## 2 Formulation in Banach Space  
Let \(X\) be a complex Banach space of functions and \(X^{*}\) the adjoint space of \(X\), i.e., the set of all complex-valued linear bounded functionals \(\Psi_{j}:X\rightarrow\mathbb{C}\), \(j\in\mathbb{N}\). Let \(\Psi=\text{col}(\Psi_{1},\Psi_{2},...,\Psi_{m})\) be a vector of linear bounded functionals \(\Psi_{j}\), \(j=1,2,...,m\), and \(G=\lambda(g_{1},g_{2},...,g_{m})\) a vector of functions \(g_{j}\in X\), \(j=1,2,...,m\). Then Equation (2) may be written in the form  
\[Tu=f,\quad f\in X, \tag{3}\]  
where the linear operator \(T:X\to X\) is defined by  
\[Tu=u-G\Psi(u). \tag{4}\]  
In (3) and (4), the components of the vectors \(G\) and \(\Psi\) are known, \(f\) is given and \(u\) has to be determined. To examine the solvability and find the unique solution of (3), we state and prove the theorem below, but first, we explain some formulae and notations which we will use. It is understood that \(\Psi(u)\) and \(\Psi(G)\) denote the \(m\times 1\) column vector and the \(m\times m\) matrix  
\[\Psi(u)=\left( \begin{array}{c}\Psi_{1}(u)\\ \Psi_{2}(u)\\ \vdots\\ \Psi_{m}(u)\end{array} \right),\quad\Psi(G)=\left[ \begin{array}{cccc}\Psi_{1 }(g_{1})&\Psi_{1}(g_{2})&\cdots&\Psi_{1}(g_{m})\\ \Psi_{2}(g_{1})&\Psi_{2}(g_{2})&\cdots&\Psi_{2}(g_{m})\\ \vdots&\vdots&\cdots&\vdots\\ \Psi_{m}(g_{1})&\Psi_{m}(g_{2})&\cdots&\Psi_{m}(g_{m})\end{array} \right],\]  
respectively. It can be easily verified that  
\[\Psi(G\mathbf{c})=(\Psi(G))\mathbf{c},\]  
where \(\mathbf{c}=\mathrm{col}(c_{1},c_{2},\ldots,c_{m})\) is a constant vector. Bold lowercase and capital letters denote vectors and matrices, respectively, whose elements are numbers. By \(\mathbf{0}\) and \(\mathbf{I}_{m}\), we mark the zero column vector and the identity matrix of order \(m\), respectively. Theorem 3.1: _Let the linear operator \(T:X\to X\) be defined by (4). Then \(T\) is injective on \(D(T)\subset X\) if and only if_  
\[\det\mathbf{V}=\det[\mathbf{I}_{m}-\Psi(G)]\
eq 0. \tag{5}\]  
_In this case, the unique solution of Equation (3) is given by_  
\[u=T^{-1}f=f+G\mathbf{V}^{-1}\Psi(f), \tag{6}\]  
_where \(T^{-1}:X\to X\) denotes the inverse operator of \(T\)._  
Proof: Suppose \(\mathbf{V}\) is a non-singular matrix, i.e., \(\det\mathbf{V}\
eq 0\), and let \(z\in\ker T\). Then,  
\[Tz=z-G\Psi(z)=0. \tag{7}\]  
Acting by the vector \(\Psi\) on both sides of (7), we obtain  
\[\Psi(z-G\Psi(z)) = \Psi(z)-\Psi(G\Psi(z))\] \[= \Psi(z)-(\Psi(G))\Psi(z)\] \[= [\mathbf{I}_{m}-\Psi(G)]\Psi(z)\] \[= \mathbf{V}\Psi(z)=\mathbf{0},\]  
which implies that \(\Psi(z)=\mathbf{0}\). Then, from (7) follows that \(z=0\) and hence \(\ker T=\{0\}\), which means that the operator \(T\) is injective. Conversely, we will prove that if \(T\) is an injective operator then \(\det\mathbf{V}\
eq 0\) or equivalently, if \(\det\mathbf{V}=0\) then \(T\) is not injective. Let \(\det\mathbf{V}=0\). Then, there exists a vector \(\mathbf{c}=\mathrm{col}(c_{1},c_{2},\ldots,c_{m})\
eq\mathbf{0}\) such that \(\mathbf{Vc}=\mathbf{0}\). Let the element \(z=G\mathbf{c}\) and note that \(z\
eq 0\); otherwise, \(\mathbf{Vc}=[\mathbf{I}_{m}-\Psi(G)]\mathbf{c}=\mathbf{c}-(\Psi(G))\mathbf{c}= \mathbf{c}-\Psi(G\mathbf{c})=\mathbf{c}=\mathbf{0}\), which contradicts the hypothesis. Substituting \(z\) into (7), we obtain  
\[Tz = G\mathbf{c}-G\Psi(G\mathbf{c})\] \[= G[\mathbf{c}-\Psi(G\mathbf{c})]\] \[\det\mathbf{V}=\det[\mathbf{I}_{m}-\Psi(G)]\
eq 0. \tag{5}\]0 \[\det\mathbf{V}=\det[\mathbf{I}_{m}-\Psi(G)]\
eq 0. \tag{5}\]1 \[\det\mathbf{V}=\det[\mathbf{I}_{m}-\Psi(G)]\
eq 0. \tag{5}\]2  
which means that \(\ker T\
eq\{0\}\), and so \(T\) is not injective. Assume now that (5) is true. Applying the vector \(\Psi\) on both sides of (3), viz. \[\det\mathbf{V}=\det[\mathbf{I}_{m}-\Psi(G)]\
eq 0. \tag{5}\]3and working as above, we have  
\[\det\mathbf{V}=\det[\mathbf{I}_{m}-\Psi(G)]\
eq 0. \tag{5}\]4  
Since \(\det\mathbf{V}\
eq 0\), it follows that \(\Psi(u)=\mathbf{V}^{-1}\Psi(f)\).