Schauder Fixed Point Theorem

We need the notion of a convex hull of a finite number of elements $x_i \in X$ , $X$ a Banach space.

Definition 4.1   Let $\{ x_1, \ldots, x_n \} := F \subset X$ . The convex hull of $F$ is defined by

$$con(F) := \{ \sum_{i = 1}^n t_i x_i : \, \sum_{i = 1}^n t_i = 1, t_i \geq 0 \}$$

Clearly $con(F)$ is compact (closed and bounded) and convex. Further,

$$con(F) = \bigcap_\alpha C_\alpha,$$

for all convex $C_\alpha$ for all convex sets containing $F$ .

Proof. Let $x, y \in con F$ . Then, there are $t_i$ and $s_i$ such that

$$x = \sum_1^n t_i x_i, \, y = \sum_1^n s_i x_i$$

Then,

$$tx + (1 - t)y = \sum_1^n (t t_i + (1 - t)s_i ) x_i$$

As $\sum_1^n t t_i + (1 - t) s_i = 1$ , we get $tx + (1 - t) y \in con F$ .
Thus, $con F = C_{\alpha_0}$ for some $\alpha_0$ . So $con F \supset C_\alpha$ . The converse $con F \subset C_\alpha$ can be shown using induction on the $x_i$ . $\qedsymbol$

Wednesday, 2-2-2005

Last time, we showed that if

$$F = \{ x_1, \ldots, x_n \} \subset X, \, X$$    - a Banach space $$$$

then $con F =$ convex hull of $F$ is the intersection of all closed convex subsets of $X$ containing $F$ .

Proposition 4.1   Shauder Projection
Let $K$ be a compact subset of the Banach space $X$ . Then given $\epsilon > 0$ , there exists a finite subset $F \subset X$ and a mapping

$$P : K \rightarrow con F$$

such that for any $x \in K$ we have

$$d(x, P(x) ) = \Vert x - P(x) \Vert < \epsilon$$

Proof. $K$ compact implies that there exists a finite $\epsilon$ -net for $K$ , $F= \{ x_1, \ldots x_n \} \subset X$ such that $\{ B(x_i, \epsilon ) \}_{i=1}^m$ is an open cover of $K$ . We define

$$\phi_i(x) = \begin{cases}\epsilon - d(x,x_i), & x \in B(x_i, \epsilon ) \\ 0, & x \notin B(x_i, \epsilon ) \end{cases}$$

Clearly, $\phi_i(x)$ is continuous for $x \in K$ and also

$$\sum_{i=1}^m \chi_i(x) > 0$$

for $x \in K$ . Next, let

$$\phi(x) := \sum_{i=1}^m \phi_i(x).$$

Then, $\phi : K \rightarrow \mathbb{R}^+$ and we define the Schauder projection

$$P(x) := \frac{ \sum_{i=1}^m \phi_i(x) x_i }{ \phi(x) }$$

So, $P : K \rightarrow X$ and is continuous. Also, since

$$\sum_{i=1}^m \frac{\phi_i(x)}{\phi(x)} = 1$$

we get $P(x) \in con F$ . Further,

$$d(P(x), x) = \left\vert \sum_{i=1}^m \frac{\phi_i(x)}{\phi_x} x_i - x \right\vert$$

$$= \left\vert \sum_{i=1}^m \frac{\phi_i(x)}{\phi(x)} (x_i - x) \right\vert$$

$$\leq \sum_{i=1}^m\frac{ \phi_i(x) }{ \phi(x) } \vert x_i - x \vert \leq \epsilon \sum_{i=1}^m \frac{ \phi_i(x)}{\phi(x)} = \epsilon$$

$\qedsymbol$

We can now prove:

Theorem 4.1   Schauder Fixed Point Theorem:
Let $C$ be a closed convex subset of the Banach space $X$ . Suppose $f: C \rightarrow C$ and $f$ is compact (i.e., bounded sets in $C$ are mapped into relatively compact sets). Then, $f$ has a fixed point in $C$ .

Proof. $f(C)$ is relatively compact, so $K = \overline{f(C)}$ is compact. For each $\epsilon > 0$ , there exists a finite $\epsilon$ -net for $K$ . Let $F = \{ x_1, \ldots, x_n \}$ be this finite $\epsilon$ -net (note that $n$ is dependent on $\epsilon$ ).

