Limit Superior and Limit Inferior

# Limit Superior and Limit Inferior

 Definition: If $(a_n)$ is a bounded sequence, then the limit superior of $(a_n)$ is a real number $x^*$ denoted $\limsup_{n \to \infty} a_n = x^*$ such that $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $a_n < x^* + \epsilon$ and there are infinitely many terms of $a_n$ in $V_{\epsilon}(x^*)$. Similarly, the limit inferior is a real number $x_*$ denoted $\liminf a_n = x_*$ such that $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $x_* - \epsilon < a_n$ and there are infinitely many terms of $a_n$ in $V_{\epsilon} (x_*)$.

Let's first look at the limit superior of a sequence:

From the definition of the limit superior of a bounded sequence, then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $a_n < x^* + \epsilon$. Therefore, given some positive $\epsilon$ we can find a natural number $N$ such that all successive terms $a_n$ are less than $x^* + \epsilon$. Therefore, for finite first few terms up until $N$ it is possible that $x^* + \epsilon < a_n$ but since there are only a finite number of terms for which this can happen, it follows that there are only a finite number of terms $a_n$ such that $x^* + \epsilon < a_n$ for any $\epsilon > 0$.

The limit superior of a sequence is analogous. For all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $x_* -\epsilon < a_n$. Therefore, given some positive $\epsilon$ we can find a natural number $N$ such that all successive terms $a_n$ are greater than $x_* - \epsilon$. Therefore, for the finite first fewterms up until $N$ it is possible that $a_n < x_* - \epsilon$ but since there are only a finite number of terms for which this can happen, it follows that there are only a finite number of terms $a_n$ such that $a_n < x_* - \epsilon$.

 Theorem 1: Let $(x_n)$ be a bounded sequence. The following statements are equivalent: a) $x^* = \limsup x_n$. b) If $A = \{ a : \mathrm{a \: is \: an \: accumulation \: point \: of \: (x_n) .} \}$ then $x^* = \sup A$. c) If $B = \{ x : x < x_n \: \mathrm{for \: at \: most \: finitely \: many \: n \in \mathbb{N}.} \}$ then $x^* = \inf B$.
• Proof: Let $(x_n)$ be a bounded sequence of real numbers.
• $a) \implies b)$ Let $x^* = \limsup x_n$, and let $s = \sup A$. We ultimately want to show that $x^* = s$. First note that $x^*$ is an accumulation point of the sequence $(x_n)$ since $\forall \epsilon > 0$, by the definition of the limit superior, $V_{\epsilon} (x^*)$ contains infinitely many terms of $(x_n)$ and eventually for some $N \in \mathbb{N}$ if $n ≥ N$ then all successive terms $x_n$ are contained in $V_{\epsilon} (x^*)$. Therefore $x^* \in A$ and so $x^* ≤ s = \sup A$ by the definition that $s$ is the supremum of $A$.
• Now we will show that it is not possible that $x^* < s$ which will force $x^* = s$.
• Suppose that $x^* < s$. Then it follows that $s - x^* > 0$ and so $\frac{s - x^*}{2} > 0$. Let $\epsilon = \frac{s - x^*}{2}$.
• Now since $s = \sup A$ (as a reminder $A$ is the set of accumulation points of the sequence $(x_n)$), then it follows that since $\frac{s - x^*}{2} < s$ that there exists an accumulation point $a \in A$ such that $\frac{s - x^*}{2} < a ≤ s$. By the definition of an accumulation point of a sequence $(x_n)$ there exists a subsequence of $(x_n)$, call it $(x_{n_k})$ such that $\forall \epsilon > 0$ and for all $N \in \mathbb{N}$ there exists an $n ≥ N$ such that $x_n$ is in $V_{\epsilon}{a}$. Let $\epsilon_1 = \mathrm{min} \{ a - \frac{s - x^*}{2}, s - a \}$. Then there exists infinitely many terms of $(x_n)$ in $V_{\epsilon_{1}} (a)$.
• But this is a contradiction to the fact that $x^* = \limsup x_n$ since then there exists infinitely many terms to the right of $x^* + \epsilon$, in other words, there does not exist an $N \in \mathbb{N}$ such that $\forall n ≥ N$ then $x_n < x^* + \epsilon$. Thus it cannot be that $x^* < s$ and so $x^* = s = \sup A$.
• $b) \implies c)$. Consider the set $B = \{ x : x < x_n \: \mathrm{for \: at \: most \: finitely \: many \: n \in \mathbb{N}.}$. Notice that for all $\epsilon > 0$ we have that $(x^* + \epsilon) \in B$ since there are only a finite number of terms $x_n$ such that $x^* + \epsilon < x_n$. Thus there are an infinite number of terms $x_n$ such that $x^* - \epsilon < x_n < x^*$, and thus $\forall \epsilon > 0$, $(x^* - \epsilon) \not \in B$. Since $x^* = \sup A$ it follows that then $x^* = \sup A = \inf B$. $\blacksquare$
 Theorem 2: A bounded sequence of real numbers $(a_n)$ is convergent to $L$ if and only if $\limsup_{n \to \infty} a_n = L = \liminf_{n \to \infty} a_n$.
• Proof 1: $\Rightarrow$ Recall that if $(a_n)$ converges that it can have at most one accumulation point. Therefore the set $A = \{ a : a \: is \: an \: accumulation \: point \: of \: (a_n) \} = \{ L \}$. Now $\sup A = L$, and so by theorem 1, $L = \sup A = \limsup_{n \to \infty} a_n$. Similarly, $L = \inf A = \liminf_{n \to \infty} a_n$ and so $\limsup_{n \to \infty} = L = \liminf_{n \to \infty} a_n$.
• $\Leftarrow$. Suppose that $x^* = \limsup_{n \to \infty} a_n$ and $x_* = \liminf_{n \to \infty} a_n$ and $x^* = L = x_*$. Then by the definition of a limit superior for $\epsilon$, $\exists N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $a_n < x^* + \epsilon = L + \epsilon$ and there are infinitely many terms of $a_n$ in $V_{\epsilon} (L)$. Similarly by the definition of a limit inferior, for $\epsilon$, $\exists N_2 \in \mathbb{N}$ such that if $n ≥ N_2$ then $L - \epsilon = x_* - \epsilon < a_n$ and there are infinitely many terms of $a_n$ in $V_{\epsilon} (L)$. Let $N = \mathrm{max} \{ N_1, N_2 \}$ and so for $n ≥ N$ we have that $L - \epsilon < a_n < L + \epsilon$ or equivalently $\mid a_n - L \mid < \epsilon$ and so $\lim_{n \to \infty} a_n = L$. $\blacksquare$