Let E be a Banach space and let g:E\to \mathbb{R} be a convex and Gâteaux differentiable function. Let C be a closed and convex subset of a real Banach space E. A mapping T:C\to E is said to be Bregman quasinonexpansive [17] if F(T)\ne \mathrm{\varnothing} and
{D}_{g}(p,Tx)\le {D}_{g}(p,x),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in C,p\in F(T).
Let C and D be nonempty subsets of a real Banach space E with D\subset C. A mapping {R}_{D}:C\to D is said to be sunny if
{R}_{D}({R}_{D}x+t(x{R}_{D}x))={R}_{D}x
for each x\in E and t\ge 0. A mapping {R}_{D}:C\to D is said to be a retraction if {R}_{D}x=x for each x\in C.
The following result was proved in [24].
Lemma 3.1 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets. Let C be a nonempty, closed and convex subset of E. Let T:C\to E be a Bregman quasinonexpansive mapping. Then F(T) is closed and convex.
Corollary 3.1 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded sets and uniformly convex on bounded sets. Let C be a nonempty, closed and convex subset of E and let T:C\to E be a Bregman nonexpansive mapping. If F(T)\ne \mathrm{\varnothing}, then it is closed and convex.
Using ideas in [43], we can prove the following result.
Theorem 3.1 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets. Let C be a nonempty, closed and convex subset of E and let T:C\to C be a mapping. Let {\{{x}_{n}\}}_{n\in \mathbb{N}} be a bounded sequence of C and let μ be a mean on {l}^{\mathrm{\infty}}. Suppose that
{\mu}_{n}{D}_{g}({x}_{n},Ty)\le {\mu}_{n}{D}_{g}({x}_{n},y)
for all y\in C. Then T has a fixed point in C.
Proof Let μ be a mean on {l}^{\mathrm{\infty}} and {\{{x}_{n}\}}_{n\in \mathbb{N}} be a bounded sequence in C. Define a mapping h:{E}^{\ast}\to \mathbb{R} by
h\left({x}^{\ast}\right)={\mu}_{n}\u3008{x}_{n},{x}^{\ast}\u3009,\phantom{\rule{1em}{0ex}}{x}^{\ast}\in {E}^{\ast}.
Since μ is linear, so is h. Observe that
\begin{array}{rl}\lefth\left({x}^{\ast}\right)\right& =\left{\mu}_{n}\u3008{x}_{n},{x}^{\ast}\u3009\right\\ \le \parallel \mu \parallel \underset{n\in \mathbb{N}}{sup}\left\u3008{x}_{n},{x}^{\ast}\u3009\right\\ \le \parallel \mu \parallel \underset{n\in \mathbb{N}}{sup}\parallel {x}_{n}\parallel \parallel {x}^{\ast}\parallel \\ =\underset{n\in \mathbb{N}}{sup}\parallel {x}_{n}\parallel \parallel {x}^{\ast}\parallel \end{array}
for all {x}^{\ast}\in {E}^{\ast}. This implies that h is a linear and continuous realvalued mapping on {E}^{\ast}. Since E is reflexive, then there exists a unique element z\in E such that
h\left({x}^{\ast}\right)={\mu}_{n}\u3008{x}_{n},{x}^{\ast}\u3009=\u3008z,{x}^{\ast}\u3009,\phantom{\rule{1em}{0ex}}{x}^{\ast}\in {E}^{\ast}.
We claim that z\in C. If not, then by the separation theorem [13] there exists {y}^{\ast}\in {E}^{\ast} such that
\u3008z,{y}^{\ast}\u3009<\underset{y\in C}{inf}\u3008y,{y}^{\ast}\u3009.
Since {\{{x}_{n}\}}_{n\in \mathbb{N}}\subset C, we conclude that
\u3008z,{y}^{\ast}\u3009<\underset{y\in C}{inf}\u3008y,{y}^{\ast}\u3009\le \underset{n\in \mathbb{N}}{inf}\u3008{x}_{n},{y}^{\ast}\u3009\le {\mu}_{n}\u3008{x}_{n},{x}^{\ast}\u3009=\u3008z,{x}^{\ast}\u3009.
