**Proof.**

(a) Since *P* ∈ CB*<sup>m</sup>*(*X*), *P* = *P*. Then from Lemma 6, *<sup>m</sup>*(*<sup>a</sup>*, *P*) = sup *x*∈*P max*. Therefore, *<sup>δ</sup>m*(*<sup>P</sup>*, *P*) = sup *a*∈*P* {*m*(*<sup>a</sup>*, *P*)} = sup *a*∈*P* {sup *x*∈*P max*}. (b)Forany*a*∈*P*,*b*∈*Q*and*c*∈*R*,wehave

$$m(a,b) - m\_{ab} \preceq m(a,c) - m\_{ac} + m(c,b) - m\_{cb}.$$

We rewrite it as

$$m(a,b) - m\_{ab} + m\_{ac} + m\_{cb} \le m(a,c) + m(c,b).$$

Since *b* is arbitrary element in *Q*, we have

$$m(a, \mathbb{Q}) - \sup\_{b \in Q} m\_{ab} + m\_{ac} + \inf\_{b \in Q} m\_{cb} \le m(a, c) + m(c, \mathbb{Q}).$$

Since *<sup>m</sup>*(*<sup>c</sup>*, *Q*) ≤ *<sup>δ</sup>m*(*<sup>R</sup>*, *Q*), we can write above inequality as

$$m(a, \mathbb{Q}) - \sup\_{b \in \mathcal{Q}} m\_{ab} + m\_{ac} + \inf\_{b \in \mathcal{Q}} m\_{cb} \le m(a, c) + \delta\_{\mathfrak{M}}(R, \mathbb{Q}).$$

As *c* is arbitrary in *R*, we have

$$m(a, \mathbb{Q}) - \sup\_{b \in Q} m\_{ab} + \inf\_{c \in R} m\_{ac} + \inf\_{c \in R} \inf\_{b \in Q} m\_{cb} \le m(a, \mathbb{R}) + \delta\_m(\mathbb{R}, \mathbb{Q}).$$

We rewrite the above inequality as

$$m(a, \mathbb{Q}) + \inf\_{c \in \mathbb{R}} \inf\_{b \in \mathbb{Q}} m\_{cb} \le m(a, \mathbb{R}) + \delta\_m(R, \mathbb{Q}) + \sup\_{b \in \mathbb{Q}} m\_{ab} - \inf\_{c \in \mathbb{R}} m\_{ac}.$$

Again, as *a* is arbitrary in *P*, we ge<sup>t</sup>

$$\delta\_{\mathfrak{m}}(P,Q) + \inf\_{c \in \mathbb{R}} \inf\_{b \in Q} m\_{cb} \le \delta\_{\mathfrak{m}}(P,R) + \delta\_{\mathfrak{m}}(R,Q) + \sup\_{a \in P} \sup\_{b \in Q} m\_{ab} - \inf\_{a \in P} \inf\_{c \in R} m\_{ac}.$$

**Proposition 2.** *For any P*, *Q*, *R* ∈ CB*<sup>m</sup>*(*X*) *following are true*

$$\begin{aligned} \text{(i)} \quad &\mathcal{H}\_{m}(P,P) = \delta\_{m}(P,P) = \sup\_{a \in P} \{ \sup\_{b \in P} m\_{ab} \}; \\\text{(ii)} \quad &\mathcal{H}\_{m}(P,Q) = \mathcal{H}\_{m}(Q,P); \\\ \text{(iii)} \quad &\mathcal{H}\_{m}(P,Q) - \sup\_{a \in P} \sup\_{b \in Q} m\_{ab} \leq \mathcal{H}\_{m}(P,R) + \mathcal{H}\_{m}(Q,R) - \inf\_{a \in P} \inf\_{c \in R} m\_{ac} - \inf\_{c \in R} \inf\_{b \in Q} m\_{cb}. \end{aligned}$$

**Proof.**


$$\begin{split} \mathcal{H}\_{\mathfrak{m}}(P,Q) &= \max\{\delta\_{\mathfrak{m}}(P,Q), \delta\_{\mathfrak{m}}(Q,P)\} \\ &\leq \max\left\{\left[\delta\_{\mathfrak{m}}(P,R) - \inf\_{a\in P}\inf\_{c\in R} m\_{ac} + \delta\_{\mathfrak{m}}(R,Q) - \inf\_{c\in R}\inf\_{b\in Q} m\_{cb} + \sup\_{a\in P}\sup\_{b\in Q} m\_{ab}\right] \right\} \\ &\leq \left[\delta\_{\mathfrak{m}}(Q,R) - \inf\_{a\in P}\inf\_{c\in R} m\_{ac} + \delta\_{\mathfrak{m}}(R,P) - \inf\_{c\in R}\inf\_{b\in Q} m\_{cb} + \sup\_{a\in P}\sup\_{b\in Q} m\_{ab}\right] \cdot \\ &\leq \max\left\{\delta\_{\mathfrak{m}}(P,R), \delta\_{\mathfrak{m}}(R,P)\right\} + \max\left\{\delta\_{\mathfrak{m}}(Q,R), \delta\_{\mathfrak{m}}(R,Q)\right\} \\ &- \inf\_{a\in P}\inf\_{c\in R} m\_{ac} - \inf\_{c\in R}\inf\_{b\in Q} m\_{cb} + \sup\sup\_{a\in P}\sup\_{b\in Q} m\_{ab} \\ &\leq \mathcal{H}\_{\mathfrak{m}}(P,R) + \mathcal{H}\_{\mathfrak{m}}(R,Q) - \inf\_{a\in P}\inf\_{c\in R} m\_{ac} - \inf\_{c\in R}\inf\_{b\in Q} m\_{cb} + \sup\sup\_{a\in P}\sup\_{b\in Q} m\_{ab}. \end{split}$$

