1
$\begingroup$

I want to prove that if $f:\mathbb R \rightarrow \mathbb R$ is continuous and $x_n$ a bounded sequence, then $\liminf_{n\rightarrow \infty}f(x_n) < f(\liminf_{n\rightarrow \infty}x_n)$.

Suppose $\liminf_{n\rightarrow \infty}x_n=a$

$$f\text{ continuous }\Rightarrow \forall \varepsilon>0,\ \exists \delta>0,\quad |x_n - a|<\delta \Rightarrow |f(x_n)-f(a)|<\varepsilon. \tag{1}$$

It follows from the definition of limit inferior that there is a sequence $x_{n_k}$ for which $x_{n_k}\rightarrow a$. In other words, $\exists k_0 \in \mathbb N$, such that $\forall k>k_0$, $|x_{n_k}-a|<\delta$. Now, by (1) we get $|x_{n_k} -a|<\delta \Rightarrow |f(x_{n_k} - f(a)|<\varepsilon$, $\forall k>k_0$.

My question is: May I infer from all this that $\liminf_{n\rightarrow \infty}f(x_n) \leq f(\liminf_{n\rightarrow \infty}x_n)$? Why?

$\endgroup$
3
  • $\begingroup$ Please, when asking questions in MSE, follow nettiquette rules(no uppercase letters) and be polite when asking questions. $\endgroup$
    – chubakueno
    Jul 2, 2013 at 2:02
  • $\begingroup$ I think this should say "$\le$" instead of "$<$". $\endgroup$
    – Michael Hardy
    Jul 2, 2013 at 3:11
  • $\begingroup$ Seconding @MichaelHardy's observation... $\endgroup$
    – copper.hat
    Jul 2, 2013 at 15:24

1 Answer 1

Reset to default
1
$\begingroup$

Let $a = \liminf_n x_n$, then there is some subsequence $x_{n_k} \to a$. Since $f$ is continuous, $f(x_{n_k}) \to f(a)$. Hence $\liminf_n f(x_n) \le \liminf_k f(x_{n_k}) = \lim_k f(x_{n_k}) = f(a)$.

Note: The only tricky part here is showing $\liminf_n f(x_n) \le \liminf_k f(x_{n_k})$. Let $I = \{ n_k \}$. Then we have $\inf_{k \ge n} f(x_k) \le \inf_{k \ge n, k \in I} f(x_k)$, and since both sides are non-decreasing, we have $\inf_{k \ge n} f(x_k) \le \lim_n \inf_{k \ge n, k \in I} f(x_k)$ followed by $\lim_n \inf_{k \ge n} f(x_k) \le \lim_n \inf_{k \ge n, k \in I} f(x_k)$. Hence

$$\liminf_n f(x_n) = \lim_n \inf_{k \ge n} f(x_k) \le \lim_n \inf_{k \ge n, k \in I} f(x_k) = \liminf_k f(x_{n_k})$$

$\endgroup$
6
  • $\begingroup$ If one knows the fact that $\liminf x_n$ is the inf of all subsequential limits, then your tricky part is immediate. $\endgroup$
    – Ted Shifrin
    Jul 2, 2013 at 3:39
  • 1
    $\begingroup$ @TedShifrin: That is a cute characterization! $\endgroup$
    – copper.hat
    Jul 2, 2013 at 3:52
  • $\begingroup$ Good morning, Copper.hat. Ok, but I want to prove other thing: $\liminf_{n\rightarrow \infty}f(x_n) < f(\liminf_{n\rightarrow \infty}x_n)$ $\endgroup$
    – Walter r
    Jul 2, 2013 at 10:18
  • $\begingroup$ The inequality cannot be strict, for example take $x_n =1$ and $f(x) =x$. At best it is $\le$, and this is what I proved above, that is, $\liminf_n f(x_n) \le f(\liminf_n x_n)$. (Note that in the above $a=\liminf_n x_n$.) $\endgroup$
    – copper.hat
    Jul 2, 2013 at 15:23
  • $\begingroup$ You've convinced me. But I think the inequality is strict. Take $f(x)=-x$ and $x_n=(-1)^n$, then $lim\ inf f(x_n)= -1$ and $f(lim\ inf x_n)=1$ $\endgroup$
    – Walter r
    Jul 3, 2013 at 1:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .