§18 Properties of Continuous Functions

Why this matters

This section proves the facts that calculus uses without comment: a continuous function on a closed interval has a largest and a smallest value, it takes every value in between, and a strictly increasing continuous function has a continuous inverse (which is how $\sqrt x$, $\sqrt[m]{x}$ and later $\log x$ are obtained). Each proof leans on completeness, through Bolzano–Weierstrass or a supremum.

A real-valued function $f$ is bounded if $\{f(x) : x \in \operatorname{dom}(f)\}$ is a bounded set, i.e. if there is a real number $M$ such that $|f(x)| \le M$ for all $x \in \operatorname{dom}(f)$.

18.1 Theorem.

Let $f$ be a continuous real-valued function on a closed interval $[a, b]$. Then $f$ is a bounded function. Moreover, $f$ assumes its maximum and minimum values on $[a, b]$: there exist $x_0, y_0$ in $[a, b]$ such that $f(x_0) \le f(x) \le f(y_0)$ for all $x \in [a, b]$.

Used in Exercise 18.1, Exercise 18.2

Proof

Assume $f$ is not bounded on $[a, b]$. Then to each $n \in \mathbb{N}$ there corresponds $x_n \in [a, b]$ with $|f(x_n)| > n$. By the Bolzano–Weierstrass Theorem 11.5, $(x_n)$ has a subsequence $(x_{n_k})$ converging to some real $x_0$, which also lies in $[a, b]$ (Exercise 8.9). Since $f$ is continuous at $x_0$, $\lim_{k\to\infty} f(x_{n_k}) = f(x_0)$; but also $\lim_{k\to\infty} |f(x_{n_k})| = +\infty$, a contradiction. So $f$ is bounded.

Now let $M = \sup\{f(x) : x \in [a, b]\}$, which is finite by the first paragraph. For each $n \in \mathbb{N}$ there is $y_n \in [a, b]$ with $M - \frac1n < f(y_n) \le M$, so $\lim f(y_n) = M$. By Bolzano–Weierstrass, a subsequence $(y_{n_k})$ converges to some $y_0 \in [a, b]$. Since $f$ is continuous at $y_0$, $f(y_0) = \lim_{k\to\infty} f(y_{n_k})$. Since $(f(y_{n_k}))_{k}$ is a subsequence of $(f(y_n))_n$, Theorem 11.3 gives $\lim_k f(y_{n_k}) = \lim_n f(y_n) = M$, so $f(y_0) = M$: $f$ assumes its maximum at $y_0$.

The last paragraph applies to $-f$, so $-f$ assumes its maximum at some $x_0 \in [a, b]$. It follows easily that $f$ assumes its minimum at $x_0$ (Exercise 18.1).

∎

Theorem 18.1 is used all the time, at least implicitly, in calculus max–min problems: it is taken for granted that a continuous function on a closed interval actually attains a maximum and a minimum. If the domain is not a closed interval, be careful (Exercise 18.3).

Theorem 18.1 is false for open intervals. For example, $f(x) = \frac1x$ is continuous but unbounded on $(0, 1)$, and $x^2$ is continuous and bounded on $(-1, 1)$ but has no maximum value there.

Intuitionwhat each hypothesis prevents

A maximum can fail to exist in three ways, and each hypothesis of Theorem 18.1 blocks one of them:

• Not closed: the top is approached at a missing endpoint, like $x^2$ on $(-1, 1)$.
• Not bounded: the function keeps rising forever, like $f(x) = x$ on $[0, \infty)$.
• Not continuous: the value at the top is “wrong”. On $[0, 1]$, let $f(x) = x$ for $x < 1$ and $f(1) = 0$. Its supremum is $1$, but it is never attained.

In the proof, closed and bounded is what lets Bolzano–Weierstrass produce a limit point that stays inside $[a, b]$, and continuity is what transfers that limit to the values of $f$.

