19

Show that the supremum of any collection of lower semicontinuous functions is lower semicontinuous.

Suppose $\{f_n\}$ is a sequence of lower semicontinuous functions on a topological space $X$. Define $$g_k=\sup_{n\ge k}f_n.$$ I could see that $$\{x: g_k(x)\gt \alpha\}=\bigcup_{n=k}^{\infty}\{x:f_n(x)\gt \alpha\}$$ from where it follows that $g_k$ is lower semicontinuous. Also $\{g_n\}$ is monotonically decreasing sequence of semicontinuous functions, and because $$\lim\sup_{n\rightarrow\infty} f_n(x)=\inf \{g_1(x), g_2(x), \cdot \cdot \cdot\}$$, $\lim\sup_{n\rightarrow\infty} f_n(x)$ is lower semicontinuous if it can be shown that a monotonically decreasing sequence of lower semicontinuous functions converges to a lower semicontinuous function. How can I show this?

Also, even if the above can be implemented, I would only prove that the supremum of a sequence of lower semicontinuous functions is lower semicontinuous. What's the way generalize this to "any collection" of lower semicontinuous functions?

Thanks.

Martin Sleziak
  • 56,060
vnd
  • 1,546
  • 1
    The problem asks for the supremum, not for $\limsup$. – Daniel Fischer Feb 19 '16 at 10:17
  • @DanielFischer Thnks for pointing. Any hint as to how could one prove that a monotonically decreasing sequence of lower semicontinuous functions converges to a lower semicontinuous function if it is true? – vnd Feb 19 '16 at 10:21
  • 1
    As for the penultimate paragraph, the limit of a monotonically decreasing sequence of lower semicontinuous functions is generally not lower semicontinuous. Recall that the characteristic function of a set $A$ is lower semicontinuous if and only if $A$ is open. Let $(U_n)$ be a decreasing sequence of open sets. Then $\lim \chi_{U_n} = \chi_{\bigcap U_n}$, and any $G_{\delta}$-set can be thus written. But not all $G_{\delta}$s are open. – Daniel Fischer Feb 19 '16 at 10:21
  • Ok, but if the convergence is uniform, then this should hold? – vnd Feb 19 '16 at 10:27
  • 1
    Yes, the limit of a (locally) uniformly convergent sequence (net) of lower [upper] semicontinuous functions is again lower [upper] semicontinuous, whether the sequence is monotonic or not. – Daniel Fischer Feb 19 '16 at 10:49

1 Answers1

27

Note that a function $f$ is lower semicontinuous iff given $x \in X$ and $r < f(x)$ there is an open neighborhood of $U$ of $x$ such that $r < f(y)$ for each $y \in U$.

Suppose that $\{ f_i \}_{i \in I}$ is any collection of lower semicontinuous functions on $X$, and let $g$ be the pointwise supremum, i.e., $g (x) = \sup_{i \in I} f_i(x)$ for all $x \in X$.

Taking $x \in X$, and any $r < g(x) = \sup_{i \in I} f_i(x)$ it must be that there is an $i \in I$ such that $r < f_i(x)$. Since $f_i$ is lower semicontinuous there is an open neighborhood $U$ of $x$ such that $r < f_i(y)$ for each $y \in U$. As $f_i(y) \leq g(y)$ for all $y$, it follows that $r < g(y)$ for each $y \in U$.

Lääne-Viru
  • 6,918