How to prove that $\limsup _{n \rightarrow \infty} f(x_n) \geq f(\limsup _{n \rightarrow \infty} (x_n))$

Question

We were given HW where it is asked to prove that, $\limsup_{n \rightarrow\infty} f(x_n) \geq f(\limsup_{n \rightarrow \infty} (x_n))$, for any bounded sequence $\{x_n\}$ and continuous function $f\colon\mathbb{R} \rightarrow \mathbb{R}$.

Could anyone give me a direction?

Thank you!

Answer 1

you can start by showing that if $x$ is an accumulation point of $\{x_n\}$, then, $f(x)$ is an accumulation point of $\{f(x_n) \}$.

Answer 2

You can start by showing contradiction, let us denote $f^* = \limsup f(x_n)$ and $s^* = \limsup (x_n)$, and $f(s^*) = f(\limsup(x_n))$

now $f^*$ is the supremum of subsequential limit of $\{f(x_n)\}$ and $f(s^*)$ belongs to the set of subsequential limit of $\{f(x_n)\}$,( since f is continuous and corresponds to the subsequence of $\{x_n\}$ which converges to $s^*$ )

hence $f^* \ge f(s^*)$ (otherwise $f^*$ is not a supremum)

Answer 3

1. Show that if $\limsup\limits_{n\to\infty} x_n=x$ then there is a subsequence $(x_{n_k})$ such that $\lim\limits_{k\to\infty} x_{n_k} = x$.

2. What can you say about $\lim\limits_{k\to\infty} f(x_{n_k})$ based on the continuity of $f$?

3. Compare $\lim\limits_{k\to\infty} f(x_{n_k})$ and $\limsup\limits_{n\to\infty} f(x_{n})$.

Answer 4

We are given two inputs: $x_{n}$ is a bounded sequence which means that $\limsup x_{n} = a$ is finite and the fact that $f: \Bbb R \to \Bbb R$ is continuous. Since $x_{n}$ is bounded and $f$ is continuous the sequence $f(x_{n})$ is also bounded and hence $\limsup f(x_{n}) = A$ exists and is finite. We need to prove that $A \geq f(a)$.

We establish this by contradiction. Let us assume $A < f(a)$ and see where we are led to. Also let $2\epsilon = f(a) - A > 0$. Since $f$ is continuous at $a$ we know that there is a $\delta > 0$ such that $|f(x) - f(a)| < \epsilon$ whenever $|x - a| < \delta$. Since $a = \limsup x_{n}$ we know that $x_{n} < a + \delta$ for all sufficiently large values of $n$ and $x_{n} > a - \delta$ for infinitely many values of $n$. Hence there are infinitely many values of $x_{n}$ such that $|x_{n} - a| < \delta$ and hence for these values of $x_{n}$ we have $|f(x_{n}) - f(a)| < \epsilon$. Thus there are infinitely many values of $x_{n}$ for which $$f(a) - \epsilon < f(x_{n}) < f(a) + \epsilon$$ Now consider the fact that $A = \limsup f(x_{n})$ which implies that "$f(x_{n}) < A + \epsilon = f(a) - \epsilon$ for all sufficiently large values of $n$" and this clearly contradicts the previous inequality "$f(a) - \epsilon < f(x_{n})$ for infinitely many values of $n$"

We have the reached the contradiction needed for our goal. This means that $A \geq f(a)$.

Update: By the way this is a very good question (at least for me). I was aware of the fact that "continuity preserves limit operation" but never bothered to check how continuity impacts the limit superior and limit inferior. The proof above has enlightened me that continuity does not preserve $\limsup$ and $\liminf$ but rather gives them a different twist. +1 for OP.