We begin the main results of this section by stating the well known Edelstein’s fixed point theorem.

#### Theorem 3.1

([23]). *Let* (*X*, *d*) *be a compact metric space and T be a mapping on X such that*
$$\begin{array}{}d(Tx,Ty)<d(x,y),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x,y\in X\phantom{\rule{1em}{0ex}}with\phantom{\rule{1em}{0ex}}x\ne y.\end{array}$$
*Then T has a unique fixed point and for any x*_{0} ∈ *X the iterate sequence* {*T*^{n} x_{0}} *converges to the fixed point of T*.

Let (*A*, *B*) be a nonempty pair in a normed linear space *X*. A mapping *T* : *A* ∪ *B* → *A* ∪ *B* is said to be noncyclic provided that *T*(*A*) ⊆ *A* and *T*(*B*) ⊆ *B*. A point (*p*, *q*) ∈ *A* × *B* is said to be a *bestproximity pair* for the noncyclic mapping *T* if
$$p=Tp,\phantom{\rule{1em}{0ex}}q=Tq\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}\parallel p-q\parallel \phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\text{dist}(A,\phantom{\rule{thinmathspace}{0ex}}B).$$

It is interesting to note that the existence of best proximity pairs for noncyclic mappings is equivalent to the existence of a solution of the following minimization nonlinear problem:
$$\underset{x\in A}{min}\parallel x-Tx\parallel ,\phantom{\rule{1em}{0ex}}\underset{y\in B}{min}\parallel y-Ty\parallel ,\phantom{\rule{1em}{0ex}}\text{and}\underset{(x,y)\in A\times B}{min}\parallel x-y\parallel .$$(3)

The existence of best proximity pairs was first studied by Eldred et al. in [24] using a geometric notion of *proximal normal structure* on nonempty, weakly compact and convex pairs in strictly convex Banach spaces for noncyclic relatively nonexpansive mappings.

#### Definition 3.2

([24]). *A convex pair* (*A*, *B*) *in a Banach space* *X* *is said to have proximal normal structure if for any bounded, closed*, *convex and proximinal pair* (*K*_{1}, *K*_{2}) ⊆ (*A*, *B*) *for which* *δ*(*K*_{1}, *K*_{2}) > dist (*K*_{1}, *K*_{2}) *and* dist (*K*_{1}, *K*_{2}) = dist (*A*, *B*), *there exits* (*x*_{1}, *x*_{2}) ∈ *K*_{1} × *K*_{2} *such that*
$$max\{{\delta}_{{x}_{1}}({K}_{2}),{\delta}_{{x}_{2}}({K}_{1})\}<\delta ({K}_{1},{K}_{2}).$$

Since every nonempty, compact and convex pair in a Banach space *X* has proximal normal structure (Proposition 2.2 of [24]), the following result concludes.

#### Theorem 3.3

(Theorem 2.2 of [24]). *Let* (*A*, *B*) *be a nonempty compact and convex pair in a strictly convex Banach space* *X* *and* *T* *be a noncyclic relatively nonexpansive mapping*, *that is*, *T* *is noncyclic and* ∥*Tx* − *Ty*∥ ≤ ∥*x* − *y*∥*for all* (*x*, *y*) ∈ *A* × *B*. *Then* *T* *has a best proximity pair*.

Motivated by Theorem 3.3, we study the convergence results of best proximity pairs for *noncyclic contractive mappings* in strictly convex Banach spaces.

#### Definition 3.4

*Let* (*A*, *B*) *be a nonempty pair in a normed linear space X*. *A mapping* *T* : *A* ∪ *B* → *A* ∪ *B* *is said to be a noncyclic contractive mapping if* *T* *is noncyclic on* *A* ∪ *B* *and*
$$\parallel Tx-Ty\parallel <\parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}all\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(x,\phantom{\rule{thinmathspace}{0ex}}y)\in A\times B,\phantom{\rule{1em}{0ex}}\mathit{w}\mathit{i}\mathit{t}\mathit{h}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\parallel x-y\parallel >\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(A,B).$$

Next lemma describes the relation between noncyclic relatively nonexpansive mappings and noncyclic contractive mappings in uniformly convex Banach spaces.

