Exact(4)
Step 1:: (Vert g_{k} Vert leqvarepsilon), stop.
Then we have begin{aligned} mathcal{G}_{q}&=max bigl{ 1, vert 1-gamma C_{1} vert bigr} =1.
Case 1: (vert svert geqalpha2^{-alpha-3}(2m+d)vert tvert ).
Step 1:: (Vert g_{k}Vert leq varepsilon), stop; Otherwise, go to the next step.
Similar(56)
So (Vert V_{k}^{-1} Vert stackrel{K}{rightarrow } Vert V_^{-1} Vert
We assume that (vert e^{t}-1 vert <1).
Let (A={xin K: vert f(x -1 vert geqvarepsilon}).
end{aligned} Obviously, we have the conclusion (Vert E^{s + 1} Vert _{infty} le Vert E^{1} Vert _{infty}).
As in the proof of Lemma 3.5, Vert x_{1} Vert leq Vert K_{P_{1}} Vert Vert N_{1}x Vert _{1}, qquad Vert y_{1} Vert leq Vert N_{2}y Vert _{1}.
Noting that ((C[0,1], Vert cdot Vert )) is dense in the ((L^{2}[0,1], Vert cdot Vert )), then X is a dense subset of ((L^{2}[0,1], Vert cdot Vert )).
Since (Vert a_{1}Vert Vert bVert + Vert a_{2}Vert le (Vert a_{1}Vert + Vert a_{2}Vert ) Vert bVert < 1), we have Vert b kVert = biglVert (1_{mathbb{A}} - a_{2})^{-1} a_{1} bbigrVert le biglVert (1_{mathbb{A}} - a_{2})^{-1}bigrVert Vert a_{1}Vert Vert bVert = sum_{i=0}^{infty} Vert a_{2}Vert ^{i} Vert a_{1}Vert Vert b Vert = frac{Vert a_{1}Vert Vert bVert }{1 - Vert a_{2}Vert } < 1.
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