0
$\begingroup$

Let $X$ be a compact and convex space and let $T=[0,1]$ be some parameter space. Let $F:X\times T\rightrightarrows X$ be a correspondence that is compact-valued, convex, and upper-hemicontinous. By Kakutani's fixed point theorem, there is a fixed point $x(t)=F(x(t),t)$ for each parameter $t\in T$.

Suppose we also know that

  1. the set $F(x, t')$ converges $F(x, t)$ as $t'\to t$ for any $x\in X$ (in the sense that the indicator functions for the two sets converge pointwise), and
  2. there is a unique fixed point $x(t)$ for each $t\in T$.

Question: Is this enough to determine that $x(t)$ is a continuous function?

Here is my very informal attempt at a proof: Consider sequences $(x_n, t_n)\to (x, t)$ such that $x_n\in F(x_n, t_n)$. Suppose $x\notin F(x, t)$. Since, $F(\cdot, t)$ and $F(\cdot, t_n)$ are close for large enough $n$, then $x\notin F(x,t_n)$. By UHC of $F$, for any open set $V$ containing $F(x, t_n)$, I can find an open set $U$ of $x$ such that for all $x'\in U$, $F(x', t_n)\in V$. I can pick $V$ such that it excludes $x$, and by consequence, any $x_n$ for $n$ large enough. However, this would imply that there is an $x_n\in U$ and $x_n\notin V$, which implies $x_n\notin F(x_n, t_n)$. Hence, $x\in F(x, t)$, and the graph $$ \{(x,t): x\in F(x, t) \} $$ is closed. This along with the uniqueness should be sufficient for showing that $x(t)$ is a continuous function.

$\endgroup$

1 Answer 1

1
$\begingroup$

I assume that you mean that $F$ is upper semicontinuous on the product space. Then in particular (since $X$ is compact, Hausdorff and $F$ has closed values), $F$ has a closed graph. This implies that also the set $\{(x,t):x\in F(x,t)\}$ is closed. This means that the multimap $$t\mapsto\{x:x\in F(x,t)\}$$ has a closed graph and assumes closed values. Since $X$ is a compact space, it follows that this multimap is upper semicontinuous.

Of course, in the single-valued case upper semicontinuity and continuity are equivalent.

In particular, your hypothesis 1 superfluous. (Well, actually it follows from the upper semicontinuity of $F$ with respect to both variables - it is the special case that you fix one variable.) Note that you really have to require the upper semicontinuity with respect to both variables, even in the metric case: Your first part of the proof becomes wrong unless you assume some sort of locally uniform convergence (which implies of course again the upper semicontinuity).

$\endgroup$
1
  • $\begingroup$ I meant for $F$ to be upper hemicontinuous only in $x$ but I see that my proof does require uniform convergence. Thanks! $\endgroup$
    – tsm
    Aug 29, 2021 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.