In this section, we study the existence of solutions to the weighted functional differential equations (1)-(2). We begin with the definition of solutions to these equations.

**Definition 4.1:** *A function y* : (−∞, *b*] → *X is said to be a solution to* (1)-(2), *if* $y|(0,b])\in C((0,b]){{\displaystyle \cap}}^{\text{}}{L}_{loc}^{1}(0,b)$, *ỹ*_{0} = ϕ *and satisfies* (1).

For the existence results on the problem (1)-(2), we need to transform the fractional differential equation into an integral equation. From Lemma 2.3 we can obtain that, if 0 < *α* < 1 and $h\in C|(0,b]){{\displaystyle \cap}}^{\text{}}{L}_{loc}^{1}(0,b)$ such that *I*^{1−α}h is absolutely continuous on [0, *b*], then the function *y* solves the fractional differential equation
$${D}^{\alpha}y(t)=h(t),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in (0,b]$$
if and only if *y* satisfies
$$y(t)=\frac{1}{\Gamma (\alpha )}\underset{0}{\overset{t}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}h(s)ds+C{t}^{\alpha -1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in (0,b]$$
for some constant *C*. See also [5, Theorem 2.23].

For the forthcoming analysis, we need the following hypothesis.

(H1) *f* : (*0, b*] × *𝓑* → *X* is continuous.

(H2) There exist a nonnegative *η* ∈ *L*^{p}(0, b] with *p > 1/α* and a continuously non-decreasing function Ω : [*0*,+∞) → [*0*,+∞) such that
$$\parallel f(t,u)\parallel \le \eta (t)\Omega (\Vert u\Vert \mathcal{B})$$(10)
for *t* ∈ *(0, b]* and every *u* ∈ *𝓑*, and
$$\beta (f(t,D))\le \eta (t)\text{\hspace{0.17em}}\underset{-\infty <\theta \le 0}{\mathrm{sup}}\text{\hspace{0.17em}}\beta (D(\theta ))$$(11)
for *t* ∈ *(0, b*] and every bounded subset *D* ⊂ 𝓑, where *D(θ)* = {*u(θ)* : *u* ∈ *D*}.

**Theorem 4.2:** *Assume the hypotheses (H1) and (H2) hold. If*
$$\underset{r\to +\infty}{\mathrm{lim}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sup}\text{\hspace{0.17em}}\frac{\Omega (r)}{r}<\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\Gamma (1+\alpha )}{{K}_{b}{b}^{2}\parallel \eta \parallel p},$$(12)
*then there exists at least a solution to* (1)-(2) *on* (−∞,*b*].

**Proof:** *From the comment above, we know that **y* is a solution to (1) if and only if *y* satisfies
$$\begin{array}{cc}y(t)=\{\begin{array}{l}\frac{1}{\Gamma (\alpha )}\underset{0}{\overset{t}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}f(s,\stackrel{~}{y}s)ds+\varphi (0){t}^{\alpha -1},\hfill \\ \varphi (t),\hfill \end{array}& \begin{array}{l}t\in (0,b],\hfill \\ t\in (-\infty ,0].\hfill \end{array}\end{array}$$For given *ϕ* : (−∞, 0] which belongs to 𝓑, let ϕ̃ be a function defined by
