Changing the order of $\lim$ and $\inf$ for point-wise converging monotonic sequence of functions

Question

Suppose $f_n: X \rightarrow \mathbb{R}$, where $X$ is some arbitrary subset of $\mathbb{R}^N$. Suppose that $$ \forall n\geq0, \forall x \in X, \; f_n(x) \leq f_{n+1}(x) $$ Let $\{f_n\}$ be such that each $f_n(x)$ is continuous and $\forall x,\; \lim_{n\rightarrow \infty}f_n(x) = f(x)$ exists, but is discontinuous. Hence, $\{f_n\}$ is point-wise convergent but not uniformly-convergent. My question is:

• Is there any such sequence for which $$ \lim_{n\rightarrow \infty} \inf_{x\in X} f_n(x) = \inf_{x\in X} \lim_{n\rightarrow \infty} f_n(x) $$ (I know this is true for uniform convergence.)

• If so, what is the best approach to prove it for that sequence?

Any solution or reference would be greatly appreciated. There have been similar questions before, like this one, however, I couldn't really relate to the answer. There's another one here.

Answer 1

It is not true in general. Let $g : \mathbb R \to \mathbb R$ such that $g(x) = 0$ if $x\leq 0$ and $g(x)=1 $ for $x\geq 1$ and increasing. Let $f_n(x) = g(x+n)$. Then $f_n$ is an increasing sequence converging to $f=1$. So the left hand side of the limit is $0$, while the right hand side is $1$.