Schauder's Fixed Point Theorem

This is a theorem for all continuous functions of a certain kind - no linearity.
This is an example of a significant research area.

Theorem 4.1.1   Brouwer's Fixed Point Theroem Let $d \in \mathbb{N}$ and let $B$ be the closed unit ball in $\mathbb{R}^d$. If $f: B \rightarrow B$ is continuous, then there is $x \in B$ such that $f(x) = x$.

Proof. If $d = 1$, this is easy. Assume there is no fixed point.

$$[-1,1] = \{ x \in [-1,1] : f(x) < x \} \cup \{ x \in [-1,1] : f(x) > x \}$$

Both sets are open and nonempty (examine 1, -1). This gives a contradiction. If $d=2$:
Assume $f$ has no fixed point. We define $g : B \rightarrow \partial B$ by letting $g(x)$ be the intersection of $\partial B$ and the ray starting at $f(x)$ and going through $x$. We can show that $g$ is continuous.
Intuitively, no such $g$ can exist. To prove this, we need to use the fundamental groups of $B$ and $\partial B$.
For general $d$, we need to use the $d^{th}$ homology group of $B$ (go take a course on algebraic topology). $\qedsymbol$

Corollary 4.1.2   $X$ is a finite dimensional normed space, $K \subset X$ is nonempty, convex, and compact. If $f : K \rightarrow K$ is continuous, then there exists $k \in K$ such that $f(k) = k$.

Proof. If $x$ is a vector space over $\mathbb{C}$, it is isomorphic to $\mathbb{R}^{2n}$ for some $n \in \mathbb{N}$. WLOG, $X$ is $\mathbb{R}^d$ for some $d$.
Second, all norms on a finite dimensional vector space are equivalent, so WLOG, the norm on $\mathbb{R}^d$ is the Euclidean norm.
If $K$ is the unit ball (or more generally, the closed ball of radius $r$), then we are done by Brouwer's theorem. In general, $K$ compact implies there is $r > 0$ such that $K \subset \overline{B_r(0)}$. Let $B = \overline{B_r(0)}$.
We construct a continuous $\phi:B \rightarrow K$. Define, for $x \in B$, $\phi(x)$ is the unique point $y$ such that

$$dist(x,K) = \Vert x - y\Vert$$

[approximation in Hilbert space - we proved this earlier]. Clearly, $\phi(x) = x$ for $x \in K$.
Claim: $\phi$ is continuous.
The rest of the proof follows as in theorem 3.3 in 933. $\qedsymbol$

Lemma 4.1.3   Let $X$ be a normed space, $K \subset X$ compact and $\epsilon > 0$. Let $A$ be a finite $\epsilon$-net for $K$, i.e., $K \subset \bigcup_{a \in A} B_\epsilon (a)$.
Define, for $a \in A$,

$$m_a : X \rightarrow [0,\epsilon ]$$

$$m_a(x) = \begin{cases}0, & x \in B_\epsilon (a) \ \epsilon - \Vert x - a\Vert, & x \in B_\epsilon (a) \end{cases}$$

If $\phi_A : K \rightarrow X$ is defined by

$$\phi_A(x) := \frac{ \sum_{a \in A} m_a(x) a }{ \sum_{a \in A} m_a(x) }$$

then $\phi_A$ is continuous ( $\phi_A(K) \subset K$) and $\Vert\phi_A(x) - x \Vert < \epsilon$ for all $x \in K$.

Proof. For $x \in K$, $\sum_{a \in A} m_a(x) > 0$ so $\phi_A$ is well-defined. As each $m_a$ is continuous, $\phi_A$ is continuous.

Monday, March 3, 2006:

Lemma 4.1.4   Let $H$ be a Hilbert space over $\mathbb{R}$. Let $K \subset H$ be a nonempty, convex, and bounded set. Define $\phi:H \rightarrow K$ by $\phi(h)$ is $y \in K$ such that $dist(h,K) = \Vert y - h\Vert$. Then, $\phi$ is continuous.

Proof. Fix$x \in H$. Translating $K$ and $x$, we may assume $\phi(x) = \vec 0$.

Claim: $\left\langle{ x, k }\right\rangle \leq 0$ for all $k \in K$.
Suppose there is $k$ such that $\left\langle{ x, k }\right\rangle > 0$. Let $\lambda \in (0,1)$. Then,

$$\Vert x - \lambda k \Vert^2 = \left\langle{ x - \lambda k, x - \lambda k }\right\rangle$$

$$= \Vert x\Vert^2 - 2 \lambda \left\langle{ x,k }\right\rangle + \lambda^2 \Vert k \Vert^2$$

There is a suitable $\lambda \in (0,1)$ such that the above is $< \Vert x\Vert^2$. But, $\vec 0, k \in K$, so $\mu k, \mu \in (0,1)$ are all in $K$. This $\lambda k$ is closer to $x$ than 0, contradicting $\phi(x) = \vec 0$.

Claim: $\left\langle{ x-\phi(x), k - \phi(x) }\right\rangle \leq 0$ for all $k \in K$.

