In this section, we show that many of the known fixed point results can be deduced from the following light version of Caristi’s theorem.

### Corollary 2.1

*Let*
\((X,d)\)
*be a complete metric space*, *and let*
\(T:X\to X\)
*be a mapping such that*

$$ d(x,y)\leq\varphi(x,y)-\varphi(Tx,Ty) $$

(5)

*for all*
\(x,y\in X\), *where*
\(\varphi:X\times X\to[0,\infty)\)
*is lower semicontinuous with respect to the first variable*. *Then*
*T*
*has a unique fixed point*.

### Proof

For each \(x\in X\), let \(y=Tx\) and \(\psi(x)=\varphi(x,Tx)\). Then for each \(x\in X\)

$$d(x,Tx)\leq\psi(x)-\psi(Tx) $$

and *ψ* is a lower semicontinuous mapping. Thus, applying Theorem 1.1 leads us to conclude the desired result. To see the uniqueness of the fixed point suppose that *u*, *v* are two distinct fixed points for *T*. Then

$$d(u,v)\leq\varphi(u,v)-\varphi(Tu,Tv)=\varphi(u,v)-\varphi(u,v)=0. $$

Thus, \(u=v\). □

### Corollary 2.2

([15], Banach contraction principle)

*Let*
\((X,d)\)
*be a complete metric space and let*
\(T:X\to X\)
*be a mapping such that for some*
\(\alpha\in[0,1)\)

$$ d(Tx,Ty)\leq\alpha d(x,y) $$

(6)

*for all*
\(x,y\in X\). *Then*
*T*
*has a unique fixed point*.

### Proof

Define \(\varphi(x,y)=\frac{d(x,y)}{1-\alpha}\). Then (6) shows that

$$ (1-\alpha)d(x,y)\leq d(x,y)-d(Tx,Ty). $$

(7)

It means that

$$ d(x,y)\leq\frac{d(x,y)}{1-\alpha}-\frac{d(Tx,Ty)}{1-\alpha} $$

(8)

and so

$$ d(x,y)\leq\varphi(x,y)-\varphi(Tx,Ty), $$

(9)

and so by applying Corollary 2.1, one can conclude that *T* has a unique fixed point. □

### Corollary 2.3

*Let*
\((X,d)\)
*be a complete metric space and let*
\(T:X\to X\)
*be a mapping such that*

$$ d(Tx,Ty)\leq\eta\bigl(d(x,y)\bigr), $$

(10)

*where*
\(\eta:[0,+\infty)\to[0,\infty)\)
*is a lower semicontinuous mapping such that*
\(\eta(t)< t\), *for each*
\(t>0\), *and*
\(\frac{\eta(t)}{t}\)
*is a nondecreasing map*. *Then*
*T*
*has a unique fixed point*.

### Proof

Define \(\varphi(x,y)=\frac{d(x,y)}{1-\frac{\eta(d(x,y))}{d(x,y)}}\), if \(x\neq y\) and otherwise \(\varphi(x,x)=0\). Then (10) shows that

$$ \biggl(1-\frac{\eta(d(x,y))}{d(x,y)}\biggr)d(x,y)\leq d(x,y)-d(Tx,Ty). $$

(11)

It means that

$$ d(x,y)\leq\frac{d(x,y)}{1-\frac{\eta(d(x,y))}{d(x,y)}}-\frac {d(Tx,Ty)}{1-\frac{\eta(d(x,y))}{d(x,y)}}. $$

(12)

Since \(\frac{\eta(t)}{t}\) is nondecreasing and \(d(Tx,Ty) < d(x,y)\),

$$ d(x,y)\leq\frac{d(x,y)}{1-\frac{\eta(d(x,y))}{d(x,y)}}-\frac {d(Tx,Ty)}{1-\frac{\eta(d(Tx,Ty))}{d(Tx,Ty)}}=\varphi(x,y)-\varphi(Tx,Ty), $$

(13)

and so by applying Corollary 2.1, one can conclude that *T* has a unique fixed point. □

The following results are the main results of this paper and play a crucial role to find the partial answers for Problem 1.1, Problem 1.2, and Problem 1.3. Comparing the partial answers for Reich’s problems, our answers include simple conditions. Also, the compactness condition on *Tx* is not needed.