Try it: break a hypothesis of the Extreme Value Theorem

Pick a scenario and drag the probe along the graph, trying to reach the top. When all hypotheses hold the probe can reach the maximum. Drop one (closed, bounded, continuous) and the supremum slips away.

graph of fsup f (when finite)your probethe maximum, attained

Probe x0.6

Jump to the highest sampled point

f(0.6) = 1.274. Distance below the supremum: 1.049. The maximum value 2.323 is attained at x ≈ 2.674, as Theorem 18.1 promises.

The Intermediate Value Theorem

Let $f(x) = x^2 - 2$, considered only on the rational numbers in $[0, 2]$. Then $f(0) = -2 < 0 < 2 = f(2)$, and $f$ is continuous. Must $f$ have a zero in this set?

Yes, by the sign changeNo, it has no rational zero

18.2 Intermediate Value Theorem.

If $f$ is a continuous real-valued function on an interval $I$, then $f$ has the intermediate value property on $I$: whenever $a, b \in I$, $a < b$, and $y$ lies between $f(a)$ and $f(b)$ (i.e. $f(a) < y < f(b)$ or $f(b) < y < f(a)$), there exists at least one $x$ in $(a, b)$ such that $f(x) = y$.

Used in §17

Proof

Assume $f(a) < y < f(b)$; the other case is similar. Let $S = \{x \in [a, b] : f(x) < y\}$. Since $a \in S$, $S$ is nonempty, so $x_0 = \sup S$ is a number in $[a, b]$. For each $n \in \mathbb{N}$, $x_0 - \frac1n$ is not an upper bound for $S$, so there is $s_n \in S$ with $x_0 - \frac1n < s_n \le x_0$. Thus $\lim s_n = x_0$ and, since $f(s_n) < y$ for all $n$,

$$f(x_0) = \lim f(s_n) \le y. \tag{1}$$

Let $t_n = \min\{b, x_0 + \frac1n\}$. Since $x_0 \le t_n \le x_0 + \frac1n$, $\lim t_n = x_0$. Each $t_n$ lies in $[a, b]$ but not in $S$, so $f(t_n) \ge y$ for all $n$. Therefore $f(x_0) = \lim f(t_n) \ge y$, and with (1), $f(x_0) = y$.

∎

Intuition

The proof picks $x_0$, the supremum of the points where $f$ is still below $y$: roughly, the last moment before $f$ reaches $y$ for good. Approaching $x_0$ from the left gives $f(x_0) \le y$, approaching from the right gives $f(x_0) \ge y$, and continuity forbids a jump in between.

Common pitfallwhat the theorem needs

• The domain must be an interval. $f(x) = \frac1x$ on $[-1, 0) \cup (0, 1]$ is continuous, with $f(-1) = -1 < 0 < 1 = f(1)$, yet it is never $0$.
• It needs completeness. Over $\mathbb{Q}$, the function $x^2 - 2$ is negative at $1$ and positive at $2$ but has no rational root. The proof breaks down at “$x_0 = \sup S$ is a number”.
• It promises existence only. There may be many solutions, and the theorem does not say where they are. Bisection, below, is one way to find one.

Try it: break a hypothesis of the Intermediate Value Theorem

Choose a value $y$ between $f(a)$ and $f(b)$. With all hypotheses in place the graph must cross the orange line. Drop continuity (a jump or an asymptote) or the interval (a gap in the domain) and the graph can pass over it without touching it.

graph of fyour value ysolution of f(x) = yf skips over y here

Value y between f(a) and f(b)1.2

f(-2) = -1, f(2) = 3, and y = 1.2 lies strictly between them. Solutions of f(x) = y: x ≈ -1.7, -0.0668, 1.76. The theorem promises at least one; there can be several.

Bisection starts with an interval of length $1$ and halves it each step. After $10$ steps, about how long is the interval that still contains a solution?

