Banach fixed-point theorem: Difference between revisions

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for all <math>x, y</math> in <math>X</math>.
 
<blockquote>'''Banach Fixed Point Theorem.''' Let <math>(X,d)</math> be a [[Empty set|non-empty]] [[complete metric space]] with a contraction mapping <math>T\colon X \to X</math>. Then <math>''T</math>'' admits a unique [[Fixed point (mathematics)|fixed-point]] <math>''x^*\'' in ''X</math>'' (i.e. <math>''T''(''x^*'') = ''x^*</math>''). Furthermore, <math>''x^*</math>'' can be found as follows: start with an arbitrary element ''x''<mathsub>x_0\in X0</mathsub> in ''X'' and define a [[sequence]] {''x<mathsub>\{x_n\}n</mathsub>''} by ''x<mathsub>x_n=T(x_{n-1})</mathsub>'' for= ''T''(''x''<mathsub>''n\geq1''−1</mathsub>) for ''n'' ≥ 1. Then {{nowrap|''x<mathsub>x_n\to x^*n</mathsub>'' → ''x*''}}.</blockquote>
 
''Remark 1.'' The following inequalities are equivalent and describe the [[Rate of convergence|speed of convergence]]:
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</math>
 
Any such value of <math>''q</math>'' is called a ''[[Lipschitz constant]]'' for <math>''T</math>'', and the smallest one is sometimes called "the best Lipschitz constant" of <math>''T</math>''.
 
''Remark 2.'' In general, <math>''d''(''T''(''x''),&nbsp;''T''(''y''))&nbsp;< &nbsp;''d''(''x'',&nbsp;''y'')</math> for all <math>''x\neq'' ≠ ''y</math>,'' is in general not enough to ensure the existence of a fixed point, as is shown by the map <math>''T\colon'' : [1,\infty)\to [1,\infty)</math>, by <math>''T''(''x'')&nbsp;=&nbsp;''x''&nbsp;+&nbsp;1/''x</math>'', which lacks a fixed point. However, if <math>''X</math>'' is [[Compact space|compact]], then this weaker assumption does imply the existence and uniqueness of a fixed point, that can be easily found as a minimizer of <math>''d''(''x'',&nbsp;''T''(''x''))</math>., Indeedindeed, a minimizer exists by compactness, and has to be a fixed point of <math>''T</math>''. It then easily follows that the fixed point is the limit of any sequence of iterations of <math>''T</math>''.
 
''Remark 3.'' When using the theorem in practice, the most difficult part is typically to define <math>''X</math>'' properly so that <math>''T''(''X'')\subseteq ⊆ ''X</math>''.
 
==Proof==