Prove $\overline{\lim\limits_{n \to \infty}} \Big(\inf f_n(x) \Big) \leq \inf \Big( \overline{\lim\limits_{n \to \infty}} f_n(x) \Big)$

Question

I have proven this problem, but I am not sure if it is correct, especially the last inequality from the set being bounded. Would appreciate tips!

Let $X \subset \mathbb{R}$ and functions $f_n: X \xrightarrow{} \mathbb{R}, n\in \mathbb{N}$, like so the set $\{f_n(x): n\in \mathbb{N}, x\in X\}$ is bounded.

Firstly we know from the definition of infimum that

$$ \underset{x \in X}{\text{inf}}f_n(x) \leq f_n(x) $$

From the limit's monotonicity we get

$$ \overline{\lim\limits_{n \to \infty}} \Big( \underset{x \in X}{\text{inf}}f_n(x) \Big) \leq \overline{\lim\limits_{n \to \infty}} f_n(x) $$

Since we know that the set is bounded, then the set has minima and maxima. Therefore infimum equals minimum, which means, that infimum is a function value. So if the inequality works for all $x$, then the right side's greatest lower bound must also be larger or equal than the left side. So we have proven that

$$ \overline{\lim\limits_{n \to \infty}} \Big( \underset{x \in X}{\text{inf}}f_n(x) \Big) \leq \underset{x \in X}{\text{inf}}\Big( \overline{\lim\limits_{n \to \infty}} f_n(x) \Big) $$

Thank you

Answer 1

Bounded subsets of $\Bbb R$ have least upper bounds and greatest lower bounds, but they need not have a maximum or a minimum. For instance, I don't see what you propose the minimum of $\{f_n(x)\,:\, n\in\Bbb N, x\in X\}$ to be for $f_n(x)=\arctan\frac1x$ and $X=(-1,0)$.

I propose:

\begin{align}\limsup_{n\to\infty}\inf_{x\in X}f_n(x)&=\inf_{n\in\Bbb N}\sup_{k\ge n}\inf_{x\in X}f_k(x)\\\inf_{x\in X}\limsup_{n\to\infty}f_n(x)&=\inf_{x\in X}\inf_{n\in\Bbb N}\sup_{k\ge n}f_k(x)\end{align}

For all $x$ and $k$, $$\inf_{t\in X}f_k(t)\le f_k(x)$$

Therefore, for all $x$ and $n$, $$\sup_{k\ge n}\inf_{t\in X}f_k(t)\le \sup_{k\ge n}f_k(x)$$

Therefore, for all $x$, $$\inf_{n\in\Bbb N}\sup_{k\ge n}\inf_{t\in X}f_k(t)\le \inf_{n\in\Bbb N}\sup_{k\ge n}f_k(x)$$

Therefore, $$\inf_{n\in\Bbb N}\sup_{k\ge n}\inf_{t\in X}f_k(t)\le \inf_{x\in X}\inf_{n\in\Bbb N}\sup_{k\ge n}f_k(x)$$

This holds verbatim under the more general convention that $\sup$, $\inf$, $\limsup$ and $\liminf$ are defined as extended real numbers.