$$\begin{array}{cc}\stackrel{~}{\varphi}(t)=\{\begin{array}{l}0,\hfill \\ \varphi (t),\hfill \end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}& \begin{array}{l}t\in (0,b],\hfill \\ t\in (-\infty ,0].\hfill \end{array}\end{array}$$Then we have ϕ̃_{0} = *ϕ*. For *z* ∈ *C*_{1−α}((0, *b*], *X*), where *C*_{1−α} (*0, b*],*X*) is the Banach space consisting of all continuous functions *f* : (*0, b*] → *X* such that lim_{t→0} *t*^{1−α} *f*(*t*) exists, endowed with the norm ∥*f*∥_{C1−α} = sup|*t*^{1−α} *f*(*t*)|; *t* ∈ *(0, b]*, we extend z̃ to (−∞, *b*], also denoted by e *z̃*, defined by
$$\begin{array}{cc}\stackrel{~}{z}(t)=\{\begin{array}{l}{t}^{1-\alpha}z(t),\hfill \\ 0,\hfill \end{array}& \begin{array}{l}t\in (0,b],\hfill \\ t\in (-\infty ,0].\hfill \end{array}\end{array}$$It is easily seen that if *y(·)* satisfies the integral equation
$$\begin{array}{cc}y(t)=\frac{1}{\Gamma (\alpha )}\underset{0}{\overset{t}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}f(s,\stackrel{~}{y}s)ds+\varphi (0){t}^{\alpha -1},& t>0,\end{array}$$
we can decompose *y(·)* as *y(t)* = *ϕ(t)* + *z(t)*, which implies that *ỹ* = *ϕ̃*_{t} + *z̃*_{t} for *t* ∈ (*0, b*], and the function *z(·)* satisfies
$$\begin{array}{cc}z(t)=\frac{1}{\Gamma (\alpha )}\underset{0}{\overset{t}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}f(s,\stackrel{~}{\varphi}s+\stackrel{~}{z}s)ds+\varphi (0){t}^{\alpha -1},& t>0.\end{array}$$(13)Set *W* = {*z* : (−∞] → *X*;*z*|(*0, b*]; ∈. *C*_{1−α}. ((*0, b*]; X), *z*_{0} = 0}. For *z* ∈ *W*, define ∥*z*∥_{W} = ∥*z*_{0}∥_{𝓑} + ∥*z*∥_{C1−α}. Then (*W*,∥*z*∥_{W}) becomes a Banach space. Define an operator *P* : *W* → *W* by
$$\begin{array}{cc}(pz)(t)=\frac{1}{\Gamma (\alpha )}\underset{0}{\overset{t}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}f(s,\stackrel{~}{\varphi}s+\stackrel{~}{z}s)ds+\varphi (0){t}^{\alpha -1},& t>0.\end{array}$$We will prove by the Schauder’s fixed point theorem that *P* has at least a fixed point *z*, and hence *z* + *ϕ̃* is a solution to (1)-(2).First note that the continuity of *P* can be derived by *(H2)* and the Lebesgue dominated convergence theorem. We now show that *P* maps bounded subsets in *W* into bounded subsets. Let *B*_{r} = {*z* ∈ *W*; ∥*z*∥_{W} ≤ *r*}. Then, for any *z* ∈ *B*_{r} and *t* ∈ *(0, b]*, we have
$$\begin{array}{ll}\Vert {t}^{1-\alpha}(pz)(t)\Vert \hfill & \le \frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{\alpha -1}\parallel f(s,\tilde{\varphi}s+\tilde{z}}s)\parallel ds+\parallel \varphi (0)\parallel \hfill \\ \hfill & \le \frac{{b}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{\alpha -1}\eta (s)\Omega (\parallel \tilde{\varphi}}s+\tilde{z}s\parallel \beta )ds+\parallel \varphi (0)\parallel .\hfill \end{array}$$Since
$$\parallel {\tilde{z}}_{s}+{\tilde{\varphi}}_{s}{\parallel}_{\mathcal{B}}\le K(s)\underset{0\le \tau \le s}{\mathrm{sup}}\parallel \tilde{z}(\tau )\parallel +M(s)\parallel {\tilde{z}}_{0}{\parallel}_{\mathcal{B}}+K(s)\mathrm{sup}\parallel \tilde{\varphi}(\tau )\parallel +M(s)\parallel \tilde{{\varphi}_{0}}{\parallel}_{\mathcal{B}}\le {K}_{b}r+{M}_{b}\parallel \varphi {\parallel}_{\mathcal{B}},$$