### Theorem 2.1

*Let*
\((X,d)\)
*be a complete metric space*, *and let*
\(T:X\to\mathcal {CB}(X)\)
*be a non*-*expansive mapping such that*, *for each*
\(x\in X\), *and for all*
\(y\in Tx\), *there exists*
\(z\in Ty\)
*such that*

$$ d(x,y)\leq\varphi(x,y)-\varphi(y,z), $$

(14)

*where*
\(\varphi:X\times X\to[0,\infty)\)
*is lower semicontinuous with respect to the first variable*. *Then*
*T*
*has a fixed point*.

### Proof

Let \(x_{0}\in X\) and let \(x_{1}\in Tx_{0}\). If \(x_{0}=x_{1}\) then \(x_{0}\) is a fixed point and we are through. Otherwise, let \(x_{1}\neq x_{0}\). By assumption there exists \(x_{2}\in Tx_{1}\) such that

$$d(x_{0},x_{1})\leq\varphi(x_{0},x_{1})- \varphi(x_{1},x_{2}). $$

Alternatively, one can choose \(x_{n}\in Tx_{n-1}\) such that \(x_{n}\neq x_{n-1}\) and find \(x_{n+1}\in Tx_{n}\) such that

$$ 0< d(x_{n-1},x_{n})\leq\varphi(x_{n-1},x_{n})- \varphi(x_{n},x_{n+1}), $$

(15)

which means that \(\{\varphi(x_{n-1},x_{n})\}_{n}\) is a non-increasing sequence, bounded below, so it converges to some \(r\geq0\). By taking the limit on both sides of (15) we have \(\lim_{n\to\infty }d(x_{n-1},x_{n})=0\). Also, for all \(m,n\in \mathbb {N}\) with \(m>n\),

$$ \begin{aligned}[b] d(x_{n},x_{m})& \leq\sum_{i=n+1}^{m} d(x_{i-1},x_{i}) \\ &\leq\sum_{i=n+1}^{m}\varphi(x_{i-1},x_{i})- \varphi (x_{i},x_{i+1}) \\ &\leq\varphi(x_{n},x_{n+1})-\varphi(x_{m},x_{m+1}). \end{aligned} $$

(16)

Therefore, by taking the limsup on both sides of (16) we have

$$\lim_{n\to\infty}\bigl(\sup\bigl\{ d(x_{n},x_{m}):m>n \bigr\} \bigr)=0. $$

It means that \(\{x_{n}\}\) is a Cauchy sequence and so it converges to \(u\in X\). Now we show that *u* is a fixed point of *T*. We have

$$ \begin{aligned}[b] d(u,Tu)&\leq d(u,x_{n+1})+d(x_{n+1},Tu) \\ &= d(u,x_{n+1})+{\mathcal{H}}(Tx_{n},Tu) \\ &\leq d(u,x_{n+1})+d(x_{n},u). \end{aligned} $$

(17)

By taking the limit on both sides of (17), we get \(d(x,Tx)=0\) and this means that \(x\in Tx\). □

The following theorem is a partial answer to Problem 1.1.

### Theorem 2.2

*Let*
\((X,d)\)
*be a complete metric space*, *and let*
\(T:X\to\mathcal{CB}(X)\)
*be a multi*-*valued function such that*

$${\mathcal{H}}(Tx,Ty)\leq\eta\bigl(d(x,y)\bigr) $$

*for all*
\(x,y \in X\), *where*
\(\eta:[0,\infty)\rightarrow[0,\infty)\)
*is a lower semicontinuous map such that*
\(\eta(t)< t\), *for all*
\(t\in (0,+\infty)\), *and*
\(\frac{\eta(t)}{t}\)
*is nondecreasing*. *Then*
*T*
*has a fixed point*.

