# Prove that oscillation function is upper semicontinous

Suppose $X$ is a Polish space and $Y$ a compact metrizable space. Denote $P_k(Y)$ to be the collection of all non-empty, compact subsets of $X$. Denote the Hausdorff metric $d_H : P_k(Y) \rightarrow P_k(Y)$ by $d_H(A,B) = \inf \{ \varepsilon > 0 : A \subseteq B_{\varepsilon} \text{ and } B \subseteq A_{\varepsilon}\}$ where $A_{\varepsilon}$ denotes the $\varepsilon$-ball around $A$.

Let $F : X \rightarrow P_k(Y)$ be a multifunction or set-valued mapping. For any closed set $P \subseteq X$, denote $F|_P$ as the restriction of $F$ to $P$, such that $P(x) = \emptyset$ for all $x \not\in P$ and $F|_P(x) = F(x)$ for all $x \in P.$ Define the oscilation function of $F|_P$ at $x$ by $osc(F|_P)(x) = \inf_{V \in N(x)} \sup_{x_1, x_2 \in V \cap P}{d_H(F(x_1), F(x_2))}.$

Question: How to prove that $osc(F|_P)$ is an upper semicontinuous function?

Definition: A multifunction $F : X \rightarrow Y$ is an upper semicontinuous function if $\{ x \in X : F(x) \subseteq U\}$ is an open subset of $X$ for every $U \subseteq Y$ open.

Remark: The oscillation function $osc(F|_P)$ is a real-valued function. This question is motivated by this paper, page $73$, Theorem $18.$

• Is $F$ upper semicontinuous? I don't think $osc(F\vert_P)$ should have any continuity properties unless it inherits it from $F$. – Theoretical Economist Mar 15 '17 at 20:48
• @TheoreticalEconomist: No, $F$ needs not be upper semicontinuous. It is just any set-valued mapping. – Idonknow Mar 16 '17 at 1:43
• Huh, interesting. Ok. Since $osc(F\vert_P)$ is a real-valued function, you want to know it is usc as a function, and not a set-valued map, correct? (Though if I remember correctly, usc functions are also usc as set-valued maps and vice-versa, so perhaps the distinction is moot.) – Theoretical Economist Mar 16 '17 at 2:39
• @TheoreticalEconomist: Yes, I want to know whether $osc(F|_P)$ is USC as a real-valued function. – Idonknow Mar 16 '17 at 2:44
• Thanks for the clarification -- was just confused by the statement of the definition of upper semicontinuity for set-valued maps when it doesn't seem the most appropriate definition to appeal to. – Theoretical Economist Mar 16 '17 at 2:50

(1) Assume that $x_n\rightarrow x$ and $g(y):={\rm osc}\ (F|P)(y)$

Then there are neighborhood $U_n$ of $x_n, y_n,\ z_n\in U_n$ and $\delta_n,\ \varepsilon_n\rightarrow 0$ s.t. $$g(x_n) +\varepsilon_n > \sup_{ y,\ z\in U_n}\ d_H(F(y),F(z)) \geq g(x_n)$$

and $$d_H(F(y_n),F(z_n)) \leq \sup_{ y,\ z\in U_n}\ d_H(F(y),F(z)) < \delta_n + d_H (F(y_n), F(z_n))$$

Hence $|d_H(F(y_n),F(z_n)) -g(x_n) |\leq \varepsilon_n+\delta_n$

(2) There is neighborhood $V_n$ of $x$ s.t. $g(x)=\lim_n\ \sup_{y,\ z\in V_n}\ d_H(F(y),F(z))$

By re-indexing we can assume that $y_n,\ z_n\in V_n$ Hence $g(x)\geq d_H(F(y_n),F(z_n))$ so that $$g(x)\geq \lim\ \sup_n\ d_H(F(y_n),F(z_n)) =\lim\ \sup_n\ g(x_n)$$

• For $(2)$, why there exists $V_n$ such that $g (x) = \lim_n \sup_{y,z \in V_n}$? I know that since $x_n$ converges to $x$, we can have a neighbourhood of $x$. But why must the equality hold? – Idonknow Mar 19 '17 at 14:16