About $0.1$About $0.001$About $0.00001$

Try it: hunting for the crossing by bisection

The proof finds the crossing as a supremum. A practical way to locate it is bisection: if $f$ changes sign on $[a, b]$, check the midpoint and keep whichever half still has a sign change. Each step halves the interval, and the Intermediate Value Theorem guarantees a solution in every interval along the way.

Halve the interval5 stepsReset

f < 0f > 0current interval and its midpoint

Step 0: f changes sign on [0.000, 2.000], of length 2.00e+0. So by the Intermediate Value Theorem a solution of x³ = 2 lies in it, and x ≈ 1.000. Each step keeps the half on which the sign still changes. The nested intervals shrink to a single point, and completeness of ℝ is what guarantees that point exists.

18.3 Corollary.

If $f$ is a continuous real-valued function on an interval $I$, then the set $f(I) = \{f(x) : x \in I\}$ is also an interval or a single point.

Used in Theorem 18.4, Exercise 18.11

Proof

The set $J = f(I)$ has the property

$$y_0, y_1 \in J \text{ and } y_0 < y < y_1 \quad\text{imply}\quad y \in J. \tag{1}$$

If $\inf J < \sup J$, such a set is an interval. In fact we show

$$\inf J < y < \sup J \quad\text{implies}\quad y \in J, \tag{2}$$

so $J$ is an interval with endpoints $\inf J$ and $\sup J$, which may or may not belong to $J$ and may or may not be finite. To prove (2), take $\inf J < y < \sup J$. Then there are $y_0, y_1 \in J$ with $y_0 < y < y_1$, so $y \in J$ by (1).

∎

Example 1.

Let $f$ be a continuous function mapping $[0, 1]$ into $[0, 1]$, i.e. $\operatorname{dom}(f) = [0, 1]$ and $f(x) \in [0, 1]$ for all $x$. Show $f$ has a fixed point: a point $x_0 \in [0, 1]$ with $f(x_0) = x_0$.

The graph of $f$ lies in the unit square, and the assertion is that it crosses the line $y = x$, which is almost obvious. A rigorous proof uses a little trick. Consider $g(x) = f(x) - x$, also continuous on $[0, 1]$. Since $g(0) = f(0) \ge 0$ and $g(1) = f(1) - 1 \le 0$, the Intermediate Value Theorem gives $g(x_0) = 0$ for some $x_0 \in [0, 1]$. Then $f(x_0) = x_0$.

Example 2.

Show that if $y > 0$ and $m \in \mathbb{N}$, then $y$ has a positive $m$th root.

The function $f(x) = x^m$ is continuous (Exercise 17.5). There is $b > 0$ with $y \le b^m$: if $y \le 1$ let $b = 1$, and if $y > 1$ let $b = y$. Thus $f(0) < y \le f(b)$, and the Intermediate Value Theorem gives $f(x) = y$ for some $x \in (0, b]$. So $y = x^m$ and $x$ is an $m$th root of $y$.

Inverse functions

Look more closely at $f(x) = x^m$ from Example 2. It is strictly increasing on $[0, \infty)$:

$$0 \le x_1 < x_2 \quad\text{implies}\quad x_1^m < x_2^m.$$

So $f$ is one-to-one and each nonnegative $y$ has exactly one nonnegative $m$th root, which makes the notation $y^{1/m}$ unambiguous. In fact $f^{-1}(y) = y^{1/m}$ is the inverse function of $f$: $f^{-1} \circ f(x) = x$ for $x \in \operatorname{dom}(f)$ and $f \circ f^{-1}(y) = y$ for $y \in \operatorname{dom}(f^{-1})$. Since $f$ is continuous, $f^{-1}(y) = y^{1/m}$ is continuous on $[0, \infty)$ by the next theorem. (For $m = 2$ this is Exercise 17.4.)

18.4 Theorem.

Let $f$ be a continuous strictly increasing function on an interval $I$. Then $f(I)$ is an interval $J$ by Corollary 18.3, and $f^{-1}$ is a function with domain $J$. The function $f^{-1}$ is continuous and strictly increasing on $J$.