*a*∈*P*

*b*∈*P*

**Remark 3.** *In general,* H*m*(*<sup>A</sup>*, *A*) = 0 *for A* ∈ CB*<sup>m</sup>*(*X*)*. It can be verified through the following example.*

**Example 4.** *Let X* = [0, ∞) *and <sup>m</sup>*(*<sup>a</sup>*, *b*) = *a*+*b* 2 *, then clearly* (*<sup>X</sup>*, *m*) *is an M-metric space. In view of* (*a*) *of Proposition 1, we have*

$$\mathcal{H}\_{\mathfrak{m}}([1,2],[1,2]) = \delta\_{\mathfrak{m}}([1,2],[1,2]) = \sup\_{p \in [1,2]} \sup\_{q \in [1,2]} m\_{pq} = \sup\_{p \in [1,2]} \sup\_{q \in [1,2]} \min\{p, q\} \neq 0.$$

In view of Proposition 2, we call H*m* : CB*<sup>m</sup>*(*X*) × CB*<sup>m</sup>*(*X*) → [0, <sup>+</sup>∞) an *M*-Pompeiu–Hausdorff type metric induced by *m*.

**Lemma 7.** *Let P*, *Q* ∈ CB*<sup>m</sup>*(*X*) *and q* > 1*. Then for every a* ∈ *P, there is at least one b* ∈ *Q such that <sup>m</sup>*(*<sup>a</sup>*, *b*) ≤ *q*H*m*(*<sup>P</sup>*, *Q*).

**Proof.** Assume that there exists an *a* ∈ *P* such that *<sup>m</sup>*(*<sup>a</sup>*, *b*) > *q*H*m*(*<sup>P</sup>*, *Q*) for all *b* ∈ *Q*. This implies that

$$\inf\_{b \in Q} \{ m(a, b) \} \ge q \mathcal{H}\_m(P, Q)\_{\prime}.$$

that is,

$$
\eta(a, Q) \ge \eta \mathcal{H}\_{\mathfrak{m}}(P, Q).
$$

Note that

$$\mathcal{H}\_{\mathfrak{m}}(P,Q) \ge \delta\_{\mathfrak{m}}(P,Q) = \sup\_{\mathbf{x} \in P} m(\mathbf{x},Q) \ge m(a,Q) \ge q \mathcal{H}\_{\mathfrak{m}}(P,Q).$$

Since H*m*(*<sup>P</sup>*, *Q*) = 0, *q* ≤ 1, which is a contradiction.

**Lemma 8.** Let *P*, *Q* ∈ CB*<sup>m</sup>*(*X*) and *r* > 0. For any *a* ∈ *P*, there is at least one *b* ∈ *Q* such that *<sup>m</sup>*(*<sup>a</sup>*, *b*) ≤ H*m*(*<sup>P</sup>*, *Q*) + *r*.

**Proof.** Assume that there exists *a* ∈ *P* such that *<sup>m</sup>*(*<sup>a</sup>*, *b*) > H*m*(*<sup>P</sup>*, *Q*) + *r* for all *b* ∈ *Q*. This implies that

$$\inf\_{b \in Q} \{ m(a, b) \} \ge \mathcal{H}\_m(P, Q) + r\_m$$

that is,

$$m(a, \mathbf{Q}) \ge \mathcal{H}\_m(P, \mathbf{Q}) + r.s$$

Now,

$$
\mathcal{H}\_m(P, \mathbb{Q}) + r \le m(a, \mathbb{Q}) \le \delta\_m(P, \mathbb{Q}) \le \mathcal{H}\_m(P, \mathbb{Q}).
$$

Thus, *r* ≤ 0, which is a contradiction.

#### **4. Fixed Point Results**

First, we state the Nadler fixed point theorem in the class of *M*-metric spaces.

