Fixed point theorem in complete metric spaces

Definition. Let (X,d) be a metric space. X is said to be complete if every Cauchy sequence in X is convergent.

Definition. A mapping f: X \rightarrow X is called contractive if there exists 0\leq k < 1 such that

d(f(x), f(y)) \leq kd(x,y) for all x, y\in X.

Remark. If f: X \rightarrow X is a contractive, then f is continuous.

Definition. A point x^* \in X is called a fixed point of f: X\rightarrow X if f(x^*) = x^*.

Theorem. Assume that (X,d) is a complete metric space, and f: X\rightarrow X is contractive. Then f has a unique fixed point.

Proof. Take any x_0\in X. For any n\geq 1, we let x_n = f(x_{n-1}). Use the assumption of f, we have

d(x_2, x_1) = d(f(x_1), f(x_0)) \leq kd(x_1,x_0)

d(x_3,x_2) = d(f(x_2),f(x_1)) \leq kd(x_2,x_1)\leq k^2d(x_1,x_0)

………. (similarly)

d(x_{n+1},x_n)\leq k^n d(x_1,x_0).

Let m<n. Using triangle inequality, we get

d(x_m,x_n) \leq d(x_m,x_{m+1}) + d(x_{m+1}, x_{m+2}) + \ldots + d(x_{n-1},x_n)

\leq (k^{m} + k^{m+1}+\ldots + k^{n-1})d(x_1,x_0)

\leq k^m(1+k+k^2+\ldots)d(x_1,x_0)

= k^m\dfrac{1}{1-k}d(x_1,x_0) \longrightarrow 0

as m\rightarrow +\infty since k<1.

This implies that \{x_n\}_{n\geq 0} is a Cauchy sequence. Because X is a complete metric space, we conclude that there exists x^* such that lim_{n\rightarrow \infty}x_n = x^*.

We claim that f(x^*) = x^*. By the continuity of f (see Remark), we find that \lim_{n\rightarrow \infty}f(x_n) = f(x^*). Take any \epsilon >0. We choose N such that

d(x_n, x^*) < \epsilon, d(f(x_n),f(x^*)) <\epsilon for all n>N.

Now we have, keep in mind that f(x_n) = x_{n+1},

d(f(x^*), x^*) \leq d(f(x^*), f(x_n)) + d(x_{n+1}, x^*) < 2\epsilon.

Since \epsilon is arbitrary, we get d(f(x^*), x^*) = , thus f(x^*) = x^*.

For the uniqueness, we assume that f(y) = y for some y\in X. Then

d(x^*, y) = d(f(x^*), f(y)) \leq kd(x^*,y)

Because 0\leq k<1, we obtain x^* = y.

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About baotangquoc

Lecturer School of Applied Mathematics and Informatics Hanoi University of Science and Technology No 1, Dai Co Viet Street, Hanoi
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One Response to Fixed point theorem in complete metric spaces

  1. NN says:

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