Used in Exercise 18.11

Proof

$f^{-1}$ is easily shown to be strictly increasing. Since $f^{-1}$ maps $J$ onto $I$, the next theorem shows $f^{-1}$ is continuous.

∎

Intuition

A strictly increasing function can only be discontinuous by jumping, and a jump leaves a gap in its range. If the range is a whole interval there are no gaps, so there are no jumps. That is the content of Theorem 18.5, and Theorem 18.4 applies it to $g = f^{-1}$.

18.5 Theorem.

Let $g$ be a strictly increasing function on an interval $J$ such that $g(J)$ is an interval $I$. Then $g$ is continuous on $J$.

Used in Exercise 18.11

Proof

Consider $x_0 \in J$, and assume $x_0$ is not an endpoint of $J$ (tiny changes are needed otherwise). Then $g(x_0)$ is not an endpoint of $I$, so there is $\varepsilon_0 > 0$ with $(g(x_0) - \varepsilon_0, g(x_0) + \varepsilon_0) \subseteq I$.

Let $\varepsilon > 0$. Since we only need the ε–δ property of Theorem 17.2 for small ε, we may assume $\varepsilon < \varepsilon_0$. Then there are $x_1, x_2 \in J$ with $g(x_1) = g(x_0) - \varepsilon$ and $g(x_2) = g(x_0) + \varepsilon$. Clearly $x_1 < x_0 < x_2$. If $x_1 < x < x_2$, then $g(x_1) < g(x) < g(x_2)$, so $g(x_0) - \varepsilon < g(x) < g(x_0) + \varepsilon$, i.e. $|g(x) - g(x_0)| < \varepsilon$. Put $\delta = \min\{x_2 - x_0, x_0 - x_1\}$. Then $|x - x_0| < \delta$ implies $x_1 < x < x_2$, hence $|g(x) - g(x_0)| < \varepsilon$.

∎

Theorem 18.5 is a partial converse to the Intermediate Value Theorem: a strictly increasing function with the intermediate value property is continuous. However, Exercise 18.12 shows a function can have the intermediate value property without being continuous.

18.6 Theorem.

Let $f$ be a one-to-one continuous function on an interval $I$. Then $f$ is strictly increasing ($x_1 < x_2$ implies $f(x_1) < f(x_2)$) or strictly decreasing ($x_1 < x_2$ implies $f(x_1) > f(x_2)$).

Proof

First we show

$$\text{if } a < b < c \text{ in } I, \text{ then } f(b) \text{ lies between } f(a) \text{ and } f(c). \tag{1}$$

If not, then, say, $f(b) > \max\{f(a), f(c)\}$. Choose $y$ with $f(b) > y > \max\{f(a), f(c)\}$. By the Intermediate Value Theorem applied to $[a, b]$ and to $[b, c]$, there are $x_1 \in (a, b)$ and $x_2 \in (b, c)$ with $f(x_1) = f(x_2) = y$, contradicting that $f$ is one-to-one.

Now choose any $a_0 < b_0$ in $I$ and suppose, say, $f(a_0) < f(b_0)$. We show $f$ is strictly increasing on $I$. By (1),

$$\begin{aligned} f(x) &< f(a_0) && \text{for } x < a_0 && [\text{since } x < a_0 < b_0], \\ f(a_0) < f(x) &< f(b_0) && \text{for } a_0 < x < b_0, \\ f(b_0) &< f(x) && \text{for } x > b_0 && [\text{since } a_0 < b_0 < x]. \end{aligned}$$

In particular,

$$\begin{aligned} f(x) &< f(a_0) \quad\text{for all } x < a_0, && (2) \\ f(a_0) &< f(x) \quad\text{for all } x > a_0. && (3) \end{aligned}$$

Consider any $x_1 < x_2$ in $I$. If $x_1 \le a_0 \le x_2$, then $f(x_1) < f(x_2)$ by (2) and (3). If $x_1 < x_2 < a_0$, then $f(x_1) < f(a_0)$ by (2), so by (1) $f(x_1) < f(x_2)$. Finally, if $a_0 < x_1 < x_2$, then $f(a_0) < f(x_2)$, so by (1) $f(x_1) < f(x_2)$.

