Superior and inferior limits with continuous functions

Question

Given two continuous functions $f$ and $g$, from $\mathbb R$ to $\mathbb R$ and with $f \leqslant g$. If we have a sequence $a_n$ such that it has a limit $a$. Is it true that lim sup $(f(a_n)$, $g(a_n))$ = lim inf $(f(a_n), g(a_n))$ = $(f(a),g(a))$ ? Is there a simple counterexample to prove that this is false?

Answer 1

Consider $f(x) = x$ and $g(x) = 1 - x$. Define $a_n = (-1/2)^n$ for all $n \geq 1$. Then $a_n \to 0$ and $f(x) \leq g(x)$ for all $x\leq 1/2$ (we don't care what happens if $x > 1/2$).

Now, $$f(a_n) = \left(-\frac{1}{2}\right)^{n}, \quad g(a_n) = 1 - \left(-\frac{1}{2}\right)^{n}$$ Define $I_n = (f(a_n), g(a_n))$. Now $\limsup I_n$ is the set of all real numbers which are contained in infinitely many $I_n$, and $\liminf I_n$ is the set of all real numbers which are contained in all but finitely many $I_n$.

Let us look at the first few $I_n$ to see the pattern:

$$I_1 = \left(-\frac{1}{2}, \frac{3}{2}\right), \quad I_2 = \left( \frac{1}{4}, \frac{3}{4} \right), \quad I_3 = \left( -\frac{1}{8}, \frac{9}{8} \right), \quad I_4 = \left( \frac{1}{16}, \frac{15}{16} \right), \quad \ldots $$ It is clear that if $\alpha \in (0,1)$, then $\alpha$ is in all except finitely many $I_n$, whereas $\alpha=0$ and $\alpha=1$ are contained in $I_n$ if and only if $n$ is odd. If $\alpha < 0$ or $\alpha > 1$ then $\alpha$ is contained in at most finitely many $I_n$.

This means that $\limsup I_n = [0,1]$ and $\liminf I_n = (0,1)$, so this is a counterexample to the claim.