We show that the equation

$$f(x) = x$$

is approximately solvable in $C$ . That is, we show there exists $x_0 \in c$ with $\vert x_0 - f(x_0)\vert < \epsilon$ . Consider the mapping

$$g := P \circ f$$

$g$ maps $C$ into $con F$ and so if we restrict $g$ to $con F$ , then since $C$ is convex, $con F \subset C$ , we have that $g: con F \rightarrow con F$ . By Brouwer's theorem and its corollaries (since $con F$ is compact, convex, and finite dimensional), there exists a $x_0 \in con F$ with $g(x_0) = x_0$ . But then

$$\vert x_0 - f(x_0)\vert = \vert g(x_0) - f(x_0)\vert = \vert P(f(x_0)) - f(x_0)\vert < \epsilon$$

where the last step is because of Prop. 4.1.
Thus, $f(x) = x$ is approximately solvable in $C$ . So, we know that there exists a fixed point $\hat x \in C$ with $\hat x = f(\hat x)$ . $\qedsymbol$

We have, as a simple consequence of Schauder:

Theorem 4.2   A compact convex subset of a Banach space $X$ has the fixed point property.

Proof. Let $C \subset X$ be compact and convex. If $f: C \rightarrow C$ , where $f$ is continuous, then $f(C)$ is compact by continuity. Therefore, there exists a $x_0 \in C$ with $x_0 = f(x_0)$ , by Schauder.
$\qedsymbol$

Example: Let $X = \ell_2 = \{ x = (x_1, x_2, \ldots ) : \,x_i \in \mathbb{R}, \sum_{i=1}^\infty x_i^2 < \infty \}$ .
Now, $X$ is a Banach space with

$$\Vert x \Vert = \sqrt{\sum_{i=1}^\infty x_i^2}$$

But there exists a closed convex subset of $X$ which does not have the fixed point property, and therefore is not compact. Let

$$C := \{ x \in \ell_2 : \Vert x \Vert \leq 1 \}.$$

We claim $C$ is not compact.
Define $f: C \rightarrow C$ by

$$f(x) = (\sqrt{1 - \Vert x \Vert^2}, x_1, x_2, \ldots ).$$

Clearly $f$ is continuous but does not have a fixed point. For any $x \in C$ ,

$$\Vert f(x) \Vert^2 = \sum_{i=1}^\infty \vert f(x_i)\vert^2 = 1 - \Vert x\Vert^2 + \sum_{i = 2}^\infty (f_i(x))^2$$

$$= 1 - \Vert x \Vert^2 + \sum_{i=1}^\infty \vert x_i\vert^2 = 1 - \Vert x\Vert^2 + \Vert x\Vert^2 = 1$$

So, $f(C) \subset S = \{ x \in \ell_2:, \Vert x \Vert = 1\}$ . If $f(x) = x$ for some $x \in C$ , then

$$\Vert f(x) \Vert = \Vert x = 1$$

and

$$(x_1, x_2, \ldots) = f(x) = (0, x_1, x_2, \ldots).$$

This implies that $x_i = 0$ for all $n$ , and we get $\Vert f(x) \Vert = 0$ , a contradiction.

It is easy to give a sequence $\{ x^{(n)} \} \subset \ell_2$ which has no convergent subsequence.

However, the Hilbert cube:

$$K := \{ x \in \ell_2 :\, \vert x_i \vert \leq \frac{1}{i} \}$$

is compact (we leave the verification up to the reader).

Sometimes we need the following version (or consequence) of the Schauder theorem. We first need the notion of the Leray-Schauder boundary condition:

Definition 4.2   Suppose that $f: X \rightarrow X$ , $X$ a Banach space. Then $f$ is said to satisfy the Leray-Schauder boundary condition if there exists a $r > 0$ such that $\Vert x \Vert = r$ implies $f(x) \neq \lambda x, \, \forall \lambda > 1$ .

In practice, one can often show that if $\Vert x \Vert = r$ , then $\Vert f(x)\Vert \leq r$ and so clearly $f$ satisfies the Leray boundary condition.

Theorem 4.3   Leray-Schauder Alternative:
Let $f: X \rightarrow X$ be completely continuous (compact) and assume that $f$ satisfies the Leray-Schauder boundary condition. Then $f$ has a fixed point.

This result is called an alternative since it says that either the equation

$$f(x) = \lambda x$$

has a solution for some $\lambda > 1$ and $\Vert x \Vert$ large or

$$f(x) = x$$

has a solution. Both could hold, of course.

Proof. Since $f$ satisfies the Leray-Schauder boundary condition, there exists $r > 0$ such that $\Vert x \Vert = r$ implies $f(x) \neq \lambda x$ for all $\lambda > 1$ . We let $B_r := \{ x \in X : \Vert x \Vert \leq r \}$ . Then, $B_r$ is closed and convex. Since $f$ is completely continuous, $f(B_r)$ is compact. We only consider $f \vert _{B_r}$ :

$$f: B_r \rightarrow X.$$

We claim there exists $x_0 \in B_r$ with $f(x_0) = x_0$ . We define the retraction $\rho: X \rightarrow B_r$ by

$$\rho(x) = \begin{cases}x, & x \in B_r \\ \frac{rx}{\Vert x\Vert}, & \Vert x \Vert > r \end{cases}$$

Then $\rho$ is continuous and the composition $\rho \circ f: B_r \rightarrow B_r$ is completely continuous (as it is the composition of a continuous map with a completely continuous map). Define $f^* := \rho \circ f$ . By the Schauder theorem, there exists a $x_0 \in B_r$ with $f^*(x_0) = x_0$ . Then, $\Vert f^*(x) \Vert = \Vert x_0 \Vert \leq r$ .
Claim: $\Vert f(x_0) \Vert \leq r$ .
If $\Vert f(x_0) > r$ , then

$$f^*(x_0) = \rho(f(x_0)) = \frac{r}{\Vert f(x_0) \Vert} f(x_0) = x_0$$

This implies $\Vert x_0 \Vert = r$ and

$$f(x_0) = \frac{\Vert f(x_0) \Vert}{r} x_0$$

and so with $\lambda = \frac{\Vert f(x_0)\Vert}{r} > 1$ , this violates the Leray-Schauder boundary condition.
$\qedsymbol$

Application of Leray-Schauder Alternative
This holds if there exists a $r > 0$ such that $\Vert x\Vert = r \implies \Vert f(x) \Vert \leq r$ .

Remark:
There is no restriction on the size of $r$ . In particular, if

$$\limsup_{\Vert x\Vert \rightarrow \infty} \frac{\Vert f(x)\Vert}{\Vert x\Vert} = \mu < 1,$$

then the Leray-Schauder boundary condition holds.

Theorem 4.4   Banach Fixed Point Theorem:
Let $X$ be a complete metric space $M \subset X$ be closed, $M \neq \emptyset$ . Assume $T: M ra M$ is a contraction (i.e., $d(Tx,Ty) \leq k d(x,y)$ for some $0 \leq k < 1$ $\forall x,y \in M$ ). Then, we have the following conclusions:

1. Existence and Uniqueness - the equation $x = Tx$ has exactly one solution $\hat x \in M$ .
   
2. For any $x_0 \in M$ , the sequence of iterates $x_{n+1} = Tx_n$ converges to the solution $\hat x$ .
   
3. A priori estimates - for all $n \geq 0$ , we have
   $$d(x_n, \hat x) \leq \frac{k^n}{1 - k} d(x_0, x_1).$$
   Moreover, we also have
   $$d(x_{n+1},\hat x) \leq \frac{k}{1 - k} d(x_n, x_{n+1})$$

Proof. We've seen this proof before.
Given $x_0 \in M$ , $x_{n+1} = Tx_n$ is Cauchy because $T$ is a contraction mapping. Since $X$ is complete and $x_n$ is Cauchy, there exists $\hat x \in M$ (as $M$ is closed) with $x_n \rightarrow x$ . $\qedsymbol$

As an application of the Schauder and/or the Banach Fixed Point Theorem, consider the nonlinear integral operator

$$x(t) = \mu \int_a^b F(t,s,x(s)) ds + \int_a^b G(t,s,x(s)) ds + \alpha g(t)$$

where $F$ . This includes many prototypes of equations. We assume:

1. $\mu$ and $\alpha$ are real parameters (generally, these are small).
2. $G$ is a nonlinear term which has order higher than first order in $x$ .
   

We now state two theorems, which will give existence of a solution of

$$x = Tx$$ (3)

where

$$Tx = \mu \int_a^b F(t,s,x(s)) ds + \int_a^b G(t,s,x(s)) ds + \alpha g(t).$$

Let

$$Q := \{ (t,s,x) \in \mathbb{R}^3 : \, a \leq t, s \leq b, \vert x\vert \leq r_0 \},$$

where $a,b,r_0$ are fixed positive numbers.

