Let (X,d) be a metric space. Then a map T:X→X is called a contraction map if it exists 0≤k<1 such that d(T(x),T(y))≤kd(x,y) for all x,y∈X. According to Banach fixed-point theorem, if (X,d) is a complete metric space and T a contraction map, then T admits a fixed-point x∗∈X, i.e. T(x∗)=x∗.
We look here at counterexamples to the Banach fixed-point theorem when some hypothesis are not fulfilled.
First, let’s consider f:R⟶Rx⟼x+1 For all x,y∈R we have |f(x)−f(y)|=|x−y|. f is not a contraction, but an isometry. Obviously, f has no fixed-point.
We now prove that a map satisfying d(g(x),g(y))<d(x,y) might also not have a fixed-point. A counterexample is the following map g:[0,+∞)⟶[0,+∞)x⟼√1+x2 Since g′(ξ)=ξ√1+ξ2<1 for all ξ∈[0,+∞), by the mean value theorem |g(x)−g(y)|=|g′(ξ)||x−y|<|x−y| for all x,y∈[0,+∞). However g has no fixed-point. Finally, let's have a look to a space (X,d) which is not complete. We take a,b∈R with 0<a<1 and for (X,d) the space X=R∖{b1−a} equipped with absolute value distance. X is not complete. Consider the map h:X⟶Xx⟼ax+b h is well defined as for x≠b1−a, h(x)≠b1−a. h is a contraction map as for x,y∈R |h(x)−h(y)|=a|x−y| However, h doesn't have a fixed-point in X as b1−a is the only real for which h(x)=x.