I want to show that if $\{ f_n \}_{n\in\mathbb{N}}$ is a sequence of upper semi-continuous functions defined, for example, on a set $G\subset\mathbb{R}$ which converges uniformly on $G$ to a function $f:G\to\mathbb{R}$, then $f$ is upper semi-continuous.

My attempt: From the upper semi-continuity of $f_n$ we know that, for every $n\in\mathbb{N}$, $$\limsup_{x\to x_0}f_n(x)\leq f_n(x_0), \ \forall x\in G.$$ So, if we could interchange the limits, by reason of uniform convergence for example, we would have $$\limsup_{x\to x_0}f(x)\leq f(x_0), \ \forall x\in G$$ and then $f$ would be upper semi-continuous.

The problem is that I cannot prove the limits' interchange. I also think that I can show that $$\limsup_{x\to x_0}f_n(x)\leq \limsup_{x\to x_0}f(x), \ \forall n\in\mathbb{N},$$ but I cannot use it to show that $$\lim_{n\to+\infty}\big(\limsup_{x\to x_0}f_n(x)\big)=\limsup_{x\to x_0}f(x).$$ Any help it would be appreciated.

EDIT: I(idonknow) would like to know a proof of the above statement by using any equivalent definition.

  • 1
    $\begingroup$ It's easier if you use a different definition/characterisation of semicontinuity. A function $g$ is upper semicontinuous at $x_0$ if and only for every $c > g(x_0)$ there is a neighbourhood $U$ of $x_0$ such that $g(x) < c$ for all $x \in U$. $\endgroup$ – Daniel Fischer Nov 23 '16 at 19:11

Let $x_0\in G$, $\alpha > f(x_0)$, and $\varepsilon := \frac{\alpha-f(x_0)}{3}$. Since $(f_n)$ is uniformly converging to $f$, there exists $N\in\mathbb{N}$ such that $|f_N(x) - f(x)| < \varepsilon$ for every $x\in G$. Since $f_N$ is u.s.c., there exists $r>0$ such that $f_N(x) < f_N(x_0) + \varepsilon$ for every $x \in B_r(x_0) \cap G$.

Hence, for every $x\in B_r(x_0) \cap G$, $$ f(x) < f_N(x) + \varepsilon < f_N(x_0) + 2\varepsilon < f(x_0) + 3\varepsilon= \alpha, $$ so that $f$ is u.s.c. at $x_0$.

  • $\begingroup$ You should correct the last equality. $\endgroup$ – Peter Elias May 8 '17 at 19:09
  • $\begingroup$ @PeterElias: Thank you. $\endgroup$ – Rigel May 8 '17 at 19:15

It suffices to show the following:

For every $\varepsilon>0$, there exists a $\delta>0$, such that $\sup_{|x-x_0|<\delta}f(x)\le f(x_0)+\varepsilon$

Fix $\varepsilon>0$. Uniform convergence of $\{f_n\}$ guarantees the existence of an $n_0\in\mathbb N$, such that $$ n\ge n_0\quad\Longrightarrow\quad |\,f_n(x)-f(x)|<\frac{\varepsilon}{3},\quad \text{for all $x\in G$}, $$ or equivalently $$ n\ge n_0\quad\Longrightarrow\quad f(x)-\frac{\varepsilon}{3}<f_n(x)<f(x)+\frac{\varepsilon}{3},\quad \text{for all $x\in G$}, $$

For the same $n_0$, the upper semi-continuity of $f_{n_0}$ guarantees the existence of a $\delta>0$, such that $\sup_{|x-x_0|<\delta}f_{n_0}(x)\le f(x_0)+\dfrac{\varepsilon}{3}$.

Altogether, we now have $$ \sup_{|x-x_0|<\delta}f(x)\le \sup_{|x-x_0|<\delta}f_{n_0}(x)+\frac{\varepsilon}{3}\le f_{n_0}(x_0)+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}\le f(x_0)+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}= f(x_0)+\varepsilon. $$

  • $\begingroup$ sorry, the upper semi-continuity of $f_{n_0}$ guarantees the existence of a $\delta >0$ such that $\sup_{|x-x_0|<\delta}f_{n_0}(x) \le f_{n_0}(x_0)+\varepsilon/3$ right? $\endgroup$ – mate89 May 30 at 7:28

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