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Consider a locally-bounded function $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ and a continuous function $g: \mathbb{R}^n \rightarrow \mathbb{R}_{> 0}$.

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

$ F(x) = \text{closure} f( x + g(x) \bar{\mathbb{B}}) + g(x) \bar{\mathbb{B}},$

where $\bar{\mathbb{B}}$ is the closed unit ball of $\mathbb{R}^n$.

Question: is $F$ Outer SemiContinuous?

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

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    Equivalently, one can prove that $\text{graph}(F) := \{ (x,y) \in \mathbb{R}^n \times \mathbb{R}^m \mid y \in F(x) \}$ is closed.2012-09-07

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Counterexample in one dimension: $f(x)=0$ if $|x|\le1$ and $f(x)=5$ otherwise. Also $g(x)\equiv1$. Now $F(0)=[-1,1]$ but for all $x\ne0$ the set $F(x)$ contains $5$.