Banach fixed-point theorem: Difference between revisions

Revision as of 13:32, 29 October 2023

Revision as of 01:09, 11 November 2023

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for all <math>x, y \in X.</math>

<blockquote>

<blockquote>'''Banach Fixed Pointfixed-point Theoremtheorem.''' Let <math>(X, d)</math> be a non-[[Empty set|non-empty]] [[complete metric space]] with a contraction mapping <math>T : X \to X.</math> Then ''T'' admits a unique [[Fixed point (mathematics)|fixed-point]] <math>x^*</math> in ''X'' (i.e. <math>T(x^*) = x^*</math>). Furthermore, <math>x^*</math> can be found as follows: start with an arbitrary element <math>x_0 \in X</math> and define a [[sequence]] <math>(x_n)_{n\in\mathbb N}</math> by <math>x_n = T(x_{n-1})</math> for <math>n \geq 1.</math> Then <math>\lim_{n \to \infty} x_n = x^*</math>.</blockquote>

''Remark 1.'' The following inequalities are equivalent and describe the [[Rate of convergence|speed of convergence]]:

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''Remark 2.'' <math>d(T(x),T(y))<d(x,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 : [1,\infty) \to [1,\infty), \,\, T(x)=x+\tfrac{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,T(x))</math>, indeed, 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>