end{aligned} The condition (zeta<1) causes φ to be a strict contraction.
Let (X, d) be a complete generalized metric space and let A : X → X be a strict contraction with the Lipschitz constant k such that d x0, A x0)) < +∞ for some x0 ∈ X.
Let E be a Banach space, ({D}subset {E}) be closed, and (F:{D}rightarrow {D}) be a strict contraction, i.e., (vert Fx-Fyvert leq kvert x-yvert ) for some (kin (0,1)) and all (x,yin {D}).
Let ((X, d)) be a complete metric space and (F : X longrightarrow X) be a strict contraction, i.e., a map satisfying d(Fx, Fy leq ad x,y), quad textit{for all } x, y in X, where (0 leq a < 1) is constant.
Let X be a nonempty complete metric space and let (T: Xrightarrow X) be a strict contraction, i.e., there is (0< k<1) such that (d(T x ,T y))leq kd x,y)), (forall x,yin X), then S has a unique fixed point (u=T u)).
One can easily see that T t u is a strict contraction with a contractive constant L = 1 − t.
If there is no contractive self-mapping on X not being a strict contraction in the switching law, the above holds if ( X, d ) is just a complete metric space.
Then Ψ is a strict contraction.
Φ 1 is a strict contraction, and.
and hence is a strict contraction whenever is small enough.
If T is a strict contraction, then it is trivially a proper contraction.
