82. The limits of indetermination of a bounded function

Suppose that $ϕ (n)$ is a bounded function, and $M$ and $m$ its upper and lower bounds. Let us take any real number $ξ$ , and consider now the relations of inequality which may hold between $ξ$ and the values assumed by $ϕ (n)$ for large values of $n$ . There are three mutually exclusive possibilities:

(1) $ξ \geq ϕ (n)$ for all sufficiently large values of $n$ ;

(2) $ξ \leq ϕ (n)$ for all sufficiently large values of $n$ ;

(3) $ξ < ϕ (n)$ for an infinity of values of $n$ , and also $ξ > ϕ (n)$ for an infinity of values of $n$ .

In case (1) we shall say that $ξ$ is a superior number, in case (2) that it is an inferior number, and in case (3) that it is an intermediate number. It is plain that no superior number can be less than $m$ , and no inferior number greater than $M$ .

Let us consider the aggregate of all superior numbers. It is bounded below, since none of its members are less than $m$ , and has therefore a lower bound, which we shall denote by $Λ$ . Similarly the aggregate of inferior numbers has an upper bound, which we denote by $λ$ .

We call $Λ$ and $λ$ respectively the upper and lower limits of indetermination of $ϕ (n)$ as $n$ tends to infinity; and write $Λ = lim^{―} ϕ (n), λ = lim_{―} ϕ (n) .$ These numbers have the following properties:

(1) $m \leq λ \leq Λ \leq M$ ;

(2) $Λ$ and $λ$ are the upper and lower bounds of the aggregate of intermediate numbers, if any such exist;

(3) if $ϵ$ is any positive number, then $ϕ (n) < Λ + ϵ$ for all sufficiently large values of $n$ , and $ϕ (n) > Λ - ϵ$ for an infinity of values of $n$ ;

(4) similarly $ϕ (n) > λ - ϵ$ for all sufficiently large values of $n$ , and $ϕ (n) < λ + ϵ$ for an infinity of values of $n$ ;

(5) the necessary and sufficient condition that $ϕ (n)$ should tend to a limit is that $Λ = λ$ , and in this case the limit is $l$ , the common value of $λ$ and $Λ$ .

Of these properties, (1) is an immediate consequence of the definitions; and we can prove (2) as follows. If $Λ = λ = l$ , there can be at most one intermediate number, viz. $l$ , and there is nothing to prove. Suppose then that $Λ > λ$ . Any intermediate number $ξ$ is less than any superior and greater than any inferior number, so that $λ \leq ξ \leq Λ$ . But if $λ < ξ < Λ$ then $ξ$ must be intermediate, since it is plainly neither superior nor inferior. Hence there are intermediate numbers as near as we please to either $λ$ or $Λ$ .

To prove (3) we observe that $Λ + ϵ$ is superior and $Λ - ϵ$ intermediate or inferior. The result is then an immediate consequence of the definitions; and the proof of (4) is substantially the same.

Finally (5) may be proved as follows. If $Λ = λ = l$ , then $l - ϵ < ϕ (n) < l + ϵ$ for every positive value of $ϵ$ and all sufficiently large values of $n$ , so that $ϕ (n) \to l$ . Conversely, if $ϕ (n) \to l$ , then the inequalities above written hold for all sufficiently large values of $n$ . Hence $l - ϵ$ is inferior and $l + ϵ$ superior, so that $λ \geq l - ϵ, Λ \leq l + ϵ,$ and therefore $Λ - λ \leq 2 ϵ$ . As $Λ - λ \geq 0$ , this can only be true if $Λ = λ$ .

Example XXXII

1. Neither $Λ$ nor $λ$ is affected by any alteration in any finite number of values of $ϕ (n)$ .

2. If $ϕ (n) = a$ for all values of $n$ , then $m = λ = Λ = M = a$ .

3. If $ϕ (n) = 1 / n$ , then $m = λ = Λ = 0$ and $M = 1$ .

4. If $ϕ (n) = (- 1)^{n}$ , then $m = λ = - 1$ and $Λ = M = 1$ .

5. If $ϕ (n) = (- 1)^{n} / n$ , then $m = - 1$ , $λ = Λ = 0$ , $M = \frac{1}{2}$ .

6. If $ϕ (n) = (- 1)^{n} {1 + (1 / n)}$ , then $m = - 2$ , $λ = - 1$ , $Λ = 1$ , $M = \frac{3}{2}$ .

7. Let $ϕ (n) = \sin n θ π$ , where $θ > 0$ . If $θ$ is an integer then $m = λ = Λ = M = 0$ . If $θ$ is rational but not integral a variety of cases arise. Suppose, , that $θ = p / q$ , $p$ and $q$ being positive, odd, and prime to one another, and $q > 1$ . Then $ϕ (n)$ assumes the cyclical sequence of values $\sin (p π / q), \sin (2 p π / q), \dots, \sin {(2 q - 1) p π / q}, \sin (2 q p π / q), \dots .$ It is easily verified that the numerically greatest and least values of $ϕ (n)$ are $\cos (π / 2 q)$ and $- \cos (π / 2 q)$ , so that $m = λ = - \cos (π / 2 q), Λ = M = \cos (π / 2 q) .$ The reader may discuss similarly the cases which arise when $p$ and $q$ are not both odd.

The case in which $θ$ is irrational is more difficult: it may be shown that in this case $m = λ = - 1$ and $Λ = M = 1$ . It may also be shown that the values of $ϕ (n)$ are scattered all over the interval $[- 1, 1]$ in such a way that, if $ξ$ is any number of the interval, then there is a sequence $n_{1}$ , $n_{2}$ , … such that $ϕ (n_{k}) \to ξ$ as $k \to \infty$ .¹

The results are very similar when $ϕ (n)$ is the fractional part of $n θ$ .

A number of simple proofs of this result are given by Hardy and Littlewood, “Some Problems of Diophantine Approximation”, Acta Mathematica, vol. xxxvii.↩︎

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81. The bounds of a bounded function

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83-84 The general principle of convergence

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