime }(x_{0})\|},\]  
provided that \(\|I-\mathfrak{F}^{\prime}(x_{0})\|<1\); namely,  
\[|\varpi|pS\|x_{0}\|^{p-1}<1. \tag{6}\]  
In addition, we denote \(\beta=\frac{1}{1-|\varpi|pS\|x_{0}\|^{p-1}}\) and do  
\[\|[\mathfrak{F}^{\prime}(x_{0})]^{-1}\mathfrak{F}(x_{0})\|\leq\|\mathfrak{F} (x_{0})\|\beta=\eta.\]  
Now, we consider \(p\geq 3\) and denote \(\omega_{k}(t;t_{0})=|\varpi|(^{p}_{k})k!\,S\,(\|x_{0}\|+t-t_{0})^{p-k}\), for \(k=2,3,\ldots,[p]\). Then, as a consequence of the latter, we can find scalar functions \(y(t)\) such that \(y^{(k)}(t)=\omega_{k}(t;t_{0})\), for \(k=2,3,\ldots,[p]\), to construct a majorizing sequence \(\{t_{n}\}\) as that given in (3) by solving the following initial value problem (see [18]):  
\[\left\{ \begin{array}{l}y^{(k)}(t)=|\varpi|\binom{p}{k}k!\,S\,(\|x_{0}\|+t-t _{0})^{p-k},\\ y(t_{0})=\frac{\eta}{\beta},\quad y^{\prime}(t_{0})=-\frac{1}{\beta},\\ y^{\prime\prime}(t_{0})=M_{2},\quad y^{\prime\prime\prime}(t_{0})=M_{3},\quad \ldots,\quad y^{(k-1)}(t_{0})=M_{k-1}.\end{array} \right.\]  
It is easy to see that there exists only one solution for the last initial value problem, that is:  
\[\varphi(t)=\int_{t_{0}}^{t}\int_{t_{0}}^{\theta_{k-1}}\cdots\int_{t_{0}}^{ \theta_{1}}\omega_{k}(z;t_{0})\,dz\,d\theta_{1}\cdots\,d\theta_{k-1}+\sum_{i= 2}^{k-1}\frac{M_{i}}{i!}(t-t_{0})^{i}-\frac{t-t_{0}}{\beta}+\frac{\eta}{\beta}\]  
\[=|\varpi|S\,(\|x_{0}\|+t-t_{0})^{p}-(t-t_{0})+\|\mathfrak{F}(x_{0})\|-|\varpi |S\|x_{0}\|^{p}. \tag{7}\]  
Notice that the scalar function defined in (7) and used to construct the scalar sequence \(\{t_{n}\}\) given in (3) with \(\varphi(t)\) defined in (7), that majorizes \(\{x_{n}\}\) in \(\Omega\), is independent of \(k\), so we can choose any \(k\), such that \(k=2,3,\ldots,[p]\), to construct the last initial value problem that gives us \(\varphi(t)\). If \(p\in[2,3)\) and using only condition (5), we consider the initial value problem  
\[\left\{ \begin{array}{l}y^{\prime\prime}(t)=|\varpi|\binom{p}{2} 2!\,S\,(\|x_{0}\|+t-t_{0})^{p-2},\\ y(t_{0})=\frac{\eta}{\beta},\quad y^{\prime}(t_{0})=-\frac{1}{\beta}, \end{array} \right.\]  
whose unique solution also is (7). Once such a majorizing sequence \(\{t_{n}\}\) is determined from \(\varphi(t)\), we have then to prove its convergence.