Brouwer’s fixed point theorem and continuous functional dependence on the fixed point.

analysisfixed-point-theoremsgeneral-topology

The famous Brower's fixed-point theorem states that any $ f $ function that maps a compact and convex set itself has a fixed point.

I would like to know if minor disturbances in the function $ f $ could only cause minor disturbances in the fixed point $ x $ of $ f $. In other words the question would be as follows. If a function $ g $ is close to $ f $ then will the fixed points of $ f $ be close to the fixed points of $ g $?

One problem with this question is that the number of fixed points of $ g $ may be greater or less than the number of fixed points of $ f $. Thus, there may be a fixed point $ x_f $ of $ f $ such that $ g (x) \neq x $ to $ x $ in some neighborhood of $ x_f $.

However the Brouwer's fixed-point theorem guarantees that the number of fixed points of $ g $ is always greater than or equal to $ 1 $. Thus, the question could be improved and put in the following terms. In the set $ \mathrm{Fix}(f) $ of fixed points of $ f $ there would be $ x_f \in \mathrm{Fix} (f) $ such that if any $ g $ application is close to $ f $ there would be $ x_g \in \mathrm{Fix }(g) $ tal which $ x_f $ is close to $ x_g $?

Technically the question would be as follows. Let $ \Omega \subset \mathbb{R}^n$ be a compact and convex set. Rig the set $ C^0(\Omega, \mathbb{R}^n) $ with the supreme norm $ \| f \|_{\infty}: = \sup \{| f (x) |: x \in \Omega \} $. Take $ f \in C^0(\Omega, \mathbb{R}^n)$ such that $ f (\Omega) \subseteq \Omega $. Given $ \epsilon> 0 $ and $ g $ satisfying the condition $ \| f-g \|_\infty <\epsilon $, with $ g(\Omega) \subseteq \Omega $, there is $ x_f \in \mathrm{Fix} (f) $, $ \delta> 0 $ and $ x_g \in \mathrm{Fix} (g) $ such that $ \| x_f-x_g \| <\delta $? How to prove it?

Best Answer

Let $\Omega=[0,1]$. Fix $\varepsilon\in(0,1]$ and define, for each $x\in[0,1]$, \begin{align*} f(x)&\equiv(1-\varepsilon)x,\\ g(x)&\equiv(1-\varepsilon)x+\varepsilon. \end{align*} Clearly, $\|f-g\|_{\infty}=\varepsilon$. However, the only fixed point of $f$ is $0$ and the only fixed point of $g$ is $1$, which are as far from each other as possible.

ADDED: That said, one can establish a continuity property of sorts that is worth exploring. For some $n\in\mathbb N$, let $\Omega$ be a non-empty, convex, compact subset of $\mathbb R^n$. Let $\mathcal C$ denote the set of continuous functions mapping $\Omega$ into itself. Define a correspondence $\Phi:\mathcal C\rightrightarrows\Omega$ as $$\Phi(f)\equiv\text{set of fixed points of $f$}\quad\text{for each $f\in\mathcal C$}.$$ By Brouwer’s theorem, $\Phi(f)$ is not empty for any $f\in\mathcal C$.

Endowing $\mathcal C$ with the supremum norm $\|\cdot\|_{\infty}$ and $\Omega$ with the Euclidean norm $\|\cdot\|_n$, we can establish the following:

THEOREM: The correspondence $\Phi$ is upper hemicontinuous in the sense that if $O$ is an open subset of $\Omega$, then the “inverse image” $$\Phi^{-1}(O)\equiv\{f\in\mathcal C\,|\,\Phi(f)\subseteq O\}$$ is open in $\mathcal C$.

Proof: For the sake of contradiction, suppose that $\Phi^{-1}(O)$ is not open. Then, one can find some $f\in\Phi^{-1}(O)$ and two sequences $(f_m)_{m\in\mathbb N}$ and $(x_m)_{m\in\mathbb N}$ in $\mathcal C$ and $\Omega$, respectively, such that for each $m\in\mathbb N$,

  • $\|f_m-f\|_{\infty}<1/m$;
  • $x_m\in\Phi(f_m)$; but
  • $x_m\in\Omega\setminus O$.

Since $\Omega\setminus O$ is compact, one can take some subsequence $(x_{m_k})_{k\in\mathbb N}$ converging to some $x\in\Omega\setminus O$. For each $k\in\mathbb N$, the following holds: \begin{align*} \|x-f(x)\|_n&\leq \|x-x_{m_k}\|_n+\|x_{m_k}-f_{m_k}(x_{m_k})\|_n\\ &+\|f_{m_k}(x_{m_k})-f(x_{m_k})\|_n+\|f(x_{m_k})-f(x)\|_n. \end{align*} The first, third, and fourth terms converge to $0$ as $k\to\infty$ because of convergence in $\Omega$, convergence in $\mathcal C$, and continuity, respectively. The second term vanishes because $x_{m_k}$ is a fixed point of $f_{m_k}$ for every $k\in\mathbb N$. It follows that $\|x-f(x)\|_n=0$, that is, $x$ is a fixed point of $f$. Since $f\in\Phi^{-1}(O)$, the conclusion is that $x\in \Phi(f)\subseteq O$, which contradicts $x\in\Omega\setminus O$. $\quad\blacksquare$

The above upper-hemicontinuity property of $\Phi$ can be given an equivalent sequential characterization as follows:

THEOREM: Let

  • $(f_m)_{m\in\mathbb N}$ be a sequence in $\mathcal C$ converging to $f\in\mathcal C$; and
  • $(x_m)_{m\in\mathbb N}$ a sequence in $\Omega$ converging to $x\in\Omega$; such that
  • $x_m$ is a fixed point of $f_m$ for each $m\in\mathbb N$, that is, $x_m\in\Phi(f_m)$.

Then, $x$ is a fixed point of $f$, that is, $x\in\Phi(f)$.

Proof: For any $m\in\mathbb N$, \begin{align*} \|x-f(x)\|_n&\leq\|x-x_m\|_n+\|x_m-f_m(x_m)\|_n\\ &+\|f_m(x_m)-f(x_m)\|_n+\|f(x_m)-f(x)\|_n. \end{align*}

Proceed as before. $\quad\blacksquare$

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