∎

Reading alongside Abbott

Abbott’s Section 4.4 proves the maximum–minimum theorem by showing that a continuous function maps compact sets to compact sets, and that a compact subset of $\mathbb{R}$ contains its supremum and infimum. Theorem 18.1’s proof is the same argument written out for $[a, b]$ with sequences. Section 4.5 gives two proofs of the Intermediate Value Theorem: one through a supremum, as in Theorem 18.2, and one through repeated halving and the nested interval property, which is the bisection widget above made rigorous. Abbott also presents the theorem as an instance of a more general fact, that continuous functions map connected sets to connected sets. Ross returns to that viewpoint in §22.

Practice: put the proof in order

The Intermediate Value Theorem comes from taking the supremum of the set where $f$ is still below $y$. Arrange the argument.

Put the proof in order: Intermediate Value Theorem (Theorem 18.2)

These are the steps of the proof, shuffled. Use the buttons to move each step into place, then check.

1. Assume $f(a) < y < f(b)$. (If instead $f(a) > y > f(b)$, apply the argument to $-f$ and $-y$.) Let $S = \{x \in [a, b] : f(x) < y\}$.
2. The set $S$ contains $a$ and is bounded above by $b$, so $x_0 = \sup S$ exists, and $a \le x_0 \le b$.
3. First case, to be ruled out: $f(x_0) < y$. Then $x_0 \ne b$, and continuity keeps $f < y$ on a short interval to the right of $x_0$, which puts points of $S$ beyond $x_0$. That contradicts $x_0$ being an upper bound.
4. Second case, to be ruled out: $f(x_0) > y$. Then $x_0 \ne a$, and continuity keeps $f > y$ on some interval $(x_0 - \delta, x_0]$, so $x_0 - \delta$ is a smaller upper bound of $S$. That contradicts $x_0$ being the least one.
5. So $f(x_0) = y$. Since $f(a) < y < f(b)$, the point $x_0$ is neither $a$ nor $b$, so $x_0 \in (a, b)$.

Practice: fill in the gaps

The first half of Theorem 18.1: a continuous function on $[a, b]$ is bounded. The gaps are the points where the argument uses completeness and closedness.

Fill in the gaps: Continuous on [a, b] implies bounded (Theorem 18.1)

This is the skeleton of the proof. In each gap choose the option that makes the argument work, then check.

1. Suppose $f$ is continuous on $[a, b]$ but not bounded. Then for each $n \in \mathbb{N}$ there is $x_n \in [a, b]$ with $|f(x_n)| > n$$|f(x_n)| < n$$f(x_n) = n$.
2. The sequence $(x_n)$ lies in $[a, b]$, so it is bounded. By the Bolzano–Weierstrass Theorem 11.5the Monotone Convergence Theorem 10.2the Intermediate Value Theorem 18.2 some subsequence $x_{n_k}$ converges, say to $x_0$.
3. Since $a \le x_{n_k} \le b$ for all $k$, the limit satisfies $a \le x_0 \le b$$a < x_0 < b$, so $x_0$ is in the domain of $f$.
4. By continuity at $x_0$, $f(x_{n_k}) \to f(x_0)$, so these values are bounded$f(x_{n_k})$ may diverge. But $|f(x_{n_k})| > n_k \ge k$ is unbounded. This contradiction proves that $f$ is bounded.

Practice: spot the error

The Intermediate Value Theorem is stated for an interval. Find where this argument forgets that.

Spot the error: a function that changes sign

One line of this argument is wrong. Press the button on the line you think is the mistake.