**Theorem 1.** *Let M-metric space* (*<sup>X</sup>*, *m*) *be M-complete and F* : *X* → CB*<sup>m</sup>*(*X*) *be a multivalued mapping. Suppose there exists λ* ∈ (0, 1) *such that*

$$
\mathcal{H}\_m(Fa, Fb) \le \lambda m(a, b), \tag{1}
$$

*for all a*, *b* ∈ *X. Then F admits a fixed point.*

**Proof.** Choose *q* = √1*λ* and *r* = √*<sup>λ</sup>*. Clearly, *q* > 1 and *r* < 1. Let *a*0 ∈ *X* be arbitrary and *a*1 ∈ *Fa*0. From Lemma 7, for *q* = √1*λ* , there exists *a*2 ∈ *Fa*1 such that

$$m(a\_1, a\_2) \le \frac{1}{\sqrt{\lambda}} \mathcal{H}\_{\mathfrak{m}}(Fa\_{0\prime}Fa\_1). \tag{2}$$

As H*m*(*Fa*0, *Fa*1) ≤ *<sup>λ</sup>m*(*<sup>a</sup>*0, *<sup>a</sup>*1), so from (2) we have

$$m(a\_1, a\_2) \le \frac{1}{\sqrt{\lambda}} \lambda m(a\_0, a\_1) = \sqrt{\lambda} m(a\_0, a\_1) = rm(a\_0, a\_1).$$

Now, from Lemma 7, there exists *a*3 ∈ *Fa*2 such that

$$m(a\_2, a\_3) \le rm(a\_1, a\_2).$$

Continuing in this way, we ge<sup>t</sup> a sequence {*ak*} of points in *X* such that *ak*+<sup>1</sup> ∈ *Fak* and for *k* ≥ 1,

$$rm(a\_k, a\_{k+1}) \le rm(a\_{k-1}, a\_k),\tag{3}$$

that is,

$$
\mu m(a\_k, a\_{k+1}) \le r^k m(a\_0, a\_1). \tag{4}
$$

By Lemma 5, we have

$$\lim\_{k \to \infty} m(a\_k, a\_{k+1}) = 0,\tag{5}$$

$$\lim\_{k \to \infty} m(a\_k, a\_k) = 0,\tag{6}$$

and

$$\lim\_{k,j \to \infty} m(a\_k, a\_j) = 0.\tag{7}$$

Also the sequence {*ak*} is *M*-Cauchy. Thus, *M*-completeness of *X* yields existence of *a* ∈ *X* such that

$$\lim\_{k \to \infty} (m(a\_k, a) - m\_{a\_k a}) = 0.$$

Since lim *k*→∞*<sup>m</sup>*(*ak*, *ak*) = 0, we have

$$\lim\_{k \to \infty} m(a\_k, a) = 0.\tag{8}$$

From (1) and (8), we have

$$\lim\_{k \to \infty} \mathcal{H}\_{\mathfrak{m}}(Fa\_k, Fa) = 0. \tag{9}$$

Now, since *ak*+<sup>1</sup> ∈ *Fak*, *<sup>m</sup>*(*ak*+1, *Fa*) ≤ H*m*(*Fak*, *Fa*). Taking limit as *k* → ∞ and using (8), we ge<sup>t</sup>

$$\lim\_{k \to \infty} m(a\_{k+1}, Fa) = 0.\tag{10}$$

As *mak*+1*Fa* ≤ *<sup>m</sup>*(*ak*+1, *Fa*), so we have

$$\lim\_{k \to \infty} m\_{a\_{k+1}Fa} = 0.\tag{11}$$

Using (*<sup>m</sup>*4), we have

$$\begin{aligned} m(a, Fa) - \sup\_{b \in Fa} m\_{ab} &\leq m(a, Fa) - m\_{aFa} \\ &\leq m(a, a\_{k+1}) - m\_{aa\_{k+1}} + m(a\_{k+1}, Fa) - m\_{a\_{k+1}Fa}. \end{aligned}$$

Varying limit as *k* → ∞ and using (8)–(11), we ge<sup>t</sup>

$$m(a, Fa) \le \sup\_{b \in Fa} m\_{ab}.\tag{12}$$

Since *mab* ≤ *<sup>m</sup>*(*<sup>a</sup>*, *b*) for every *b* ∈ *Fa*, this implies that

$$m\_{ab} - m(a, b) \le 0.$$

Thus

> sup{*mab* − *<sup>m</sup>*(*<sup>a</sup>*, *b*) : *b* ∈ *Fa*} ≤ 0,

that is,

$$\sup\_{b \in Fa} m\_{ab} - \inf\_{b \in Fa} m(a, b) \le 0.$$

This gives

$$\sup\_{a \in Fa} m\_{ab} \le m(a, Fa). \tag{13}$$

From (12) and (13), we have

$$m(a, Fa) = \sup\_{b \in Fa} m\_{ab} \dots$$

Thus, by Lemma 6, *a* ∈ *Fa* = *Fa*. **Example 5.** Let *X* = [0, 2] be endowed with *m*-metric *<sup>m</sup>*(*<sup>a</sup>*, *b*) = |*a* − *b*| + *a*+*b* 2 . Then (*<sup>X</sup>*, *m*) is an *M*-complete *M*-metric space (as in Example 3). Let *F* : *X* → CB*<sup>m</sup>*(*X*) be a mapping defined as

$$F(a) = \left[0, \frac{1}{7}a^2\right] \text{ for all } \ a \in X.$$

We shall show that for *λ* ∈ (0, <sup>1</sup>), H*m*(*Fa*, *Fb*) ≤ *<sup>λ</sup>m*(*<sup>a</sup>*, *b*), i.e., (1) holds for all *a*, *b* ∈ *X*. We have following three possible cases:

