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Let $f:[a,b]\to \mathbb{R}$ be a bounded function. Let $$\bar{f}(x) = \inf_{\delta>0}\sup_{|x-y|<\delta} f(y)$$ $$\underline{f}(x) = \sup_{\delta>0}\inf_{|x-y|<\delta} f(y)$$

(small side question, does anyone know the latex code for a small bar underneath an item?)

The problem (well it's got 4 parts, but I'll just put one and hopefully figure things out on the other 3 with a little nudge)

Show that $\bar{f}$ is upper semi-continuous on $[a,b]$

The definition of upper semi-continuous we are using is: A function $f$ is upper semi-continuous at $x$ if whenever $f(x)<\alpha$ there is an open set $O$, containing $x$, such that $f(y)<\alpha$ for $y\in O$.

I've tried showing this directly and by contradiction, but in both cases I'm simply finding I don't fully understand how to deal with these functions. For example, my direct argument:

We must show that when $\bar{f}(x) < \alpha$, there is an open set $O\subset [a,b]$, containing $x$, such that $\bar{f}(z) < \alpha$ for all $z\in O$. But this means we must show if $$\inf_{\delta>0}\sup_{|x-y| < \delta} f(y) < \alpha,$$ then there is an open set $O\subset[a,b]$ containing $x$ such that $$\inf_{\delta>0}\sup_{|z-y| < \delta} f(y) < \alpha,$$ for all $z\in O$. Fix some $\delta'>0$. For all $\delta > \delta'$, $$ \inf \sup \{y:|z-y|<\delta\}f(y) \leq \inf \sup \{y:|z-y|<\delta'\} f(y).$$ so...?

Some other thoughts:

We note if we have $$ f(y) > \alpha$$ for all $y\neq x$ on some open neighbourhood of $x$, with $f(x) < \alpha$, then $\bar{f}(x) \geq \alpha$, as $\bar{f}(x)$ does not actually use the value of $f$ at $x$. Hence if $f$ has a removable discontinuity or singularity, $\bar{f}$ will not notice. Hence we may assume our function $f$ does not have a removable singularity. But then $f$ can only have jump discontinuity or an essential discontinuity. If $f$ has a jump discontinuity, perhaps we can find some open neighbourhood $U$ of $f$ containing $x$ such that $f(U)$ is homeomorphic to something connected apart from the jump, and well behaved. If $f$ has an essential discontinuity at $x$, the limit must not exist (as our function is bounded), thought $\bar{f}$ still exists. Again, we could `zoom in' to some connected component (apart from the discontinuity) and things should be nicer there? I'm not sure how to pick such an open neighbourhood of our point though...

Even the slightest nudge would be appreciated :) Thanks!

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up vote 4 down vote accepted

Suppose that $\bar{f}$ is not upper semicontinuous at $x$; then there is an $\alpha>\bar{f}(x)$ such that for each $\epsilon>0$ there is a $y_\epsilon \in (x-\epsilon,x+\epsilon) \cap [a,b]$ such that $\bar{f}(y_\epsilon)\ge \alpha$. This implies that for every $\epsilon>0$, $$\sup_{|y-x|<\epsilon}\bar{f}(y)\ge \alpha\;.\tag{1}$$ On the other hand, you know that $$\bar{f}(x) = \inf_{\epsilon>0}\;\;\;\sup_{|y-x|<\epsilon}f(y)<\alpha\;,$$ where $\sup_{|y-x|<\epsilon}f(y)$ cannot increase as $\epsilon$ gets smaller. Pick any $\beta\in (\bar{f}(x),\alpha)$; then there must be some $\epsilon_0>0$ such that $f(y)<\beta$ whenever $|y-x|<\epsilon_0$. What does this tell you about $$\bar{f}(y) = \inf_{\epsilon>0}\;\;\;\sup_{|z-y|<\epsilon}f(z)$$ when $|y-x|<\epsilon_0$? Now compare with $(1)$.

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Hm, well I suppose if $f(y)<\beta$ for any $y$ with $|y-x|<\epsilon_0$, then you could pick any particular $y$, and let $\epsilon$ be small enough so that the ball of all $z$ with $|z-y|<\epsilon$ is contained in the set of all $y$ with $|x-y|<\epsilon_0$. Given this, we have that $\bar{f}(y) < \beta$. But this contradicts (1) as the supremum, as you mentioned, cannot decrease when $\epsilon$ decreases. – Alex Oct 22 '11 at 4:58
@Alex: Looks good. – Brian M. Scott Oct 22 '11 at 7:24
Thanks! Also nice how this gives us lower semi-continuity of $\underline{f}$ as $\underline{f} = \overline{-f}$! – Alex Oct 22 '11 at 15:58

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