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Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a locally bounded, discontinuous, function and let $\delta: \mathbb{R}^n \rightarrow \mathbb{R}_{\geq 0}$ be a continuous function.

Define the set-valued mapping $ F: \mathbb{R}^n \rightrightarrows \mathbb{R}^m $ as

$$ F(x) := \bigcap_{r >0} \text{closure} \left( f( x+ \delta(x)\mathbb{B} + r \delta(x) \mathbb{B} ) \right), $$

where $\mathbb{B}$ denotes the closed ball.

Question: is $F$ Outer SemiContinuous?


1) It is known that the set-valued mapping $\bar{F}(x):= \bigcap_{r>0} \text{closure} \left(f(x+r \mathbb{B}) \right)$ is Outer SemiContinuous.

2) Definition of Outer SemiContinuity: 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 $\limsup_{x \rightarrow \bar x} S(x) = S(\bar x)$.

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For continuous $d: \mathbb{R}^n \rightarrow \mathbb{R}_{>0}$, define $G(x) := \bar{F}(x+d(x)\mathbb{B}) = \bigcap_{r>0} \text{closure} (f(x+d(x)\mathbb{B}+r\mathbb{B} ))$. Is $F_1$ equal to $F$ with $\delta$ strictly-positive valued? – Adam Sep 14 '12 at 13:37
up vote 2 down vote accepted

Yes. First, notice it suffices to prove that for every $y$ the set $\{x: y\notin F(x)\}$ is open. Now, if $y\notin F(x)$ then there exists $r>0$ for which the corresponding set in the intersection does not contain $y$. Finally, if $|x'-x|$ is small enough, the $r/2$-set in the definition of $F(x')$ does not contain $y$, proving the claim.

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Why it is sufficient to prove that $\{x\mid y \in F(x)\}$ is open? – Adam Sep 15 '12 at 9:17
We must prove the following: if $y$ is not in $S(\bar x)$, then $y$ is not in the $\limsup $. The openness condition I stated implies $y\notin S(x)$ for all $x$ in some neighborhood of $\bar x$, – user31373 Sep 15 '12 at 14:44
But I'm not clear about what happens when $\delta(x)=0$. I mean for those $x$ such that $\delta(x)=0$ we have $F(x) = \{f(x)\}$, while for $x$ such that $\delta(x)>0$ we have $F(x) \supset \{f(x)\}$. Is this fine for $F$ being OSC? – Adam Sep 16 '12 at 8:21
@Adam I did not notice that you allow $\delta$ to vanish. If $\delta$ is zero, then there is no reason for $F$ to be continuous. The counterexample I gave you elsewhere applies here too: let $\delta$ be identically zero, $f(0)=1$, and $f=0$ elsewhere. – user31373 Sep 16 '12 at 17:40
So $F$ is outer semicontinuous in the set of $x$ such that $\delta(x)>0$, right? – Adam Sep 16 '12 at 20:52

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