F. Equicontinuity Thm

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Equicontinuity

A function f(x) is said to be equicontinuous on an interval [a,b] if and only if for each e > 0, there exist at least one d > 0 such that

|f(x₂)-f(x₁)| < e

whenever x₁ and x₂ are both in the interval [a,b] and |x₁-x₂| < d.

Thereom F.5 [Equicontinuity Theorem] Suppose b > a. Suppose f(x) is continuous at each point in the interval [a,b]. Then for every e > 0 there exists a d > 0 such that for every pair of points x₁ and x₂ in the interval [a,b],

|x₂-x₁| < d implies |f(x₂)-f(x₁)| < e

The proof given below holds in the case of a function of one variable. It can be generalized to provide the demonstration of a theorem on the equicontinuity of functions of several variables.

Proof of Equicontinuity Theorem.

|f(x₁)-f(x₂)|

|f(x₁)-f(x)+[f(x)-f(x₂)]|

|f(x₁)-f(x)|+|f(x)-f(x₂)|

e =

Therefore the set U(x) =

{d Î [0,1] | x₁ & x₂ Î [x-d,x+d]Ç[a,b] Þ |f(x₁)-f(x₂)| £½e}

contains at least one positive number d £ 1. Thus it is a non-empty subset of the interval [0,1]. Recall the arrow Þ means implies. Therefore we can let g(x) = supU(x). Then g(x) is the largest d > 0 such that |f(x₁)-f(x₂)| £ [1/2]e whenever x₁ and x₂ are in the interval [a,b] within a distance d of the point x. Since U(x) contains at least one positive number, we conclude the function value g(x) = d > 0.

Let x₀ be in the interval [a,b]. We wish to show that g(x) is continuous at x₀. Put d₀ = g(x₀) > 0. Suppose the element c of [a,b] satisfies |c-x₀| £ [1/2]d₀. Then c belongs to the interval [x₀-d₀,x₀+d₀]Ç[a,b]. Further the smallest distance of c to the endpoints of this interval is £ d₀-|x₀-c| = d. From the definition of U(x), we can conclude that d > 0 belongs to U(c). Therefore g(c) ³ d = d₀-|x₀-c| = g(x₀)-|x₀-c|.

Now |x-c| £ [1/2] g(x₀) also implies the largest distance of c to the endpoints of the interval [x₀-d₀,x₀+d₀]Ç[a,b] is d₀+|x-c|. Now suppose for the sake of contradiction that g(c) > d₀+|x-c|. Then d = g(c)-|x₀-c| > d₀ belongs to U(x₀) and hence g(x₀) ³ d > d₀ = g(x₀). But the latter is impossible. Therefore g(c) £ d₀+|x-c| = g(x)+|x-c| when |x-c| £ [1/2] g(x₀) and c is in the interval [a,b].

The above inequalities imply that

g(x₀)-|x₀-c| £ g(c) £ g(x₀)+|x₀-c|

when |x-c| £ [1/2] g(x₀). The inequality is equivalent to |g(x₀)-g(c)| £ |x₀-c|. This latter implies that g(x) is continuous at x₀. And thus g(x) is continuous at each point x = x₀ in the interval [a,b]. By definition, g(x) > 0.

From the finite covering theorem, there exists a whole number n > 0 and a set of numbers x₁,x₂,¼,x_n in [a,b] with the property that if x is a number in the interval [a,b] then there exists an whole number j with 1 £ j £ n and |x-x_j| < g(x_j).

Put (eta) h = [1/2]min_{1 £
j £ n}g(x_j). Now suppose x and y are both in the interval [a,b] with |x-y| < h. Then there exists a x_j with |x_j-x| £ g(x_j). This implies |f(x_j)-f(x)| < [1/2]e. It further implies by the triangle inequality that |y-x_j| = |(y-x)+(x-x_j)| £ |y-x|+|x-x_j| < h+g(x_j) £ 2g(x_j) and hence |f(y)-f(x_j)| £ [1/2]e. Finally, we obtain that

|f(x)-f(y)|

=

|f(x)-f(x_j)+(f(x_j)-f(y))|

£

|f(x)-f(x_j)|+|f(x_j)-f(y)|

£

1
2 e+ 1
2 e = e.

Thus |x-y| < h = d implies |f(x)-f(y)| £ e.

The previous proof has a consequence not stated. In the proof, let e = 1. Then there exists x₁,¼,x_n in [a,b] such that if x is in [a,b] then for some j, |x-x_j| £ d and |f(x)-f(x_j)| £ e = 1. The last inequality is equivalent to

-1 £ f(x)-f(x_j) £ 1

The latter implies

-1+f(x_j) £ f(x) £ 1+f(x_j)

Put K = -1+min_{1 £
q £ n} f(x_q) and M = +1+max_{1 £ q
£ n}f(x_q). Then K £ f(x) £ M if x is in the interval [a,b]. This holds since there exist a whole number j ³ 1 such that

K £ -1+f(x_j) £ f(x) £ 1+f(x_j) £ M

Thus the range W of the function f(x) is contained in the finite interval [K,M]. This implies the first conclusion in the following assertion.