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Let $\phi: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a locally bounded function.

Consider the following set-valued map $\Phi: \mathbb{R}^n \rightrightarrows \mathbb{R}^m$ (i.e. $x \in \mathbb{R}^n$ is mapped to a subset of $\mathbb{R}^m$) defined as: $$ \Phi(x) := \bigcap_{\rho > 0} \overline{ \phi(x + \rho \mathbb{B}) } $$

Prove that $\Phi(\cdot)$ is outer semicontinuous, i.e.:

for any sequence $\{(f_i,y_i)\}_{i=1}^{\infty}$, $(f_i,y_i) \in \mathbb{R}^m \times \mathbb{R}^n$ and $(f_i,y_i) \rightarrow (f,y)$, such that $f_i \in \Phi(y_i)$, we have $f \in \Phi(y)$.

Notes: $\overline{S}$ denotes the closure of the set $S$. Therefore $\overline{ \phi(x + \rho \mathbb{B}) }$ denotes the closure of the set $\{\phi(x+\epsilon): \ \epsilon \in \delta \mathbb{B}\}$, where $\mathbb{B} \subset \mathbb{R}^n$ is the unitary closed ball.

Asaf Karagila
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user693
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1 Answers1

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This seems a little suspicious, since I haven't use local boundedness anywhere. I'll post this anyway; since I was not able to find a mistake. (In case you know an example showing that the above is not true without the assumption that $\phi$ is locally bounded, please, let me know. Including the example would be nice.) $\newcommand{\ol}[1]{\overline{#1}}\newcommand{\ve}\varepsilon\newcommand{\norm}[1]{\|{#1}\|}\newcommand{\mc}[1]{\mathcal{#1}}$

First, let's have a look at the meaning of your function $\Phi(x)$. I will denote the ball of radius $\ve$ around $x$ by $B(x,\ve)$. I.e., $B(x,\ve)=x+\ve\mathbb B$. $$y\in\Phi(x)=\bigcap_{\varepsilon>0} \ol{\phi(B(x,\ve))}$$ means that, for each $\ve>0$, the point $y$ belongs to $\ol{\phi(B(x,\ve))}$, i.e. every neighborhood of $y$ intersects $\phi(B(x,\ve))$. $$(\forall \delta>0)(\forall \ve>0) \phi(B(x,\ve))\cap B(y,\delta)\ne\emptyset$$ $$(\forall \delta>0)(\forall \ve>0) (\exists z) \norm{z-x}<\ve \land \norm{\phi(z)-y}<\ve$$

We know that $f_i\in\Phi(y_i)$, $y_i\to y$, $f_i\to f$.

Suppose we are given $\delta,\ve>0$.

First we choose $i$ such that $\norm{y_i-y}<\ve/2$ and $\norm{f_i-f}<\delta/2$.

We know that there exists $z$ such that $\norm{z-y_i}<\ve/2$ and $\norm{\phi(z)-f_i}<\delta/2$.

Using triangle inequality we get $\norm{z-y}<\ve$ and $\norm{\phi(z)-f}<\delta$. So for $y$ and $f$ (and any given positive $\delta$ and $\ve$) we have found $z$ that verifies the above condition.

Note: You multifunction $\Phi(x)$ can be equivalently defined as the intersection $\bigcap \ol{\phi(U)}$ over all neighborhoods $U$ of $x$. As I've learned recently from Dave L. Renfro's answer here, this means that $\Phi(x)$ is the cluster set of the function $\phi$ at the point $x$. The references in this answer suggest that cluster sets have been studied quite extensively, so it is quite probable that cluster multifunction have been studied too and you'll be able to find some facts about it in the literature.

Added: Your question (at least if my reinterpretation of it is correct) in fact asks whether the multifunction assigning to each point cluster set of a given function at this point has closed graph. I've tried to Google a little to see whether I found some known results on this function. What I was able to find was this:

This paper studies cluster function of $f$ defined as $C_f(x)=\bigcap_{U\in\mc N(x)} \overline{f(U)}$. By $\mc N(x)$ I mean the set of all neighborhoods of $x$. However, it does not mention the property that this multifunction has closed graph.

  • Christian Richter: The Cluster Function of Single-Valued Functions; Set-Valued Analysis, Volume 14, Number 1 (2006), 25-40, DOI: 10.1007/s11228-005-5135-y

In this paper it is mentioned that a multifunction $F$ is the cluster function of some $f$ if $F$ has a closed graph and admits a selection $f$ whose graph is dense in the graph of $F$.

This paper studies a more general notion of $\mathcal E$-cluster points. It is shown that the cluster multifunction $\mathcal E_F$ has closed graph.

Maybe these papers and the references given there can be useful for you if you plan to study the multifunction $\Phi$ further.

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Martin Sleziak
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  • I'm not clear about the equivalence between $y \in \Phi(x)$ and the two conditions you provided. For instance, $B(x,\epsilon)$ is a neighborhood of $x$, I don't see why you address neighborhoods of $y$. Instead, I would say that "$\phi(x)$ + its limit points" contain $y$. – user693 Apr 19 '12 at 16:03
  • $B(x,\varepsilon)$ is the same thing as $x+\varepsilon\mathbb B$. – Martin Sleziak Apr 19 '12 at 16:11
  • Yes, so I don't see why every neighborhood of $y$ intersects $\phi(B(x,\epsilon))$ – user693 Apr 19 '12 at 16:17
  • $y$ belongs to closure of a set $A$ if an only if every neighborhood of $y$ intersects $A$. See also limit point at Wikipedia. – Martin Sleziak Apr 19 '12 at 16:25
  • We know that $f_i \in \bigcap_{\rho > 0} \overline{ \phi(y_i + \rho \mathbb{B}) } $. Taking the limit we have: $f = \lim_{i \rightarrow \infty} f_i \in \lim_{i \rightarrow \infty} \bigcap_{\rho > 0} \overline{ \phi(y_i + \rho \mathbb{B}) }$. Does such RHS contain (or is equal to) $\Phi(x)$? – user693 Apr 19 '12 at 16:28
  • @Adam: Maybe it would be reasonable to take this to chat instead of clustering the comments here. – Martin Sleziak Apr 19 '12 at 16:31
  • Dear Martin, I'm not clear why "$y$ belongs to each $\overline{ \phi(B(x,\epsilon))}$". Intuitively, why this should be true if $\epsilon$ is "very large"? For instance, if $\phi(\cdot)$ is continuous, then $\Phi(x) = \phi(x)$ is single-valued and $y \notin \phi(x+\epsilon)$. – user693 Apr 19 '12 at 16:33