#### Lemma 3.5

*Let* (*A*, *B*) *be a nonempty compact and convex pair in a strictly convex Banach space* *X* *and* *T* : *A* ∪ *B* → *A* ∪ *B* *be a noncyclic contractive mapping*. *Then* *T* *is noncyclic relatively nonexpansive*.

#### Proof

We only have to prove that ∥*Tx* − *Ty*∥ = dist (*A*, *B*) whenever ∥*x* − *y*∥ = dist (*A*, *B*). So let ∥*x* − *y*∥ = dist (*A*, *B*). Choose a sequence ({*x*_{n}}, {y_{n}}) in *A* × *B* such that ∥*x*_{n} − *y*_{n}∥ > dist (*A*, *B*) and *x*_{n} ≠ *x*, *y*_{n} ≠ *y* for any *n* ∈ *N*. By the compactness condition of the pair (*A*, *B*), we may assume that lim_{n→∞}x_{n} = *x* ∈ *A* and lim_{n → ∞}y_{n} = *y* ∈ *B*. Then lim_{n → ∞}∥*x*_{n} − *y*_{n}∥ = dist(*A*; *B*). Notice that if ∥*x*_{n0} − *y*∥ = dist (*A*, *B*) for some *n*_{0} ∈ ℕ, then by the strictly convexity of *X* we must have x_{n0} =x which is a contradiction. Thus
$$\text{dist}(A,\phantom{\rule{thinmathspace}{0ex}}B)\le \parallel {\mathcal{P}}_{A}(Ty)-Ty\parallel \le \parallel T{x}_{n}-Ty\parallel <\parallel {x}_{n}-y\parallel .$$

Therefore, ∥Tx_{n}−*Ty*∥ → dist(*A*; *B*). Since ∥𝓟_{A}(*Ty*)−*Ty*∥≤∥*Tx*_{n} − *Ty*∥ and *T**y* ∈ *B*_{0},
$$T{x}_{n}\to {\mathcal{P}}_{A}(Ty).$$

Similarly we can see that *Ty*_{n} → 𝓟_{B}(*Tx*). In view of the fact that ∥*Tx*_{n} − *Ty*_{n}∥ → dist(*A*; *B*), we obtain ∥𝓟_{A}(*Ty*) − 𝓟_{B}(*Tx*)∥ = dist (*A*, *B*). Again, using the strict convexity of *X*,
$$Tx={\mathcal{P}}_{A}(Ty),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Ty={\mathcal{P}}_{B}(Tx).$$

Thereby, ∥*Tx* − *Ty*∥ = dist (*A*, *B*) and the result follows.□

Next example shows that the strictly convexity of the Banach space *X* in Lemma 3.5 is a necessary condition.

#### Example 3.6

*Let* *X* = {ℝ^{2},∥.∥_{∞}} *and let* *A* = {(0, *s*) : 0 ≤ *s* ≤ 2} *and* *B* = {(1, *t*) : 0 ≤ *t* ≤ 2}. *It is clear that* dist (*A*, *B*) = 1. *Define the noncyclic mapping* *T* : *A* ∪ *B* → *A* ∪ *B* *by*
$$T(0,\phantom{\rule{thinmathspace}{0ex}}s)=(0,\phantom{\rule{thinmathspace}{0ex}}\frac{s}{1+s}),\phantom{\rule{1em}{0ex}}T(1,\phantom{\rule{thinmathspace}{0ex}}t)=(1,\phantom{\rule{thinmathspace}{0ex}}\frac{t}{1+t}).$$

*For* (*x*, *y*) ∈ *A* × *B* *if* ∥*x* − *y*∥_{ ∞} > dist (*A*, *B*), *then*
$$\begin{array}{}\parallel Tx-Ty{\parallel}_{\mathrm{\infty}}=\parallel (0,\frac{s}{1+s})-(1,\frac{t}{1+t}){\parallel}_{\mathrm{\infty}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=max\{1,\frac{|s-t|}{(1+s)(1+t)}\}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}<|s-t|\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\parallel x-y{\parallel}_{\mathrm{\infty}},\end{array}$$

*which implies that* *T* *is noncyclic contractive*. *Besides*, ∥*Tx* − *Ty*∥_{∞} = dist (*A*, *B*), *when* ∥*x* − *y*∥_{∞} = dist (*A*, *B*). *Hence*, *T* *is noncyclic relatively nonexpansive*.

