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Exercise from book:

Let $\left(f_{k}\right)_{k=1}^{\infty}$ be a sequence of functions and suppose that they are all upper semi-continuous at $x_{0}$. Define the function $g$ by $g(x)=\inf _{1 \leq k<\infty} f_{k}(x)$. Show that $g$ is upper semi-continuous at $x_{0}$.

My attempt: By definition of upper semi continuity we have

$\lvert x-x_0\rvert \Longrightarrow f_{k}(x) < f_{k}(x_0)+\varepsilon$

$ \inf f_{k}(x) < f_{k}(x_0)+\varepsilon$

I think it completes the proof. Is it right or I missing something?????

I am not familiar with topology. so I am interested in only real analysis terms

Daniele Tampieri
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emil agazade
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    To show the upper semi-continuity of $g$ at $x_0$ you have to show that $\forall \varepsilon > 0$, there exists $\delta > 0$ such that $|x - x_0| < \delta$ implies $g(x) < g(x_0) + \varepsilon$. It seems to me that your proof is not complete. – Falcon May 23 '21 at 13:08
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    You have the main idea but some details are missing. – Falcon May 23 '21 at 13:12
  • @Falcon thank you. I will try – emil agazade May 23 '21 at 13:15

1 Answers1

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Let $\varepsilon > 0$ and $g(x) = \inf_{k \in \mathbb N} f_k(x)$. Therefore, there exists $k_0 \in \mathbb N$ with $$f_{k_0}(x_0) - g(x_0) < \frac{\varepsilon}{2}.$$ Moreover, for such a $k_0$, there is $\delta_{k_0}> 0$ with the property that $$|x - x_0| < \delta_{k_{0}} \quad \Rightarrow \quad f_{k_0}(x) < f_{k_0}(x_0) + \frac{\varepsilon}{2}.$$ We deduce that, if $|x - x_0| < \delta_{k_0}$, $$g(x) \le f_{k_0}(x) < f_{k_0}(x_0) + \frac{\varepsilon}{2} < g(x_0) + \varepsilon.$$

Falcon
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