### Proof

Let \(x\in X\) and \(y\in Tx\). If \(y=x\) then *T* has a fixed point and the proof is complete, so we suppose that \(y\neq x\). Define

$$\theta(t)=\frac{\eta(t)+t}{2}\quad \mbox{for all }t\in(0,+\infty). $$

We have \({\mathcal{H}}(Tx,Ty)\leq\eta(d(x,y))<\theta(d(x,y))<d(x,y)\). Thus there exists \(\epsilon_{0}>0\) such that \(\theta(d(x,y))={\mathcal{H}}(Tx,Ty)+\epsilon_{0}\). So there exists \(z\in Ty\) such that

$$ d(y,z)< {\mathcal{H}}(Tx,Ty)+\epsilon_{0}=\theta\bigl(d(x,y) \bigr)< d(x,y). $$

(18)

We again suppose that \(y\neq z\); therefore \(d(x,y)-\theta(d(x,y))\leq d(x,y)-d(y,z)\) or equivalently

$$d(x,y)< \frac{d(x,y)}{1-\frac{\theta(d(x,y))}{d(x,y)}}-\frac {d(y,z)}{1-\frac{\theta(d(x,y))}{d(x,y)}}, $$

since \(\frac{\theta(t)}{t}\) is also a nondecreasing function and \(d(y,z)< d(x,y)\) we get

$$d(x,y)< \frac{d(x,y)}{1-\frac{\theta(d(x,y))}{d(x,y)}}-\frac {d(y,z)}{1-\frac{\theta(d(y,z))}{d(y,z)}}. $$

Define \(\Phi(x,y)=\frac{d(x,y)}{1-\frac{\theta(d(x,y))}{d(x,y)}}\) if \(x\neq y\), otherwise 0 for all \(x,y\in X\). It means that

$$d(x,y)< \Phi(x,y)-\Phi(y,z). $$

Therefore, *T* satisfies (14) of Theorem 2.1 and so we conclude that *T* has a unique fixed point *u* and the proof is completed. □

The following theorem is a partial answer to Problem 1.2.

### Corollary 2.4

([12], Mizoguchi-Takahashi’s type)

*Let*
\((X,d)\)
*be a complete metric space and let*
\(T:X\to\mathcal{CB}(X)\)
*be a multi*-*valued mapping such that*

$$ {\mathcal{H}}(Tx,Ty)\leq\eta\bigl(d(x,y)\bigr)d(x,y) $$

(19)

*for all*
\(x,y\in X\), *where*
\(\eta:[0,+\infty)\to[0,1)\)
*is a lower semicontinuous and nondecreasing mapping*. *Then*
*T*
*has a fixed point*.

### Proof

Let \(\theta(t)=\eta(t)t\), \(\theta(t)< t\) for all \(t\in R_{+}\), and \(\frac{\theta(t)}{t}=\eta(t)\) is a nondecreasing mapping. By the assumption \(d(Tx,Ty)\leq\eta(d(x,y))d(x,y)=\theta(d(x,y))\) for all \(x,y\in X\), therefore by Theorem 2.2
*T* has a fixed point. □

Note that if \(\eta:[0,\infty)\rightarrow[0,1)\) is a nondecreasing map then for all \(s\in[0,+\infty)\)

$$\begin{aligned} \limsup_{t\to s^{+}}\eta(t)&=\inf _{\delta>0}\sup_{s\leq t< s+\delta}\eta(t) \\ &=\lim_{\delta\to0}\sup_{s\leq t< s+\delta}\eta(t)\leq \eta(s+ \delta)< 1. \end{aligned} $$

It means that Corollary 2.4 comes from the Mizoguchi-Takahashi’s results directly and here we deduce it from our results [12].

The following theorem is a partial answer to Problem 1.3.

### Corollary 2.5

*Let*
\((X,d)\)
*be a complete metric space*, *and let*
\(T:X\to\mathcal{CB}(X)\)
*be a multi*-*valued function such that*

$${\mathcal{H}}(Tx,Ty)\leq d(x,y)-\theta\bigl(d(x,y)\bigr) $$

*for all*
\(x,y \in X\), *where*
\(\theta:(0,\infty)\rightarrow(0,\infty)\)
*is an upper semicontinuous map such that*, *for all*
\(t\in(0,+\infty)\), \(\frac{\theta(t)}{t}\)
*is non*-*increasing*. *Then*
*T*
*has a fixed point*.

### Proof

Let \(\eta(t)=t-\theta(t)\), for each \(t>0\). Then \(\eta(t)< t\), for each \(t>0\), and \(\frac{\eta(t)}{t}=1-\frac{\theta(t)}{t}\) is nondecreasing. Thus, the desired result is obtained by Theorem 2.2. □