*But if we modify* *T* : *A* → *A* *as*
$T(0,\phantom{\rule{thinmathspace}{0ex}}s)=(0,\phantom{\rule{thinmathspace}{0ex}}\frac{s}{1+s})\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0<s\le 2$
*and* T(0, 0) = (0, 1.1), *then* *T* *is still noncyclic contractive*. *To see* *T* *this*, *let* **x** : = (0,0). *Since*
$$\{\mathrm{y}\in B\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}\parallel \mathrm{x}-\mathrm{y}{\parallel}_{\mathrm{\infty}}>1\}=\{(1,\phantom{\rule{thinmathspace}{0ex}}t)\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}1<t\le 2\},$$

*we have*
$$\parallel T\mathrm{x}-T\mathrm{y}\parallel =max\{1,\phantom{\rule{thinmathspace}{0ex}}1.1-\frac{t}{1+t}\}=1<\parallel \mathrm{x}-\mathrm{y}{\parallel}_{\mathrm{\infty}},$$

*whenever* 1 < *t* ≤ 2. *On the other hand*, *if* **v** : = (1,0), *then*
$$\parallel T\mathrm{x}-T\mathrm{v}\parallel =max\{1.1,\phantom{\rule{thinmathspace}{0ex}}1\}>1=\parallel \mathrm{x}-\mathrm{v}{\parallel}_{\mathrm{\infty}},$$

*that is*, *T* *is not a noncyclic relatively nonexpansive mapping*.

The following theorem is an extension of Edelstein’s fixed point theorem in strictly convex Banach spaces.

#### Theorem 3.7

*Let* (*A*, *B*) *be a nonempty compact and convex pair in a strictly convex Banach space* *X* *and* *T* : *A* ∪ *B* → *A* ∪ *B* *be a noncyclic contractive mapping*. *Then* *T* *has a unique best proximity pair Moreover for any* (*x*_{0}, *y*_{0}) ∈ *A*_{0} × *B*_{0} *if we define* *x*_{n+1} := *Tx*_{n} *and* *y*_{n+1} := *Ty*_{n} *then the sequence* {(*x*_{n}, *y*_{n})} *converges to the best proximity pair of T*.

#### Proof

It follows from Lemma 3.5 that *T* is a noncyclic relatively nonexpansive mapping. Since the pair (*A*, *B*) is compact and convex, the existence of a best proximity pair for the mapping *T* is concluded from Theorem 3.3. Suppose (*p*, *q*) ∈ *A* × *B* is a best proximity pair of the mapping *T*. Then *p* = *Tp*, *q* = *Tq* and ∥*p*−*q*∥ = dist (*A*, *B*). It is worth noticing that the fixed point sets of *T* in *A*_{0} and *B*_{0} are singleton. Indeed, if *p*′ ∈ *A*_{0} such that *p*′ = *T**p*′ and *p* ≠ *p*′ then from the strictly convexity of *X* we have ∥*p*′ − *q*∥ > dist (*A*, *B*). Therefore,
$$\parallel {p}^{\prime}-q\parallel =\parallel T{p}^{\prime}-Tq\parallel <\parallel {p}^{\prime}-q\parallel ,$$

which is impossible. Equivalently, we can see that the fixed point set in *B*_{0} is singleton. This implies that *T* has a unique best proximity pair in *A* × *B*. Let *x*_{0} ∈ *A*_{0} and *x*_{n+1} = *Tx*_{n}. Assume that {*x*_{nk}} is a subsequence of {*x*_{n}} such that *x*_{nk} → *z* ∈ *A*_{0}. Thus
$$\begin{array}{}d({x}_{n},\phantom{\rule{thinmathspace}{0ex}}{\mathcal{P}}_{B}p)=d(T{x}_{n-1},\phantom{\rule{thinmathspace}{0ex}}{\mathcal{P}}_{B}(Tp))\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=d(T{x}_{n-1},\phantom{\rule{thinmathspace}{0ex}}T({\mathcal{P}}_{B}p))<d({x}_{n-1},\phantom{\rule{thinmathspace}{0ex}}{\mathcal{P}}_{B}p)\end{array}$$

