The Limit Superior and Limit Inferior of Functions of Real Numbers
Recall from The Limit Superior and Limit Inferior of Sequences of Real Numbers that if $(a_n)_{n=1}^{\infty}$ is a sequence of real numbers then we defined:
(1)We now define the limit superior and limit inferior as $x \to \infty$ for a function $f : (a, \infty) \to \mathbb{R}$ where $a \in \mathbb{R}$:
Definition: Let $f : (a, \infty) \to \mathbb{R}$. The Limit Superior as $x \to \infty$ of $f$ is defined as $\displaystyle{\limsup_{x \to \infty} f(x) = \lim_{x \to \infty} \sup_{t \geq x} \{ f(t) \} = \inf_{x \geq a} \left \{ \sup_{t \geq x} \{ f(t) \} \right \}}$. The Limit Inferior as $x \to \infty$ of $f$ is defined as $\displaystyle{\liminf_{x \to \infty} f(x) = \lim_{x \to \infty} \inf_{t \geq x} \{ f(t) \} = \inf_{x \geq a} \left \{ \sup_{t \geq x} \{ f(t) \} \right \} }$. |
We can similarly define the limit superior and limit inferior as $x \to -\infty$ for a function $f : (-\infty, b) \to \mathbb{R}$ where $b \in \mathbb{R}$:
Definition: Let $f : (-\infty, b) \to \mathbb{R}$. The Limit Superior as $x \to -\infty$ of $f$ is defined as $\displaystyle{\limsup_{x \to -\infty} f(x) = \lim_{x \to -\infty} \sup_{t \leq x} \{ f(t) \} = \inf_{x \leq b} \left \{ \sup_{t \leq x} \{ f(t) \} \right \}}$. The Limit Inferior as $x \to \infty$ of $f$ is defined as $\displaystyle{\liminf_{x \to -\infty} f(x) = \lim_{x \to -\infty} \inf_{t \leq x} \{ f(t) \} = \inf_{x \leq b} \left \{ \sup_{t \leq x} \{ f(t) \} \right \} }$. |
We now prove some fundamental results regarding the limit superior and limit inferior as $x \to \infty$ of a function a function $f : (a, \infty) \to \mathbb{R}$. Analogous results can be proven for the limit superior and limit inferior as $x \to -\infty$ of a function $f : (-\infty, b) \to \mathbb{R}$.
Theorem 1: Let $f : (a, \infty) \to \mathbb{R}$ where $a \in \mathbb{R}$. Then $\displaystyle{\lim_{x \to \infty} f(x) = L}$ if and only if $\displaystyle{\limsup_{x \to \infty} f(x) = L = \liminf_{x \to \infty} f(x)}$. |
- Proof: $\Rightarrow$ Suppose that $\displaystyle{\lim_{x \to \infty} f(x) = L}$. Then for all $\epsilon > 0$ there exists an $M \in \mathbb{R}$, $M \geq a$ such that if $x \geq M$ then $\: f(x) - L \: < \epsilon$. So if $x \geq M$ then:
- So for $x \geq M$ we have that $L + \epsilon$ is an upper bound for $f$ and $L - \epsilon$ is a lower bound for $f$. Thus for $x \geq M$ we have that:
- Hence $\displaystyle{\liminf_{x \to \infty} f(x) = \lim_{x \to \infty} \inf_{t \geq x} \{ f(t) \} = L}$ and $\displaystyle{\limsup_{x \to \infty} f(x) = \lim_{x \to \infty} \sup_{t \geq x} \{ f(t) \} = L}$. $\blacksquare$
- $\Leftarrow$ Suppose that $\displaystyle{\limsup_{x \to \infty} f(x) = L = \liminf_{x \to \infty} f(x)}$. Let $\epsilon > 0$ be given.
- Since $\displaystyle{\limsup_{x \to \infty} f(x) = L}$ we have that there exists an $M_1 \in \mathbb{R}$, $M_1 \geq a$ such that if $x \geq M_1$ then:
- Also, since $\displaystyle{\liminf_{x \to \infty} f(x) = L}$ we have that there exists an $M_2 \in \mathbb{R}$, $M_2 \geq a$ such that if $x \geq M_2$ then:
- Let $M = \max \{ M_1, M_2 \}$. Clearly $M \geq a$ since $M_1, M_2 \geq a$. Then if $x \geq M$ we have that $(*)$ and $(**)$ hold so:
- Hence $\displaystyle{\lim_{x \to \infty} f(x) = L}$. $\blacksquare$