7.6: Fixed point theorem and Picard’s theorem again

In this section we prove a fixed point theorem for contraction mappings. As an application we prove Picard’s theorem. We have proved Picard’s theorem without metric spaces in . The proof we present here is similar, but the proof goes a lot smoother by using metric space concepts and the fixed point theorem. For more examples on using Picard’s theorem see .

Let $(X,d)$ and $(X',d')$ be metric spaces. $F \colon X \to X'$ is said to be a contraction (or a contractive map) if it is a $k$-Lipschitz map for some $k < 1$, i.e. if there exists a $k < 1$ such that $$d'\bigl(F(x),F(y)\bigr) \leq k d(x,y) \ \ \ \ \text{for all } x,y \in X.$$

If $T \colon X \to X$ is a map, $x \in X$ is called a fixed point if $T(x)=x$.

[Contraction mapping principle or Fixed point theorem] [thm:contr] Let $(X,d)$ be a nonempty complete metric space and $T \colon X \to X$ is a contraction. Then $T$ has a fixed point.

Note that the words complete and contraction are necessary. See .

Pick any $x_0 \in X$. Define a sequence $\{ x_n \}$ by $x_{n+1} := T(x_n)$. $$d(x_{n+1},x_n) = d\bigl(T(x_n),T(x_{n-1})\bigr) \leq k d(x_n,x_{n-1}) \leq \cdots \leq k^n d(x_1,x_0) .$$ So let $m \geq n$ $$\begin{split} d(x_m,x_n) & \leq \sum_{i=n}^{m-1} d(x_{i+1},x_i) \\ & \leq \sum_{i=n}^{m-1} k^i d(x_1,x_0) \\ & = k^n d(x_1,x_0) \sum_{i=0}^{m-n-1} k^i \\ & \leq k^n d(x_1,x_0) \sum_{i=0}^{\infty} k^i = k^n d(x_1,x_0) \frac{1}{1-k} . \end{split}$$ In particular the sequence is Cauchy. Since $X$ is complete we let $x := \lim_{n\to \infty} x_n$ and claim that $x$ is our unique fixed point.

Fixed point? Note that $T$ is continuous because it is a contraction. Hence $$T(x) = \lim T(x_n) = \lim x_{n+1} = x .$$

Unique? Let $y$ be a fixed point. $$d(x,y) = d\bigl(T(x),T(y)\bigr) = k d(x,y) .$$ As $k < 1$ this means that $d(x,y) = 0$ and hence $x=y$. The theorem is proved.

Note that the proof is constructive. Not only do we know that a unique fixed point exists. We also know how to find it. Let us use the theorem to prove the classical Picard theorem on the existence and uniqueness of ordinary differential equations.

Consider the equation $$\frac{dx}{dt} = F(t,x) .$$ Given some $t_0, x_0$ we are looking for a function $f(t)$ such that $f'(t_0) = x_0$ and such that $$f'(t) = F\bigl(t,f(t)\bigr) .$$ There are some subtle issues. Look at the equation $x' = x^2$, $x(0)=1$. Then $x(t) = \frac{1}{1-t}$ is a solution. While $F$ is a reasonably “nice” function and in particular exists for all $x$ and $t$, the solution “blows up” at $t=1$.

Let $I, J \subset {\mathbb{R}}$ be compact intervals and let $I_0$ and $J_0$ be their interiors. Suppose $F \colon I \times J \to {\mathbb{R}}$ is continuous and Lipschitz in the second variable, that is, there exists $L \in {\mathbb{R}}$ such that $$\left\lvert {F(t,x) - F(t,y)} \right\rvert \leq L \left\lvert {x-y} \right\rvert \ \ \ \text{ for all $x,y \in J$, $t \in I$} .$$ Let $(t_0,x_0) \in I_0 \times J_0$. Then there exists $h > 0$ and a unique differentiable $f \colon [t_0 - h, t_0 + h] \to {\mathbb{R}}$, such that $f'(t) = F\bigl(t,f(t)\bigr)$ and $f(t_0) = x_0$.