where *M*_{b} = sup_{0≤s≤b} *M(s)*. It follows from (H2) and Holder’s inequality that
$$\begin{array}{ll}\parallel {t}^{1-\alpha}(pz)(t)\parallel \hfill & \le \frac{{b}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{\alpha -1}\eta (s)ds\Omega ({K}_{b}r+{M}_{b}\parallel \varphi \parallel \mathcal{B})+\parallel \varphi (0)\parallel}\hfill \\ \hfill & \le \frac{{b}^{1-\alpha}}{\Gamma (\alpha )}\Omega ({K}_{b}r+{M}_{b}\parallel \varphi \parallel \mathcal{B})({\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{(\alpha -1)q}ds{)}^{1/q}\parallel \eta \parallel p+\parallel \varphi (0)\parallel .}\hfill \\ \hfill & \le \Omega ({K}_{b}r+{M}_{b}\parallel \varphi \parallel \mathcal{B})\frac{{b}^{2}\parallel \eta \parallel p}{\Gamma (1+\alpha )}+\parallel \varphi (0)\parallel \text{\hspace{0.17em}}:=l,\hfill \end{array}$$
where ${\Vert \eta \Vert}_{p}={({\displaystyle {\int}_{o}^{b}|\eta (s){|}^{p}}ds)}^{1/p}$ and 1/*p* + 1/*q* = 1; (*α* − 1)*q* > −1. Therefore, ∥*Pz*∥_{W} ≤ *l* for every *z* ∈ *B*_{r}, which implies that *P* maps bounded subsets into bounded subsets in *W*.Next we prove that *PB̃* is equicontinuous for every bounded subsets *B* ⊂ *W*. Let *z* ∈ *B*_{r} and *t*_{1}, *t*_{2} ∈ *(0, b]* with *t*_{1} < *t*_{2}, then we have
$$\begin{array}{l}\parallel {t}_{2}^{1-\alpha}(pz)({t}_{2})-{t}_{1}^{1-\alpha}(pz)({t}_{1})\parallel \\ \le \frac{{t}_{2}^{1-\alpha}-{t}_{2}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{{t}_{2}}{\int}}{({t}_{2}-s)}^{\alpha -1}\parallel f(s,\tilde{\varphi}}s+\tilde{z}s)\parallel ds+\frac{{t}_{1}^{1-\alpha}}{\Gamma (\alpha )}[{\displaystyle \underset{0}{\overset{{t}_{1}}{\int}}({({t}_{2}-s)}^{\alpha -1}-{({t}_{1}-s)}^{\alpha -1})\parallel f(s,\tilde{\varphi}s+\tilde{z}s)\parallel ds}\\ +{\displaystyle \underset{{t}_{1}}{\overset{{t}_{2}}{\int}}{({t}_{2}-s)}^{\alpha -1}\parallel f(s,\tilde{\varphi}}s+\tilde{z}s)\parallel ds]\\ \le \frac{{t}_{2}^{1-\alpha}-{t}_{1}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{{t}_{2}}{\int}}{({t}_{2}-s)}^{\alpha -1}}\eta (s)\Omega (\parallel \tilde{\varphi}s+\tilde{z}s\parallel \mathcal{B})ds\\ +\frac{{t}_{1}^{1-\alpha}}{\Gamma (\alpha )}[{\displaystyle \underset{0}{\overset{{t}_{1}}{\int}}({({t}_{2}-s)}^{\alpha -1}-{({t}_{1}-s)}^{\alpha -1})\eta (s)\Omega (\parallel \tilde{\varphi}s+\tilde{z}s\parallel \mathcal{B})ds+{\displaystyle \underset{{t}_{1}}{\overset{{t}_{2}}{\int}}{({t}_{2}-s)}^{\alpha -1}\eta (s)\Omega (\parallel \tilde{\varphi}s+\tilde{z}s\parallel \mathcal{B})ds]}}\\ \le \frac{({t}_{z}^{1-\alpha}-{t}_{1}^{1-\alpha})\Omega (ro)}{\Gamma (\alpha )}({\displaystyle \underset{0}{\overset{{t}_{2}}{\int}}{({t}_{2}-s)}^{{(\alpha -1)}^{q}ds{)}^{1/q}}({\displaystyle \underset{0}{\overset{{t}_{2}}{\int}}{\eta}^{p}(s)ds{)}^{1/p}}}\\ +\frac{{t}_{1}^{1-\alpha}\Omega (ro)}{\Gamma (\alpha )}[({\displaystyle \underset{0}{\overset{{t}_{1}}{\int}}{({({t}_{2}-s)}^{\alpha -1}-{({t}_{1}-s)}^{\alpha -1})}^{q}ds{)}^{1/q}({\displaystyle \underset{0}{\overset{{t}_{1}}{\int}}{\eta}^{p}(s)ds{)}^{1/p}}}\\ +({\displaystyle \underset{{t}_{1}}{\overset{{t}_{2}}{\int}}{({t}_{2}-s)}^{(\alpha -1)q}ds{)}^{1/q}{({\displaystyle \underset{0}{\overset{{t}_{1}}{\int}}{\eta}^{p}}(s)ds)}^{1/p}}\\ \le \frac{\parallel \eta \parallel p\Omega (ro)}{{r}_{1}\Gamma (\alpha )}({b}^{{r}_{2}}({t}_{z}^{1-\alpha}-{t}_{1}^{1-\alpha})+(2{(}^{{t}_{2}}+({t}_{2}^{{r}_{2}}-{t}_{1}^{{r}_{2}}))),\end{array}$$
where *r*_{0} = *K*_{b}r + *M*_{b}||*Φ*||_{B}, *r*_{1} = ((*α* − 1) *q* + 1)^{1/q} and *r*_{2} = [(*α*−1)*q*+1]/*q* > 0. It follows that $\Vert {t}_{2}^{1-\alpha}(pz)({t}_{2})-{t}_{1}^{1-\alpha}(pz)({t}_{1})\Vert \to 0\text{\hspace{0.05em}}\text{\hspace{0.17em}}as\text{\hspace{0.17em}}{t}_{2}\to {t}_{1}\to 0$, and the convergence is independent of *z* ∈ *B*_{r}, which implies that the set *PB*_{r}̃ is equicontinuous.Now we have to verify that there exists a closed convex bounded subset *B*_{r0}, such that *PB*_{r0} ⊂ *B*_{r0}. We derive from inequlity (12), i.e., ${\mathrm{lim}}_{r\to +\infty}\frac{\Omega (r)}{r}<\frac{\Gamma (1+\alpha )}{{k}_{b}{b}^{2}{\Vert \eta \Vert}_{p}}$, that there exists a constant *r*_{0} > 0 such that
$$\frac{{b}^{2}\parallel \eta \parallel p}{\Gamma (1+\alpha )}\Omega ({K}_{b}ro+{M}_{b}\parallel \varphi \parallel \mathcal{B})+\parallel \varphi (0)\parallel \text{\hspace{0.17em}}\text{\hspace{0.17em}}<\text{\hspace{0.17em}}ro.$$Define *B*_{r0} = {*z* ∈ *W*; ∥*z*∥_{W} ≤ *r*_{0}}. Then *B*_{r0} is closed, convex and bounded in *W*. Then, for every *z* ∈ *B*_{r0} and *t* ∈ *(0, b]*, similar to the proof of *P* maps bounded subsets in *W* into bounded subset, we have
$$\begin{array}{ll}\parallel {t}^{1-\alpha}({p}_{1}z)(t)-{t}^{1-\alpha}({p}_{1}{z}^{*})(t)\parallel \hfill & \le \frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{\alpha -1}}\parallel f(s,\tilde{\varphi}s+\tilde{z}s)-f(s,\tilde{\varphi}s+{\tilde{z}}_{s}^{*}\parallel ds\hfill \\ \hfill & \le \frac{L{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{\alpha -1}\parallel \tilde{z}s-{\tilde{z}}_{s}^{*}\parallel \mathcal{B}ds}\hfill \end{array}$$It then follows that ∥*Pz*∥_{W} ≤ *r*_{0} for all *z* ∈ *Br*_{0}, and hence *PB*_{r0} ⊂ *B*_{r0}.Now we prove that there exists a compact subset *M* ⊂ *B*_{r0} such that *PM* ⊂ *M*. We first construct a series of sets {*M*_{n}} ⊂ *B*_{r0} by
$$\begin{array}{ccc}{M}_{0}={B}_{ro},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{M}_{1}=\overline{conv}pMo,& {M}_{n+1}=\overline{conv}p{M}_{n},& n=1,2,\mathrm{....}\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\end{array}$$From the above proof it is easy to see that *M*_{n+1} ⊂ *M*_{n} for *n* = *1, 2*, … and each *M̃*_{n} is equi-continuous. Further, from Lemma 2.5, 2.6 and 2.7 we can derive that
