4.9 Continuous functions: three big theorems

Let $a$ , $b\in\mathbb{R}$ with $a<b$ so that $[a,b]$ is a closed, bounded interval. Here we study continuous functions $f\colon[a,b]\to\mathbb{R}$ and show that they have a number of special properties. These properties are described by three big theorems:

•

Intermediate value theorem;
•

Boundedness theorem;
•

Extreme value theorem.

The intermediate value theorem

Let $a$ , $b\in\mathbb{R}$ with $a<b$ and suppose $f\colon[a,b]\to\mathbb{R}$ is continuous and satisfies $f(a)<f(b)$ . As illustrated in Figure 4.23, we can think of the graph of $f$ as a continuous path starting at $(a,f(a))$ and ending at $(b,f(b))$ . Now chose some ‘intermediate value’ $f(a)<y_{0}<f(b)$ along the $y$ -axis. We draw the horizontal straight line $y=y_{0}$ , which separates the points $(a,f(a))$ and $(b,f(b))$ . It is intuitively clear that the path traced out by the graph of $f$ must cross the line $y=y_{0}$ at some point. That is, there must exist some $x_{0}\in[a,b]$ such that $f(x_{0})=y_{0}$ . This simple intuition is made precise by the intermediate value theorem.

Figure 4.23: The intermediate value theorem. For

f\colon[a,b]\to\mathbb{R}

continuous, given

f(a)\leq y_{0}\leq f(b)

, there exists some

x_{0}\in[a,b]

such that

f(x_{0})=y_{0}

Theorem 4.91 (Intermediate value theorem).

Let $a$ , $b\in\mathbb{R}$ with $a<b$ and suppose $f\colon[a,b]\to\mathbb{R}$ is continuous with $f(a)\leq f(b)$ . For all $y_{0}\in\mathbb{R}$ such that $f(a)\leq y_{0}\leq f(b)$ , there exists some $x_{0}\in[a,b]$ such that $f(x_{0})=y_{0}$ .

From the sketch in Figure 4.23, the idea behind the intermediate value theorem is intuitively clear. However, it is in fact a deep result which relies on the completeness axiom. To see this is the case, the following example shows how the theorem implies the existence of numbers such as $\sqrt{2}$ which are missing from $\mathbb{Q}$ (recall from Chapter 1 that the existence of these numbers relies on the completeness axiom).

Example 4.92.

We can use the intermediate value theorem to give another proof of the existence of $\sqrt{2}$ . Consider the function $p_{2}\colon[0,2]\to[0,4]$ given by $p_{2}(x):=x^{2}$ for all $x\in[0,2]$ . Then $p_{2}$ is continuous and satisfies $p_{2}(0)=0<2<4=p_{2}(2)$ . Hence, by the intermediate value theorem, we see that there exists some $s\in[0,2]$ such that $p_{2}(s)=2$ . In other words, $s^{2}=2$ so that $s=\sqrt{2}$ .

Exercise 4.93.

Use the intermediate value theorem to show the following.

(i)

For all $x_{0}\geq 0$ there exists some $s\geq 0$ such that $s^{2}=x_{0}$ .
(ii)

More generally, for all $x_{0}\geq 0$ and all $n\in\mathbb{N}$ there exists some $s\geq 0$ such that $s^{n}=x_{0}$ .

In order to prove Theorem 4.91, we shall make use of the following simple lemma, which exploits the continuity hypothesis.

Lemma 4.94.

Let $I\subseteq\mathbb{R}$ be an interval, $f\colon I\to\mathbb{R}$ be continuous at $x_{0}\in I$ and $y_{0}\in\mathbb{R}$ .

(a)

If $f(x_{0})>y_{0}$ , then there exists some $\delta>0$ such that for all $x\in I$ with $|x-x_{0}|<\delta$ , we have $f(x)>y_{0}$ .
(b)

If $f(x_{0})<y_{0}$ , then there exists some $\delta>0$ such that for all $x\in I$ with $|x-x_{0}|<\delta$ , we have $f(x)<y_{0}$ .

Figure 4.24: Illustration of the proof of Lemma 4.94 (a).

Proof.

We shall only prove part (a); part (b) follows either by a similar argument or by reflection (replacing $f$ with $-f$ ).

Since $f(x_{0})>y_{0}$ , we have

\varepsilon:=f(x_{0})-y_{0}>0.

Applying the $\varepsilon$ - $\delta$ definition of continuity with this value of $\varepsilon$ , there exists some $\delta>0$ such that for all $x\in I$ with $|x-x_{0}|<\delta$ we have

|f(x)-f(x_{0})|<\varepsilon.

Observe that

f(x)=f(x_{0})-(f(x)-f(x_{0}))\geq f(x_{0})-|f(x)-f(x_{0})|.

Therefore, for all $x\in I$ with $|x-x_{0}|<\delta$ , we have

f(x)>f(x_{0})-\varepsilon=f(x_{0})-(f(x_{0})-y_{0})=y_{0}

as required. ∎

Proof (of Theorem 4.91).

We may assume $f(a)<y_{0}<f(b)$ , since otherwise we can take $x_{0}$ to be either $a$ or $b$ .

Consider the set

E:=\big{\{}x\in[a,b]:f(x)<y_{0}\big{\}}.

We illustrate this set in Figure 4.25.

Figure 4.25: The set

E:=\big{\{}x\in[a,b]:f(x)<y_{0}\big{\}}

and

x_{0}:=\sup E

from the proof of the intermediate value theorem, drawn along the

x

-axis.

Since we assumed $y_{0}>f(a)$ , it follows that $a\in E$ and so $E$ is nonempty. On the other hand, $x\leq b$ for all $x\in E$ so that $b$ is an upper bound for the set $E$ . Hence, $E$ is nonempty and bounded above, so $x_{0}:=\sup E$ exists by the completeness axiom and $x_{0}\in[a,b]$ .

Intuitively, it should be clear that $f(x_{0})=y_{0}$ and so $x_{0}$ is the point we are looking for: see Figure 4.25. In order to prove this is the case, we use a Goldilocks approach: we show $f(x_{0})\not>y_{0}$ (not too hot) and $f(x_{0})\not<y_{0}$ (not too cold), and thereby conclude $f(x_{0})=y_{0}$ (just right).

Arguing by contradiction, suppose $f(x_{0})>y_{0}$ . By Lemma 4.94 (a), there exists some $\delta>0$ such that $f(x)>y_{0}$ for all $x\in[a,b]$ with $|x-x_{0}|<\delta$ . On the other hand, since $x_{0}$ is the supremum of $E$ , there exists some $c\in E$ such that $x_{0}-\delta<c\leq x_{0}$ (see Lemma 1.31). Since $c\in E$ , we have $f(c)<y_{0}$ . However, $|c-x_{0}|<\delta$ and so we must have $f(c)>y_{0}$ . This is a contradiction. Therefore, we must have $f(x_{0})\leq y_{0}$ . Note, in particular, this tells us that $x_{0}\neq b$ , since $f(b)>y_{0}$ .

Now suppose $f(x_{0})<y_{0}$ . By Lemma 4.94 (b), there exists some $\delta>0$ such that $f(x)<y_{0}$ for all $x\in[a,b]$ with $|x-x_{0}|<\delta$ . Since $x_{0}<b$ , it follows that there exists some $d\in[a,b]$ satisfying $x_{0}<d<\min\{x_{0}+\delta,b\}$ : see Figure 4.26.

Figure 4.26: The choice of

d

from the proof of the intermediate value theorem.

Since $|d-x_{0}|<\delta$ , it follows that $f(d)<y_{0}$ so that $d\in E$ . However, $x_{0}<d$ , which contradicts the fact that $x_{0}$ is the supremum of $E$ . Therefore, we must have $f(x_{0})\geq y_{0}$ .

Since we have shown $f(x_{0})\leq y_{0}$ and $f(x_{0})\geq y_{0}$ , it follows that $f(x_{0})=y_{0}$ , as required. ∎

We have already seen that the intermediate value theorem gives us a quick and easy way to define radicals such as $\sqrt{2}$ . However, it has many other important uses. We give one further application here and explore many more applications in Worksheet 8.

Corollary 4.95.

Let $I\subseteq\mathbb{R}$ be an interval and $f\colon I\to\mathbb{R}$ be continuous. Then the image $\mathrm{Im}(f):=\{f(x):x\in I\}$ is an interval.

We illustrate Corollary 4.95 in Figure 4.27 . This is another manifestation of the idea that the graphs of continuous functions do not have any ‘breaks’ or ‘jumps’.

Figure 4.27: Examples illustrating Corollary 4.95. In both cases,

f\colon[a,b]\to\mathbb{R}

is continuous and the image

\mathrm{Im}(f)

is an interval.

To prove Corollary 4.95 we shall use a simple characterisation of intervals. In particular, the following are equivalent:

1.

$E\subseteq\mathbb{R}$ is an interval;
2.

If $y\in\mathbb{R}$ such that $y_{1}<y<y_{2}$ for some $y_{1}$ , $y_{2}\in E$ then $y\in E$ .

Condition 2 says that any point $y$ sandwiched in between two points $y_{1}$ , $y_{2}$ belonging to $E$ also belongs to $E$ . This characterisation should be intuitive and can be proved by considering the supremum and infimum of $E$ , treating the cases where $E$ is bounded or unbounded above or below separately.

Proof (of Corollary 4.95).

Pick $y_{1}$ , $y_{2}\in\mathrm{Im}(f)$ with $y_{1}<y_{2}$ and let $y_{1}<y<y_{2}$ , By the intermediate value theorem, there exists some $x\in I$ such that $f(x)=y$ . Hence $y\in\mathrm{Im}(f)$ . It therefore follows from the above characterisation that $\mathrm{Im}(f)$ is an interval. ∎

Exercise 4.96.

Show that there exists a function $f\colon[-1,1]\to\mathbb{R}$ such that the image $\mathrm{Im}(f)$ is not an interval. Thus, the hypothesis that $f$ is continuous is necessary in Corollary 4.95.

Example 4.97.

Consider the restriction of the exponential function $\exp\colon[0,\infty)\to[1,\infty)$ to the non-negative real line.

Recall (from Proposition 4.6) that it is immediate from the formula (4.2) that $\exp$ is increasing on $[0,\infty)$ and $\exp(x)\geq\exp(0)=1$ for all $x\in\mathbb{R}$ . Thus, $\exp$ is an injective function on $[0,\infty)$ .

We claim $\exp\colon[0,\infty)\to[1,\infty)$ is also a surjective function. We know from Exercise 4.82 (iii) that $\lim_{x\to\infty}\exp(x)=\infty$ and so given any $y\geq 1$ there exists some $b>0$ such that $\exp(0)=1\leq y\leq\exp(b)$ . Since $\exp$ is continuous, by the intermediate value theorem there exists some $x\in[1,b]\subset[1,\infty)$ such that $\exp(x)=y$ . This shows $\exp\colon[0,\infty)\to[1,\infty)$ is surjective, as claimed.

As mentioned earlier, these results continue to hold for $\exp\colon\mathbb{R}\to(0,\infty)$ defined on the entire real line, but we shall postpone the proof until the next chapter.

Exercise 4.98.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be continuous and satisfy $\displaystyle\lim_{x\to\infty}f(x)=\infty$ and $\displaystyle\lim_{x\to-\infty}f(x)=-\infty$ . Show that $f$ is surjective.

The boundedness theorem

There are many examples of unbounded continuous functions: for instance, $x^{2}$ is unbounded on $\mathbb{R}$ . This example is clearly related to the fact that $\mathbb{R}$ is an unbounded domain. Another kind of example is $1/x$ , which is continuous and unbounded on the bounded domain $(0,1)$ .

Intuitively, the unboundedness of $1/x$ occurs because the open interval $(0,1)$ has a ‘hole’ at $0$ , which allows the function to escape off to infinity. If we consider a continuous function $f\colon[a,b]\to\mathbb{R}$ on a closed, bounded interval $[a,b]$ , then there are no such means for the function to escape to infinity.

Exercise 4.99.

Observe that $x^{2}$ is bounded on any closed, bounded interval $[a,b]\subset\mathbb{R}$ and $1/x$ is bounded on any closed, bounded interval $[a,b]\subset(0,1)$ .

From the above, we might begin to expect that any continuous function $f\colon[a,b]\to\mathbb{R}$ is bounded. The following theorem shows this is indeed the case.

Theorem 4.100 (Boundedness theorem).

Let $a$ , $b\in\mathbb{R}$ with $a<b$ . If $f\colon[a,b]\to\mathbb{R}$ is continuous then $f$ is bounded.

As we have already observed, in Theorem 4.100 it is important to consider continuous functions defined in a closed bounded interval $[a,b]$ . Indeed, the functions $x^{2}$ and $1/x$ show that the result can fail for continuous functions defined over other kinds of intervals.

Exercise 4.101.

Show that the conclusion of Theorem 4.100 can also fail if $f$ is not continuous. In particular, find an example of an unbounded discontinuous function $f\colon[0,1]\to\mathbb{R}$ .

To prove Theorem 4.100, our first step is to establish the following ‘local’ version.³³ 3 We essentially already saw this in Exercise 4.48.

Lemma 4.102 (Local boundedness).

Let $I\subseteq\mathbb{R}$ be an interval and $f\colon I\to\mathbb{R}$ be continuous at $x_{0}\in I$ . Then there exists some $\delta>0$ such that $f$ is bounded on $I\cap(x_{0}-\delta,x_{0}+\delta)$ .

Proof.

Since $f$ is continuous at $x_{0}$ , by taking $\varepsilon:=1$ in the $\varepsilon$ - $\delta$ definition of continuity, we see that there exists some $\delta>0$ such that

|f(x)-f(x_{0})|<1\qquad\text{for all $x\in I$ with $|x-x_{0}|<\delta$.}

Let $M:=1+|f(x_{0})|$ . Then, by the triangle inequality, we see that

$\displaystyle\|f(x)\|$	$\displaystyle=\|f(x)-f(x_{0})+f(x_{0})\|$
	$\displaystyle\leq\|f(x)-f(x_{0})\|+\|f(x_{0})\|$
	$\displaystyle<1+(M-1)$
	$\displaystyle=M\qquad\text{for all $x\in I\cap(x_{0}-\delta,x_{0}+\delta)$.}$

Thus, $f$ is bounded on $I\cap(x_{0}-\delta,x_{0}+\delta)$ , as required. ∎

Armed with Lemma 4.102, we now turn to the proof of Theorem 4.100. The argument also relies on the least upper bound axiom. This should perhaps not come as a surprise: we have already noted that boundedness of continuous functions is related to the presence or lack of ‘holes’ in the domain.

Proof (of Theorem 4.100).

Let

E:=\big{\{}x\in[a,b]:\text{ $f$ is bounded on $[a,x]$}\big{\}}.

Our goal is to show that $b\in E$ . Indeed, this means precisely that $f$ is bounded on the whole interval $[a,b]$ , which is what we want to show.

Note that $a\in E$ so that $E$ is nonempty and that $E$ is bounded above by $b$ . Thus, by the completeness axiom, $s:=\sup E$ exists.

Claim: $s\in E$ and $s=b$ .

Once we have the claim, then we are done since this shows $b\in E$ .

Since $a\leq s\leq b$ , the function $f$ is defined at $s$ and, moreover is continuous at that point. By Lemma 4.102, there exists some $\delta>0$ such that $f$ is bounded on $[a,b]\cap(s-\delta,s+\delta)$ . In particular, there exists some $M_{1}>0$ such that

(4.16) (4.16)

|f(x)|\leq M_{1}\qquad\text{for all $x\in[a,b]\cap(s-\delta,s+\delta)$.}

On the other hand, by the approximation property of the supremum, there exists some $c\in E$ such that $s-\delta<c<s$ . In particular, since $c\in E$ , we know that $f$ is bounded on $[a,c]$ and so there exists some $M_{2}>0$ such that

(4.17) (4.17)

|f(x)|\leq M_{2}\qquad\text{for all $x\in[a,c]$.}

Combining (4.16) and (4.17), for $M:=\max\{M_{1},M_{2}\}$ we see that

(4.18) (4.18)

|f(x)|\leq M\qquad\text{for all $x\in[a,b]\cap[a,s+\delta)$.}

It immediately follows from (4.18) that $f$ is bounded on $[a,s]$ and so $s\in E$ . This establishes the first part of the claim.

It remains to show $s=b$ . We already know that $s\leq b$ and so it suffices to show $s\geq b$ .

We argue by contradiction, assuming $s<b$ . Under this assumption, we can find some $d$ such that $s<d<\min\{s+\delta,b\}$ . By (4.18), the function $f$ is bounded on $[a,d]$ and so $d\in E$ . However, $d>s$ so this contradicts the fact that the supremum is an upper bound for $E$ . Hence, we must have $s=b$ as claimed. ∎

The extreme value theorem

By Theorem 4.100, any continuous function $f\colon[a,b]\to\mathbb{R}$ is automatically bounded. Our goal is now to prove an upgrade of this theorem, by showing $f$ attains both a maximum and minimum value.

Definition 4.103.

Let $E\subseteq\mathbb{R}$ and $f\colon E\to\mathbb{R}$ .

1

We say $x_{M}\in E$ is a maximum point for $f$ if $f(x)\leq f(x_{M})$ for all $x\in E$ ;
2

We say $x_{m}\in E$ is a minimum point for $f$ if $f(x)\geq f(x_{m})$ for all $x\in E$ .

If a function $f\colon E\to\mathbb{R}$ has a maximum (respectively, minimum) point, then it is bounded above (respectively, below). However, the converse of this statement is not true.

Exercise 4.104.

