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Proposition 6.2.5: Algebra with Continuous Functions

  • The identity function f(x) = x is continuous in its domain.
  • If f(x) and g(x) are both continuous at x = c, so is f(x) + g(x) at x = c.
  • If f(x) and g(x) are both continuous at x = c, so is f(x) * g(x) at x = c.
  • If f(x) and g(x) are both continuous at x = c, and g(x) # 0, then f(x) / g(x) is continuous at x = c.
  • If f(x) is continuous at x = c, and g(x) is continuous at x = f(c), then the composition g(f(x)) is continuous at x = c.

Proof:

Suppose f(x) = x. Then, given any > 0 choose = / 2. Then, if

| x - c | <
it implies that
| f(x) - f(c) | = | x - c | < = / 2 <
. Hence, the identity function is indeed continuous. Was it really necessary to take = / 2 ?

The sum of continuous functions is continuous follows directly from the triangle inequality. Take any > 0. There exists 1 > 0 such that whenever

| x - c | < 1
we know that
| f(x) - f(c) | <
(because f is continuous at c). There also exists 2 > 0 such that whenever
| x - c | < 2
we know that
| g(x) - g(c) | <
(because g is continuous at c). But then, if we let = min(1 , 2), we have: if
| x - c | <
then
| (f(x) + g(x)) - (f(c) + g(c)) | ≤
      | f(x) - f(c) | + | g(x) - g(c) | <
      + = 2
That finishes the proof. (That we don't get a simple should not bother us any more).

The product of two continuous functions is again continuous, which follows from a simple trick. We will only look at the trick involved, and leave the details to the reader:

| f(x) g(x) - f(c) g(c) |
      = | f(x) g(x) - f(x) g(c) + f(x) g(c) - f(c) g(c) |
      ≤ | f(x) | | g(x) - g(c) | + | g(c) | | f(x) - f(c) |
With this trick the rest of the proof should not be too difficult.

A similar trick works for the quotient. Here is the idea:

| f(x) / g(x) - f(c) / g(c) | = | 1 / g(x) g(c) | | f(x) g(c) - f(c) g(x) |
Can you see how to continue ? Adding and subtracting will help again.

As for composition of functions, we have to proceed somewhat different: We know that f(x) is continuous at c, and g(x) is continuous at f(c). Therefore, given any > 0 there exists 1 > 0 such that whenever

| t - d | < 1
then
| g(t) - g(d) | <
There also exists 2 > 0 such that if
| x - c | < 2
then
| f(x) - f(c) | < 1
. (Note that we have replaced the usual by 1 here) Now let
= min(1 , 2)
and substitute t = f(x) and d = f(c). We have: if
| x - c | <
then
| f(x) - f(c) | < 1
and then
| g(f(x)) - g(f(c)) | <
In other words, f(g(x)) is continuous at x = c.
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