Case I: *a* = *b* = *p*. Then *Fa* = [0, 17*p*2] = *Fb*. Here, for *λ* ≥ 27,

$$
\mathcal{H}\_m(Fa, Fb) = \frac{1}{7}p^2 \le \lambda p = \lambda m(p, p) = \lambda m(a, b).
$$

Case II: *a* < *b*. Then *Fa* = [0, 17 *<sup>a</sup>*<sup>2</sup>], *Fb* = [0, 17 *b*2] and *Fa* ⊆ *Fb*. In this case,

$$\mathcal{H}\_{\mathfrak{M}}(Fa, Fb) = \max\left\{\frac{1}{7}a^2, |\frac{1}{7}a^2 - \frac{1}{7}b^2| + \frac{\frac{1}{7}a^2 + \frac{1}{7}b^2}{2}\right\}.$$

Since *a* < *b*, 17 *a*2 < | 17 *a*2 − 17 *b*2| + 17 *a*2<sup>+</sup> 17 *b*2 2 . So we ge<sup>t</sup>

$$\mathcal{H}\_m(Fa, Fb) = |\frac{1}{7}a^2 - \frac{1}{7}b^2| + \frac{\frac{1}{7}a^2 + \frac{1}{7}b^2}{2}$$

and *<sup>m</sup>*(*<sup>a</sup>*, *b*) = |*a* − *b*| + (*a*+*b*) 2 . Then one can see that

$$\begin{aligned} \mathcal{H}\_{\text{m}}(Fa, Fb) &= |\frac{1}{7}a^2 - \frac{1}{7}b^2| + \frac{\frac{1}{7}a^2 + \frac{1}{7}b^2}{2} \\ &= \frac{1}{7}|(a-b)(a+b)| + \frac{1}{7}\frac{a^2 + b^2}{2} \\ &= \frac{1}{7}\left[|a-b|(a+b) + \frac{(a+b)^2 - 2ab}{2}\right] \\ &\le \frac{1}{7}\left[|a-b| + \frac{(a+b)}{2}\right](a+b) \\ &= \frac{(a+b)}{7}m(a,b). \end{aligned}$$

Case III: *a* > *b*. Then *Fa* = [0, 17 *<sup>a</sup>*<sup>2</sup>], *Fb* = [0, 17 *b*2] and *Fb* ⊆ *Fa*. In this case,

$$\mathcal{H}\_m(Fa, Fb) = \max\left\{\frac{1}{7}b^2, |\frac{1}{7}a^2 - \frac{1}{7}b^2| + \frac{\frac{1}{7}a^2 + \frac{1}{7}b^2}{2}\right\}.$$

Since *b* < *a*, 17 *b*2 < | 17 *a*2 − 17 *b*2| + 17 *a*2<sup>+</sup> 17 *b*2 2 . So, we ge<sup>t</sup>

$$\mathcal{H}\_m(Fa, Fb) = |\frac{1}{7}a^2 - \frac{1}{7}b^2| + \frac{\frac{1}{7}a^2 + \frac{1}{7}b^2}{2}$$

and *<sup>m</sup>*(*<sup>a</sup>*, *b*) = |*a* − *b*| + (*a*+*b*) 2 . Following Case II, one can easily show that

$$\mathcal{H}\_{\mathfrak{m}}(Fa, Fb) \le \frac{(a+b)}{\mathcal{T}} m(a, b).$$

From above three cases, it is clear that (1) is satisfied for *λ* ≥ 47 . Thus, all the required conditions of Theorem 1 are satisfied. Hence *F* admits a fixed point, which is *a* = 0.

Next, we present our fixed point result corresponding to multivalued Kannan contractions in *M*-metric spaces.

**Theorem 2.** *Let M-metric space* (*<sup>X</sup>*, *m*) *be M-complete and F* : *X* → CB*<sup>m</sup>*(*X*) *be a multivalued mapping. Suppose there exists λ* ∈ (0, 12) *such that*

$$\mathcal{H}\_m(Fa, Fb) \le \lambda [m(a, Fa) + m(b, Fb)],\tag{14}$$

*for all a*, *b* ∈ *X. Then F admits a fixed point in X.*

**Proof.** Let *a*0 ∈ *X* be arbitrary. Fix an element *a*1 ∈ *Fa*0. We can now choose *a*2 ∈ *Fa*1 such that

$$m(a\_1, a\_2) = m(a\_1, Fa\_1) \le \mathcal{H}\_m(Fa\_{0"}, Fa\_1) \dots$$

Again, we can choose *a*3 ∈ *Fa*2 such that

$$m(a\_2, a\_3) \le \mathcal{H}\_m(Fa\_1, Fa\_2).$$

Continuing in this way, we ge<sup>t</sup> a sequence {*ak*} such that *ak*+<sup>1</sup> ∈ *Fak* with

$$m(a\_k, a\_{k+1}) \le \mathcal{H}\_m(Fa\_{k-1}, Fa\_k). \tag{15}$$

Using (14) in (15), we ge<sup>t</sup>

$$\begin{aligned} m(a\_k, a\_{k+1}) &\leq \lambda \left[ m(a\_{k-1}, Fa\_{k-1}) + m(a\_k, Fa\_k) \right] \\ &\leq \lambda \left[ m(a\_{k-1}, a\_k) + m(a\_k, a\_{k+1}) \right] .\end{aligned}$$

