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I have been stumped for long by this exercise (3.12(d)) from Stokey and Lucas's Recursive Methods in Economic Dynamics. Would greatly appreciate any hints.

Let $\phi: X \to Y$ and $\psi: X \to Y$ be lower hemicontinuous correspondences (set-valued functions), and suppose that for all $x \in X$

$$\Gamma(x)=\{y \in Y: y \in \phi(x) \cap \psi(x)\}\neq \emptyset$$

Show that if $\phi$ and $\psi$ are both convex valued, and if $\mathrm{int} \phi(x) \cap \mathrm{int} \psi(x) \neq \emptyset$, then $\Gamma(x)$ is lower hemicontinuous at $x$.

[A correspondence $\Gamma: X \to Y$ is said to be lower hemicontinuous at $x \in X$ if $\Gamma(x)$ is nonempty and if, for every $y \in \Gamma(x)$ and every sequence $x_n \to x$, there exists $N \geq 1$ and a sequence $\{y_n\}_{n=N}^\infty$ such that $y_n \to y$ and $y_n \in \Gamma(x_n)$, all $n \geq N$.

Intuitively this means that the graph of $\Gamma(x)$ cannot suddenly broaden out.]

EDIT: We can assume that $X$ and $Y$ are subsets of $\mathbf{R}^n$.

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By "int", do you mean the interior of the set? – Niel de Beaudrap Oct 8 '10 at 14:42
@Niel de Beaudrap. Yes. – Jyotirmoy Bhattacharya Oct 8 '10 at 15:04
A paper proves this as a byproduct of proving something similar for infinite dimensionsional $X$ and $Y$. Could there be a simpler approach just for the finite-dimensional case?… – Jyotirmoy Bhattacharya Oct 8 '10 at 15:18
And convex-valued, I assume, means that $\phi(x)$ and $\psi(x)$ are convex sets for all $x$? – Paul VanKoughnett Oct 8 '10 at 16:19
@Paul VanKoughnett. Yes. – Jyotirmoy Bhattacharya Oct 8 '10 at 16:36
up vote 2 down vote accepted

Here is a somewhat detailed outline of an argument that I think works. As this is a homework problem, some of the pieces of the argument do need to be filled in.

Assume we're in $\mathbb{R}^m$. For fixed $y \in \Gamma(x)$, the fact that $\Gamma(x)$ has a nonempty interior means that there are $m$ points $z_1, z_2, \ldots, z_m$ in the interior of $\Gamma(x)$ such that these $m$ points and $y$ together are affinely independent. Thus you can take sufficiently small balls around each of these $m+1$ points such that the balls do not intersect and that any set consisting of one point from each ball is also affinely independent. Let $z_0 = y$. Since $\phi$ is lower hemicontinuous, for each $z_i$ there exists a sequence $z_{i_n} \to z_i$ and an $N_i$ such that $z_{i_n} \in \phi(x_n)$ and $z_{i_n}$ is inside that small ball around $z_i$ for all $n \geq N_i$. For each $n \geq \max \{N_i\}$, construct the convex hull $C_n$ of $\{z_{0_n}, z_{1_n}$, $z_{2_n}, \ldots, z_{i_n}\}$. Since $\phi$ is convex-valued, $C_n$ is a subset of $\phi(x_n)$. Do the same thing for each $n$ for the $\psi$ function to obtain sets $D_n$. Let $S_n = C_n \cap D_n$. The intersection of the convex hulls of two sets of $m+1$ affinely independent points in $\mathbb{R}^m$ that are pairwise close to each other must be nonempty. (Consider the supporting hyperplanes.) Let $y_n$ be the point in $S_n$ closest to $y$. Since the extreme points of $S_n$ converge to the extreme points of the convex hull of $\{z_0, z_1, z_2, \ldots, z_m\}$, $S_n$ converges as a set to the convex hull of $\{z_0, z_1, z_2, \ldots, z_m\}$. Thus the point in $S_n$ closest to $y$ $(= z_0)$ must converge to $y$; i.e., $y_n \to y$.

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Thanks! I haven't checked all the details yet but this is a much better starting point from what I had before. – Jyotirmoy Bhattacharya Oct 14 '10 at 7:58
If you would like more detail on the answer, just ask. – Mike Spivey Oct 14 '10 at 13:32

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