This is a contradiction. Thus we have z\in C. In view of (2.5), for any y\in C and n\in \mathbb{N}, we deduce that
{D}_{g}({x}_{n},y)={D}_{g}({x}_{n},Ty)+{D}_{g}(Ty,y)+\u3008{x}_{n}Ty,\mathrm{\nabla}g(Ty)\mathrm{\nabla}g(y)\u3009.
Thus we have, for any y\in C, that
\begin{array}{rl}{\mu}_{n}{D}_{g}({x}_{n},y)& ={\mu}_{n}{D}_{g}({x}_{n},Ty)+{\mu}_{n}{D}_{g}(Ty,y)+{\mu}_{n}\u3008{x}_{n}Ty,\mathrm{\nabla}g(Ty)\mathrm{\nabla}g(y)\u3009\\ ={\mu}_{n}{D}_{g}({x}_{n},Ty)+{D}_{g}(Ty,y)+\u3008zTy,\mathrm{\nabla}g(Ty)\mathrm{\nabla}g(y)\u3009.\end{array}
By the assumption, we have that
{\mu}_{n}{D}_{g}({x}_{n},Ty)\le {\mu}_{n}{D}_{g}({x}_{n},y)
for all y\in C. This implies that
{\mu}_{n}{D}_{g}({x}_{n},y)\le {\mu}_{n}{D}_{g}({x}_{n},y)+{D}_{g}(Ty,y)+\u3008zTy,\mathrm{\nabla}g(Ty)\mathrm{\nabla}g(y)\u3009
(3.1)
for all y\in C. Putting y=z in (3.1) and taking into account (2.6), we see that
\begin{array}{rl}0& \le {D}_{g}(Tz,z)+\u3008zTz,\mathrm{\nabla}g(Tz)\mathrm{\nabla}g(z)\u3009\\ ={D}_{g}(z,Tz)+\u3008zTz,\mathrm{\nabla}g(z)\mathrm{\nabla}g(Tz)\u3009+\u3008zTy,\mathrm{\nabla}g(Tz)\mathrm{\nabla}g(z)\u3009\\ ={D}_{g}(z,Tz).\end{array}
Then we have 0\le {D}_{g}(z,Tz), which implies that {D}_{g}(z,Tz)=0. In view of Lemma 2.2, we conclude that Tz=z, which completes the proof. □
Remark 3.1 Let g and T be as in Example 1.1. Let x\in [0,0.9] be fixed. Then {\{{T}^{n}x\}}_{n\in \mathbb{N}} is a bounded sequence in [0,0.9]. Set {x}_{n}:={T}^{n}x for n=1,2,\dots . It is obvious that T satisfies all the aspects of the hypothesis of Theorem 3.1, so it has a fixed point.
Corollary 3.2 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets. Let C be a nonempty, closed and convex subset of E and let T:C\to C be a mapping. Suppose that there exist x\in C and a Banach limit μ such that {\{{T}^{n}x\}}_{n\in \mathbb{N}} is bounded and
{\mu}_{n}{D}_{g}({T}^{n}x,Ty)\le {\mu}_{n}{D}_{g}({T}^{n}x,y)
for all y\in C. Then T has a fixed point.
Corollary 3.3 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets. Let C be a nonempty, closed and convex subset of E and let T:C\to C be a Bregman nonexpansive mapping. Suppose that there exists x\in C such that {\{{T}^{n}x\}}_{n\in \mathbb{N}} is bounded. Then T has a fixed point.
Proof Let μ a Banach limit on {l}^{\mathrm{\infty}} and x\in C be such that {\{{T}^{n}x\}}_{n\in \mathbb{N}} is bounded. Then we have
{\mu}_{n}{D}_{g}({T}^{n}x,Ty)={\mu}_{n}{D}_{g}({T}^{n+1}x,Ty)\le {\mu}_{n}{D}_{g}({T}^{n}x,y)
for all y\in C. In view of Corollary 3.2, we deduce that F(T)\ne \mathrm{\varnothing}, which completes the proof. □
Corollary 3.4 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded sets and uniformly convex on bounded sets. Let C be a nonempty, bounded, closed and convex subset of E and let T:C\to C be a Bregman nonexpansive mapping. Then T has a fixed point.
Definition 3.1 Let A and C be nonempty subsets of a real Banach space E with A\subset C. We say that A is a Bregman nonexpansive retract of C if there exists a Bregman nonexpansive map R:C\to A such that R(a)=a for every a\in A.