Thus,

$$m(a\_k, a\_{k+1}) \le \frac{\lambda}{1 - \lambda} m(a\_{k-1}, a\_k).$$

Let *r* = *λ*1−*λ* . Since *λ* < 12 , we have *r* < 1. So,

$$m(a\_k, a\_{k+1}) \le rm(a\_{k-1}, a\_k). \tag{16}$$

Thus, from Lemma 5, we have

$$\lim\_{k \to \infty} m(a\_k, a\_{k+1}) = 0,\tag{17}$$

$$\lim\_{k \to \infty} m(a\_k, a\_k) = 0,\tag{18}$$

and

$$\lim\_{k,j \to \infty} m(a\_k, a\_j) = 0.\tag{19}$$

Moreover, the sequence {*ak*} is a *M*-Cauchy. *M*-completeness of *X* yields existence of *a*<sup>∗</sup> ∈ *X* such that

$$\lim\_{k \to \infty} (m(a\_{k'}a^\*) - m\_{a\_k a^\*}) = 0 \text{ and } \lim\_{k \to \infty} (M\_{a\_k a^\*} - m\_{a\_k a^\*}) = 0.$$

Due to (18), we ge<sup>t</sup>

$$\lim\_{k \to \infty} m(a\_k, a^\*) = 0 \text{ and } \lim\_{k \to \infty} M\_{a\_k a^\*} = 0.$$

Thus, we have

$$\lim\_{k \to \infty} [M\_{a\_k a^\*} + m\_{a\_k a^\*}] = 0.1$$

This implies that

$$m(a^\*, a^\*) = 0 \text{ and hence } \; m\_{a^\* \to a^\*} = 0. \tag{20}$$

We shall show that *a*<sup>∗</sup> ∈ *Fa*<sup>∗</sup>. Since

$$m(a\_{k+1}, Fa^\*) \le \mathcal{H}\_m(Fa\_k, Fa^\*) \le \lambda [m(a\_k, Fa\_k) + m(a^\*, Fa^\*)].$$

Taking limit as *k* → <sup>∞</sup>, we ge<sup>t</sup>

$$\lim\_{k \to \infty} m(a\_{k+1}, Fa^\*) = 2\lambda m(a^\*, Fa^\*). \tag{21}$$

Suppose *<sup>m</sup>*(*a*<sup>∗</sup>, *Fa*∗) > 0, then we have

$$m(a^\*, Fa^\*) - m\_{a^\* \to a^\*} \le m(a^\*, a\_{k+1}) - m\_{a^\* a\_{k+1}} + m(a\_{k+1}, Fa^\*) - m\_{a\_{k+1} \to a^\*}$$

Taking limit as *k* → ∞ and using (21), we ge<sup>t</sup> *<sup>m</sup>*(*a*<sup>∗</sup>, *Fa*∗) ≤ <sup>2</sup>*λm*(*a*<sup>∗</sup>, *Fa*<sup>∗</sup>), which is a contradiction (as 2*λ* < 1). So

$$m(a^\*, Fa^\*) = 0.\tag{22}$$

.

Also, using (20), we have

$$\sup\_{b \in Fa} m\_{a^\*b} = \sup\_{b \in Fa} \min \{ m(a^\*, a^\*), m(b, b) \} = 0. \tag{23}$$

From (22) and (23), we ge<sup>t</sup>

$$m(a^\*, Fa^\*) = \sup\_{b \in Fa} m\_{a^\*b}.$$

Thus, from Lemma 6, we ge<sup>t</sup> *a*<sup>∗</sup> ∈ *Fa*∗ = *Fa*<sup>∗</sup>.

**Example 6.** *Let X* = [0, 1] *and m* : *X* × *X* → [0, ∞) *be defined as*

$$m(a,b) = \frac{a+b}{2}.$$

*Then* (*<sup>X</sup>*, *m*) *is an M-complete M-metric space. Let F* : *X* → CB*<sup>m</sup>*(*X*) *be a mapping defined as*

$$F(a) = \begin{cases} \begin{array}{ll} [0, a^2] & \text{if } a \in [0, \frac{1}{2}], \\\\ \left[\frac{a}{3}, \frac{a}{2}\right] & \text{if } a \in [\frac{1}{2}, 1]. \end{array} \end{cases}$$

*Then one can easily verify that there exists some λ in* (0, 12) *such that*

$$\mathcal{H}\_m(Fa, Fb) \le \lambda \left[ m(a, Fa) + m(b, Fb) \right].$$

*Thus F satisfies all the conditions in Theorem 2 and hence it has a fixed point (namely* 0*) in X.*