Hence, *d*(*z*,*P*_{B}p) = lim_{k → ∞}*d*(*x*_{nk}, 𝓟_{B}p). From Proposition 3.4 of [25] *T* is continuous on *A*_{0} ∪ *B*_{0}. Suppose *d*(*z*,𝓟_{B}*p*) > dist(*A*, *B*). We now have
$$\begin{array}{}d(z,{\mathcal{P}}_{B}\phantom{\rule{thinmathspace}{0ex}}p)=\underset{k\to \mathrm{\infty}}{lim}d({x}_{{n}_{k+1},}{\mathcal{P}}_{B}\phantom{\rule{thinmathspace}{0ex}}p)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\underset{k\to \mathrm{\infty}}{lim}d(T{x}_{{n}_{k},}{\mathcal{P}}_{B}\phantom{\rule{thinmathspace}{0ex}}p)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=d(Tz,{\mathcal{P}}_{B}\phantom{\rule{thinmathspace}{0ex}}p)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=d(Tz,{\mathcal{P}}_{B}(Tp))\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=d(Tz,T({\mathcal{P}}_{B}\phantom{\rule{thinmathspace}{0ex}}p))\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}<d(z,{\mathcal{P}}_{B}\phantom{\rule{thinmathspace}{0ex}}p),\end{array}$$
which is a contradiction and so we must have *d*(*z*,𝓟_{B}*p*) = dist(*A*, *B*). Then *z* = *p*. Since any convergent subsequence of {*x*_{nk}} converges to *p*, the sequence itself converges to *p*. Similarly we can prove the convergence of {*y*_{n}} to the point *q* and this competes the proof.□

#### Lemma 3.9

*Let* (*A*, *B*) *be a nonempty closed*, *and convex pair in a normed linear space X and* (*T*;*S*) *be a cyclic pair defined on A* ∪ *B such that*

*S*(*A*) ⊆ *T*(*A*) ⊆ *B and S*(*B*) ⊆ *T*(*B*) ⊆ *A*,

*T*(*A*) *and T*(*B*) *are compact subsets of B and A respectively*.

∥*Sx* − *Sy*∥ < ∥*Tx* − *Ty*∥, *for all* (*x*, *y*)∈ *A* × *B such that* ∥*Sx* − *Sy*∥ > dist(*A*, *B*),

*Then*
$$\begin{array}{}\text{dist}(S(A),S(B))=\text{dist}(T(A),T(B))=\text{dist}(A,B).\end{array}$$

#### Proof

Clearly
$$\begin{array}{}\text{dist}(S(A),S(B))\ge \text{dist}(T(A),T(B))\ge \text{dist}(A,B).\end{array}$$(4)

If dist(*S*(*A*), *S*(*B*)) = dist(*A*, *B*), then there is nothing to prove. Suppose dist(*S*(*A*), *S*(*B*)) > dist(*A*, *B*). By the assumption (*iii*),
$$\begin{array}{}\text{dist}(S(A),S(B))\le \text{dist}(T(A),T(B)).\end{array}$$

Therefore, dist(*S*(*A*), *S*(*B*)) = dist(*T*(*A*),*T*(*B*)) and so dist(*T*(*A*),*T*(*B*)) > dist(*A*, *B*). Let *a*′ ∈ *T*(*B*), *b*′ ∈ *T*(*A*) be such that dist(*T*(*A*),*T*(*B*)) = ∥*a*′ − *b*′∥ > dist(*A*, *B*). Assume *a*′ = *T*(*b*) and *b*′ = *T*(*a*) for some (*a*,*b*)∈ *A* × *B*. Since
$$\begin{array}{}\parallel S(a)-S(b)\parallel \ge \text{dist}(S(A),S(B))>\text{dist}(A,B),\end{array}$$
we have
$$\begin{array}{}\parallel S(a)-S(b)\parallel <\parallel T(a)-T(b)\parallel =\text{dist}(T(A),T(B)),\end{array}$$
and this is a contradiction with (4) and the result follows.□

#### Theorem 3.10

*Let* (*A*, *B*) *be a nonempty, closed and convex pair in a strictly convex Banach space X*. *Let* (*T*;*S*) *be a cyclic pair defined on* *A* ∪ *B such that*

*S*(*A*) ⊆ *T*(*A*) ⊆ *B and S*(*B*) ⊆ *T*(*B*) ⊆ *A*

*T*(*A*) *and T*(*B*) *are compact and convex subsets of B and A respectively*.