1. Let $f(x) = \frac1x$ on its domain $\mathbb{R} \setminus \{0\}$. This function is continuous at every point of its domain.
2. We have $f(-1) = -1 < 0 < 1 = f(1)$.
3. By the Intermediate Value Theorem there is $x$ between $-1$ and $1$ with $f(x) = 0$.

Check your understanding

Take the quiz (3 questions)

$f(x) = \frac1x$ is continuous on $(0, 1]$ but unbounded. Why doesn’t Theorem 18.1 apply?

The interval $(0, 1]$ is not closed.$f$ is not continuous at $x = 1$.It does apply: $f$ is bounded by $1$ on $(0, 1]$.

Let $p(x) = x^5 + x - 1$. What does the Intermediate Value Theorem tell you?

$p$ has a root in $(0, 1)$, since $p(0) = -1 < 0 < 1 = p(1)$.Nothing, because $p$ has degree $5$.$p(\frac12) = 0$

Can a continuous one-to-one function on $[0, 2]$ have $f(0) = 0$, $f(1) = 2$ and $f(2) = 1$?

No: by Theorem 18.6 it would have to be strictly increasing or strictly decreasing.Yes, as long as it is continuous.Only if $f$ is a polynomial.

Exercises

Exercise 18.1

Let $f$ be as in Theorem 18.1. Show that if $-f$ assumes its maximum at $x_0 \in [a, b]$, then $f$ assumes its minimum at $x_0$.

Solution

Saying $-f$ assumes its maximum at $x_0$ means $-f(x_0) \ge -f(x)$ for all $x \in [a,b]$. Multiplying by $-1$ reverses the inequality: $f(x_0) \le f(x)$ for all $x \in [a,b]$. So $f$ assumes its minimum at $x_0$.

Tags: Warm-upIVT & EVT

Used in Theorem 18.1

Exercise 18.2

Reread the proof of Theorem 18.1 with $[a, b]$ replaced by $(a, b)$. Where does it break down? Discuss.

Solution

The limit $x_0$ of the Bolzano–Weierstrass subsequence lies in $[a, b]$ but may be an endpoint, which is not in $(a, b)$, so continuity at $x_0$ cannot be used.

Tags: Warm-upExplain why

Exercise 18.3

Use calculus to find the maximum and minimum of $f(x) = x^3 - 6x^2 + 9x + 1$ on $[0, 5)$.

Solution

$f'(x) = 3(x - 1)(x - 3)$. Candidates: $f(0) = 1$, $f(1) = 5$, $f(3) = 1$, and $f(x) \to 21$ as $x \to 5$. The minimum is $1$ (at $0$ and $3$); there is no maximum, since $\sup = 21$ is not attained on $[0, 5)$.

Tags: Warm-upComputation

Exercise 18.4

Let $S \subseteq \mathbb{R}$ and suppose there is a sequence $(x_n)$ in $S$ converging to a number $x_0 \notin S$. Show there exists an unbounded continuous function on $S$.

Hint

Try $f(x) = \frac1{x - x_0}$.

Solution

Define $f(x) = \dfrac{1}{x - x_0}$ on $S$. Since $x_0 \notin S$, the denominator never vanishes, so $f$ is a quotient of continuous functions with nonzero denominator and is continuous on $S$. It is unbounded: given $M > 0$, because $x_n \to x_0$ and $x_n \neq x_0$ we can pick $n$ with $0 < |x_n - x_0| < 1/M$, and then $|f(x_n)| > M$.

Tags: StandardExamples & counterexamplesSubsequences

Exercise 18.5

1. Let $f$ and $g$ be continuous on $[a, b]$ with $f(a) \ge g(a)$ and $f(b) \le g(b)$. Prove $f(x_0) = g(x_0)$ for at least one $x_0$ in $[a, b]$.
2. Show Example 1 can be viewed as a special case of part (a).

Hint

Apply the Intermediate Value Theorem to $f - g$.

