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

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.

• A counter example is $f_n(x)=1-1/n$ for $0< x<1/n$ and $f_n(x)=2$ for $x>1/n$. Then $\inf\lim f_n = 2$ while $\lim\inf f_n = 1$ Oct 15, 2013 at 6:55
• Btw, the above example can be modified to work on any $X$ having at least two points. The case $X=\{a\}$ for some real $a$ is fine. Oct 15, 2013 at 7:07
• Thanks, however, I wanted to know if there is some sequence of point-wise converging functions that does satisfy this. Can we create an example for that? Oct 15, 2013 at 7:16
• An example when it works is easy. By the way, the sentence starting with "Hence" is wrong, you might change that to "Hence $f_n$ converges, but we do not know if the convergence is uniform.". Oct 15, 2013 at 7:40
• Thanks for pointing out! I actually forgot to state that $\{f_n\}$ are continuous and point-wise convergent, and $f(x)$ is discontinuous. This should imply that $\{f_n\}$ is not uniformly convergent (right?). Also, unfortunately, I'm unable to figure out an example as you said (silly me!), could you please give me one when it is satisfied? Oct 15, 2013 at 8:15

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 $1$.
• Thanks, but earlier I had forgotten to mention that the $\{f_n(x)\}$ are continuous functions while $f(x)$ is discontinuous. I have modified the questions, but I guess there is a counter-example for that too. Oct 15, 2013 at 22:01