Then, for all u,v, we obtain P u - P v ≤ L ( μ + 1 ) 1 - α B ( 1, α ) Γ u - v. Thus, the operator P is a contraction mapping then in view of Banach fixed point theorem, P has a unique fixed point which corresponds to the solution of the initial value problem (15).
Hence by the assumption of the theorem we have that P is a contraction mapping then in view of the Banach fixed point theorem, P has a unique fixed point which is corresponding to the solution of Equation (1).
In view of the Banach fixed point theorem and Lemma 2.1, has a unique fixed point and (2.2).
Also, in view of (3.13), for fixed and, is a bounded function of and.
Now g ( z, μ ( z ) ) is a contraction mapping whenever ρ ∈ ( 0, 1 ) ; therefore, in view of the Banach fixed point theorem, Eq. (15) has a unique analytic solution in the unit disk and consequently the problem (14).
Now, to show that G ( z, μ ( z ), z μ ′ ( z ) ) is a contraction mapping, Thus, in view of the Banach fixed point theorem, Eq. (20) has a unique analytic solution in the unit disk and consequently the problem (19).
end{aligned}Therefore, if (K < Xi ^{-1},) then in view of the Banach fixed point theorem, Eq. (1) has a unique solution, which clearly is the only m-P(Lambda Lambda)M solution.
In view of Proposition 2.5, Fix ( Q T ) = Fix ( T ), thus, we get x = T Q x = Q x, which turns out that x = T x.
This concept is of interest in view of the Earle-Hamilton fixed point theorem [21].
In view of the Avery-Peterson fixed point theorem, this paper investigates the existence of three positive solutions for the second-order boundary value problem with integral boundary conditions left { textstylebegin{array}l} u t)+h(t)f t,u(t),u'(t))=0,quad 0< t< 1, u(0 -alpha u'(0 -alpha0}^{1}g_{1}(s)u(s),ds, u(1)+beta u'(1)=int_{0}^{1}g_{2}(s)u(s),ds.
In view of (C.1), for any fixed positive integer number m, there exists N > 0 such that | O n | ≤ d m for every n > N. As n is sufficiently large, we have ∑ j = 1 N a j 2 + ( m − 1 m ) 2 d 2 ∑ j = N + 1 n j 2 α ≤ ∑ j = 1 n a j 2 ≤ ∑ j = 1 N a j 2 + ( m + 1 m ) 2 d 2 ∑ j = N + 1 n j 2 α. (3.7).
