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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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1 Answer 1

up vote 3 down vote accepted

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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