∥*Sx* − *Sy*∥ > ∥*Tx* − *Ty*∥ *for all* (*x*, *y*) ∈ *A* × *B such that* ∥*Sx* − *Sy*∥ > dist(*A*; *B*)

*S and T commute*.

*Then* (*T*;*S*) *has a unique common best proximity point in B*.

#### Proof

Let
$$\begin{array}{}[T(A){]}_{0}=\{y\in T(A):d(x,y)=\text{dist}(T(A),T(B)),\text{\hspace{0.17em}for some}\phantom{\rule{1em}{0ex}}x\in T(B)\},\end{array}$$
and
$$\begin{array}{}[T(B){]}_{0}=\{x\in T(B):d(x,y)=\text{dist}(T(A),T(B)),\text{\hspace{0.17em}for some}\phantom{\rule{1em}{0ex}}y\in T(A)\}.\end{array}$$

Notice that from Lemma 3.9, dist(*T*(*A*),*T*(*B*)) = dist(*A*, *B*). To show *ST*^{− 1} is singleton, let *x* ∈[*T*(*B*)]_{0}. Then there exists *y* ∈[*T*(*A*)]_{0} such that ∥*x* − *y*∥ = dist(*A*; *B*). For any *z* ∈ *ST*^{− 1} *x* ⊆ *T*(*B*) and *w* ∈ *ST*^{− 1}*y* ⊆ *T*(*A*),
$$\begin{array}{}\parallel z-w\parallel =\parallel S{x}^{\prime}-S{y}^{\prime}\phantom{\rule{thinmathspace}{0ex}}\parallel \end{array}$$

where *z* = *Sx*′ and *w* = *Sy*′ for some *x*′ ∈ *T*^{−1}*x* and *y*′ ∈ *T*^{−1}*y*. If ∥*Sx*′ − *Sy*′ ∥> dist (*A*, *B*), then
$$\begin{array}{}\parallel S{x}^{\prime}-S{y}^{\prime}\parallel <\parallel T{x}^{\prime}-T{y}^{\prime}\parallel \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\parallel T{T}^{-1}x-T{T}^{-1}y\parallel \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\parallel x-y\parallel \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\text{dist}(A,\phantom{\rule{thinmathspace}{0ex}}B),\end{array}$$
which is impossible. So
$$\begin{array}{}\parallel z-w\parallel =\parallel S{x}^{\prime}-S{y}^{\prime}\parallel =\text{dist}(A,\phantom{\rule{thinmathspace}{0ex}}B),\end{array}$$
for any *z* ∈ *ST*^{−1}*x* and *w* ∈ *ST*^{−1}*y*. It now follows from the strict convexity of *X* that *ST*^{−1}*x* and *ST*^{−1}*y* are singleton. Also, it is clear that *ST*^{−1}([*T*(*A*)]_{0}) ⊆[*T*(*A*)]_{0} and *ST*^{−1}([*T*(*B*)]_{0}) ⊆[*T*(*B*)]_{0}, that is, *ST*^{−1} : [*T*(*A*)]_{0}∪[*T*(*B*)]_{0}→[*T*(*A*)]_{0}∪[*T*(*B*)]_{0} is a noncyclic mapping. Let (*x*, *y*)∈[*T*(*A*)]_{0} ×[*T*(*B*)]_{0} be such that ∥*x* − *y*∥ > dist (*T*(*A*),*T*(*B*))(= dist(*A*, *B*)). If ∥ *ST*^{−1}*x* − *ST*^{−1}*y* ∥ > dist (*A*, *B*), then
$$\begin{array}{}\parallel S{T}^{-1}x-S{T}^{-1}y\parallel <\parallel T{T}^{-1}x-T{T}^{-1}y\parallel =\parallel x-y\parallel ,\end{array}$$
which implies that *ST*^{−1} is a noncyclic contractive mapping on a compact and convex pair ([*T*(*A*)]_{0}, [*T*(*B*)]_{0}). Now using Theorem 3.7 for any *x* ∈[*T*(*A*)]_{0}∪[*T*(*B*)]_{0} the sequence {(*ST*^{−1})^{n}*x*}_{n≥1} converges to the unique fixed point *z* of *ST*^{−1}. Since *S* and *T* commutes,
$$\begin{array}{}Tz=T(S{T}^{-1}z)=Sz=S(S{T}^{-1}z)\end{array}$$
Suppose ∥*Sz*−*T*(*Sz*)∥ > dist (*A*, *B*). Thus
$$\begin{array}{}\parallel Sz-T(Sz)\parallel =\parallel Sz-S(Tz)\parallel \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}<\parallel Tz-T(Tz)\parallel \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\parallel Sz-T(Sz)\parallel ,\end{array}$$
which is a contradiction. Hence ∥*Sz*−*T*(*Sz*)∥= dist (*A*, *B*). On the other hand, if ∥*Sz*−*S*(*Sz*)∥ > dist (*A*, *B*), then
$$\begin{array}{}\parallel Sz-S(Sz)\parallel <\parallel Tz-T(Sz)\parallel \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\parallel Sz-T(Sz)\parallel \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\parallel Sz-S(Tz)\parallel \\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\parallel Sz-S(Sz)\parallel ,\end{array}$$
which is impossible. Thereby ∥*Sz*−*S*(*Sz*)∥= dist (*A*, *B*) and so the point *Sz* is a common best proximity point for the cyclic pair (*T*; *S*). Uniqueness of the common best proximity point follows as in the proof of Theorem 2.1. □