**Example 7.** *Let X* = [0, 1] *be endowed with m-metric <sup>m</sup>*(*<sup>x</sup>*, *y*) = *x*+*y* 2 . *Then* (*<sup>X</sup>*, *m*) *is an M-complete M-metric space. We define the mapping F* : *X* → CB*<sup>m</sup>*(*X*) *as*

$$F(a) = \begin{cases} \{\frac{1}{5}\} & \text{if } a = 0, \\\\ \left[\frac{a}{8(1+a^2)}, \frac{a}{4(1+a^2)}\right] & \text{if } a > 0. \end{cases}$$

*For a* = 0 *and b* = 110*, there does not exist any λ in* (0, 12) *such that*

$$\mathcal{H}\_m(F(0), F(\frac{1}{10})) \le \lambda \left[ m(0, F(0)) + m(\frac{1}{10'}, F(\frac{1}{10})) \right].$$

*Thus F does not satisfy* (14) *in Theorem 2. Evidently, F has no fixed point in X.*

#### **5. Homotopy Results in** *M***-Metric Spaces**

The following result is required in the sequel while proving a homotopy result in *M*-metric spaces.

**Proposition 3.** *Let F* : *X* → CB*<sup>m</sup>*(*X*) *be a multivalued mapping satisfying* (1) *for all a*, *b in M-metric space* (*<sup>X</sup>*, *<sup>m</sup>*)*. If c* ∈ *Fc for some c* ∈ *X, then <sup>m</sup>*(*<sup>a</sup>*, *a*) = 0 *for a* ∈ *Fc.*

**Proof.** Let *c* ∈ *Fc*. Then *<sup>m</sup>*(*<sup>c</sup>*, *Fc*) = sup *b*∈*Fc mc*,*<sup>b</sup>* = sup *b*∈*Fc mbb*. Also

$$\mathcal{H}\_m(Fc, Fc) = \delta\_m(Fc, Fc) = \sup\_{b \in Fc} m\_{bb}.$$

Assume that *<sup>m</sup>*(*<sup>c</sup>*, *c*) > 0. We have

$$\sup\_{b \in Fc} m\_{bb} = \mathcal{H}\_m(Fc, Fc) \le \lambda m(c, c)\_\prime$$

that is,

$$\sup\_{b \in Fc} m\_{bb} \le \lambda m(c, c).$$

Since *c* ∈ *Fc*, it is a contradiction. So *<sup>m</sup>*(*<sup>a</sup>*, *a*) = 0 for every *a* ∈ *Fc*.

**Theorem 3.** *Let* O *(resp.* C *) be an open (resp. closed) subset in an M-complete M-metric space* (*<sup>X</sup>*, *m*) *such that* O⊂C*. Let* G : C × [*μ*, *ν*] → CB*<sup>m</sup>*(*X*) *be a mapping satisfying the following conditions:*

*(a) a* ∈ G / (*a*, *t*) *for all a* ∈C\O *and each t* ∈ [*μ*, *ν*]*;*

*(b) there exists λ* ∈ (0, 1) *such that for every t* ∈ [*μ*, *ν*] *and all a*, *b* ∈ C *we have*

$$\mathcal{H}\_m(\mathcal{G}(a,t), \mathcal{G}(b,t)) \le \lambda m(a,b);$$

*(c) there exists a continuous mapping ψ* : [*μ*, *ν*] → R *satisfying*

$$\mathcal{H}\_m(\mathcal{G}(a,t), \mathcal{G}(a,s)) \le \lambda |\psi(t) - \psi(s)|\varphi$$

*(d) if c* ∈ G(*<sup>c</sup>*, *t*) *then* G(*<sup>c</sup>*, *t*) = {*c*}.

*If* G(., *<sup>t</sup>*1) *admits a fixed point in* C *for at least one t*1 ∈ [*μ*, *ν*]*, then* G(., *t*) *admits a fixed point in* O *for all t* ∈ [*μ*, *ν*]*. Moreover, the fixed point of* G(., *t*) *is unique for any fixed t* ∈ [*μ*, *ν*].

**Proof.** Consider, the set

$$\mathcal{W} = \{ t \in [\mu, \nu] \, | \, a \in \mathcal{G}(a, t) \text{ for some } a \in \mathcal{O} \}.$$

Then W is nonempty, because G(., *<sup>t</sup>*1) has a fixed point in C for at least one *t*1 ∈ [*μ*, *<sup>ν</sup>*], that is, there exists *a* ∈ C such that *a* ∈ G(*<sup>a</sup>*, *<sup>t</sup>*1) and as (*a*) holds, we have *a* ∈ O.

We will show that W is both closed and open in [*μ*, *ν*]. First, we show that it is open.

Let *t*0 ∈ W and *a*0 ∈ O with *a*0 ∈ G(*<sup>a</sup>*0, *<sup>t</sup>*0). As O is open subset of *X*, *Bm*(*<sup>a</sup>*0,*<sup>r</sup>*) ⊆ O for some *r* > 0. Let *ε* = *r* + *maa*0− *<sup>λ</sup>*(*r* + *maa*0) > 0. As *ψ* is continuous on [*μ*, *<sup>ν</sup>*], there exists *δ* > 0 such that

$$|\psi(t) - \psi(t\_0)| < \epsilon, \text{ for all } t \in \mathcal{S}\_\delta(t\_0),$$

where *<sup>S</sup>δ*(*<sup>t</sup>*0)=(*<sup>t</sup>*0 − *δ*, *t*0 + *<sup>δ</sup>*).

