Define the operator T by
$$\begin{aligned} Tu(t)={}& \int^{t}_{0}\frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)}f\bigl(s, u(s), D^{\alpha-1}u(s)\bigr)\,ds \\ &{}+\frac{- \int^{\infty}_{0}f(s, u(s), D^{\alpha-1}u(s))\,ds+ \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}\frac{(\xi_{i}-s)^{\alpha -1}}{\Gamma(\alpha)}f(s, u(s), D^{\alpha-1}u(s))\,ds}{\Gamma(\alpha )- \sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha-1}}t^{\alpha-1}. \end{aligned}$$
Theorem 3.1
Assume that
\(f: J\times R\times R\rightarrow R\)
is continuous. Then problem (1.1)-(1.2) has at least one solution under the assumption that
-
(H)
there exist nonnegative functions
\(a(t)(1+t^{\alpha-1}), b(t), c(t)\in L^{1}(J)\), such that
$$\bigl\| f(t,x,y)\bigr\| \leq a(t)|x|+b(t)|y|+c(t), $$
where
\(\int^{\infty}_{0}c(t)\,dt<+\infty\).
Proof
First of all, in view of
$$\begin{aligned}& \begin{aligned}[b] Tu'(t)={}& \int^{t}_{0}\frac{(t-s)^{\alpha-2}}{\Gamma(\alpha )}f\bigl(s, u, D^{\alpha-1}u\bigr)\,ds \\ &{}+\frac{- \int^{\infty}_{0}f(s, u, D^{\alpha-1}u)\,ds+ \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}\frac{(\xi_{i}-s)^{\alpha -1}}{\Gamma(\alpha)}f(s, u, D^{\alpha-1}u)\,ds}{\Gamma(\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha-1}}(\alpha-1)t^{\alpha-2}, \end{aligned} \\& \begin{aligned}[b] &D^{\alpha-1} Tu(t)\\ &\quad=\int^{t}_{0}f\bigl(s, u(s), D^{\alpha-1}u(s) \bigr)\,ds\\ &\qquad{}+\frac{- \int^{\infty}_{0}f(s, u(s), D^{\alpha -1}u(s))\,ds+ \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}\frac{(\xi _{i}-s)^{\alpha-1}}{\Gamma(\alpha)}f(s, u(s), D^{\alpha -1}u(s))\,ds}{\Gamma(\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi _{i}^{\alpha -1}}\Gamma(\alpha), \end{aligned} \end{aligned}$$
together with the continuity of f, we see that \(T'u(t)\) and \(D^{\alpha -1}Tu(t)\) are continuous on J.
In the following we divide the proof into several steps.
Step 1 Choose the positive number
$$R>\max\{R_{1}, R_{2}, R_{3}\}, $$
where
$$\begin{aligned}& R_{1}= \frac{ \frac{1}{\Gamma(\alpha)}\int^{1}_{0}c(t)\,dt +\frac{1}{\Gamma(\alpha)\Lambda} \sum^{m-2}_{i=1}\beta_{i}\int ^{\xi _{i}}_{0} (\xi_{i}-t)^{\alpha-1}c(t)\,dt +\frac{1}{\Lambda}\int^{\infty}_{0}c(t)\,dt}{1- \frac{1}{\Gamma (\alpha )}\int^{1}_{0}(a(t)+b(t))\,dt -\frac{1}{\Gamma(\alpha)\Lambda} \sum^{m-2}_{i=1}\beta_{i}\int ^{\xi _{i}}_{0} (\xi_{i}-t)^{\alpha-1}(a(t)+b(t))\,dt -\frac{1}{\Lambda}\int^{\infty}_{0}(a(t)+b(t))\,dt}, \\& R_{2}= \frac{ \frac{1}{\Gamma(\alpha)}\int^{1}_{0}c(t)\,dt +\frac{\alpha-1}{\Gamma(\alpha)\Lambda} \sum^{m-2}_{i=1}\beta _{i}\int ^{\xi_{i}}_{0} (\xi_{i}-t)^{\alpha-1}c(t)\,dt +\frac{\alpha-1}{\Lambda}\int^{\infty}_{0}c(t)\,dt}{1- \frac {1}{\Gamma (\alpha)}\int^{1}_{0}(a(t)+b(t))\,dt -\frac{\alpha-1}{\Gamma(\alpha)\Lambda} \sum^{m-2}_{i=1}\beta _{i}\int ^{\xi_{i}}_{0} (\xi_{i}-t)^{\alpha-1}(a(t)+b(t))\,dt -\frac{\alpha-1}{\Lambda}\int^{\infty}_{0}(a(t)+b(t))\,dt}, \\& R_{3}=\frac{ \int^{1}_{0}c(t)\,dt -+ \frac{1}{\Lambda} \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0} (\xi_{i}-t)^{\alpha-1}c(t)\,dt +\frac{\Gamma(\alpha)}{\Lambda}\int^{\infty}_{0}c(t)\,dt}{1- \int ^{1}_{0}(a(t)+b(t))\,dt - \frac{1}{\Lambda} \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0} (\xi_{i}-t)^{\alpha-1}(a(t)+b(t))\,dt -\frac{\Gamma(\alpha)}{\Lambda}\int^{\infty}_{0}(a(t)+b(t))\,dt}, \end{aligned}$$
and \(\Lambda=\Gamma(\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi _{i}^{\alpha -1}\).
Let set
$$U=\bigl\{ u(t)\in Y:\bigl\| u(t)\bigr\| _{Y}\leq R\bigr\} . $$
Then, \(A:U\rightarrow U\). In fact, for any \(u(t)\in U\), we have
$$\begin{aligned} &\frac{|Tu(t)|}{1+t^{\alpha-1}}\\ &\quad= \biggl|\int ^{t}_{0}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)(1+t^{\alpha -1})}f\bigl(s,\ u(s), D^{\alpha-1}u(s)\bigr)\,ds\\ &\qquad{}+\frac{- \int^{\infty}_{0}f(s, u(s), D^{\alpha -1}u(s))\,ds+ \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}\frac{(\xi _{i}-s)^{\alpha-1}}{\Gamma(\alpha)}f(s, u(s), D^{\alpha -1}u(s))\,ds}{\Gamma(\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi _{i}^{\alpha -1}}\frac{t^{\alpha-1}}{(1+t^{\alpha-1})} \biggr|\\ &\quad\leq\frac{1}{\Gamma(\alpha)} \int ^{t}_{0}\bigl(a(s)\bigl|u(s)\bigr|+b(s)\bigl|D^{\alpha-1}u(s)\bigr|+c(s) \bigr)\,ds \\ &\qquad{}+\frac{1}{\Lambda} \int^{\infty}_{0}\bigl(a(s)\bigl|u(s)\bigr|+b(s)\bigl|D^{\alpha -1}u(s)\bigr|+c(s) \bigr)\,ds\\ &\qquad{}+\frac{ \sum^{m-2}_{i=1}\beta_{i}}{\Lambda} \int^{\xi _{i}}_{0}\frac{(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha )}\bigl(a(s)\bigl|u(s)\bigr|+b(s)\bigl|D^{\alpha-1}u(s)\bigr|+c(s) \bigr)\,ds\\ &\quad\leq\frac{\|u\|_{Y}}{\Gamma(\alpha)} \int ^{1}_{0}\bigl(a(t)+b(t)\bigr)\,dt+ \frac{1}{\Gamma(\alpha)} \int^{1}_{0}c(t)\,dt \\ &\qquad{}+\frac{\|u\|_{Y}}{\Lambda} \int^{\infty}_{0}\bigl(a(t)+b(t)\bigr)\,dt+ \frac {1}{\Lambda} \int^{\infty}_{0}c(t)\,dt\\ &\qquad{}+\frac{\|u\|_{Y}}{\Lambda} \sum^{m-2}_{i=1}\beta _{i} \int^{\xi_{i}}_{0} \frac{(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha)}\bigl(a(t)+b(t) \bigr)\,dt+\frac{ \sum^{m-2}_{i=1}\beta_{i}}{\Lambda} \int^{\xi_{i}}_{0} \frac{(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha)}c(t)\,dt\\ &\quad\leq R, \\ & \frac{|T'u(t)|}{1+t^{\alpha-2}}\\ &\quad= \int^{t}_{0}\frac {(t-s)^{\alpha-2}}{\Gamma(\alpha)(1+t^{\alpha-2})}f\bigl(s, u(s), D^{\alpha -1}u(s)\bigr)\,ds\\ &\qquad{}+\frac{- \int^{\infty}_{0}f(s, u(s), D^{\alpha -1}u(s))\,ds+ \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}\frac{(\xi _{i}-s)^{\alpha-1}}{\Gamma(\alpha)}f(s, u(s), D^{\alpha -1}u(s))\,ds}{\Gamma(\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi _{i}^{\alpha -1}}\\ &\qquad{}\times\frac{(\alpha-1)t^{\alpha-2}}{(1+t^{\alpha-2})}\\ &\quad\leq\frac{1}{\Gamma(\alpha)} \int ^{t}_{0}\bigl(a(s)\bigl|u(s)\bigr|+b(s)\bigl|D^{\alpha-1}u(s)\bigr|+c(s) \bigr)\,ds\\ &\qquad{}+\frac{\alpha -1}{\Lambda} \int^{\infty}_{0}\bigl(a(s)\bigl|u(s)\bigr|+b(s)\bigl|D^{\alpha-1}u(s)\bigr|+c(s) \bigr)\,ds\\ &\qquad{}+(\alpha-1)\frac{ \sum^{m-2}_{i=1}\beta_{i}}{\Lambda} \int^{\xi_{i}}_{0}\frac{(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha )}\bigl(a(s)\bigl|u(s)\bigr|+b(s)\bigl|D^{\alpha-1}u(s)\bigr|+c(s) \bigr)\,ds\\ &\quad\leq \frac{\|u\|_{Y}}{\Gamma(\alpha)} \int ^{1}_{0}\bigl(a(t)+b(t)\bigr)\,dt+ \frac{1}{\Gamma(\alpha)} \int ^{1}_{0}c(t)\,dt\\ &\qquad{}+\frac{(\alpha-1)\|u\|_{Y}}{\Lambda} \int^{\infty }_{0}\bigl(a(t)+b(t)\bigr)\,dt+ \frac{\alpha-1}{\Lambda} \int^{\infty}_{0}c(t)\,dt\\ &\qquad{}+\frac{(\alpha-1)\|u\|_{Y}}{\Lambda} \sum^{m-2}_{i=1} \beta_{i} \int^{\xi_{i}}_{0} \frac{(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha)}\bigl(a(t)+b(t) \bigr)\,dt \\ &\qquad{}+\frac{(\alpha-1) \sum^{m-2}_{i=1}\beta_{i}}{\Lambda} \int^{\xi_{i}}_{0} \frac{(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha)}c(t)\,dt\\ &\quad\leq R, \\ & \bigl|D^{\alpha-1}Tu(t)\bigr|\\ &\quad\leq \int^{t}_{0}\bigl|f\bigl(s, u(s), D^{\alpha-1}u(s) \bigr)\bigr|\,ds +\frac{\Gamma(\alpha)}{\Gamma(\alpha)- \sum^{m-2}_{k=1}\beta _{i}\xi _{i}^{\alpha-1}} \int^{t}_{0}\bigl|f\bigl(s, u(s), D^{\alpha-1}u(s) \bigr)\bigr|\,ds\\ &\qquad{}+\frac{ \sum^{m-2}_{i=1}\beta_{i}}{\Gamma(\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha-1}} \int^{\xi _{i}}_{0}(\xi _{i}-s)^{\alpha-1}\bigl|f \bigl(s, u(s), D^{\alpha-1}u(s)\bigr)\bigr|\,ds\\ &\quad\leq R* \int^{1}_{0}\bigl(a(t)+b(t)\bigr)\,dt+ \int^{1}_{0}c(t)\,dt \\ &\qquad{}+\frac{\Gamma(\alpha)R}{\Lambda} \int^{\infty}_{0}\bigl(a(t)+b(t)\bigr)\,dt+\frac{\Gamma(\alpha)}{\Lambda} \int^{\infty}_{0}c(t)\,dt \\ &\qquad{}+\frac{ \sum^{m-2}_{i=1}\beta_{i}R}{\Lambda} \int^{\xi _{i}}_{0}(\xi _{i}-s)^{\alpha-1} \bigl(a(t)+b(t)\bigr)\,dt +\frac{ \sum^{m-2}_{i=1}\beta_{i}}{\Lambda} \int^{\xi_{i}}_{0}(\xi _{i}-s)^{\alpha-1}c(t)\,dt\\ &\quad\leq R. \end{aligned}$$
Hence, \(\|Tu(t)\|_{Y}\leq R\), which shows that \(A:U\rightarrow U\).
Step 2 Let V be a nonempty subset of U. We will show that TV is relative compact. Let \(I\subset J\) be a compact interval, \(t_{1}, t_{2}\in I\) and \(t_{1}< t_{2}\). Then for any \(u(t)\in V\), we have
$$\begin{aligned} & \biggl|\frac{Tu(t_{2})}{1+t_{2}^{\alpha-1}}-\frac {Tu(t_{1})}{1+t_{1}^{\alpha-1}} \biggr| \\ &\quad= \biggl| \int^{t_{2}}_{0}\frac {(t_{2}-s)^{\alpha-1}}{\Gamma(\alpha)(1+t_{2}^{\alpha-1})}f\bigl(s, u,\ D^{\alpha-1}u\bigr)\,ds \\ &\qquad{}+\frac{- \int^{\infty}_{0}f(s, u, D^{\alpha -1}u)\,ds+ \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}\frac{(\xi _{i}-s)^{\alpha-1}}{\Gamma(\alpha)}f(s, u, D^{\alpha-1}u)\,ds}{\Gamma (\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha-1}}\frac {t_{2}^{\alpha-1}}{(1+t_{2}^{\alpha-1})} \\ &\qquad{}- \int^{t_{1}}_{0}\frac{(t_{1}-s)^{\alpha-1}}{\Gamma (\alpha)(1+t_{1}^{\alpha-1})}f\bigl(s, u, D^{\alpha-1}u\bigr)\,ds \\ &\qquad{}-\frac{- \int^{\infty}_{0}f(s, u, D^{\alpha -1}u)\,ds+ \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}\frac{(\xi _{i}-s)^{\alpha-1}}{\Gamma(\alpha)}f(s, u, D^{\alpha-1}u)\,ds}{\Gamma (\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha-1}}\frac {t_{1}^{\alpha-1}}{(1+t_{1}^{\alpha-1})} \biggr| \\ &\quad\leq \int^{t_{1}}_{0}\biggl|\frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)(1+t_{2}^{\alpha-1})}- \frac{(t_{1}-s)^{\alpha-1}}{\Gamma (\alpha )(1+t_{1}^{\alpha-1})}\biggr|\bigl|f\bigl(s, u, D^{\alpha-1}u\bigr)\bigr|\,ds\\ &\qquad{} + \int^{t_{2}}_{t_{1}}\frac{(t_{2}-s)^{\alpha-1}}{\Gamma(\alpha )}\bigl|f\bigl(s,\ u, D^{\alpha-1}u\bigr)\bigr|\,ds \\ &\qquad{}+\biggl|\frac{- \int^{\infty}_{0}f(s, u(s), D^{\alpha -1}u(s))\,ds+ \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}\frac{(\xi _{i}-s)^{\alpha-1}}{\Gamma(\alpha)}f(s, u, D^{\alpha-1}u)\,ds}{\Gamma (\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha-1}}\biggr| \\ &\qquad{}\times\biggl|\frac{t_{2}^{\alpha-1}}{1+t_{2}^{\alpha-1}}-\frac {t_{1}^{\alpha -1}}{1+t_{1}^{\alpha-1}} \biggr|, \\ & \biggl|\frac{T'u(t_{2})}{1+t^{\alpha-2}}-\frac {T'u(t)_{1}}{1+t^{\alpha -2}} \biggr| \\ &\quad\leq \biggl| \int^{t_{2}}_{0}\frac{(t_{2}-s)^{\alpha-2}}{\Gamma (\alpha )(1+t_{2}^{\alpha-2})}f\bigl(s, u, D^{\alpha-1}u\bigr)\,ds- \int ^{t_{1}}_{0}\frac{(t_{1}-s)^{\alpha-2}}{\Gamma(\alpha )(1+t_{1}^{\alpha -2})}f\bigl(s, u, D^{\alpha-1}u\bigr)\,ds \biggr| \\ &\qquad{}+(\alpha-1) \biggl|\frac{- \int^{\infty}_{0}f(s, u, D^{\alpha -1}u)\,ds+ \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}\frac{(\xi _{i}-s)^{\alpha-1}}{\Gamma(\alpha)}f(s, u, D^{\alpha-1}u)\,ds}{\Gamma (\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha-1}} \biggr| \\ &\qquad{}\times\biggl|\frac{t_{2}^{\alpha-1}}{1+t_{2}^{\alpha-1}}- \frac{t_{1}^{\alpha -1}}{1+t_{1}^{\alpha-1}} \biggr| \end{aligned}$$
