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)

\begin{align} \quad \limsup_{n \to \infty} a_n = \lim_{n \to \infty} \sup_{k \geq n} \{ a_k \} = \inf_{n \geq 1} \left \{ \sup_{k \geq n} \{ a_k \} \right \} \end{align}

(2)

\begin{align} \quad \liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf_{k \geq n} \{ a_k \} = \sup_{n \geq 1} \left \{ \inf_{k \geq n} \{ a_k \} \right \} \end{align}

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:

(3)

\begin{align} \quad L - \epsilon < f(x) < L + \epsilon \end{align}

• 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:

(4)

\begin{align} \quad L - \epsilon \leq \inf_{t \geq x} \{ f(t) \} \leq \sup_{t \geq x} \{ f(t) \} \leq L + \epsilon \quad \Leftrightarrow \quad \biggr \lvert \inf_{t \geq x} \{ f(t) \} - L \biggr \rvert < \epsilon \quad , \quad \biggr \lvert \sup_{x \geq t} \{ f(t) \} - L \biggr \rvert < \epsilon \end{align}

• 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:

(5)

\begin{align} \quad \biggr \lvert \sup_{t \geq x} \{ f(t) \} - L \biggr \rvert < \epsilon \quad \Leftrightarrow \quad L -\epsilon < \sup_{t \geq x} \{ f(t) \} < L + \epsilon \quad (*) \end{align}

• 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:

(6)

\begin{align} \quad \biggr \lvert \inf_{t \geq x} \{ f(t) \} - L \biggr \rvert < \epsilon \quad \Leftrightarrow \quad L - \epsilon < \inf_{t \geq x} \{ f(t) \} < L + \epsilon \quad (**) \end{align}

• 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:

(7)

\begin{align} \quad L - \epsilon \leq \inf_{t \geq x} \{ f(t) \} \leq f(x) \leq \sup_{t \geq x} \{ f(t) \} \leq L + \epsilon \quad \Leftrightarrow \quad \biggr \lvert f(x) - L \biggr \rvert < \epsilon \end{align}

• Hence $\displaystyle{\lim_{x \to \infty} f(x) = L}$. $\blacksquare$