Continuity of fixed-points?

Question

A version of Brouwer's fixed-point theorem states, given a compact, convex set $S$, that $\forall f \colon S \xrightarrow{cont.} S, \exists x_0 \in S \colon f(x_0) = x_0$. Thus, if $F \colon S \times (0,1) \xrightarrow{cont.} S$, then $\forall t \in (0,1), \exists x_t \in S \colon F(x_t,t) = x_t$.

It seems "intuitively obvious", that the function $t \rightarrow x_t$ is continuous, but I can't quite nail the argument; I haven't got as far as my first coffee yet. Is it back to $\epsilon$ and $\delta$ again?

Answer 1

First note that the set

$$G = \{ (t, x) \in (0, 1) \times S : F(x, t) = x \}$$

is a closed subset of $S \times (0, 1)$. Indeed, if $F(x, t) \neq x$, slightly moving $x$ and $t$ changes both sides just a little, so they're still not equal (it's easy to formalize this argument).

Now if $G$ is a function $(0, 1) \to S$, that is, if for each $t \in (0, 1)$ there is a unique $x \in S$ such that $(t, x) \in G$, then $G$ is a continuous function: let $G(t) = x$ and let $x \in V \subseteq S$ be open. Then $\{ t \} \times (S \setminus V)$ is compact and for each $y \in S \setminus V$ since $(t, y) \notin G$, there is an open neighborhood $(t, y) \in U_y \times V_y \subseteq \big( (0, 1) \times S \big) \setminus G$. The family

$$\{ U_y \times V_y : y \in S \setminus V \}$$

is an open cover of $\{ t \} \times (S \setminus V)$. Pick a finite subcover $\{ U_i \times V_i : i = 1, \ldots, n \}$ and let

$$U = \bigcap_{i=1}^n U_i.$$

Then $U \times (S \setminus V)$ is disjoint from $G$, hence for $s \in U$ we have $G(s) \in V$.

In general, however, $G$ does not have to be a function. Moreover, it's possible that there is no $G_0 \subseteq G$ which is a continuous function $(0, 1) \to S$. For example, for clarity of presentation replace $(0, 1)$ with $(-1, 1)$, take $S = [-1, 1]$ and let

$$G = \left\{ (t, x) \in (-1, 1) \times [-1, 1] : t = 5 x \left( x - \frac{1}{2} \right) \left(x + \frac{1}{2} \right) \right\}$$

It's simple to find a continuous $F : [-1, 1] \times (-1, 1) \to [-1, 1]$ such that $F(x, t) = x \iff (t, x) \in G$. But obviously there is no continuous function $G_0 \subseteq G$.

Answer 2

Your function is not well defined, since $x_t$ isn't necessarily unique. But if the fixed point is always unique, the following argument works.

Take some $t \in (0,1)$ and $t_n \in (0,1)$ with $t_n \to t$. Let $x$ be some subsequential limit of $x_{t_n}$, say $x_{t_{n_k}} \to x$. Then, $x = \lim_k x_{t_{n_k}} = \lim_k F(x_{t_{n_k}},t_{n_k}) = F(x,t)$, where the last equality holds by continuity of $F$ (since, of course, $t_{n_k} \to t$). Hence, we showed that any subsequential limit of $(x_{t_n})_n$ is a fixed point of $F(\cdot,t)$. By uniqueness of the fixed point, we must have that $x$ is the only subsequential limit. But, by compactness, this means that $x_{t_n} \to x$.