Inclusion of uppersemicontinuous maps in continuous maps

Question

I have a question please. Is there a result about the inclusion of uppersemicontinuous maps in continuous maps ? I mean if F is uppersemicontinuous, is there always a map G continuous such that $F\subset G$.

Thank you in advance for your answers,

Answer 1

Consider $F : \Bbb{R} \rightrightarrows \Bbb{R}$, defined by $$F(x) = \begin{cases} \{1\} & \text{if } x > 0 \\ \{-1\} & \text{if }x < 0 \\ \{-1, 1\} & \text{if } x = 0. \end{cases}$$ Then $F$ is USC with full domain. Locally around $x_0 < 0$ and $x_0 > 1$, the function is single-valued and constant, making it continuous. At $x_0 = 0$, $F(0) + [-\varepsilon, \varepsilon]$ will always contain both $\{1\}$ and $\{-1\}$, and hence $F(x)$ for any $x \in \Bbb{R}$, making $F$ USC at $0$.

Now, suppose $G \subseteq F$, also with full domain. We have exactly $3$ options: $G(x) = F(x)$ for $x \neq 0$ and either:

• $G(0) = \{-1\}$,
• $G(0) = \{1\}$, or
• $G(0) = \{-1, 1\} = F(0)$.

It's not difficult to see that the first two options are not USC. If we add $[0.5, 0.5]$ to $G(0)$, then in the first case, $G(x) \not\subseteq G(0) + [0.5, 0.5]$ for any $x > 0$. Similarly, in the second case, this is true for $x < 0$. Either way, this is a failure of USC.

In the third case, $G = F$, which is not LSC. If we choose $y = 1 \in F(0)$, and $\varepsilon = 1$, then any $x < 0$ the only possible $t \in F(x)$ is $t = -1$, which is not within $\varepsilon$ of $1$.

So, no, it's not always possible to choose a continuous function from a USC function.