Throughout this section, we denote the set

Y of all continuous functions on

J by

**Proof.** We introduce a function

${d}_{1}:$ $Y\times Y\to [0,\infty ]$, where

Y defined by (

5) with

where

$K=\frac{(L+1)|{b}_{\gamma}|}{1-(L+1)|{a}_{\gamma}|}>0$ and

${a}_{\gamma},{b}_{\gamma}$ are given in (

2)

Let $\psi (\xb7)={e}^{K(\xb7-{\tau}_{0})}$ in Lemma 1, we obtain $(Y,{d}_{1})$ is a generalized complete metric space.

Next, we consider the operator

$P:Y\to Y$ as follows:

for any

$f,g\in Y$, where

${f}_{0}=f({\tau}_{0})$. Please note that any fixed point of

P solves (

7). Indeed, the function

$u-{a}_{\gamma}g(\tau ,u)=v$ in (

10) is invertible, it is increasing. We denote its inverse

$u=G(\tau ,v)$, and

G is globally Lipschitz in

v and locally Lipschitz in

$\tau $ by our assumptions. So, any fixed point of (

10) satisfies

Now clearly the function

$\tau \to {b}_{\gamma}{\int}_{{\tau}_{0}}^{\tau}g(s,f(s))ds+{f}_{0}$ is locally Lipschitz in

$\tau $, we see that the composition function

$\tau \to G(\tau ,{b}_{\gamma}{\int}_{{\tau}_{0}}^{\tau}g(s,f(s))ds+{f}_{0})$ is also locally Lipschitz in

$\tau $. So, any fixed point

$f(\tau )$ of (

10) is a locally Lipschitz function, and thus it is locally absolute continuous on

J. So really (

10) gives solutions of (

7). As a matter of fact, we need just that

$u-{a}_{\gamma}g(\tau ,u)=v$ is invertible, i.e.,

$u-{a}_{\gamma}g(\tau ,u)$ is strictly monotonic in

u, and we can extend our results for more general case. We shall consider (

11) instead of (

10).

We prove that

$Pf$ is continuous. Let

${\tau}_{1},{\tau}_{2}\in J$, and

${\tau}_{1}<{\tau}_{2}$, we have

Then, for all $f\in Y$, as ${\tau}_{1}\to {\tau}_{2}$, the right-hand side of the above inequality tends to zero (due to $\left[{A}_{1}\right]$ and $f\in Y$). Thus, $Pf$ is continuous, i.e., $Pf\in Y$ for all $f\in Y$.

Then, we have

for all

${f}_{0}\in Y,$ and

$\tau \in \phantom{\rule{4pt}{0ex}}J$. Therefore, by (

9), we obtain

${d}_{1}(P{f}_{0},{f}_{0})<\infty ,{f}_{0}\in Y.$Similarly, we have

for all

$f\in Y,$ and

$\tau \in \phantom{\rule{4pt}{0ex}}J$, which implies that

that is

$\{f\in Y\phantom{\rule{4pt}{0ex}}|\phantom{\rule{4pt}{0ex}}{d}_{1}({f}_{0},f)<\infty \}=Y.$Next, we show that

P is strictly contractive on

Y. For any

$l,n\in Y$, we get

for all

$\tau \in J$. Thus, for any

$l,n\in Y$ and all

$\tau \in J$, we have

Therefore, P is strictly contractive on Y.

When $k=1$ and $Y={\mathsf{\Omega}}^{*}$, the operator P satisfies all the conditions of Theorem 1.

On the other hand, by (

6), we have

Similar to the approach in ([

4], Theorem 2), we can obtain

for all

$\tau \in J$. From (

10), (

12) is equivalent to

Multiply both sides of (

13) by

${e}^{-K(\tau -{\tau}_{0})}$,

for all

$\tau \in J$. Then

By Theorem 1, there exists a unique solution

$f:$ $J\to \mathbb{R}$ of (

7) satisfying

by (

9), we have

which implies that (

8) holds. □

**Proof.** We introduce a function

${d}_{2}:$ $Y\times Y\to [0,\infty ]$, where

Y defined by (

5) with

Let $\psi (\xb7)={e}^{K(\xb7-{\tau}_{0})}G(\xb7)$ in the Lemma 1, $(Y,{d}_{2})$ is a generalized complete metric space.

Consider

$P:Y\to Y$ defined in (

10). Similar to the method of Theorem 2, we can conclude that

${d}_{2}(P{f}_{0},f)<\infty $ for each

${f}_{0}\in X$ and

$\{f\in Y\phantom{\rule{4pt}{0ex}}|\phantom{\rule{4pt}{0ex}}{d}_{2}({f}_{0},f)<\infty \}=Y$.

Next, we prove that

P is strictly contractive on

Y. Note

for all

$\tau \in J.$For any

$l,n\in Y$, let

${M}_{l,n}\in [0,\infty ]$ be an arbitrary constant with

${d}_{2}(l,n)\le {M}_{l,n}$, by (

17), we obtain

Then, for each

$l,n\in Y$, we have

for all

$\tau \in J$. Thus, for any

$l,n\in Y$ and all

$\tau \in J$, we have

that is,

${d}_{2}(Pl,Pn)\le \frac{L}{L+1}{M}_{l,n},\phantom{\rule{4pt}{0ex}}\forall \phantom{\rule{4pt}{0ex}}\tau \in J.$ Hence, we obtain

Therefore, P is strictly contractive on Y. When $k=1$ and $Y={\mathsf{\Omega}}^{*}$, the operator P satisfies all the conditions of Theorem 1.

On the other hand, by (

14), we have

By simple computation, we can obtain

Multiply both sides of (

18) by

${e}^{-K(\tau -{\tau}_{0})}$, then,

Then

By Theorem 1, there exists a unique solution

$f:$ $J\to \mathbb{R}$ of (

7) satisfying

By (

17), we have

which implies (

16) holds. The proof is complete. □