Solution

1. Let $h = f - g$, continuous on $[a,b]$, with $h(a) \ge 0 \ge h(b)$. If $h(a) = 0$ or $h(b) = 0$ take $x_0 = a$ or $b$. Otherwise $h(a) > 0 > h(b)$, and the Intermediate Value Theorem with $y = 0$ gives $x_0 \in (a,b)$ with $h(x_0) = 0$, i.e. $f(x_0) = g(x_0)$.
2. Take $[a,b] = [0,1]$, $g(x) = x$. Since $f$ maps into $[0,1]$, $f(0) \ge 0 = g(0)$ and $f(1) \le 1 = g(1)$, so (a) yields $x_0$ with $f(x_0) = x_0$.

Tags: StandardIVT & EVT

Exercise 18.6

Prove $x = \cos x$ for some $x$ in $(0, \frac\pi2)$. (Find it numerically in the widget above.)

Solution

Let $h(x) = x - \cos x$, continuous on $[0, \frac\pi2]$. Then $h(0) = -1 < 0$ and $h(\frac\pi2) = \frac\pi2 > 0$. By the Intermediate Value Theorem there is $x \in (0, \frac\pi2)$ with $h(x) = 0$, that is, $x = \cos x$.

Tags: Warm-upIVT & EVT

Exercise 18.7

Prove $x e^x = 2$ for some $x$ in $(0, 1)$.

Solution

Let $f(x) = xe^x$, continuous on $[0,1]$ as a product of continuous functions. Since $f(0) = 0 < 2 < e = f(1)$, the Intermediate Value Theorem gives $x \in (0,1)$ with $f(x) = 2$.

Tags: Warm-upIVT & EVT

Exercise 18.8

Suppose $f$ is a real-valued continuous function on $\mathbb{R}$ and $f(a)f(b) < 0$ for some $a, b \in \mathbb{R}$. Prove there exists $x$ between $a$ and $b$ such that $f(x) = 0$.

Solution

Since $f(a)f(b) < 0$, the numbers $f(a)$ and $f(b)$ are nonzero with opposite signs, so $0$ lies between them, and $a \neq b$. Apply the Intermediate Value Theorem to $f$ on the interval with endpoints $a$ and $b$, with $y = 0$. It gives $x$ strictly between $a$ and $b$ with $f(x) = 0$.

Tags: Warm-upIVT & EVT

Used in Exercise 18.9

Exercise 18.9

Prove that a polynomial function $f$ of odd degree has at least one real root.

Hint

It may help to consider first the case of a cubic, $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3$ with $a_3 \ne 0$. For large $|x|$ the leading term dominates, so $f$ takes both signs.

Solution

Let $f(x) = a_0 + a_1x + \cdots + a_nx^n$ with $n$ odd and $a_n \neq 0$. Replacing $f$ by $-f$ if necessary (same roots) we may assume $a_n > 0$. Let $S = |a_0| + \cdots + |a_{n-1}|$ and $R = \max\{1, 2S/a_n\} + 1$, so $R > 1$ and $a_n R > 2S$. For $|x| = R$, $|a_0 + \cdots + a_{n-1}x^{n-1}| \le S R^{n-1} < \tfrac12 a_n R^n .$ Hence $f(R) \ge a_nR^n - SR^{n-1} > \tfrac12 a_nR^n > 0$. Because $n$ is odd, $(-R)^n = -R^n$, so $f(-R) \le -a_nR^n + SR^{n-1} < 0$. Now $f$ is continuous (Exercise 17.5), and by Exercise 18.8 (or the Intermediate Value Theorem) $f$ has a root in $(-R, R)$.

Tags: StandardIVT & EVT

Exercise 18.10

Suppose $f$ is continuous on $[0, 2]$ and $f(0) = f(2)$. Prove there exist $x, y$ in $[0, 2]$ such that $|y - x| = 1$ and $f(x) = f(y)$.