$$\begin{array}{l}B({\tilde{M}}_{n+1}(t))=\beta ({t}^{1-\alpha}{M}_{n+1}(t))=\beta ({t}^{1-\alpha}P{M}_{n}(t))\\ =\beta \left[\frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle {\int}_{0}^{t}{(t-s)}^{\alpha -1}f(s,\tilde{\varphi}s+{\tilde{M}}_{ns})}ds+\varphi (0)\right]\\ \le \beta \left[\frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle {\int}_{0}^{t}{(t-s)}^{\alpha -1}f(s,\tilde{\varphi}s+{\tilde{M}}_{ns})}ds\right]\\ \le \frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle {\int}_{0}^{t}{(t-s)}^{\alpha -1}f(s,\tilde{\varphi}s+{\tilde{M}}_{ns}))}ds]\\ \le \frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle {\int}_{0}^{t}{(t-s)}^{\alpha -1}\eta (s)su{p}_{-\infty <\theta \le 0}}\beta (\tilde{\varphi}s(\theta )+{\tilde{M}}_{ns}(\theta ))ds\\ \le \frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle {\int}_{0}^{t}{(t-s)}^{\alpha -1}\eta (s)su{p}_{-\infty <\theta \le 0}}[\beta (\tilde{\varphi}s(\theta )+\beta {\tilde{M}}_{ns}(\theta ))]ds\\ \le \frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle {\int}_{0}^{t}{(t-s)}^{\alpha -1}\eta (s)su{p}_{-\infty <\theta \le 0}}\beta ({\tilde{M}}_{ns}(\theta ))ds\\ =\frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle {\int}_{0}^{t}{(t-s)}^{\alpha -1}\eta (s)su{p}_{-\infty <\theta \le 0}}\beta ({s}^{1-\alpha}{(}^{\theta}{(}^{\theta}{M}_{n}(\theta +s))ds\\ =\frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle {\int}_{0}^{t}{(t-s)}^{\alpha -1}\eta (s)su{p}_{-\infty <\theta \le 0}}\beta ({s}^{1-\alpha}{(}^{\theta}{(}^{\theta}{M}_{n}(\theta +s))ds\end{array}$$(14)From the proof above we know that {(*θ* + *s*)^{1−α} *M*_{n}(*θ*+*s*)} is equicontinuous at *θ* and *θ*+*s*)^{1−α} *M*_{n}(*θ* + *s*)|_{θ=−s} = {*ϕ*(0)}. So from Lemma 2.5 we know that the function *θ* ↦ *β*((θ+*s*)^{1−α} *M*_{n}(*θ*+*s*))is continuous, and *β*((θ+*s*)^{1−α} *M*_{n}. (*θ*+*s*))|_{θ=−s} = 0. Therefore, if the supermum sup_{−s≤θ≤0} *β*(*s*^{1−α}(θ + *s*)^{α−1}(*θ* + *s*^{1−α} *M*_{n} (*θ* + *s*)) is taken at *θ* = −*s*, then sup_{−s≤θ≤0} *β*(*s*^{1−α}(*θ* + *s*)^{α−1} (*θ* + *s*)^{1−α} *M*_{n} (*θ* + *s*)) ≡ 0. Otherwise, there exist δ_{n} > 0, such that
$$\begin{array}{ll}\hfill & {\mathrm{sup}}_{-s\le \theta \le 0}\beta ({s}^{1-\alpha}{(\theta +s)}^{\alpha -1}{(\theta +s)}^{1-\alpha}{M}_{n}(\theta +s))\hfill \\ =\hfill & {\mathrm{sup}}_{-s+{\delta}_{n}\le \theta \le 0}\beta ({s}^{1-\alpha}{(\theta +s)}^{\alpha -1}({(\theta +s)}^{1-\alpha}{M}_{n}(\theta +s))\hfill \\ =\hfill & {\mathrm{sup}}_{{\delta}_{n}\le \tau \le s}\beta ({s}^{1-\alpha}{\tau}^{\alpha -1}{\tau}^{1-\alpha}{M}_{n}(\tau ))\hfill \\ \le \hfill & {\mathrm{sup}}_{{\delta}_{n}\le \tau \le s}\beta ({s}^{1-\alpha}{\delta}_{n}^{{}^{\alpha -1}}{\tau}^{1-\alpha}{M}_{n}(\tau ))\hfill \\ =\hfill & {s}^{1-\alpha}{\delta}_{n}^{\alpha -1}{\mathrm{sup}}_{0\le \tau \le s}\beta ({\tilde{M}}_{n}(\tau )).\hfill \end{array}$$(15)Subsitting (15) into (14) we obtain that
$$\beta ({\tilde{M}}_{n+1}(t))\le \frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{\alpha -1}\eta (s){s}^{1-\alpha}{\delta}_{n}^{\alpha -1}}\underset{0\le \tau \le s}{\mathrm{sup}}\beta ({\tilde{M}}_{n}(\tau ))\mathrm{ds}$$(16)
for *n* = *1, 2*, …. Define the functions *f*_{n}(t) = sup_{τ ∈ [0, t]} *β*(*M̃*_{n}(*τ*)) for *n* = 1,2,… and take supermum on both sides of (16), We get due to Lemma 3.9 that
$${f}_{n+1}(t)\le \frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{\alpha -1}\eta (s){s}^{1-\alpha}{\delta}_{n}^{\alpha -1}}{f}_{n}(s)\mathrm{ds}$$(17)
for *n* = *1, 2*, …. The fact that *M*_{n+1} ⊂ *M*_{n} implies that *f*_{n+1}*(t)* ≤ *f*_{n}*(t)* for all *t* ∈ [0, *b*] and *n* = *1, 2*, …, and each *f*_{n} is continuous on [0, *b*]. Therefore, the limit lim_{n} *f*_{n}*(t)* = *f(t)* exists for *t* ∈ [0, *b*]. We claim that *f(t)* = 0 for all *t* ∈ [0, *b*]. In fact, take inf_{n≤1} *δ*_{n} = *δ*. If *δ* > 0, then inequality (17) implies that
$${f}_{n+1}(t)\le \frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{\alpha -1}\eta (s){s}^{1-\alpha}{\delta}^{\alpha -1}f(s)s}$$(18)
for all *t* ∈ [0,*b*] and *n* = 1, 2, …. Taking limit as *n* → ∞ in (18) we get
$$f(t)\le \frac{{t}^{1-\alpha}}{\Gamma (\alpha )}\underset{0}{\overset{t}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}\eta {(s)}^{1-\alpha \text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta \alpha -1}f(s)ds$$
for all *t* ∈ [0, *b*]. An application of Lemma 3.7 yields *f(t)* = 0 for all *t* ∈ [0, *b*].If *δ* = 0, then there are two cases. In case when there exists an *n*_{0} such that *δ*_{n0} = 0, *f*_{n0}*(t)* = 0 for all *t* ∈ [0, *b*] according to the definition of δ_{n}, the inequality (17) implies that *f*_{n}*(t)* = 0 for all *n* > *n*_{0} and *t* ∈ [0, *b]*, and hence *f(t)* = lim_{n} *f*_{n}*(t)* = 0 for all *t* ∈ [0, *b]*. In case when δ_{n} ≠ 0 for all *n* ≥ 1, lim_{n} δ_{n} = 0, the definition of *f*_{n} and *δ*_{n} imply that lim_{n} *f*_{n}*(t)* = 0 = *f(t)* for all *t* ∈ [0, *b*]. So we have *f(t)* = 0 for all *t* ∈ [0, *b]* in each case, as claimed. Therefore, ${\cap}_{n=1}^{\infty}{M}_{n}=M$ is nonempty and compact in *W* due to Lemma 2.5, and *PM* ⊂ *M* by the definition of *M*_{n}.Up to now we have verified that there exists a nonempty bounded convex and compact subset *M* ⊂ *W* such that *PM* ⊂ *M*. An employment of Schauder’s fixed point theorem shows that there exists at least a fixed point *z* of *P* in *M*. Then *y* = *z* + *ϕ̃* is the solution to (1)-(2) on [0, *b*], which completes the proof. ⎕

Below we consider the existence result which is based on the Lipschitz condition. We need the following hypothesis.

(H3) There exists a constant *L* > 0 such that
$$\parallel f(t,u)-f(t,u)\parallel \le L\parallel u-v\parallel \mathcal{B}$$
for *t* ∈ *(0, b]* and every *u, v* ∈ 𝓑.

**Theorem 4.3:** *Assume that (H1) and (H3) hold. Then there exists a unique solution to* (1)-(2) *on* (−∞, *b*].

**Proof:** *As in the proof of Theorem 4.2, we define the operator **P* : *W* → *W* by
$$\begin{array}{ll}(PZ)(t)=\frac{1}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{\alpha -1}f(s,\tilde{\varphi}s+\tilde{z}s})s+\varphi (0){t}^{\alpha -1},\hfill & t\text{\hspace{0.17em}}>0.\hfill \end{array}$$(19)If *z* ∈ *W* is a fixed point of *P*, then *y* = *z* + *ϕ̃* is a solution of (1)-(2).Let *K(b)* = sup{*K(t)*; *t* ∈ *(0, b]*}, where *K(·)* is the function that appeared in Definition 2.4. Let *N* = [*b*(2*LK*_{b}*b*^{1−α}=Γ(1+*α*))^{1/α}], and *h*_{i} = *ib/N*. Then 0 < *h*_{1} < *h*_{2} < … < *h*_{N} = *b* and
$$\frac{L{K}_{b}{b}^{1-\alpha}{({h}_{i+1}-{h}_{i})}^{\alpha}}{\Gamma (1+\alpha )}<\frac{1}{2}$$(20)
for *i* = *1, 2, …, N:*We first focus on the interval(0, *h*_{1}]. Let *W*_{1} = {*z* : (−∞, *h*_{1}] → *X* ; *z* |_{0,h1}] ∈ *C*_{1−α}((0, *h*_{1}]; *X*), *z*|_{0} = 0} and define ∥*z*∥_{W1} = ∥z_{0}∥_{𝓑} + sup{|*t*^{1−α}*z*(*t*)|; 0 < *t* < *h*_{1}} = sup{|*t*^{1−α}*z*(*t*)|; 0 < t < *h*_{1}} for *z* ∈ *W*_{1}.Then (*W*_{1}, ∥*z*∥_{W1}) is a Banach space. Define the operator *P*_{1} : *W*_{1} → *W*_{1} by
$$\begin{array}{cc}({p}_{1}z)(t)=\frac{1}{\Gamma (\alpha )}\underset{0}{\overset{t}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}f(s,\stackrel{~}{\varphi}s+\stackrel{~}{z}s)ds+\varphi (0){t}^{\alpha -1}.& t\in (0,{h}_{1}].\end{array}$$(21)For *z, z** ∈ *W*_{1} and *t* ∈ (0, *h*_{1}], we have
$$\begin{array}{ll}\parallel {t}^{1-\alpha}({p}_{1}z)(t)-{t}^{1-\alpha}({p}_{1}{z}^{*})(t)\parallel \hfill & \le \frac{{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{\alpha -1}}\parallel f(s,\tilde{\varphi}s+\tilde{z}s)-f(s,\tilde{\varphi}s+{\tilde{z}}_{s}^{*}\parallel ds\hfill \\ \hfill & \le \frac{L{t}^{1-\alpha}}{\Gamma (\alpha )}{\displaystyle \underset{0}{\overset{t}{\int}}{(t-s)}^{\alpha -1}\parallel \tilde{z}s-{\tilde{z}}_{s}^{*}\parallel \mathcal{B}ds}\hfill \end{array}$$Since
$$\begin{array}{ll}\parallel \tilde{z}s-{\tilde{z}}_{s}^{*}\parallel \mathcal{B}\text{\hspace{0.17em}}\hfill & \le K(s)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{0\le \tau \le s}{\mathrm{sup}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left\{\parallel \tilde{z}(\tau )-{\tilde{z}}^{*}(\tau )\parallel \right\}+M(s)\parallel \tilde{z}0-{\tilde{z}}_{0}^{*}\parallel \mathcal{B}\hfill \\ \hfill & \le {K}_{b}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{0\le \tau \le s}{\mathrm{sup}}\text{\hspace{0.17em}}\left\{\parallel {\tau}^{1-\alpha}z(\tau )-{\tau}^{1-\alpha}{z}^{*}(\tau )\parallel \right\}\le {K}_{b}\parallel z-{z}^{*}\parallel {W}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \end{array}$$
we have
$$\parallel {t}^{1-\alpha}({p}_{1}z)(t)-{t}^{1-\alpha}({p}_{1}{z}^{*})(t)\parallel \le \frac{L{K}_{b}{t}^{1-\alpha}}{\Gamma (\alpha )}\underset{0}{\overset{t}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}ds\parallel z-{z}^{*}\parallel {W}_{1}\le \frac{L{K}_{b}{b}^{1-\alpha}{h}_{1}^{\alpha}}{\Gamma (1+\alpha )}\parallel z-{z}^{*}\parallel {W}_{1},$$
and hence
$$\parallel \text{\hspace{0.17em}}{p}_{1}z\text{\hspace{0.17em}}-\text{\hspace{0.17em}}{p}_{1}z\ast \parallel {w}_{1}\le \frac{L{K}_{b}{b}^{1-\alpha}{h}_{1}^{\alpha}}{\Gamma (1+\alpha )}\parallel z-{z}^{\ast}\parallel {W}_{1}.$$(22)From (20) and the Banach contraction principle we know that there exists a unique *z* ∈ *W*_{1} satisfying
$$z(t)=\frac{1}{\alpha}\underset{0}{\overset{t}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}f(s,\stackrel{~}{{\phi}_{s}}+{\stackrel{~}{z}}_{s})ds\text{\hspace{0.17em}}+\phi (0){t}^{\alpha -1}$$(23)
for *t* ∈(0, *h*_{1}], which is the unique solution to the integral equation (13) on the interval (0, *h*_{1}].Next we consider the interval (*h*_{1}, *h*_{2}]. Restrict the function *z* ∈ *W* on the interval (*h*_{1}, *h*_{2}] to construct *W*_{2} and define ∥*z*∥_{W2} = ∥z_{0}∥_{𝓑} + sup{|*t*^{1−α} *z*(*t*)|; *h*_{1} < *t* ≤ *h*_{2}} = sup{|*t*^{1−α} z(t);| *h*_{1} < *t* ≤ h_{2}} for *z* ∈ *W*_{2}. Then (*W*_{2}, ∥*z*∥ _{W2}) is Banach space. For *t* ∈ (*h*_{1}, *h*_{2}], rewrite equation (13) as
$$z(t)=\frac{1}{\Gamma (\alpha )}\underset{{h}_{1}}{\overset{t}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}f(s,\stackrel{~}{\varphi}s+\stackrel{~}{z}s)ds+\frac{1}{\Gamma (\alpha )}\underset{0}{\overset{{h}_{1}}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}f(s,\stackrel{~}{\varphi}s+\stackrel{~}{z}s)ds+\varphi (0){t}^{\alpha -1}$$(24)Since the function *z* is uniquely defined on (0, *h*_{1}], the second integral can be considered as a known function. Using the same arguments as above, we can obtain that there exists a unique function *z* ∈ *W*_{2} satisfying
$$z(t)=\frac{1}{\Gamma (\alpha )}\underset{{h}_{1}}{\overset{t}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}f(s,\stackrel{~}{\varphi}s+\stackrel{~}{z}s)ds+\frac{1}{\Gamma (\alpha )}\underset{0}{\overset{{h}_{1}}{{\displaystyle \int}}}{(t-s)}^{\alpha -1}f(s,\stackrel{~}{\varphi}s+\stackrel{~}{z}s)ds+\varphi (0){t}^{\alpha -1}$$(25)
for *t* ∈ (*h*_{1}, *h*_{2}], which is the unique solution to the integral equation (13) on the interval (*h*_{1}, *h*_{2}]. Taking the next interval (*h*_{2}, *h*_{3}], repeating this process, we conclude that there exists a unique solution to the integral equation (13) on the interval (0, *h*_{N}] = *(0, b]*. Set *y* = *z* + *ϕ̃*, then *y* is the unique solution to the fractional differential equation (1)-(2). ⎕

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