Proving a result with limit inf

Question

If a function $f: \mathbb R \to \mathbb R$ is continuous in $x$, $f$ is sequentially continuous in $x$: $$ \lim_{n\rightarrow +\infty} x_n = x \Rightarrow \lim_{n\rightarrow +\infty} f(x_n)=f(x)$$

I guess it is not possible in general if one consider $\liminf$ or $\limsup$ instead of $\lim$
$$ \liminf_{n\rightarrow +\infty} x_n = x \Rightarrow \liminf_{n\rightarrow +\infty} f(x_n)=f(x)$$

Also, I guess that if $f$ is also monotone increasing, in addition to continuous, then the result is true with $\liminf$ or $\limsup$. However, I am not able to prove this.

Answer 1

We know that $$\liminf_{n\rightarrow \infty} x_n = x$$ iff for all $\varepsilon>0$, all but finitely many terms of the sequence are more than $x-\varepsilon$ and infinitely many terms are less than $x+\varepsilon$.

Now, if $f$ is an increasing continuous function, fix $\varepsilon>0$. By continuity of $f$,

$$\exists\delta>0:|x-x_n|<\delta\implies|f(x)-f(x_n)|<\varepsilon$$

Consider the sequence $\{f(x_i)\}_{i\in\mathbb N}$. Suppose that infinitely many terms of this sequence are less than $f(x)-\varepsilon$. Then infinitely many terms of original sequence are less than $x-\delta$, which is a contradiction. Similarly, if only finitely many terms of this sequence are less than $f(x)+\varepsilon$, that means all but finitely many terms of this sequence are more than $f(x)+\varepsilon$, which in turn implies that all but finitely many terms of the original sequence are more than $x+\delta$, which is a contradiction again.

Similarly you can prove the limsup part.

EDIT

If $f$ is decreasing, then $-f$ is increasing. Hence, liminf of $\{-f(x_i)\}_{i\in\mathbb N}$ is $-f(x)$. Now, we know that on multiplication by $-1$, image of limsup becomes liminf and vice versa. Hence, limsup of $\{f(x_i)\}_{i\in\mathbb N}$ is $f(x)$. Similarly, you can do for limsup of the sequence.

Hope it helps:)

Answer 2

Attempt:

Let $x_n$ be a bounded sequence.

Want to show that if $f$ is increasing, continuous then

$\liminf \limits_{n \rightarrow \infty} x_n =L$ implies $\liminf \limits _{n \rightarrow \infty} f(x_n)=f(L)$.

Recall: $L:=\liminf \limits _{n \rightarrow \infty} (x_n)$ is a limit point of $(x_n)$, and

$L= \inf ${$H| H$ limit point of $(x_n)$}.

There exists a subsequence $(x_{n_k})$ with

$\lim_{k \rightarrow \infty}x_{n_k} =L$;

$\lim_{k \rightarrow \infty} f(x_{n_k})=f(L)$;

$f(L)$ is a limit point of $f(x_n)$.

Need to show that $f(L) =\liminf \limits _{n \rightarrow \infty }f(x_n)$.

Assume $D$ is a limit point of $f(x_n)$; there exists a subsequence $f(x_{n_l})$ with

$\lim_{l \rightarrow \infty}f(x_{n_l})=D$;

$x_{n_l}$ is bounded, since $x_n$ is bounded (assumption).

Bolzano Weierstrass:

There exists a convergent subsequence $b_s$ of $x_{n_l}$ with limit $M$.

$M$ is a limit point of $x_n$ and

$L \le M$.

Continuity:

$\lim_{s \rightarrow \infty} f(b_s)=f(M)=$

$\lim_{l \rightarrow \infty}f(x_{n_l})=D$;

Since $f$ is increasing:

$f(L)\le f(M)=D$, i e.

$f(L)= \liminf \limits _{ n \rightarrow \infty} f(x_n)$.