5.2 Continuity and differentiability

Differentiability makes precise the notion of a ‘smooth’ graph. If a function is differentiable, so that its graph is smooth, then it is reasonable to expect that the graph should not have any breaks. In other words, any differentiable function should be continuous. This is indeed the case for all the examples of differentiable functions we saw in Section 5.1 and we now show it is true in general.

Theorem 5.20.

Let $I\subseteq\mathbb{R}$ be an open interval and suppose $f\colon I\to\mathbb{R}$ is differentiable at $a\in I$ . Then $f$ is continuous at $a$ .

By the contrapositive of Theorem 5.20, if $f$ is not continuous, then $f$ is not differentiable. We illustrate this in Figure 5.7, which shows the failure of differentiability in the presence of a jump discontinuity.

Figure 5.7: An illustration of Theorem 5.20. The function is not continuous at

0

. For

h>0

, we see that the secant line through

(0,f(0))

and

(h,f(h))

has steeper and steeper gradient as

h

gets small. We can use this to show

f

is not differentiable at

0

To prove Theorem 5.20, we introduce the following useful lemma, which is essentially a mild reformulation of the definition of the derivative.

Lemma 5.21.

Let $I\subseteq\mathbb{R}$ be an open interval and $f\colon I\to\mathbb{R}$ be differentiable at $a$ . Then

(5.9) (5.9)

F\colon I\to\mathbb{R};\qquad F(x):=\begin{cases}\displaystyle\frac{f(x)-f(a)}% {x-a}&\text{if $x\in I\setminus\{a\}$,}\\ f^{\prime}(a)&\text{if $x=a$}\end{cases}

is continuous at $a$ and satisfies $f(x)=F(x)(x-a)+f(a)$ for all $x\in I$ .

Exercise 5.22.

Give an interpretation of the definition of $F$ from (5.9): in particular, for $x\in I$ , what does the value $F(x)$ measure? Illustrate your interpretation with a figure.

Proof.

Since $f$ is differentiable at $a$ , using the formulation of the definition of the derivative from Remark 5.3, it follows that

\lim_{x\to a}F(x)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f^{\prime}(a)=F(a).

Thus, by definition, $F$ is continuous at $a$ .

Moreover, the equation $f(x)=F(x)(x-a)+f(a)$ is satisfied for all $x\in I$ . When $x=a$ it reduces to $f(a)=f(a)$ which is trivially true. For $x\neq a$ the equation is simply a rearranged form of the definition of $F$ in that case. ∎

Exercise 5.23.

Compare Lemma 5.21 to our discussion of the function $\mathrm{sinc}$ from Example 4.44. How are they related?

Proof (of Theorem 5.20).

By Lemma 5.21, there exists a function $F\colon I\to\mathbb{R}$ which is continuous at $a$ with $F(a)=f^{\prime}(a)$ and satisfies

f(x)=F(x)(x-a)+f(a)\qquad\text{for all $x\in I$.}

By the limit laws, it follows that

\lim_{x\to a}f(x)=\lim_{x\to a}F(x)(x-a)+f(a)=F(a)\cdot 0+f(a)=f(a).

Thus, by definition $f$ is continuous at $a$ . ∎

It is important to note that the converse of Theorem 5.20 does not hold. That is, not every continuous function is differentiable. Indeed, we have already seen multiple examples of continuous functions which fail to be differentiable, such as the function $f(x):=|x|$ from Example 5.7.