Let (X,d) be a metric space and let \mathcal{CB}(X) denote the family of nonempty, closed and bounded subsets of X. For A,B\in \mathcal{CB}(X) let \rho (A,B)={sup}_{x\in A}dist(x,B) and \rho (B,A)={sup}_{x\in B}dist(x,A). The usual Hausdorff distance H(A,B) between A and B is defined as
H(A,B)=max\{\rho (A,B),\rho (B,A)\}.
A mapping T:X\to \mathcal{CB}(X) is called a multivalued contraction mapping if there exists a constant k\in (0,1) such that
H(Tx,Ty)\le kd(x,y),\phantom{\rule{1em}{0ex}}x,y\in X.
A point x\in X is said to be a fixed point of T if x\in Tx. Our point of departure in this section is the following celebrated theorem of Nadler [9].
Theorem 3.1 Let (X,d) be a complete metric space, and suppose T:X\to \mathcal{CB}(X) be a multivalued contraction mapping. Then T has a fixed point.
Our purpose in this section is to extend Nadler’s theorem by replacing the Hausdorff metric with other metrics on \mathcal{CB}(X) which are either metrically or sequentially equivalent to H.
One example of a metric on \mathcal{CB}(X) which is metrically equivalent to the Hausdorff metric H is the metric {H}^{+}, which was introduced in [10]. {H}^{+} is defined by setting
{H}^{+}(A,B)=\frac{1}{2}(\rho (A,B)+\rho (B,A)),\phantom{\rule{1em}{0ex}}A,B\in \mathcal{CB}(X).
Clearly, {H}^{+} is metrically equivalent to the Hausdorff metric:
\frac{1}{2}H(A,B)\le {H}^{+}(A,B)\le H(A,B).
A multivalued mapping T:X\to \mathcal{CB}(X) is called an {H}^{+}contraction if

(1)
there exists k\in (0,1) such that
{H}^{+}(Tx,Ty)\le kd(x,y)\phantom{\rule{1em}{0ex}}\text{for every}x,y\in X;
and

(2)
for every x\in X and y\in Tx,
dist(y,Ty)\le {H}^{+}(Tx,Ty).
It follows immediately from the definition of the Hausdorff metric H that if A,B\in \mathcal{CB}(X) and if x\in X. Then for each \epsilon >0 there exists y\in Y such that
d(x,y)\le H(A,B)+\epsilon .
Thus, condition (2) is always true if {H}^{+} is replaced with the usual Hausdorff metric H. This is precisely the fact about the Hausdorff metric that Nadler used in his proof. In fact, Nadler’s proof yields the following result. This theorem implies that every {H}^{+}contraction of a complete metric space X into \mathcal{CB}(X) has a fixed point [10].
Theorem 3.2 Let (X,d) be a complete metric space, and let D be any metric on \mathcal{CB}(X) which is sequentially equivalent to the Hausdorff metric H. Suppose T:X\to \mathcal{CB}(X) satisfies

(1)
there exists
k\in (0,1)
such that
D(Tx,Ty)\le kd(x,y)\phantom{\rule{1em}{0ex}}\mathit{\text{for every}}x,y\in X;
and

(2)
if x\in X and y\in Tx,
Then T has a fixed point, i.e., there exists x\in X such that x\in Tx.
Proof (cf. [9]) By saying D is sequentially equivalent to H, we mean that for A\in \mathcal{CB}(X) and \{{A}_{n}\}\subset \mathcal{CB}(X),
\underset{n\to \mathrm{\infty}}{lim}D({A}_{n},A)=0\phantom{\rule{1em}{0ex}}\iff \phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty}}{lim}H({A}_{n},A)=0.
Suppose T:X\to \mathcal{CB}(X). Select {x}_{0}\in X and {x}_{1}\in Tx. By (2) and (1), there exists {x}_{2}\in T{x}_{1} such that
\begin{array}{rl}d({x}_{1},{x}_{2})& \le D(T{x}_{0},T{x}_{1})+k\\ \le kd({x}_{0},{x}_{1})+k.\end{array}
Similarly, there exists {x}_{3}\in T{x}_{2} such that
\begin{array}{rl}d({x}_{2},{x}_{3})& \le D(T{x}_{1},T{x}_{2})+{k}^{2}\\ \le kd({x}_{1},{x}_{2})+{k}^{2}\\ \le k[kd({x}_{0},{x}_{1})+k]+{k}^{2}\\ ={k}^{2}d({x}_{0},{x}_{1})+2{k}^{2}.\end{array}
In general, for each i\in \mathbb{N} there exists {x}_{i+1}\in T{x}_{i} such that
\begin{array}{rl}d({x}_{i},{x}_{i+1})& \le D(T{x}_{i1},{x}_{i})+{k}^{i}\\ \le kd({x}_{i1},{x}_{i})+{k}^{i}\\ \le k[D(T{x}_{i2},T{x}_{i1})+{k}^{i1}]+{k}^{i}\\ \le {k}^{2}d({x}_{i2},{x}_{i1})+2{k}^{i}\\ \le \cdots \\ \le {k}^{i}d({x}_{0},{x}_{1})+i{k}^{i}.\end{array}
Therefore,
\sum _{i=0}^{\mathrm{\infty}}d({x}_{i},{x}_{i+1})\le d({x}_{0},{x}_{1})\sum _{i=0}^{\mathrm{\infty}}{k}^{i}+\sum _{i=0}^{\mathrm{\infty}}i{k}^{i}<\mathrm{\infty}.
Hence, \{{x}_{n}\} is a Cauchy sequence, so there exists x\in X such that {lim}_{n\to \mathrm{\infty}}{x}_{n}=x. It follows from (1) that {lim}_{n\to \mathrm{\infty}}D(T{x}_{n},Tx)=0. Since D and H are equivalent, {lim}_{n\to \mathrm{\infty}}H(T{x}_{n},Tx)=0. Since {x}_{n+1}\in T{x}_{n}, it follows from the definition of Hausdorff metric that {lim}_{n\to \mathrm{\infty}}dist({x}_{n},Tx)=0, and since Tx is closed, x\in Tx. □
Remark 3.3 The point valued analog of Theorem 3.2 is rather trivial. Let (X,d) be a complete metric space, and let ρ be any metric on X which is sequentially equivalent to the d. Suppose T:X\to X satisfies

