Banach fixed-point theorem: Difference between revisions

Revision as of 14:27, 4 September 2024

Revision as of 07:18, 16 September 2024

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:<math>d(x_{n+1}, x_n) \le q^n d(x_1, x_0).</math>

This follows by [[Principle of mathematical induction|induction]] on ''<math>n''</math>, using the fact that ''<math>T''</math> is a contraction mapping. Then we can show that <math>(x_n)_{n\in\mathbb N}</math> is a [[Cauchy sequence]]. In particular, let <math>m, n \in \N</math> such that <math>m > n </math>:

: <math>\begin{align}

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\end{align}</math>

Let ''ε''<math>\varepsilon > 0</math> be arbitrary. Since <math>q \in [0,1)</math>, we can find a large <math>N \in \N</math> so that

:<math>q^N < \frac{\varepsilon(1-q)}{d(x_1, x_0)}.</math>

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:<math>d(x_m, x_n) \leq q^n d(x_1, x_0) \left ( \frac{1}{1-q} \right ) < \left (\frac{\varepsilon(1-q)}{d(x_1, x_0)} \right ) d(x_1, x_0) \left ( \frac{1}{1-q} \right ) = \varepsilon.</math>

This proves that the sequence <math>(x_n)_{n\in\mathbb N}</math> is Cauchy. By completeness of (''X'',''d''), the sequence has a limit <math>x^* \in X.</math> Furthermore, <math>x^*</math> must be a [[Fixed point (mathematics)|fixed point]] of ''<math>T''</math>:

:<math>x^*=\lim_{n\to\infty} x_n = \lim_{n\to\infty} T(x_{n-1}) = T\left(\lim_{n\to\infty} x_{n-1} \right) = T(x^*). </math>

As a contraction mapping, ''<math>T''</math> is continuous, so bringing the limit inside ''<math>T''</math> was justified. Lastly, ''<math>T''</math> cannot have more than one fixed point in <math>(''X'','' d'')</math>, since any pair of distinct fixed points ''p''<submath>1p_1</submath> and ''p''<submath>2p_2</submath> would contradict the contraction of ''<math>T''</math>:

:<math> d(T(p_1),T(p_2)) = d(p_1,p_2) > q d(p_1, p_2).</math>