Proposition: 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 > 0 such that whenever | x - c | < we know that | f(x) - f(c) | < (because f is continuous at c)
• there also exists > 0 such that whenever | x - c | < we know that | g(x) - g(c) | < (because g is continuous at c).

But then, if we let = min( , ), 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 continuos 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 > 0 such that whenever | t - d | < then | g(t) - g(d) | < .
• there also exists > 0 such that if | x - c | < then | f(x) - f(c) | < .

(Note that we have replaced the usual by here) Now let = min( , ), and substitute t = f(x) and d = f(c). We have:

• if | x - c | < then | f(x) - f(c) | < .
• and then | g(f(x)) - g(f(c)) | < .

In other words, f(g(x)) is continuous at x = c.