Definition 3.2 Let C be a nonempty, closed and convex subset of a real Banach space E. The mapping T:C\to C is called Bregman NRmap if Fix(T) is a Bregman nonexpansive retract of C.
Theorem 3.2 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded sets and uniformly convex on bounded sets. Let C be a nonempty, bounded, closed and convex subset of E. Let T:C\to C be a continuous Bregman NRmap. Let S:C\to C be a Bregman nonexpansive mapping such that (S,T) is a Banach operator pair. Then F(S,T) is not empty.
Proof Since the retract of a nonempty space is nonempty, Fix(T) is nonempty and is closed as T is continuous. Since T is a Bregman NRmap, then there exists a Bregman nonexpansive retract R:C\to Fix(T). Since (S,T) is a Banach operator pair, then S(Fix(T))\subset Fix(T). Hence S\circ R:C\to C is a Bregman nonexpansive map such that S\circ R(C)\subset Fix(T). Corollary 3.4 implies the existence of a fixed point of S\circ R. Clearly, such a fixed point is a fixed point of S which belongs to Fix(T). Hence Fix(T)\cap Fix(S)=F(S,T) is not empty. □
Example 3.1 Let E be a reflexive and smooth Banach space and let C be a closed and convex subset of E such that 0\in C. Let T:C\to E be defined as
T(x)=x,\phantom{\rule{1em}{0ex}}x\in C.
Then T is a Bregman quasinonexpansive mapping with g(x)=\frac{1}{2}{\parallel x\parallel}^{2}, \mathrm{\nabla}g(x)=Jx for all x\in C and F(T)=\{0\}. Indeed, it is clear that
\parallel Tx\parallel \le \parallel x\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in C.
This implies that
g(Tx)g(x)=\frac{1}{2}{\parallel Tx\parallel}^{2}\frac{1}{2}{\parallel x\parallel}^{2}\le \u30080,\mathrm{\nabla}g(Tx)\mathrm{\nabla}g(x)\u3009=\u3008p,\mathrm{\nabla}g(Tx)\mathrm{\nabla}g(x)\u3009
for all p\in F(T). Then we have
{\parallel p\parallel}^{2}+{\parallel Tx\parallel}^{2}2\u3008p,\mathrm{\nabla}g(Tx)\u3009\le {\parallel p\parallel}^{2}+{\parallel x\parallel}^{2}2\u3008p,\mathrm{\nabla}g(x)\u3009.
This means that
{D}_{g}(p,Tx)\le {D}_{g}(p,x),
for all p\in F(T) and x\in C. Hence, T is a Bregman quasinonexpansive mapping. Define a mapping R:C\to \{0\} by
R(x)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in C.
Then T is a Bregman NRmap.
Assume now that h:E\to \mathbb{R} is a lower semicontinuous function satisfying the following conditions:

(i)
h is totally convex on bounded sets;

(ii)
h, as well as its Fenchel conjugate {h}^{\ast}, are defined and (Gâteaux) differentiable on E and {E}^{\ast}, respectively;

(iii)
{h}^{\prime} is uniformly continuous and {h}^{\ast} is bounded on bounded sets.
Let A:domA\to {E}^{\ast} be an operator and Ω be a nonempty subset of domA such that 0\in \mathrm{\Omega}, A(0)=0 and C\subset domA. For any \alpha \in (0,\mathrm{\infty}), we define the operator {A}_{\alpha}^{h}:domA\to E by
{A}_{\alpha}^{h}x={{h}^{\ast}}^{\prime}({h}^{\prime}(x)\alpha Ax).
It is worth mentioning that Ax=0 if and only if x\in domA is a fixed point of {A}_{\alpha}^{h}. The operator A is said to be inversestronglymonotone relative to h on the set Ω if there exist a real number \alpha >0 and a vector z\in \mathrm{\Omega} such that
\u3008Ay,{A}_{\alpha}^{h}yz\u3009\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in \mathrm{\Omega}.
If we set S:={A}_{\alpha}^{h}, then S is a Bregman nonexpansive mapping (for more details, see [41]). It is clear that T and S satisfy all the aspects of the hypothesis of Theorem 3.2 and T and S have a common fixed point.