#### Example 3.11

*Let X* = {ℝ^{2}, ∥.∥_{2}} *and let*
$\begin{array}{}A=\{(0,s):0\le s\le \frac{1}{2}\}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}B=\{(1,t):0\le t\le \frac{1}{2}\}.\end{array}$
*Thus* dist(*A*, *B*) = 1. *Define the cyclic pair* (*T* ; *S*) *on A* ∪ *B as follows*:
$$\begin{array}{}S(0,s)=(1,{s}^{3}),\phantom{\rule{thinmathspace}{0ex}}S(1,t)=(0,{t}^{3})\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}T(0,s)=(1,{s}^{2}),T(1,t)=(0,{t}^{2}).\end{array}$$
*Then*
$$\begin{array}{}S(A)={\displaystyle \{(1,s):0\le s\le \frac{1}{8}\},\phantom{\rule{1em}{0ex}}S(B)=\{(0,t):0\le t\le \frac{1}{8}\},}\\ T(A)={\displaystyle \{(1,s):0\le s\le \frac{1}{4}\},\phantom{\rule{1em}{0ex}}T(B)=\{(0,t):0\le t\le \frac{1}{4}\}.}\end{array}$$

*Thereore*, *S*(*A*) ⊆ *T*(*A*) ⊆ *B and S*(*B*) ⊆ *T*(*B*) ⊆ *A*. *Also*, *T*(*A*) *and* *T*(*B*) *are compact and convex subsets of B and A*, *respectively*. *Moreover*,
$$\begin{array}{}ST(0,s)=S(1,{s}^{2})=(0,{s}^{6})=T(1,{s}^{3})=TS(0,s),\\ \\ ST(1,\phantom{\rule{thinmathspace}{0ex}}t)=S(0,\phantom{\rule{thinmathspace}{0ex}}{t}^{2})=(1,\phantom{\rule{thinmathspace}{0ex}}{t}^{6})=T(0,{t}^{3})=TS(1,\phantom{\rule{thinmathspace}{0ex}}t),\end{array}$$
*and so*, *S* *and T are commuting*. *Finally if* ∥*S*(0, *s*)−*S*(1, *t*)∥ > dist(*A*, *B*), *we conclude that s* ≠ *t*. *Hence*,
$$\begin{array}{}\parallel S(0,s)-S(1,t)\parallel =\sqrt{1+(|{s}^{3}-{t}^{3}|{)}^{2}}<\sqrt{1+(|{s}^{2}-{t}^{2}|{)}^{2}}=\parallel T(0,s)-T(1,t)\parallel ,\end{array}$$
*whenever s*, *t* ≥0 *and s* + *t* ≤ 1. *Thereby all of the assumptions of Theorem 3.10 hold and the cyclic pair* (*T*; *S*) *has a unique common best proximity point in B and this point is p* = (0,0).

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