Without loss of generality assume $t_0 =0$. Let $M := \sup \{ \left\lvert {F(t,x)} \right\rvert : (t,x) \in I\times J \}$. As $I \times J$ is compact, $M < \infty$. Pick $\alpha > 0$ such that $[-\alpha,\alpha] \subset I$ and $[x_0-\alpha, x_0 + \alpha] \subset J$. Let $$h := \min \left\{ \alpha, \frac{\alpha}{M+L\alpha} \right\} .$$ Note $[-h,h] \subset I$. Define the set $$Y := \{ f \in C([-h,h]) : f([-h,h]) \subset [x_0-\alpha,x_0+\alpha] \} .$$ Here $C([-h,h])$ is equipped with the standard metric $d(f,g) := \sup \{ \left\lvert {f(x)-g(x)} \right\rvert : x \in [-h,h] \}$. With this metric we have shown in an exercise that $C([-h,h])$ is a complete metric space.

Show that $Y \subset {\mathbb{C}}([-h,h])$ is closed.

Define a mapping $T \colon Y \to C([-h,h])$ by $$T(f)(t) := x_0 + \int_0^t F\bigl(s,f(s)\bigr)~ds .$$

Show that $T$ really maps into $C([-h,h])$.

Let $f \in Y$ and $\left\lvert {t} \right\rvert \leq h$. As $F$ is bounded by $M$ we have $$\begin{split} \left\lvert {T(f)(t) - x_0} \right\rvert &= \left\lvert {\int_0^t F\bigl(s,f(s)\bigr)~ds} \right\rvert \\ & \leq \left\lvert {t} \right\rvert M \leq hM \leq \alpha . \end{split}$$ Therefore, $T(Y) \subset Y$. We can thus consider $T$ as a mapping of $Y$ to $Y$.

We claim $T$ is a contraction. First, for $t \in [-h,h]$ and $f,g \in Y$ we have $$\left\lvert {F\bigl(t,f(t)\bigr) - F\bigl(t,g(t)\bigr)} \right\rvert \leq L\left\lvert {f(t)- g(t)} \right\rvert \leq L \, d(f,g) .$$ Therefore, $$\begin{split} \left\lvert {T(f)(t) - T(g)(t)} \right\rvert &= \left\lvert {\int_0^t F\bigl(s,f(s)\bigr) - F\bigl(s,g(s)\bigr)~ds} \right\rvert \\ & \leq \left\lvert {t} \right\rvert L \, d(f,g) \\ & \leq h L\, d(f,g) \\ & \leq \frac{L\alpha}{M+L\alpha} \, d(f,g) . \end{split}$$ We can assume $M > 0$ (why?). Then $\frac{L\alpha}{M+L\alpha} < 1$ and the claim is proved.

Now apply the fixed point theorem () to find a unique $f \in Y$ such that $T(f) = f$, that is, $$f(t) = x_0 + \int_0^t F\bigl(s,f(s)\bigr)~ds .$$ By the fundamental theorem of calculus, $f$ is differentiable and $f'(t) = F\bigl(t,f(t)\bigr)$.

We have shown that $f$ is the unique function in $Y$. Why is it the unique continuous function $f \colon [-h,h] \to J$ that solves $T(f)=f$? Hint: Look at the last estimate in the proof.

Exercises

Suppose $X = X' = {\mathbb{R}}$ with the standard metric. Let $0 < k < 1$, $b \in {\mathbb{R}}$. a) Show that the map $F(x) = kx + b$ is a contraction. b) Find the fixed point and show directly that it is unique.

Suppose $X = X' = [0,\nicefrac{1}{4}]$ with the standard metric. a) Show that the map $F(x) = x^2$ is a contraction, and find the best (largest) $k$ that works. b) Find the fixed point and show directly that it is unique.

[exercise:nofixedpoint] a) Find an example of a contraction of non-complete metric space with no fixed point. b) Find a 1-Lipschitz map of a complete metric space with no fixed point.

Consider $x' =x^2$, $x(0)=1$. Start with $f_0(t) = 1$. Find a few iterates (at least up to $f_2$). Prove that the limit of $f_n$ is $\frac{1}{1-t}$.

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