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.


1 Answer 1


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).

  • $\begingroup$ 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

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