## 2 Statement of the Problem
We have \(-0.2(t^{-0.5}+1)=\alpha(t)\leq x\leq y\leq\beta(t)=-0.1t-0.05\leq-0.05\), \(-0.49\leq\alpha(0.5)\leq\sup_{s\in[t-0.5,t]}\alpha(s)\leq u\leq v\leq\sup_{s \in[t-0.5,t]}\beta(s)\leq-0.0105\) and therefore, \(\frac{y+0.2t^{-0.5}+0.25}{t+1}\geq\frac{-0.2(t^{-0.5}-0.2+0.2t^{-0.5}+0.25}{t+ 1}=\frac{0.05}{t+1}\geq 0.025\) for \(t\in[0,1],\;x,y,u,v\in\mathbb{R}\) and  
\[F(t,x,u)-F(t,y,v) =(x+0.2t^{-0.5}+0.25)\frac{u}{t+1}-(y+0.2t^{-0.5}+0.25)\frac{v}{t+ 1}\] \[=\frac{u}{t+1}(x-y)+\frac{y+0.2t^{-0.5}+0.25}{t+1}(u-v)\leq-0.49( x-y)+0.025(u-v).\]  
Thus, \(M=-0.49,\;L=0.025>0\). Denote \(\alpha^{(0)}(t)=\alpha(t)\) and \(\beta^{(0)}(t)=\beta(t)\) for \(t\in[-0.5,1]\). Figure 5: Graphs of the function \(\alpha(t)\) and its upper bound defined by (9). Figure 6: Graphs of the functions \(\beta(t)\) and \(\alpha(t)\). _Based on Step 4 in Algorithm I the \(n\)-th lower successive approximation, \(n=1,2,\ldots\), is defined by_  
\[\alpha^{(n)}(t)= \begin{cases}-0.15-\frac{0.041}{n},\ \ \ \ \ t\in[-0.5,0]\\ (-0.15-\frac{0.041}{n})\Gamma(0.5)E_{0.5,0.5}(-0.49t^{0.5})t^{-0.5}\\ \ \ \ +\int_{0}^{t}(t-s)^{-0.5}E_{0.5,0.5}(-0.49(t-s)^{0.5})\times\\ \ \ \ \ \ \times\Big{(}\alpha^{(n-1)}(s)+0.2s^{-0.5}+0.25\Big{)}\frac{ \sigma\in[s-0.5,s]}{s+1}ds\\ \ \ \ -\int_{0}^{t}(t-s)^{-0.5}E_{0.5,0.5}(-0.49(t-s)^{0.5})\times\\ \ \ \ \ \ \ \ \ \times\Big{(}0.49\alpha^{(n-1)}(s)+0.025\Big{(}\sup_{ \sigma\in[s-0.5,s]}\alpha^{(n-1)}(\sigma)-\sup_{\sigma\in[s-0.5,s]}\alpha^{(n )}(\sigma)\Big{)}\Big{)}ds,\\ \ \ \ \ \ \ \ t\in(0,1],\end{cases}  \tag{10}\]  
_and according to Step 5 in Algorithm I the \(n\)-th upper successive approximation, \(n=1,2,\ldots\), is defined by_  
\[\beta^{(n)}(t)=\begin{cases}-0.15+\frac{0.05}{n},\ \ \ \ \ t\in[-0.5,0]\\ (-0.15+\frac{  
_As it can be seen in Figure 10 the inequality \(\alpha(t)\leq\beta(t)\) is valid on \([-0.5,1]\), i.e., the considered lower and upper solutions of IVP (11) are properly defined._  
_To apply the iterative scheme, suggested in Algorithm II we need to check the validity of the corresponding conditions given in the theoretical part in Section 5, Theorem 2. We get_  
\[F(t,u)-F(t,v)=f(t)u-f(t)v=f(t)(u-v)\leq 0.04(u-v)\ \ \mbox{for}\ t\in[0,1],\ u,v\in \mathbb{R}.\]  
_Thus \(L=0.04>0\) and \(\frac{0.04}{\Gamma(0.5)}=0.56419<1\)._  
_Define the zero approximations \(\alpha^{(0)}(t)=\alpha(t)\) and \(\beta^{(0)}(t)=\beta(t)\) for \(t\in[-0.5,1]\)._  
Figure 8: Example 2. Graph of the function \(\alpha(t)\). Figure 9: Example 2. Graphs of the bound \(-0.1\) and the integral. _For any \(t\in(0,T]\) the inequality_  
\[F\left(t,u(t),\sup_{s\in[t-r,t]}u(s)\right)-F\left(t,\alpha(t),\sup_ {s\in[t-r,t]}\alpha(s)\right)\] \[\leq M\big{(}u(t)-\alpha(t)\big{)}+L\left(\sup_{s\in[t-r,t]}u(s)- \sup_{s\in[t-r,t]}\alpha(s)\right)\] _holds with_ \(M\in\mathbb{R}\)_,_ \(L>0\)_._
3. _The function_ \(\beta\in PC([-r,T],\mathbb{R})\) _be such that_ \(\beta(t)=\Omega(v,\beta,\varepsilon+\delta)(t),\ t\in[-r,T]\) _with_ \(\delta>0\)_._  
_Then the inequality \(\alpha(t)<\beta(t)\) holds for all \(t\in[-r,T]\)._  
Proof. : Assume the contrary. Since \(\alpha(t)<\beta(t)\) for \(t\in[-r,0]\) and \(t^{1-q}\alpha(t)|_{t=0}=\phi(0)+\varepsilon<\phi(0)+\varepsilon+\delta=t^{1-q} \beta(t)|_{t=0}\) there exists a point \(t^{*}\in(0,T]\) such that \(\alpha(t)<\beta(t)\) for \(t\in[-r,t^{*})\) and \(\alpha(t^{*})=\beta(t^{*})\).