2
$\begingroup$

I'll briefly state the problem I've been working on and what i've tried and got stuck on.

Let's $f: X \to [-\infty, \infty]$ be any function on an arbitrary topological space $X$. I proved this result and let's use this as given:

$\underline{f}(x) = \displaystyle\sup_{N \in N_x}\inf_{\xi \in N}f(\xi) = \displaystyle\lim_{n \to \infty}\left(\displaystyle\inf_{\xi \in I_n}f(\xi)\right)$

where $N_x$ is a collection of neighborhoods of a point $x \in X$ and for any nonincreasing sequence of intervals

$I_0 \supseteq I_1 \supseteq \dots$ where the lengths decreases to zero, $l(I_n) \to 0$, and $x \in \mathrm{interior}(I_n)$ for each $n \in \mathbb{Z}_{\geq 0}$

Problem: Show that $\underline{f}$ is lower-semicontinuous.

So far, the definitions I"m using for lower-semicontinuous (LSC), is $\underline{f}^{-1}(a,\infty]$ is open for any $a \in \overline{\mathbb{R}}$, $\underline{f}^{-1}[-\infty,a]$ is closed, and for any $x \in X, \epsilon > 0$, there exists a neighborhood of $x$, $N$, such that \underline{f}(x) - \underline{f(}x') < \epsilon for all x' \in N.

How i started the proof, is if $\underline{f}^{-1}(-\infty)$ is closed, then $\underline{f}^{-1}(a, \infty]$ is open. If $\underline{f}^{-1}(\infty)$ is closed, then $\underline{f}^{-1}[-\infty,\alpha)$ is open for any $\alpha \in \overline{\mathbb{R}}$. Take the intersection of these guys and you get $\underline{f}^{-1}(\beta, \alpha)$ is open for any $\beta < \alpha \in \overline{\mathbb{R}}$. So $\underline{f}^{-1}(\beta, \infty]$ is open.

Consider $\underline{f}(x) \in \mathbb{R}$. Then using that $\underline{f}(x) = \lim_{n \to \infty}(\inf_{\xi \in I_n}f(\xi))$, we have the following

Let $\epsilon > 0$, there is an $N \in \mathbb{Z}_{\geq 0}$ such that $\underline{f}(x) - \inf_{\xi \in I_N}{f}(\xi) < \epsilon$, for a fixed $x$.

Likewise, we have that for $\epsilon = \inf_{\xi \in I_{N_1}}f(\xi) - \inf_{\xi \in I_N}f(\xi)$,

\inf_{\xi \in I_{N_1}}{f}(\xi) - \underline{f}(x') < \inf_{\xi \in I_{N_1}}f(\xi) - \inf_{\xi \in I_N}f(\xi) which gives us -f(x') < -\inf_{\xi \in I_N}f(\xi). Then we have \underline{f}(x) - \underline{f}(x') < \epsilon.

At this point, I got stuck how to find a neighborhood of $x$ such that it satisfies \underline{f}(x) - \underline{f}(x') < \epsilon. I only got it for one point.

  • 1
    I'm almost certain for the rest of the problem set $X = \mathbb{R}$. See the definition of an interval in [the supplementary notes](http://math.berkeley.edu/~wodzicki/H104.F10/Integral.pdf), and the subsequent questions on pages 2 and 3 of the homework. But in reality the best way to get a clarification of what an interval is supposed to mean in this context is to *ask your professor*.2012-04-05

0 Answers 0