Theorem 4.5   Assume $F, G: Q \rightarrow \mathbb{R}$ are continuous and $g: [a,b] \rightarrow \mathbb{R}$ . Further, for all $(t,s,x) \in Q$ ,

$$\vert G(t,s,x)\vert \leq K \vert x\vert^\rho$$

for fixed $\rho > 1$ and some $K > 0$ . Then, there exists a $\mu_0$ and $\alpha_0 > 0$ such that the equation $x = Tx$ has a solution for fixed $\mu$ and $\alpha$ with $\vert\mu\vert \leq \mu_0$ and $\vert\alpha\vert \leq \alpha_0$ .

Proof. The Banach space we are interested in is $X = \mathcal{C}[a,b]$ with norm $\Vert x \Vert = \displaystyle \max_{a \leq t \leq b} \vert x(t)\vert$ , and

$$M := \{ x \in X : \, \Vert x \Vert \leq r \}.$$

Clearly, $M$ is closed bounded, and convex. We want to show that $T:M \rightarrow X$ actually maps $T : M \rightarrow M$ , where $r$ is small. In order to apply Schauder's Theorem, we also need to show that $T$ is a compact operator. To show compactness, we prove the following:

Lemma 4.1   Let $K : [a,b] \times [a,b] \times [-R, R] \rightarrow \mathbb{R}$ be continuous, $-\infty < a < b < \infty, 0 < R < \infty$ . Define the integral operators $S$ and $\hat T$ by

$$(Sx)(t) = \int_a^t K(t,s,x(s)) ds$$

and

$$(\hat T x)(t) = \int_a^b K(t,s,x(s)) ds$$

for $x \in \hat M := \{ x \in \mathcal{C}[a,b] : \Vert x \Vert \leq R \}$ . Then, the integral operators are compact (and continuous).

Proof. Since the set $A := [a,b] \times [a,b] \times [-R,R]$ is compact, the function $K:A \rightarrow \mathbb{R}$ is bounded and uniformly continuous. Hence, there exists a $\alpha > 0$ such that

$$\vert K(t,s,x) \vert \leq \alpha, \, \forall(t,s,x) \in A$$

and for each $\epsilon > 0$ , there exists a $\delta_\epsilon > 0$ such that

$$\vert K(t_1,s_1,x_1) - K(t_2,s_2,x_2)\vert < \epsilon$$

if $\vert t_1 - t_2\vert + \vert s_1 - s_2\vert + \vert x_1 - x_2\vert < \delta_\epsilon$ for all $(t_i, s_i, x_i) \in A$ . So let $x \in \hat M$ and $z = Sx$ , then

$$\vert z(t)\vert = \vert(Sx)(t)\vert \leq \left\vert\int_a^t K(t,s... ...ds \right\vert \leq \alpha \left\vert \int_a^b ds \right\vert = \alpha (b - a)$$

Hence, $S$ is bounded on $\hat M$ . Similarly, $\hat T$ is bounded on $\hat M$ . Then for $\vert t_1 - t_2\vert \leq \min \{ \epsilon , \delta_\epsilon \}$ , we have

$$\vert z(t_1) - z(t_2) \vert = \vert(Sx)(t_1) - (Sx)(t_2) \vert = ... ...t\vert \int_a^{t_1} K(t,s,x(s)) ds - \int_a^{t_2} K(t_2,s,x(s)) ds \right\vert$$

$$= \left\vert \int_a^{t_1} [K(t,s,xs) - K(t_2,s,x(s))]ds + \int_a^t K(t_2,s,x(s)) ds - \int_a^{t_2} K(t_2,s,x(s)) ds \right\vert$$

$$\leq \left\vert \int_a^b \epsilon ds \right\vert + \left\vert \in... ...silon (b - a) + \vert t_1 - t_2\vert \alpha \leq \epsilon ( (b - a) + \alpha )$$

This holds for any $x \in \hat M$ and $z = Sx$ . Therefore, $S(\hat M)$ is equicontinuous and so by Ascoli-Arzela, $S$ is a compact operator.

Also, $S$ is continuous on $\hat M$ since if $x_n \in \hat M$ is such that $x_n \rightarrow \hat x$ , then with $z_n = S x_n$ , we get $\hat z = S \hat x$ . Since $x_n(t) \rightarrow \hat x(t)$ uniformly on $[a,b]$ and by uniform continuity of $K$ , $S$ is continuous on $\hat M$ . Therefore, $S$ is compact and continuous; similarly, the proof for $\hat T$ follows. $\qedsymbol$

Lemma 4.1 shows that the operator $T$ is compact and continuous on $M$ and maps $M$ into $\mathcal{C}[a,b]$ .

