The Limit Superior and Limit Inferior of Functions of Real Numbers

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