Prove that $\varliminf_{n \to \infty}\left(\inf_{x\in X} f_n(x)\right) \leq \inf_{x\in X}\left( \varliminf_{n \to \infty} f_n (x)\right)$

Question

We also know that $f_n(x)$ is bounded for every $n\in\mathbb{N}, x\in X$.

Firstly using liminf definition I rewrote

$\varliminf_{n \to \infty} f_n(x) = \lim_{n \to \infty} \inf_{k\geq n} f_k(x)$

And we know that this inequality holds:

$\varliminf_{n \to \infty} f_n(x) = \lim_{n \to \infty} \inf_{k\geq n} f_k(x) \geq \inf_{k\geq n} f_k(x)$

Then i took infinum from both sides:

$\inf_{x\in X}\left( \varliminf_{n \to \infty} f_n (x) \right) \geq \inf_{x\in X}\left( \inf_{k\geq n} f_k(x) \right)$

And finally using monotocity I got:

$\varliminf_{n \to \infty}\left( \inf_{x\in X}\varliminf_{n\to\infty} f_n(x) \right)\geq \varliminf_{n \to \infty}\left( \inf_{k\geq n} \left( \inf_{x\in X} f_k(x) \right) \right)$

And now i want to get rid of the inner liminf from the left side, but I can't figure out how.

Could someone help me out with the proof? I really need this proven for my real analysis course :)

Thanks!

Answer 1

Using perhaps a cleaner notation makes it fall into place:

Since $$\tag{1} \inf_{y\in X}f_n(y)\le f_n(x)\,,\quad \forall x\in X\,, $$ it is trivial that $$\tag{2} \varliminf_{n \to \infty}\Big( \inf_{y\in X}f_n(y)\Big)\le \varliminf_{n \to \infty} f_n(x)\,,\quad \forall x\in X\,. $$ Now take the inf on the RHS and you are done.

Your formula $$\varliminf_{n \to \infty}\left( \inf_{x\in X}\varliminf_{n\to\infty} f_n(x) \right)\geq \varliminf_{n \to \infty}\left( \inf_{k\geq n} \left( \inf_{x\in X} f_k(x) \right) \right)$$ does not make sense because on the LHS you have two limits of $n\to\infty\,.$