and
$$\begin{aligned} & \bigl|D^{\alpha-1}Tu(t_{2})-D^{\alpha-1}Tu(t_{1}) \bigr| \\ &\quad\leq \biggl| \int^{t_{2}}_{0}f\bigl(s, u(s), D^{\alpha-1}u(s) \bigr)\,ds- \int ^{t_{1}}_{0}f\bigl(s, u(s), D^{\alpha-1}u(s) \bigr)\,ds \biggr| \\ &\quad\leq \int^{t_{2}}_{t_{1}} \bigl|f\bigl(s, u(s), D^{\alpha -1}u(s) \bigr)\,ds \bigr|. \end{aligned}$$
Note that for any \(u(t)\in V\), we have \(f(t, u(t), D^{\alpha-1}u(t))\) is bounded on I. Then it is easy to see that \(\frac{|Tu(t)|}{1+t^{\alpha-1}}\), \(\frac{|T'u(t)|}{1+t^{\alpha-2}}\), and \(D^{\alpha-1}Tu(t)\) are equicontinuous on I.
Considering the condition H, for given \(\varepsilon>0\), there exists a constant \(L>0\) such that
$$\int^{+\infty}_{L}\bigl|f\bigl(t, u(t), D^{\alpha-1}u(t) \bigr)\bigr|< \varepsilon. $$
On the other hand, since \(\lim_{t\rightarrow+\infty}\frac {t^{\alpha -1}}{1+t^{\alpha-1}}=1\), there exists a constant \(T_{1}>0\) such that \(t_{1}, t_{2}\geq T_{1}\),
$$\biggl|\frac{t_{1}^{\alpha-1}}{1+t_{1}^{\alpha-1}}-\frac {t_{2}^{\alpha -1}}{1+t_{2}^{\alpha-1}} \biggr|< \varepsilon. $$
Similarly, in view of \(\lim_{t\rightarrow+\infty}\frac {(t-L)^{\alpha-1}}{1+t^{\alpha-1}}=1\), there exists a constant \(T_{2}>L>0\) such that \(t_{1}, t_{2}\geq T_{2}\) and \(0\leq s\leq L\),
$$\biggl|\frac{(t_{1}-s)^{\alpha-1}}{1+t_{1}^{\alpha-1}}-\frac {(t_{2}-s)^{\alpha -1}}{1+t_{2}^{\alpha-1}}\biggr|< \varepsilon. $$
In view of \(\lim_{t\rightarrow+\infty}\frac{(t-L)^{\alpha -2}}{1+t^{\alpha-2}}=1\), there exists a constant \(T_{3}>L>0\) such that \(t_{1}, t_{2}\geq T_{3}\), and \(0\leq s\leq L\),
$$\biggl|\frac{(t_{1}-s)^{\alpha-2}}{1+t_{1}^{\alpha-2}}-\frac {(t_{2}-s)^{\alpha -2}}{1+t_{2}^{\alpha-2}}\biggr|< \varepsilon. $$
Now choose \(T>\max\{T_{1}, T_{2}, T_{3}\}\). Then for \(t_{1}, t_{2}\geq T\), we have
$$\begin{aligned} & \biggl|\frac{Tu(t_{2})}{1+t_{2}^{\alpha-1}}-\frac {Tu(t_{1})}{1+t_{1}^{\alpha-1}} \biggr| \\ &\quad\leq\frac{\max_{t\in [0,L],u\in V}|f(t, u, D^{\alpha-1}u)|}{\Gamma(\alpha)}L\varepsilon +\frac{2}{\Gamma(\alpha)}\varepsilon \\ &\qquad{}+ \biggl|\frac{- \int^{\infty}_{0}f(s, u, D^{\alpha -1}u)\,ds+ \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}\frac{(\xi _{i}-s)^{\alpha-1}}{\Gamma(\alpha)}f(s, u, D^{\alpha-1}u)\,ds}{ \Gamma(\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha -1}} \biggr|\varepsilon, \\ & \biggl|\frac{T'u(t_{2})}{1+t_{2}^{\alpha-1}}-\frac {T'u(t_{1})}{1+t_{1}^{\alpha-1}} \biggr| \\ &\quad\leq\frac{\max_{t\in [0,L],u\in V}|f(t, u, D^{\alpha-1}u)|}{\Gamma(\alpha)}L\varepsilon +\frac{2}{\Gamma(\alpha)}\varepsilon \\ &\qquad{}+(\alpha-1) \biggl|\frac{- \int^{\infty}_{0}f(s, u, D^{\alpha-1}u)\,ds+ \sum^{m-2}_{i=1}\beta_{i}\int^{\xi_{i}}_{0}\frac {(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha)}f(s, u, D^{\alpha-1}u)\,ds}{ \Gamma(\alpha)- \sum^{m-2}_{i=1}\beta_{i}\xi_{i}^{\alpha -1}} \biggr|\varepsilon, \end{aligned}$$
and
$$\bigl|D^{\alpha-1}Tu(t_{2})-D^{\alpha-1}Tu(t_{1}) \bigr|\leq \int ^{t_{2}}_{t_{1}} \bigl|f\bigl(s, u(s), D^{\alpha-1}u(s) \bigr)\,ds \bigr| < \varepsilon. $$
Consequently, Lemma 2.5 shows that TV is relative compact.
Step 3 \(T:U\rightarrow U\) is a continuous operator.
Let \(u_{n}, u\in U\), \(n=1,2,\ldots\) , and \(\|u_{n}-u\| _{Y}\rightarrow0\) as \(n\rightarrow\infty\). Then, we have
$$\begin{aligned} & \biggl|\frac{Tu_{n}(t)}{1+t^{\alpha-1}}-\frac {Tu(t)}{1+t^{\alpha-1}} \biggr| \\ &\quad\leq \int^{t}_{0}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )(1+t^{\alpha -1})} \bigl|f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr)-f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr) \bigr|\,ds \\ &\qquad{}+\frac{t^{\alpha-1}}{(1+t^{\alpha-1})\Lambda} \sum^{m-2}_{i=1} \beta _{i} \int_{0}^{\xi_{i}}\frac{(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha )} \bigl|f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr)\\ &\qquad{}-f\bigl(s, u_{n}(s), D^{\alpha -1}u_{n}(s)\bigr) \bigr|\,ds \\ &\qquad{}+\frac{t^{\alpha-1}}{(1+t^{\alpha-1})\Lambda} \int^{\infty }_{0} \bigl|f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr)-f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr) \bigr|\,ds \\ &\quad\leq\biggl( \frac{2}{\Gamma(\alpha)}+\frac{4}{\Lambda}\biggr) \int ^{\infty}_{0}\bigl|f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr)-f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr) \bigr|\,ds \\ &\quad\leq\biggl( \frac{4}{\Gamma(\alpha)}+\frac{8}{\Lambda }\biggr)R \int^{\infty}_{0}\bigl[\bigl(1+t^{\alpha-1} \bigr)a(t)+b(t)\bigr]\,dt\\ &\qquad{}+\biggl( \frac{4}{\Gamma (\alpha)}+\frac{8}{\Lambda}\biggr) \int^{\infty}_{0}c(t)\,dt, \\ & \biggl|\frac{T'u_{n}(t)}{1+t^{\alpha-2}}-\frac {T'u(t)}{1+t^{\alpha-2}} \biggr| \\ &\quad\leq \int^{t}_{0}\frac{(t-s)^{\alpha-2}}{\Gamma(\alpha )(1+t^{\alpha -1})} \bigl|f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr)-f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr) \bigr|\,ds \\ &\qquad{}+\frac{(\alpha-1)t^{\alpha-2}}{(1+t^{\alpha-2})\Lambda} \sum^{m-2}_{i=1} \beta_{i} \int_{0}^{\xi_{i}}\frac{(\xi{i}-s)^{\alpha -1}}{\Gamma(\alpha)} \bigl|f\bigl(s, u_{n}(s), D^{\alpha -1}u_{n}(s)\bigr)\\ &\qquad{}-f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr) \bigr|\,ds \\ &\qquad{}+\frac{(\alpha-1)t^{\alpha-2}}{(1+t^{\alpha-2})\Lambda } \int^{\infty}_{0} \bigl|f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr)-f\bigl(s,\ u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr) \bigr|\,ds \\ &\quad\leq\biggl( \frac{2}{\Gamma(\alpha)}+\frac{4(\alpha -1)}{\Lambda}\biggr) \int^{\infty}_{0}\bigl|f\bigl(s, u_{n}(s), D^{\alpha -1}u_{n}(s)\bigr)-f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr) \bigr|\,ds \\ &\quad\leq\biggl( \frac{4}{\Gamma(\alpha)}+\frac{8(\alpha -1)}{\Lambda}\biggr)R \int^{\infty}_{0}\bigl[\bigl(1+t^{\alpha-1} \bigr)a(t)+b(t)\bigr]\,dt\\ &\qquad{}+\biggl( \frac {4}{\Gamma(\alpha)}+\frac{8(\alpha-1)}{\Lambda}\biggr) \int^{\infty}_{0}c(t)\,dt, \\ &\bigl|D^{\alpha-1}Tu_{n}(t)-D^{\alpha-1}Tu(t)\bigr| \\ &\quad\leq \int ^{\infty}_{0} \bigl|f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr)-f\bigl(s,\ u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr) \bigr|\,ds \\ &\qquad{}+\frac{\Gamma(\alpha)}{\Lambda} \sum^{m-2}_{k=1} \beta _{i} \int_{0}^{\xi_{i}}\frac{(\xi_{i}-s)^{\alpha-1}}{\Gamma(\alpha )} \bigl|f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr)-f\bigl(s, u_{n}(s), D^{\alpha -1}u_{n}(s)\bigr) \bigr|\,ds \\ &\qquad{}+\frac{\Gamma(\alpha)}{\Lambda} \int^{\infty}_{0} \bigl|f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr)-f\bigl(s, u_{n}(s), D^{\alpha -1}u_{n}(s)\bigr) \bigr|\,ds \\ &\quad\leq\biggl( \frac{2}{\Gamma(\alpha)}+\frac{2+2\Gamma(\alpha )}{\Lambda}\biggr) \int^{\infty}_{0}\bigl|f\bigl(s, u_{n}(s), D^{\alpha -1}u_{n}(s)\bigr)-f\bigl(s, u_{n}(s), D^{\alpha-1}u_{n}(s)\bigr) \bigr|\,ds \\ &\quad\leq\biggl( \frac{4}{\Gamma(\alpha)}+\frac{4+4\Gamma(\alpha )}{\Lambda}\biggr)R \int^{\infty}_{0}\bigl[\bigl(1+t^{\alpha-1} \bigr)a(t)+b(t)\bigr]\,dt\\ &\qquad{}+\biggl( \frac {4}{\Gamma(\alpha)}+\frac{4+4\Gamma(\alpha)}{\Lambda}\biggr) \int ^{\infty}_{0}c(t)\,dt. \end{aligned}$$
Then the operator T is continuous in view of the Lebesgue dominated convergence theorem. Thus by Schauder’s fixed point theorem we conclude that the problem (1.1)-(1.2) has at least one solution in U and the proof is completed. □