Remark 3.2 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets. Let C be a nonempty, bounded, closed and convex subset of E and let T:C\to C be a Bregman nonexpansive mapping. Then, in view of Corollary 3.4 and Lemma 3.1, Fix(T) is not empty and closed convex which implies that Fix(T) is a Bregman nonexpansive retract of C. Thus T is a Bregman NRmap.
Theorem 3.3 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets. Let T and S be two Bregman nonexpansive selfmappings defined on a closed and convex subset C of E. If (S,T) is a Banach operator pair and T(C) is bounded, then Fix(T)\cap Fix(S)\ne \mathrm{\varnothing}.
Proof Let K=\overline{conv}(T(C)). Then T:K\to K and K is nonempty and bounded. In view of Corollary 3.4, the fixed point set Fix(T) of T is nonempty and bounded. Since (S,T) is a Banach operator pair, S:Fix(T)\to Fix(T). By Corollary 3.4, S has a fixed point in Fix(T) as required. □
The following slight extension of Theorem 3.3 can be proved easily.
Theorem 3.4 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets. Let C be a nonempty, closed and convex subset of E. Let X be a normed space and T and S be two Bregman nonexpansive selfmappings defined on a closed convex set C\subset E. If (S,T) is a Banach operator pair, and if \overline{{T}^{n}(C)} is bounded for some n\in \mathbb{N}, then Fix(T)\cap Fix(S)\ne \mathrm{\varnothing}.
Corollary 3.5 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets. Let C be a nonempty, bounded, closed and convex subset of E. Let T:C\to C be Bregman nonexpansive. Let S:C\to C be a Bregman nonexpansive mapping such that (S,T) is a Banach operator pair. Then F(S,T) is not empty.
Corollary 3.6 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets. Let C be a nonempty, closed and convex subset of E. Let T:C\to C be a Bregman nonexpansive map such that T(C) is bounded and T(C)\subset Fix(T). Let S:C\to C be a Bregman nonexpansive mapping such that (S,T) is nontrivially a Banach operator pair. Then Fix(S)\cap Fix(T) is not empty.
Corollary 3.7 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets. Let C be a nonempty, closed and convex subset of E. Let S,T:C\to C be a nontrivially Banach operator pair such that Fix(T) is bounded and S is a Bregman nonexpansive map. Assume that T:C\to Fix(T) is a Bregman nonexpansive map. Then Fix(S)\cap Fix(T) is not empty.
Theorem 3.5 Let E be a reflexive Banach space and let g:E\to \mathbb{R} be a convex, continuous, strongly coercive and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets. Let C be a nonempty, closed and convex subset of E which has the property that every Bregman nonexpansive mapping of C\to C is Bregman NRmap. Suppose T:C\to C is a mapping for which {T}^{n} is Bregman nonexpansive for some n\in \mathbb{N}, and suppose the restriction of T to Fix({T}^{n}) is also Bregman nonexpansive. Then Fix(T) is a nonempty Bregman nonexpansive retract of C. Consequently, if S:C\to C is Bregman nonexpansive and if (S,T) is a Banach operator pair, then Fix(T)\cap Fix(S) is a nonempty Bregman nonexpansive retract of C.
Proof By assumption, there exists a Bregman nonexpansive retraction {R}_{1} of C onto Fix({T}^{n}). Consequently, T\circ {R}_{1} is a Bregman nonexpansive mapping of C into C, so Fix(T\circ {R}_{1}) is a nonempty Bregman nonexpansive retract of C. But x\in Fix(T\circ {R}_{1})\iff x\in Fix(T)\cap Fix({T}^{n}), and by Lemma 1 [44]
x\in Fix(T)\cap Fix\left({T}^{n}\right)\phantom{\rule{1em}{0ex}}\iff \phantom{\rule{1em}{0ex}}x\in Fix\left({T}^{n+1}\right)\cap Fix\left({T}^{n}\right)=Fix(T).
Therefore there is a Bregman nonexpansive retraction {R}_{2} of C onto Fix(T). So, S\circ {R}_{2} is a Bregman nonexpansive mapping of C into Fix(T). Therefore Fix(S\circ {R}_{2})=Fix(S)=Fix(S)\cap Fix(T) is a nonempty Bregman nonexpansive retract of C. □
We might observe that in the above theorem it is not necessary that T be Bregman nonexpansive. The only facts needed for the proof is that Fix({T}^{n}) be a Bregman nonexpansive retract of C.