Recall the function $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x):=1-(1+x^{2})^{-1}$ for all $x\in\mathbb{R}$ from Exercise 4.11. Show that $f$ is bounded above by $1$ but has no maximum point.

Often, whether or not a function has a maximum or a minimum point depends on the domain of the function. For instance, if we let $R>0$ and consider the restriction $f|_{[-R,R]}$ of the function in Exercise 4.104 to the interval $[-R,R]$ , then $f|_{[-R,R]}$ has maximum points at $-R$ and $R$ .

Exercise 4.105.

Show that the function $p_{2}\colon[0,1]\to\mathbb{R}$ given by $p_{2}(x):=x^{2}$ for all $x\in[0,1]$ has a maximum point, but the restriction $p_{2}|_{(0,1)}\colon(0,1)\to\mathbb{R}$ has no maximum point.

The contrasting behaviour of the functions in Exercise 4.105, which only differ in whether their domain is a closed interval or an open interval, hints at another special behaviour of continuous functions on closed, bounded intervals.

Theorem 4.106 (Extreme value theorem).

Let $a$ , $b\in\mathbb{R}$ with $a<b$ and $f\colon[a,b]\to\mathbb{R}$ be continuous. Then there exist $x_{m}$ , $x_{M}\in[a,b]$ such that

(4.19) (4.19)

f(x_{m})\leq f(x)\leq f(x_{M})\qquad\text{for all $x\in[a,b]$.}

In particular, $f$ has a minimum point and a maximum point on $[a,b]$ .

Theorem 4.106 is an upgrade of Theorem 4.100: not only does (4.19) imply that $f$ is bounded, but it also tells us that $f$ attains its (least upper and greatest lower) bounds.

In Theorem 4.106 it is important to consider continuous functions defined on a closed, bounded interval $[a,b]$ . Indeed, Exercise 4.105 shows that a bounded, continuous function on the bounded interval $(0,1)$ can fail to have a maximum point – so the “closed” assumption is essential. Likewise, Exercise 4.104 shows that a bounded, continuous function on an unbounded interval can fail to have a maximum point – so the “bounded” assumption is also essential. Hence, we need both the closed and the bounded hypothesis to ensure the conclusion holds.

Proof (of Theorem 4.106).

We show the existence of a maximum point. The existence of a minimum point can be shown in a similar manner, or can be derived from the existence of a maximum point using a reflection argument (replacing $f$ with $-f$ ).

Since $f\colon[a,b]\to\mathbb{R}$ is continuous, by Theorem 4.100 it is bounded. In particular, the image

E:=\mathrm{Im}(f):=\{f(x):x\in[a,b]\}

is nonempty and bounded above. Hence, by the completeness axiom $s:=\sup E$ exists.

We claim that there exists some $x_{M}\in[a,b]$ such that $f(x_{M})=s$ . Once this is established, it follows that $f(x)\leq f(x_{M})$ for all $x\in[a,b]$ , since $s$ is an upper bound for $E=\mathrm{Im}(f)$ .

It remains to prove the claim. We argue by contradiction, assuming that there does not exist any $x\in[a,b]$ such that $f(x)=s$ . In this case, the function

g\colon[a,b]\to\mathbb{R};\qquad g(x):=s-f(x)\qquad\text{for all $x\in[a,b]$}

satisfies $g(x)>0$ for all $x\in[a,b]$ . Thus, by Corollary 4.51, the function $1/g\colon[a,b]\to\mathbb{R}$ is well-defined and continuous. By the boundedness theorem applied to $1/g$ , there exists some $M>0$ such that $0<1/g(x)\leq M$ for all $x\in[a,b]$ and so

(4.20) (4.20)

s-f(x)=g(x)\geq 1/M\qquad\text{for all $x\in[a,b]$.}

However, since $s$ is the supremum of $E$ , by taking $\varepsilon:=1/M$ in the approximation property, there must exist some $y\in E$ such that $s-1/M<y<s$ . In particular, $y=f(x)$ for some $x\in[a,b]$ and rearranging gives $s-f(x)<1/M$ , which contradicts (4.20). Consequently, there must exist some $x_{M}\in[a,b]$ such that $f(x_{M})=s$ , as required. ∎

$\displaystyle\|f(x)\|$	$\displaystyle=\|f(x)-f(x_{0})+f(x_{0})\|$
	$\displaystyle\leq\|f(x)-f(x_{0})\|+\|f(x_{0})\|$
	$\displaystyle<1+(M-1)$
	$\displaystyle=M\qquad\text{for all $x\in I\cap(x_{0}-\delta,x_{0}+\delta)$.}$