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

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?

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 $$n\in\mathbb R$$, 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. $$\quad\blacksquare$$

• Great answer. Have a reference to these theorems? – MathOverview Oct 18 '19 at 21:43
• @MathOverview I constructed them on my own based on the framework laid out in Chapter E of Efe Ok’s Real Analysis with Economic Applications. I highly recommend this excellent book if you are interested in correspondences (set-valued functions), hemicontinuity, and the equivalence (under some regularity conditions) between the inverse-image and sequential characterizations of upper hemicontinuity. – triple_sec Oct 18 '19 at 23:59