(1)
there exists k\in (0,1) such that
\rho (Tx,Ty)\le kd(x,y)\phantom{\rule{1em}{0ex}}\text{for every}x,y\in X;
and

(2)
if x\in X,
d(Tx,{T}^{2}x)\le \rho (Tx,{T}^{2}x).
Then T has a fixed point.
Proof
Combining (1) and (2), we have
d(Tx,{T}^{2}x)\le kd(x,Tx),\phantom{\rule{1em}{0ex}}x\in X.
This implies that ({T}^{n}x) is a Cauchy sequence in (X,d), so {lim}_{n\to \mathrm{\infty}}{T}^{n}x={x}_{0} exists. Since (1) implies T is continuous, T{x}_{0}={x}_{0}. □
As noted above, (1) alone is sufficient if D=H because (2) is redundant in this case. However, the following example shows that (1) alone is not sufficient if D={H}^{+}.
Example Take X=[0,\mathrm{\infty}) with the metric: d(x,y)=\frac{xy}{xy+1} \mathrm{\forall}x,y\in X. Define T:X\to \mathcal{CB}(X) by setting T(x)=[x+1,\mathrm{\infty}). Clearly, T has no fixed point. However, if x>y then T(x)\subseteq T(y), and
\begin{array}{rl}{H}^{+}(T(x),T(y))& =\frac{1}{2}d(T(x),T(y))\\ =\frac{1}{2}\frac{x+1(y+1)}{x+1(y+1)+1}\\ =\frac{1}{2}\frac{xy}{xy+1}\\ =\frac{1}{2}d(x,y).\end{array}
We now turn to an analog of Theorem 2.2 for setvalued mappings.
A mapping T:X\to \mathcal{CB}(X) is said to be an (\epsilon ,k)uniform local multivalued contraction (where \epsilon >0 and k\in (0,1)) if for x,y\in X, d(x,y)<\epsilon \Rightarrow H(Tx,Ty)\le kd(x,y). This definition, given in [9], is modelled after a concept introduced by Edelstein in [11].
Theorem 3.4 Let (X,d) be a metric space and T:X\to \mathcal{CB}(X). Suppose there exists a metric transform ϕ and k\in (0,1) such that the following conditions hold:

(a)
For each x,y\in X,
\varphi (H(Tx,Ty))\le kd(x,y).

(b)
There exists c\in (0,1) such that for t>0 sufficiently small,
Then for \epsilon >0 sufficiently small, T is an (\epsilon ,c)uniform local multivalued contraction on (X,d).
Proof Let x,y\in X, and observe that
\varphi (H(Tx,Ty))\le kd(x,y).
Now suppose there exists c\in (0,1) such that for t sufficiently small,
Then for d(x,y) sufficiently small,
\varphi (H(Tx,Ty))\le kd(x,y)\le \varphi (cd(x,y))
and since ϕ is strictly increasing this in turn implies
Thus, for \epsilon >0 sufficiently small, T is an (\epsilon ,c)uniform local multivalued contraction on (X,d). □
A metric space (X,d) is said to be εchainable (where \epsilon >0 is fixed) if given a,b\in X there is an εchain joining a and b. This means there exists a finite set of points {\{{x}_{i}\}}_{i=1}^{n} in X such that a={x}_{1}, b={x}_{n} and d({x}_{i},{x}_{i+1})<\epsilon for all i=1,\dots ,n1. The following result is also due to Nadler.
Theorem 3.5 ([[9], Theorem 6])
Let (X,d) be a complete εchainable metric space. If T:X\to \mathcal{CB}(X) is an (\epsilon ,k) uniform local multivalued contraction, then T has a fixed point.
By combining the above result with Theorem 3.4 we obtain the following.
Theorem 3.6 If, in addition to the assumptions of Theorem 3.4, X is complete and connected, then T has a fixed point.
Proof A connected metric space is εchainable for any \epsilon >0. □