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Consider an outer semicontinuous set-valued mapping $S: \mathbb{R}^n \rightrightarrows \mathbb{R}^m$.

(Namely the sets $S(x) \subseteq \mathbb{R}^m$ do not "change continuously" as a function of $x$)

1) Find a continuous function $f: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}_{\geq 0}$ such that

$f(x,y) = 0 \Longleftrightarrow y \in S(x) $

2) Can we make $f$ such that $\lim_{|(x,u)| \rightarrow \infty} f(x,y) = \infty$?

For instance the Euclidean distance of $y$ to the set $S(x)$, i.e. $f(x,y) = |y|_{S(x)}$, satisfies the property but it is not continuous.

Note: definition of outer semicontinuity for a set-valued map:

A set-valued mapping $S: \mathbb{R}^n \rightrightarrows \mathbb{R}^m $ is outer semicontinuous at $\bar x$ if

$ \limsup_{x \rightarrow \bar x} S(x) \subset S(\bar x) $

or equivalently $\limsup_{x \rightarrow \bar x} S(x) = S(\bar x)$.

1 Answers 1

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Your idea was almost right – the Euclidean distance of $(x,y)$ from $T=\{(a,b)\mid b\in S(a)\}$ works.

Clearly, if $y\in S(x)$, then $(x,y)\in T$, and $|(x,y)|_T=0$. Conversely, if $|(x,y)|_T=0$, then there is a sequence of points $(x_n,y_n)$ with $y_n\in S(x_n)$ and $\lim_{n\to\infty}(x_n,y_n)=(x,y)$, thus $\lim_{n\to\infty}x_n=x$ and $\lim_{n\to\infty}y_n=y$, and hence $y\in\limsup_{x'\to x}S(x')=S(x)$.

Another way of saying the same thing is that $S$ is outer semicontinuous iff $T$ is closed. See also this.

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    @Adam: You've deleted question (2) in the meantime; the above comments no longer make sense for someone reading the thread now. Please keep in mind that the content of the site is for everyone, not just for those posting. To answer your question: If $T$ is unbounded, then there are points $(x,y)$ with $y\in S(x)$ with $|(x,y)|$ arbitrarily large. If $y\in S(x)$ implies $f(x,y)=0$, that contradicts the condition that $f$ tends to $\infty$ for $|(x,y)|\to\infty$.2012-06-20