Ex4-8

Question:
(a) Let f and g be two functions that are usc at x_0. Show that f+g is usc at x_0. Is f-g usc at x_0? When is fg usc at x_0?
(b) If \{f_k\} is a sequence of functions that are usc at x_0, show that \inf_k f_k(x) is usc at x_0.
(c) If \{f_k\} is a sequence of functions that are usc at x_0 and converge uniformly near x_0, show that \lim f_k is usc at x_0.
Solutions: 
Recall: Let f be defined on E,and let x_0 be a limit point of E that lies in E. Then is said to be upper semicontinuous (usc) at x_0 if
\limsup_{x\to x_0;x\in E} f(x)\leq f(x_0).
(a) Given f and g be two functions that are usc at x_0. This implies that \limsup_{x\to x_0} f(x)\leq f(x_0) and \limsup_{x\to x_0} g(x)\leq g(x_0). If \limsup_{x\to x_0} f(x)+\limsup_{x\to x_0} g(x) is well-defined, that is not (\infty-\infty) or (-\infty+\infty), then 
\begin{align}
\limsup_{x\to x_0} [f(x)+g(x)]&\leq \limsup_{x\to x_0}f(x)+\limsup_{x\to x_0}g(x)\\
&\leq f(x_0)+g(x_0)
\end{align}Thus, f+g is usc at x_0.
For the function f-g, it is not usc at x_0. One counterexample is:
Let f(x)=0\quad \text{for all}~x and g(x) = \begin{cases} 0, & \text{if }x<0 \\ 1, & \text{if }x\geq 0 \end{cases}, then f and g are usc at x_0=0. However, 
f(x)-g(x) = \begin{cases} 0, & \text{if }x<0 \\ -1, & \text{if }x\geq 0 \end{cases}
, which is not usc at x_0=0.
For the product function fg, it is usc at x_0 if
\limsup_{x\to x_0} [f(x)g(x)]\leq (\limsup_{x\to x_0}f(x))(\limsup_{x\to x_0}g(x)).
This inequality can only be true when both functions f and g are nonnegative.
(b) Given \{f_k\} is a sequence of functions that are usc at x_0. So, \limsup_{x\to x_0} f_k(x)\leq f_k(x_0)\quad\text{for all}~k\in\mathbb{N}. Now, for all k,
\begin{align}
\limsup_{x\to x_0}[\inf_k f_k(x)]&\leq \limsup_{x\to x_0} f_k(x)\\
&\leq f_k(x_0).
\end{align}
Since the above inequality is true for all k, we have
\limsup_{x\to x_0}\inf_k f_k(x)\leq \inf_k f_k(x_0)and thus \inf_k f_k(x) is usc at x_0.
(c) Given \{f_k\} is a sequence of functions that are usc at x_0. So, \limsup_{x\to x_0} f_k(x)\leq f_k(x_0)\quad\text{for all}~k\in\mathbb{N}.
Taking limits on both sides, 
\lim_{k\to \infty}\limsup_{x\to x_0} f_k(x)\leq \lim_{k\to \infty}f_k(x_0).Since \{f_k\}converge uniformly near x_0, we can interchange lim and limsup, then
\limsup_{x\to x_0}\lim_{k\to \infty} f_k(x)\leq \lim_{k\to \infty}f_k(x_0)
and thus \lim f_k is usc at x_0.