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I have been asked to prove If $f$ is bounded, then $g(x)= \overline{\lim}_{y\to x} f(y)$ is upper semi continuous.

This means somehow I have to show that for some $x_0$ $$\overline{\lim}_{x\to x_0} g(x)\leq g(x_0)$$ Now $$\overline{\lim}_{x\to x_0}g(x)= \lim_{\delta\to 0}\sup_{0\leq|x-x_0|\leq \delta} g(x)$$ $$=\lim_{\delta\to 0}\sup_{0\leq|x-x_0|\leq \delta}( \overline{\lim}_{y\to x} f(y))$$

So for $g$ to be upper semi continuous, I have to show that

$$\lim_{\delta\to 0}\sup_{0\leq|x-x_0|\leq \delta}( \overline{\lim}_{y\to x} f(y))\leq \overline{\lim}_{y\to x_0} f(y)$$

But how this is true?? Basically my question is why $g$ is upper semi continuous

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Let $\epsilon > 0$. Then by the definition of $g$, there is an $\delta > 0$ such that \[ f(x) < g(x_0) + \epsilon, \quad \text{all $x \in B_\delta(x_0)$ (the open ball around $x_0$)} \] Now let $x \in B_\delta(x_0)$ by given, and set $\eta := \delta - |x-x_0|$. Then we have \[ g(x) =\varlimsup_{y\to x} f(y) \le \sup_{y \in B_\eta(x)} f(y) \le g(x_0) + \epsilon. \] As $x \in B_\delta(x_0)$ was arbitrary, we have \[ \varlimsup_{x \to x_0} g(x) \le \sup_{x\in B_\delta(x_0)} g(x) \le g(x_0) + \epsilon. \] As $\epsilon$ was arbitrary, $\varlimsup_{x\to x_0} g(x) \le g(x_0)$.

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    If i set $f(x)=\mathbb I_{x=x_0}(x)*1000 $, does your arguments always work ? Because in this particular case we have : $g(x_0)=0 $ and $f(x) $, for $x $ around $x_0 $, is not less than epsilon.. Someone has an idea ?2016-09-29
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    martini, do you have an answer for @the-owner ?2016-11-06
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    @user46944 I have. the-owner's example is upper semi-continuous, he is wrong in stating that $g(x_0) = 0$. We have $g(x_0) = 1000$, as $g = f$.2016-11-09
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    @the-owner See the above comment2016-11-09
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    @user46944 Because I was little bit confused about the lim sup definition. Take a glance at the wikipedia's [definition](https://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#Functions_from_metric_spaces_to_metric_spaces), **it excludes the point where the limit goes $x_0$** (or $a$. on wiki). But I have seen **another definition that does not exclude the point $x_0$** (check the book of rockafellar that does not exclude the point (check the book of rockafellar 1970 "convex analysis" section 7 page 51 - [here](http://convexoptimization.com/TOOLS/ConvexAnalysis.pdf) ).2016-11-10
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    @user46944 ... In wikipedia's article the limit at $x_0$ could involve problems (of $\infty$) so we exclude it. Whereas In Rockafellar the limit always exists as long as we work on $[-\infty,+\infty]$, so there is no reason for excluding $x_0$ in this framework.2016-11-10
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    So for wikipedia $g(x_0)=0$ and Rockafellar $g(x_0)=1000$.2016-11-10