Since *a*0 ∈ G(*<sup>a</sup>*0, *<sup>t</sup>*0), by Proposition 3, *<sup>m</sup>*(*<sup>c</sup>*, *c*) = 0 for every *c* ∈ G(*<sup>a</sup>*0, *<sup>t</sup>*0). Keeping this fact in view, we have

$$m\_{p\mathbb{C}} = 0,\text{ for every } p \in X. \tag{24}$$

Now, using (*iii*) of Proposition 2 and (24), we have

$$\begin{split} m(\mathcal{G}(a,t),a\_{0}) &= \mathcal{H}\_{\mathfrak{m}}(\mathcal{G}(a,t),\mathcal{G}(a\_{0},t\_{0})) \\ &\leq \mathcal{H}\_{\mathfrak{m}}(\mathcal{G}(a,t),\mathcal{G}(a,t\_{0})) + \mathcal{H}\_{\mathfrak{m}}(\mathcal{G}(a,t\_{0}),\mathcal{G}(a\_{0},t\_{0})) \\ &\quad - \inf\_{p\in\mathcal{G}(a,t)} \inf\_{q\in\mathcal{G}(a,t\_{0})} m\_{pq} - \inf\_{q\in\mathcal{G}(a,t\_{0})} \inf\_{c\in\mathcal{G}(a\_{0},t\_{0})} m\_{q\epsilon} + \sup\_{p\in\mathcal{G}(a,t)} \sup\_{c\in\mathcal{G}(a\_{0},t\_{0})} m\_{pc} \\ &\leq \mathcal{H}\_{\mathfrak{m}}(\mathcal{G}(a,t),\mathcal{G}(a,t\_{0})) + \mathcal{H}\_{\mathfrak{m}}(\mathcal{G}(a,t\_{0}),\mathcal{G}(a\_{0},t\_{0})) \\ &\leq \lambda |\psi(t) - \psi(t\_{0})| + \lambda m(a,a\_{0}) \\ &\leq \lambda \varepsilon + \lambda \left(m\_{\mathrm{an}\_{0}} + r\right) \\ &= \lambda (r + m\_{\mathrm{an}\_{0}} - \lambda(r + m\_{\mathrm{an}\_{0}})) + \lambda \left(m\_{\mathrm{an}\_{0}} + r\right) \\ &\leq r + m\_{\mathrm{an}\_{0}} - \lambda \left(r + m\_{\mathrm{an}\_{0}}\right) + \lambda \left(m\_{\mathrm{an}\_{0}} + r\right) \\ &\leq r + m\_{\mathrm{an}\_{0}}.\end{split}$$

Thus for each fixed *t* ∈ *<sup>S</sup>δ*(*<sup>t</sup>*0), G(., *t*) : *Bm*(*<sup>a</sup>*0,*<sup>r</sup>*) → CB*<sup>m</sup>*(*Bm*(*<sup>a</sup>*0,*<sup>r</sup>*)) satisfies all the hypotheses of Theorem 1 and so G(., *t*) admits a fixed point in *Bm*(*<sup>a</sup>*0,*<sup>r</sup>*) ⊆ C. But this fixed point must be in O to satisfy (*a*). Therefore, *<sup>S</sup>δ*(*<sup>t</sup>*0) ⊆ W and hence W is open in [*μ*, *ν*].

Next, we show that W is closed in [*μ*, *ν*]. Let {*tk*} be a convergen<sup>t</sup> sequence in W to some *s* ∈ [*μ*, *ν*]. We need to show that *s* ∈ W.

The definition of the set W implies that for all *k* ∈ N \ {0}, there exists *ak* ∈ O with *ak* ∈ G(*ak*, *tk*). Then using (*d*), (*iii*) of Proposition 2 and the outcome of Proposition 3, we have

$$\begin{aligned} m(a\_{k'}a\_{\dot{j}}) &= \mathcal{H}\_m(\mathcal{G}(a\_{k'}t\_k), \mathcal{G}(a\_{\dot{j}'}t\_{\dot{j}})) \\ &\leq \mathcal{H}\_m(\mathcal{G}(a\_{k'}t\_k), \mathcal{G}(a\_{k'}t\_{\dot{j}})) + \mathcal{H}\_m(\mathcal{G}(a\_{k'}t\_{\dot{j}}), \mathcal{G}(a\_{\dot{j}'}t\_{\dot{j}})) \\ &\leq \lambda |\psi(t\_k) - \psi(t\_{\dot{j}})| + \lambda m(a\_{k'}a\_{\dot{j}}). \end{aligned}$$

This gives us

$$m(a\_{k'}a\_{\bar{j}}) \le \frac{\lambda}{1-\lambda} \quad |\psi(t\_k) - \psi(t\_{\bar{j}})|.$$