Hint

Consider $g(x) = f(x + 1) - f(x)$ on $[0, 1]$; note $g(0) = -g(1)$.

Solution

Define $g(x) = f(x+1) - f(x)$ on $[0,1]$; it is continuous and $g(0) = f(1) - f(0), \qquad g(1) = f(2) - f(1) = f(0) - f(1) = -g(0).$ If $g(0) = 0$, take $x = 0$, $y = 1$. Otherwise $g(0)$ and $g(1)$ have opposite signs, so by the Intermediate Value Theorem $g(x) = 0$ for some $x \in (0,1)$. In either case there is $x \in [0,1]$ with $f(x+1) = f(x)$; put $y = x + 1 \in [1,2]$. Then $|y - x| = 1$ and $f(x) = f(y)$.

Tags: ChallengeIVT & EVT

Exercise 18.11

1. Show that if $f$ is strictly increasing on an interval $I$, then $-f$ is strictly decreasing on $I$.
2. State and prove Theorems 18.4 and 18.5 for strictly decreasing functions.

Solution

1. If $x_1 < x_2$ in $I$ then $f(x_1) < f(x_2)$, so $-f(x_1) > -f(x_2)$. Thus $-f$ is strictly decreasing.

2. Theorem 18.4 for decreasing $f$. Let $f$ be continuous and strictly decreasing on $I$. By Corollary 18.3, $J = f(I)$ is an interval. Put $g = -f$: it is continuous and strictly increasing on $I$ by (a) and Theorem 17.3, so $g^{-1}$ is continuous and strictly increasing on $g(I) = -J$ by Theorem 18.4. Since $f^{-1}(y) = g^{-1}(-y)$ for $y \in J$, the inverse $f^{-1}$ is a composite of the continuous map $y \mapsto -y$ and $g^{-1}$, hence continuous. It is strictly decreasing: $y_1 < y_2$ gives $-y_1 > -y_2$, so $g^{-1}(-y_1) > g^{-1}(-y_2)$.

   Theorem 18.5 for decreasing $g$. Let $g$ be strictly decreasing on an interval $J$ with $g(J)$ an interval. Then $-g$ is strictly increasing on $J$ and $(-g)(J) = -g(J)$ is an interval, so $-g$ is continuous by Theorem 18.5. Thus $g = -(-g)$ is continuous by Theorem 17.3.

Tags: StandardIVT & EVT

Exercise 18.12

Let $f(x) = \sin\left(\frac1x\right)$ for $x \ne 0$ and $f(0) = 0$.

1. Observe that $f$ is discontinuous at $0$ by Exercise 17.10(b).
2. Show $f$ has the intermediate value property on $\mathbb{R}$.

Hint

For (b), if an interval $[a, b]$ contains $0$ (and $a < b$), then on the nonzero side $f$ takes every value in $[-1, 1]$.

Solution

1. This is Exercise 17.10(b).

2. Note $|f| \le 1$ everywhere. Let $a < b$ and let $y$ lie between $f(a)$ and $f(b)$; then $y \in [-1, 1]$.

   If $0 \notin [a,b]$, then $f(x) = \sin(1/x)$ is continuous on $[a,b]$ (a composite of continuous functions), and the Intermediate Value Theorem applies.

   If $0 \in [a,b]$, then $(a,b)$ contains an interval $(0, c)$ or $(-c, 0)$ for some $c > 0$. Take the first case (the second is symmetric, using $-1/x$). As $x$ runs over $(0,c)$, $1/x$ runs over $(1/c, \infty)$, which contains a closed interval of length $2\pi$, say $[t, t + 2\pi]$. On that interval $\sin$ is continuous and takes both the values $1$ and $-1$ (a full period), so by the Intermediate Value Theorem it takes the value $y$ at some $s \in [t, t+2\pi]$. Then $x = 1/s \in (0, c) \subseteq (a,b)$ and $f(x) = y$.

Tags: ChallengeIVT & EVTSequences & continuity