# limit inferior and infimum (real analysis)

I want to prove $$\liminf_{n \to \infty} \left( \inf_{x \in X} f_n(x) \right) \leq \inf_{x \in X} \left( \limsup_{n \to \infty} f_n(x)\right)$$

when $$X \subset \mathbb{R}$$ and $$f_n\colon X \rightarrow \mathbb{R}$$ so that $$\{f_n(x)\>\colon \> n \in \mathbb{N}, x \in X\}$$ is bounded.

I know that should be able to use the property that if $$x_n \leq y_n$$ then $$\limsup_{n} x_n \leq \limsup_{n} y_n$$ and that $$\inf_{n} x_n \leq \liminf_{n} x_n$$.

But how am I supposed finish this proof?

• yes thank you, it's supposed to be bounded – Hannah Apr 2 '19 at 20:08

You have that $$\inf_{x\in X}f_n(x)\le f_n(x)\:\:\:\: \forall n\in \Bbb N\iff \liminf_{n\to\infty}\left( \inf_{x \in X} f_n(x) \right)\le\liminf_{n\to\infty}f_n(x)\:\:\:\:\forall x\in X.$$ But we have also that $$\liminf_{n\to\infty}f_n(x)\le\limsup_{n\to\infty}f_n(x)\quad \forall x\in X.$$ Now, since the infimum of a bounded from below set in $$\Bbb R$$ is the largest of all lower bounds, we get $$\liminf_{n \to \infty} \left( \inf_{x \in X} f_n(x) \right) \leq \inf_{x \in X} \left( \limsup_{n \to \infty} f_n(x)\right),$$ because $$\liminf_{n\to\infty}\big( \inf_{x \in X} f_n(x) \big)$$ is a lower bound for the set $$\left\{\limsup_{n\to\infty}f_n(x)\:\big|\:x\in X\right\},$$ which in turn is bounded and thus bounded from below since $$\{f_n(x)\>\colon \> n \in \mathbb{N}, x \in X\}$$ is.

By properties of the infimum, it suffices to prove that for every $$x\in X$$, we have $$\liminf_n(\inf_{z\in X} f_n(z)) \,\le\, \limsup_n f_n(x)$$ But this easily follows, since $$\inf_z f_n(z) \le f_n(x)$$, hence $$\liminf_n(\inf_z f_n(z)) \,\le\, \liminf_n f_n(x)\,\le\, \limsup_n f_n(x)$$