Pick $y, z \in H$. Then,

$$\left\langle{ z - \phi(z), \phi(y)-\phi(z) }\right\rangle \leq 0$$

$$\left\langle{ y - \phi(y), \phi(z) - \phi(y) }\right\rangle \leq 0$$

Adding,

$$\left\langle{ z - \phi(z), \phi(y) - \phi(z) }\right\rangle + \left\langle{ \phi(y) - y, \phi(y) - \phi(z) }\right\rangle \leq 0$$

$$\left\langle{ z - \phi(z) + \phi(y) - y, \phi(y) - \phi(z) }\right\rangle \leq 0$$

$$\Vert \phi(z) - \phi(y) \Vert^2 + \left\langle{ z - y, \phi(y) - \phi(z) }\right\rangle \leq 0$$

By Cauchy-Schwartz,

$$\Vert \phi(z) - \phi(y) \Vert^2 \leq \Vert \phi(z) - \phi(y) \Vert \Vert y - z \Vert$$

$$\Vert \phi(z) - \phi(y) \Vert \leq \Vert z - y \Vert$$

$\qedsymbol$

Recall:

$$\phi_A(x) = \frac{ \sum_{a \in A} m_a(x) a }{ \sum_{a \in A} m_a(x) }$$

It remains to show that $\Vert\phi_A(x) - x \Vert < \epsilon$ for all $x \in K$. Then,

$$\phi_A(x) - x = \frac{\sum_{a \in A} m_a(x) a }{ \sum_{a \in A} m_a(x) } - \frac{ \sum_{a \in A}( m_a(x) x ) }{ \sum_{a \in A} m_a(x) }$$

$$= \frac{ \sum_{a \in A} m_a(x)(a-x) }{ \sum_{a \in A} m_a(x) }$$

Let

$$A' = \{ a \in A : \Vert x - a \Vert < \epsilon \}$$

Then, $m_a(x) = 0$ if $a \notin A'$ and so the above formula remains unchanged if we restrict the sums to those $a \in A'$.
By the definition of $A'$, for $a \in A'$, $\Vert x - a \Vert < \epsilon$. Thus,

$$\Vert \phi_A(x) - x \Vert \leq \frac{ \sum_{a \in A'} m_a(x) \Vert x - a \Vert }{ \sum_{a \in A'} m_a(x) } < \epsilon$$

$\qedsymbol$

Definition 4.1.5   $X$ normed space, $E \subset X$. Then, $f: E \rightarrow X$ is compact if $f$ is continuous and for all $A \subset E$, bounded, $\overline{f(A)}$ is compact.

Example: If $E$ is compact, every continuous $f: E \rightarrow X$ is compact.

Theorem 4.1.6   Schauder's Fixed Point Theorem
Let $X$ be a normed space and $E \subset X$ closed, bounded, and convex. If $f: E \rightarrow X$ with $f(E) \subset E$ is compact,

Proof. For $n \in \mathbb{N}$, let $A_n$ be an $\frac1n$-net for $\overline{f(E)}$, a compact set. Let $\phi_n$ be the $\phi_{A_n}$ in the previous lemma. As $E$ is convex,

$$\phi(\overline{f(E)}) \subset$$   co $$( \overline{f(E)} ) \subset E$$

Let $f_n = \phi_n \circ f$. Then, $f_n(E) \subset E$ and

$$\Vert f_n(x) - f(x) \Vert = \Vert \phi_n(f(x)) - f(x) \Vert < \frac1n$$

Let $X_n =$   span$A_n$. $E_n = X_n \cap E$. Notice that $f(E_n) \subset E_n$. Since $\dim X_n < \infty$, we can apply the corollary to $f_n$ acting on $E_n \subset X_n$ to get $e_n \in E_n$ such that $f_n(e_n) = e_n$. Let $e$ be a cluster point of $(f(e_n))$ in $K$; i.e., there exists $\{ n_k \}$ such that $f(e_{n_k}) \rightarrow e$ such that

$$\Vert e_{n_k} - e \Vert \leq \Vert f_{n_k}(e_{n_k}) - f(e_{n_k}) \Vert + \Vert f(e_{n_k}) - e \Vert$$

$$\leq \frac1{n_k} + \Vert f(e_{n_k}) - e\Vert$$

Thus, $e_{n_k} \rightarrow e$. So,

$$f(e) = \lim_{k \rightarrow \infty} f(e_{n_k}) = e$$

$\qedsymbol$

Wednesday, March 8, 2006:

Two big results about weak and weak-star topologies on Banach spaces:

Theorem 4.1.7   Krein-Smulian Theorem Let $X$ be a Banach space, $A \subset X^*$ convex. If $A \cap \overline{B_r(0)}$ is wk $^*$-closed, then $A$ is wk $^*$-closed.

The same result is true for $A \subset X$ convex and the weak topology, but the proof is easy:
It suffices to prove that $A$ is norm-closed. Consider a sequence $(x_n)$ in $A$ such that $x_n \rightarrow x_0$ in norm. Thus, there is some $r > 0$ such that $(x_n) \subset \overline{B_r(0)}$. By the hypothesis, since $(x_n) \subset A \cap \overline{B_r(0)}$, $x_0 \in A$. This proof does not work for the weak-star topology and $A \subset X^*$. There are norm closed convex sets that are not wk $^*$-closed.
For example, pick $F \in X^{**}\setminus X$. Then, $\ker F$ is norm closed as $F$ is continuous, but not wk $^*$ closed.

Theorem 4.1.8   Eberlein-Smulian Theorem
$X$ Banach space, $A \subset X$ TFAE:

1. Sequences in $A$ have weakly convergent subsequences.
2. Sequences in $A$ have weak cluster points.
3. wk -closure of $A$ is weakly compact.

The content of this theorem is that to show a set is weakly compact, we need only look at sequences