Notice that for $\Vert x \Vert \leq r$ , then

$$\max_t \left\vert\int_a^b G(t,s,x(s)) ds\right\vert \leq \left\vert\int_a^b K r^\rho ds \right\vert = (b - a) K r^\rho \leq \frac{r}2$$

if

$$r^{\rho - 1} \leq \frac{1}{2(b-a)K},$$

and this holds if

$$r \leq \left( \frac{1}{2(b - a)K} \right)^{\frac{1}{p -1}}.$$

We also need $r \leq r_0$ . Further, choose $\mu_0, \alpha_0 > 0$ such that

$$\max_t \left\vert \mu \int_a^b F(t,s,x(s)) ds \right\vert \leq \frac{r}4$$

and $\max_t \vert\alpha_0 g(t) \vert \leq \frac{r}{4}$ . Then, for fixed $\mu$ and $\alpha$ with $\vert\mu\vert \leq \mu_0, \vert\alpha\vert \leq \alpha_0$ , $T$ maps $M$ into itself; i.e.,

$$\Vert Tx\Vert = \max_t \vert(Tx)(t)\vert$$

$$\leq \max_t \left\vert \mu \int_a^bF(t,s,x(s)) ds \right\vert + \... ...left\vert \int_a^b G(t,s,x(s)) ds \right\vert + \max_t \vert \alpha g(t) \vert$$

$$\leq \frac{r}4 + \frac{r}2 + \frac{r}4 = r$$

Therefore, $T : M \rightarrow M$ , so by Schauder's Fixed Point Theorem, there exists a $\hat x$ such that $\hat x = T\hat x$ .
$\qedsymbol$

Theorem 4.6   Suppose that $F$ and $G$ both have continuous first partial derivatives with respect to $x$ and satisfy

$$G_x(t,s,0) = 0$$

for all $(t,s,0) \in Q$ . Then there exists $\mu_0, \alpha_0 > 0$ and $r > 0$ such that for fixed $\mu$ and $\alpha$ with

$$\vert\mu\vert < \mu_0, \, \vert\alpha\vert \leq \alpha_0$$

such that the equation $x = Tx$ has a unique solution $x = x(t)$ on $[a,b)$ with $\max\vert x(t)\vert \leq r$ on $[a,b]$ . $x(t)$ can be attained by successive approximations with $x_0(t) = 0$ .

Proof. Let $X = \mathcal{C}[a,b]$ be the set of all continuous functions on $[a,b]$ with the supremum norm.

$$M := \{ x \in X : \Vert x \vert \leq r_0 \}$$

We only need to show that the Banach Fixed Point Theorem applies to $T$ . Recall that for $x, y \in M$ , by the Mean Value Theorem,

$$\vert G(t,s,x(s)) - G(t,s,y(s))\vert \leq \vert G_x(t,s,\theta(s)) \vert \vert x(s) - y(s)\vert$$

where

$$\theta(s) = x(s) + \xi_1 (y(s) - x(s)), \, 0 < \xi_1 < 1.$$

Likewise for F,

$$\vert F(t,s,x(s)) - F(t,s,y(s))\vert \leq \vert F_x(t,s,\eta(s))\vert \vert x(s) - y(s)\vert$$

where

$$\eta(s) = x(s) + \xi_2(y(s) - x(s)), \, 0 < \xi_2 < 1.$$

For any $x, y \in M$ and $r > 0$ small,

$$\max_t \left\vert \int_a^b [ G(t,s,x(s)) - G(t,s,y(s)) ] ds \right \vert$$

$$\leq \max_t \left\vert \int_a^b G_x(t,s,\theta(s)) ds \right\vert \cdot \max_t \vert x(t) - y(t)\vert$$

$$\leq \Vert x - y \Vert \max_t \left\vert\int_a^b G_x(t,s,\theta(s)) ds \right\vert$$

$$\leq \frac14\Vert x - y\Vert$$

if $\vert G_x(t,s,u) \vert (b - a) < \frac14$ for all $t,s \in [a,b]$ and $\vert u\vert \leq r$ .
Concerning $F$ , we have the parameter $\mu$ multiplying this term, and since $F_x(t,s,u)$ is continuous on $Q$ , it is bounded. For small $\mu_0$ and $\alpha_0$ positive,

$$\Vert Tx - Ty \Vert \leq \frac12 \Vert x - y \Vert$$

$\qedsymbol$