5.3: Sequences of Functions
Most functions you have encountered have the property that at (almost) every point the function has a limit that agrees with its value there. This is a very useful feature of a function, and it is called continuity.
DEFINITION. Continuous Let \(f\) be a real function with domain \(X \subseteq \mathbb{R} .\) Let \(a \in X\) . Then we say \(f\) is continuous at \(a\) if \(\lim _{X \ni x \rightarrow a} f(x)=\) \(f(a)\) . We say \(f\) is continuous on \(X\) if it is continuous at every point of \(X\)
Intuitively, the idea of a continuous function on an interval is that it has no jumps. We shall make this precise in Chapter 8 when we prove the Intermediate Value Theorem 8.10, which asserts that if a continuous function on an interval takes on two distinct values \(c\) and \(d\) , it must also take on every value between \(c\) and \(d\) .
EXAMPLE 5.18. Prove that the function \(f(x)=x^{2}\) is continuous on \(\mathbb{R}\) . Discussion. How would we do this from first principles? We need to show that for every \(a \in \mathbb{R}\) , for every \(\varepsilon>0\) , we can always find a \(\delta>0\) such that for any \(x \in \mathbb{R}\) \[|x-a|<\delta \quad \Longrightarrow \quad\left|x^{2}-a^{2}\right|<\varepsilon .\] (Why don’t we need to add the hypothesis \(0<|x-a|\) ?) The easiest way to do this is to write down a formula that, given \(a\) and \(\varepsilon\) , produces a \(\delta\) satisfying (5.19).
As \(x^{2}-a^{2}=(x-a)(x+a)\) , if \(|x-a|\) is less than some number \(\delta\) (still unspecified), then \(\left|x^{2}-a^{2}\right|\) is less than \(\delta|x+a|\) . So we want \[\delta|x+a| \leq \varepsilon\] We can’t choose \(\delta=\varepsilon /|x+a|\) , because \(\delta\) cannot depend on \(x\) . But if \(|x-a|<\delta\) , then \[\begin{aligned} |x+a| & \leq|x|+|a| \\ &<|a|+\delta+|a|=2|a|+\delta, \end{aligned}\] SO \[\left|x^{2}-a^{2}\right|<\delta(2|a|+\delta) \stackrel{?}{\leq} \varepsilon .\] We must choose \(\delta\) so that the last inequality holds. By the quadratic formula, \[\delta(2|a|+\delta) \leq \varepsilon \quad \Longleftrightarrow \delta \leq \sqrt{|a|^{2}+\varepsilon}-|a| .\] So choose \(\delta=\sqrt{|a|^{2}+\varepsilon}-|a|\) and (5.19) holds.
REMARK. A formally correct proof could have been reduced to:
PROOF. Let \(a \in \mathbb{R}\) and \(\varepsilon>0\) . Then letting \(\delta=\sqrt{|a|^{2}+\varepsilon}-|a|\) we have \(|x-a|<\delta \Longrightarrow\left|x^{2}-a^{2}\right|<\varepsilon\) . Q.E.D.
However, while a diligent reader could verify that this proof is correct, pulling \(\delta\) out of a hat like this doesn’t give the reader the insight that our much longer proof does. Remember, a proof has more than one function: not only must it convince the reader that the claimed result is true, but it should also help the reader understand why the result is true. A good proof should be describable in a few English sentences, so that a knowledgeable listener can then go write down a more detailed proof fairly easily.
REMARK. One does not need to choose the largest value of \(\delta\) so that the inequality \(\stackrel{?}{\leq}\) in (5.21) holds - any positive \(\delta\) that satisfies the inequality will work. This allows one to simplify the algebra. For example, let \(\delta_{1}\) be such that \[|x-a|<\delta_{1} \Rightarrow|x+a|<2|a|+1 .\] (Such a \(\delta_{1}\) exists from the continuity of the simpler function \(x \mapsto x\) ). Then let \[\delta=\min \left(\delta_{1}, \frac{\varepsilon}{2|a|+1}\right)\] and (5.20) holds.
One could imagine repeating proofs like the above to show that \(x^{3}\) , \(x^{4}\) , and so on are continuous, but we want to take big steps. Can we show all polynomials are continuous?
First observe that because limits behave well with respect to algebraic operations (Theorem 5.9), and continuity is defined in terms of limits, then algebraic combinations of continuous functions are continuous.
Proposition 5.22. Suppose \(f: X \rightarrow \mathbb{R}\) and \(g: X \rightarrow \mathbb{R}\) are real functions that are continuous at \(a \in X\) . Let \(c\) and \(d\) be scalars \({ }^{2}\) . Then \(c f+d g\) and \(f g\) are both continuous at a, and so is \(f / g\) if \(g(a) \neq 0\) .
PROOF. Exercise.
Constant functions are continuous (Lemma 5.15), and the function \(f(x)=x\) is continuous (Exercise 5.16). So one can prove by induction on the degree of polynomial, using Proposition 5.22, that that all polynomials are continuous (Exercise 5.27). Once you have proved that all polynomials are continuous, you may prove that rational functions are continuous wherever the denominator doesn’t vanish.
This result is used so frequently that we will state it formally.
\({ }^{2}\) A scalar is just a fancy word for a number. PROPOSITION 5.23. Every polynomial is continuous on \(\mathbb{R}\) . Every rational function is continuous wherever the denominator is non-zero.
What about the exponential function \[e^{x}:=\sum_{n=0}^{\infty} \frac{x^{n}}{n !} ?\] Each partial sum is a polynomial, and hence continuous; so if we knew that the limit of a sequence of continuous functions were continuous, we would be done. This turns out, however, to be a subtle problem, which we address in the next Section.