If r = 0, it is easy to see that {T n x} converges in norm to a fixed point of T for any x ∈ K. Now we assume that r > 0. Let x be arbitrary element in K and {n i } be a subsequence of the positive integers {n}.
Conversely, if a, b are arbitrary elements in L with η ( a ) ≤ η ( b ), let c = a ∧ b.
Let x, y ∈ E 1 be arbitrary elements.
end{aligned} Since u was arbitrary element of X in (3.13), we obtain omega^{T}bigl(F X),varepsilonbigr) leq omega^{T}(p,varepsilon)+frac{1}{(n-1-k)!} bigl[varepsilon T^{n}U_{r_{0}}^{T}+phi^{T}( varepsilon TU_{r_{0}}^{T} bigr].
Let H be a Hilbert space and W be a closed subspace of H such that (dimW < infty) and ({ w_{1},ldots,w_{n},w_{n}}) is any basis for W. Let g be an arbitrary element in H and (g_{0}) be the unique best approximation to g out of W. Then |g-g_{0}|=G_{g}, where G_{g}=sqrt{frac{F g,w_{1},ldots,w_{n})}{F(w_{1},ldots,w_{n})}}, and F is introduced in [25].
Let f be an arbitrary element in H. Since Y is a finite dimensional and closed subspace, it is a complete subset of H. Therefore, f has the unique best approximation out of Y, such that ( y_{0} ) exists y_{0} in Y; forall y in Y: left| {f - y_{0} } right|_{2} le left| {f - y} right|_{2}.
Let z 0 be an arbitrary element in a Hilbert space H, the sequence of iterations { z n } is defined by finding u n ∈ C such that { G ( u n, y ) + 〈 u n − z n, y − u n 〉 ≥ 0, ∀ y ∈ C, z n + 1 = P C ( z n − β n [ z n − u n + ∑ i = 1 ∞ γ i A i ( z n ) + α n z n ] ) z n + 1 = P C ( z n − β n [ ∑ i = 1 ∞ γ i A i ( z n ) + ( 1 + α n ) z n − u n ] ), n ≥ 0, (2.1).
Let be an arbitrary element in and define.
For this purpose, let Ψ be an arbitrary element in the dual space ( L 2 ( R ) ) ′ of L 2 ( R ).
But this procedure would work only at a large scale, wherein there would be an arbitrary element in both determining the scale and the averaging procedure.
