Leray-Schauder fixed point theorem

By | January 8, 2018

First, we derive the Schauder fixed point theorem.

Theorem 1 (Schauder fixed point theorem). Let S be a compact convex set in a Banach space X and let T be a continuous mapping of S into itself. Then T has a fixed point, that is, Tx=x for some x\in S.

Proof. Fix k\in\mathbb{N}. Since S is compact, \left\{ B_{\frac{1}{k}}\left(x\right)\right\} _{x\in K} has a finite subcover which covers S. Write \left\{ B_{\frac{1}{k}}\left(x_{1}\right),\dots,B_{\frac{1}{k}}\left(x_{N}\right)\right\}.
Let S_{k}=\mathrm{conv}\left(x_{1},\dots,x_{N}\right). Define J_{k}:S\rightarrow S_{k} by

    \[ J_{k}x=\frac{\sum_{i}\mathrm{dist }\left(x,S\setminus B_{\frac{1}{k}}\left(x_{i}\right)\right)x_{i}}{\sum_{i}\mathrm{dist }\left(x,S\setminus B_{\frac{1}{k}}\left(x_{i}\right)\right)}. \]

The map is well-defined. Since \mathrm{dist } is continuous, J_{k}x is continuous. Also,

    \[ \norm{J_{k}x-x}{}\le\frac{\sum_{i}\mathrm{dist }\left(x,S\setminus B_{\frac{1}{k}}\left(x_{i}\right)\right)\norm{x_{i}-x}{}}{\sum_{i}\mathrm{dist }\left(x,S\setminus B_{\frac{1}{k}}\left(x_{i}\right)\right)}. \]

Note that for such x\in S with \mathrm{dist }\left(x,S\setminus B_{\frac{1}{k}}\left(x_{i}\right)\right)>0, x\in B_{\frac{1}{k}}\left(x_{i}\right). Thus, for such i, we have \norm{x_{i}-x}{}<\frac{1}{k}. Hence \norm{J_{k}x-x}{}<\frac{1}{k}. Now J_{k}\circ T:S_{k}\rightarrow S_{k} is continuous, S_{k} is compact and convex set. So S_{k} is homeomorphic to \mathbb{D}^{N}. Hence by the Brouwer fixed point theorem, J_{k}\circ T\left(x_{k}\right)=x_{k} for some x_{k}\in S_{k}.

Now since \left\{ x_{k}\right\} \subset S and S is compact, there exists a convergent subsequence \left\{ x_{n_{k}}\right\} which converges to x in S. We claim that x is a fixed point. Note

    \[ \norm{Tx_{n_{k}}-x_{n_{k}}}{}=\norm{J_{n_{k}}\circ T\left(x_{n_{k}}\right)-x_{n_{k}}}{}<\frac{1}{n_{k}}\rightarrow0 \]

as k\rightarrow\infty. Hence Tx=x.

Using this theorem, we obtain another version of Schauder fixed point theorem.

Theorem 2. Let S be a bounded and closed convex set in a Banach space X and let T be a compact mapping of S into itself. Then T has a fixed point.

Proof. Let A be the closed convex hull of \overline{TS}. Then A is convex and since the closed convex hull of a compact set is itself compact, A is compact. Note that A\subset S because TS\subset S and \mathcal{S} is closed. So \overline{TS}\subset S. Since S is convex and A is the convex hull of \overline{TS}, A\subset S.
Thus, by the Schauder fixed point theorem, T has a fixed point in A. So we are done.

Now we prove the Leray-Schauder fixed point theorem. Here we consider the simplest form.

Theorem 3 (Leray-Schauder). Let X be a Banach space and let T be a compact mapping of X into itself. Also, suppose there exists a constant M such that

    \[ \norm xX<M \]

for all x\in X and \sigma\in\left[0,1\right] satisfying x=\sigma Tx. Then T has a fixed point.

Proof. We may assume M=1. Define \tilde{T}:X\rightarrow X by

    \[ \tilde{T}\left(x\right)=\begin{cases} T\left(x\right) & \text{if }\norm{T\left(x\right)}{}<1\\ \frac{T\left(x\right)}{\norm{T\left(x\right)}{}} & \text{if }\norm{T\left(x\right)}{}\ge1. \end{cases} \]

Let \mathbb{D}=\left\{ x\in X:\norm x{}\le1\right\}. Then \tilde{T} maps \mathbb{D} to \mathbb{D}. We show that \tilde{T}:\mathbb{D}\rightarrow\mathbb{D}
is compact. First, \tilde{T}:\mathbb{D}\rightarrow\mathbb{D} is continuous. For those x with \norm{T\left(x\right)}{}=1, we
have

    \[ \tilde{T}\left(x\right)=\frac{T\left(x\right)}{\norm{T\left(x\right)}{}}=T\left(x\right). \]

From the continuity of T, \tilde{T} is continuous on \mathbb{D}.

Next, let \left\{ u_{n}\right\} be a sequence in the ball \mathbb{D}. There are two cases:

  •  there exists a subsequence \left\{ u_{n_{k}}\right\} satisfying \norm{\tilde{T}\left(u_{n_{k}}\right)}{}<1 for all k;
  •  there exists a subsequence \left\{ u_{n_{k}}\right\} satisfying \norm{\tilde{T}\left(u_{n_{k}}\right)}{}\ge1 for all k.

For the first case, since \tilde{T}\left(u_{n_{k}}\right)=T\left(u_{n_{k}}\right) and T is compact, there exists a subsequence \left\{ u_{\tilde{n}_{k}}\right\} such that \left\{ T\left(u_{\tilde{n}_{k}}\right)\right\} converges strongly in X. So \tilde{T}u_{\tilde{n}_{k}} converges strongly in X.

For the second case, choose a subsequence \left\{ u_{\tilde{n}_{k}}\right\} so that \frac{1}{\norm{T\left(u_{\tilde{n}_{k}}\right)}{}}\rightarrow\alpha and T\left(u_{\tilde{n}_{k}}\right)\rightarrow z as k\rightarrow\infty by Bolzano-Weierstrass theorem on real numbers and compactness of T. Hence

    \[ \tilde{T}\left(u_{\tilde{n}_{k}}\right)\rightarrow\alpha z. \]

Thus, \tilde{T} is compact.

Hence by Theorem 2, \tilde{T} has a fixed point x. We claim that x is also a fixed point of T. Suppose \norm{Tx}{}\ge1 with Tx=x. Then

    \[ x=\tilde{T}x=\sigma Tx\quad\text{with }\sigma=\frac{1}{\norm{Tx}{}} \]

and so \norm x{}=1, which contradicts the assumption. Hence \norm{Tx}{}<1 and consequently x=T^{*}x=Tx.

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