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Contraction mapping theorem
2022-04-23 14:42:00 【patrickpdx】
Definition
set up X X X Is the distance space , T : X → X T: X \rightarrow X T:X→X Is a mapping , If exist 0 ≤ λ < 1 0\leq \lambda<1 0≤λ<1, bring
d ( T x , T y ) ≤ λ d ( x , y ) , ∀ x , y ∈ X d(Tx,Ty)\leq \lambda d(x,y), \quad \forall x,y\in X d(Tx,Ty)≤λd(x,y),∀x,y∈X
said T T T It's compressed
lemma
The contractive mapping is continuous
if x n → x x_{n} \rightarrow x xn→x, be : T x n → T x Tx_{n}\rightarrow Tx Txn→Tx
prove :
d ( T x n , T x ) ≤ λ d ( x n , x ) → 0 d(Tx_{n}, Tx) \leq \lambda d(x_{n},x) \rightarrow 0 d(Txn,Tx)≤λd(xn,x)→0
Contraction mapping theorem
There is a unique fixed point for contractive mappings on complete metric spaces
prove :
set up X X X Is an arbitrary complete distance space , T T T yes X X X Compression mapping on . Take whatever you like x 0 ∈ X x_{0} \in X x0∈X, The following proves that the sequence of numbers { x n } \{x_{n}\} {
xn} :
x n = T x n − 1 x_{n} = T x_{n-1} xn=Txn−1
The limits of x = lim n → ∞ x n x = \lim\limits_{n\rightarrow \infty}x_{n} x=n→∞limxn There is , And is a fixed point
(1) First of all, prove that { x n } \{x_{n}\} {
xn} yes Cauchy Sequence :
d ( x n + 1 , x n ) = d ( T x n , T x n − 1 ) ≤ λ d ( x n , x n − 1 ) d(x_{n+1},x_{n}) = d(Tx_{n},Tx_{n-1})\leq \lambda d(x_{n},x_{n-1}) d(xn+1,xn)=d(Txn,Txn−1)≤λd(xn,xn−1)
, in turn,
d ( x n + 1 , x n ) ≤ λ n d ( x 1 , x 0 ) d(x_{n+1},x_{n})\leq \lambda^{n}d(x_{1},x_{0}) d(xn+1,xn)≤λnd(x1,x0)
For any positive integer n n n and p p p :
KaTeX parse error: No such environment: align at position 8: \begin{̲a̲l̲i̲g̲n̲}̲ d(x_{n+p},x_{n…
So for ∀ ϵ > 0 \forall \epsilon>0 ∀ϵ>0, ∃ N \exists N ∃N, Properly n > N n>N n>N when , Yes ∀ p \forall p ∀p Satisfy :
∣ d ( x n + p , x n ) ∣ ≤ ϵ |d(x_{n+p},x_{n})|\leq \epsilon ∣d(xn+p,xn)∣≤ϵ
(2) Because of space X X X It's complete , therefore { x n } \{x_{n}\} { xn} convergence
(3) , in turn, x = lim n → ∞ x n = lim n → ∞ T x n − 1 = T lim n → ∞ x n − 1 = T x x=\lim\limits_{n\rightarrow \infty}x_{n} = \lim\limits_{n\rightarrow \infty} Tx_{n-1} = T \lim\limits_{n\rightarrow \infty}x_{n-1} = Tx x=n→∞limxn=n→∞limTxn−1=Tn→∞limxn−1=Tx
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本文为[patrickpdx]所创,转载请带上原文链接,感谢
https://yzsam.com/2022/04/202204231427118779.html
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