Since *ψ* is continuous and {*tk*} converges to *s*, varying *k*, *j* → ∞ in the above inequality, we ge<sup>t</sup>

$$\lim\_{k,j \to \infty} m(a\_{k\prime} a\_j) = 0.$$

As *mak aj* ≤ *<sup>m</sup>*(*ak*, *aj*), so

$$\lim\_{k,j \to \infty} m\_{a\_k a\_j} = 0.$$

Also lim *k*→∞ *<sup>m</sup>*(*ak*, *ak*) = 0 = lim*k*→∞ *<sup>m</sup>*(*aj*, *aj*). Therefore

$$\lim\_{k,j \to \infty} (m(a\_{k'}a\_j) - m\_{a\_ka\_j}) = 0 \text{ and } \lim\_{k,j \to \infty} (M\_{a\_ka\_j} - m\_{a\_ka\_j}) = 0.$$

Thus {*ak*} is an *M*-Cauchy sequence. Using (*iii*) of Definition 3, there exists *a*<sup>∗</sup> ∈ *X* such that

$$\lim\_{k \to \infty} (m(a\_{k^\*}a^\*) - m\_{a\_k a^\*}) = 0 \quad \text{and} \quad \lim\_{k \to \infty} (M\_{a\_k a^\*} - m\_{a\_k a^\*}) = 0.$$

But lim *k*→∞ *<sup>m</sup>*(*ak*, *ak*) = 0, so

$$\lim\_{k \to \infty} m(a\_{k\prime}a^\*) = 0 \text{ and } \lim\_{k \to \infty} M\_{a\_k a^\*} = 0.$$

Thus, we ge<sup>t</sup> *<sup>m</sup>*(*a*<sup>∗</sup>, *a*<sup>∗</sup>) = 0. We shall prove *a*<sup>∗</sup> ∈ G(*a*<sup>∗</sup>, *<sup>t</sup>*<sup>∗</sup>). We have

$$\begin{split} m(a\_{k\prime}\mathcal{G}(a^\*,t^\*)) &\leq \mathcal{H}\_{\mathfrak{m}}(\mathcal{G}(a\_{k\prime}t\_k), \mathcal{G}(a^\*,t^\*)) \\ &\leq \mathcal{H}\_{\mathfrak{m}}(\mathcal{G}(a\_{k\prime}t\_k), \mathcal{G}(a\_{k\prime}t^\*)) + \mathcal{H}\_{\mathfrak{m}}(\mathcal{G}(a\_{k\prime}t^\*), \mathcal{G}(a^\*,t^\*)) \\ &\leq \lambda \left| \psi(a\_k) - \psi(t^\*) \right| + \lambda m(a\_{k\prime}a^\*). \end{split}$$

Varying *k* → ∞ in above inequality, we ge<sup>t</sup>

$$\lim\_{k \to \infty} m(a\_k, \mathcal{G}(a^\*, t^\*)) = 0.$$

Hence

$$m(a^\*, \mathcal{G}(a^\*, t^\*)) = 0.\tag{25}$$

Since *<sup>m</sup>*(*a*<sup>∗</sup>, *a*<sup>∗</sup>) = 0, we have

$$\sup\_{b \in \mathcal{G}(a^\*, t^\*)} m\_{a^\*b} = \sup\_{b \in \mathcal{G}(a^\*, t^\*)} \min \{ m(a^\*, a^\*), m(b, b) \} = 0. \tag{26}$$

From (25) and (26), we ge<sup>t</sup>

$$m(a^\*, \mathcal{G}(a^\*, t^\*)) = \sup\_{b^\* \in \mathcal{G}(a^\*, t^\*)} m\_{a^\*b^\*}$$

Therefore, from Lemma 6, we have *a*<sup>∗</sup> ∈ G(*a*<sup>∗</sup>, *<sup>t</sup>*<sup>∗</sup>). Thus *a*<sup>∗</sup> ∈ O. Hence *t*∗ ∈ W and W is closed in [*μ*, *ν*].

As [*μ*, *ν*] is connected and W is both open and closed in it, so W = [*μ*, *ν*]. Thus G(., *t*) admits a fixed point in O for all *t* ∈ [*μ*, *ν*].

For uniqueness, fix *t* ∈ [*μ*, *<sup>ν</sup>*], then there exists *a* ∈ O such that *a* ∈ G(*<sup>a</sup>*, *t*). Suppose *b* is another fixed point of G(*b*, *t*), then from (*d*) we have

$$m(a,b) = \mathcal{H}\_m(\mathcal{G}(a,t), \mathcal{G}(b,t)) \le \lambda m(a,b),$$

a contradiction. Thus, the fixed point of G(., *t*) is unique for any *t* ∈ [*μ*, *ν*].

**Author Contributions:** All authors contributed equally to this paper. All authors have read and approved the final manuscript.

**Funding:** This research received no external funding.

**Acknowledgments:** The fifth author would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) group number RG-DES-2017-01-17.

**Conflicts of Interest:** The authors declare no conflict of interest.
