53 The Mean Value Theorems

#todo/school : Read §5.3 📅 2024-12-03 ✅ 2024-12-03
#todo/school : Do problems 5.3 # 1, 2, 3, 4, 6, 7 📅 2024-12-06 ✅ 2024-12-05
#todo/school : Do problems 5.3 # 8, 10, 12 📅 2024-12-06 ✅ 2024-12-06

5.3.1

Question

A function $f : A \to R$ is Lipschitz on $A$ if $\exists M > 0$ such that:

| \frac{f (x) - f (y)}{x - y} | \leq M

for all $x \neq y \in A$ .
a. Show that if $f$ is differentiable on a closed interval $[a, b]$ and if $f^{'}$ is continuous on $[a, b]$ then $f$ is Lipschitz on $[a, b]$ .
b. Review 43 Continuity Practice#4.3.11 for the definition of contractive. If we add the assumption that $| f^{'} (x) | < 1$ on $[a, b]$ does it follow that $f$ is contractive on this set?

Proof

a. Suppose $f$ is differentiable on $[a, b]$ (and thus $f$ is continuous on at least $[a, b]$ ) and $f^{'}$ is continuous on $[a, b]$ .

By the Extreme Value Theorem then $f^{'}$ attains an absolute maximum and absolute minimum in that interval. Let's denote the maximum $m_{1}$ and the minimum $m_{0}$ , so then $\exists x_{1}, x_{0}$ such that $f^{'} (x_{0}) = m_{0}$ (the min.) and $f^{'} (x_{1}) = m_{1}$ (the max).

Now notice if $x_{0} = x_{1}$ then the max is the min, so we have a constant function. In that case $f^{'} (x) = k$ for some $k \in R$ so then by our corollary then $f (x) = k x + c$ . So letting $x \neq y \in [a, b]$ then $| \frac{f (x) - f (y)}{x - y} | = k$ so choose then $M = k$ as an obvious example.

Now if instead $x_{0} \neq x_{1}$ then the max isn't the min. Then choose $M = max {| f (x_{0}) |, | f (x_{1}) |}$ . To show $f$ is Lipschitz, assume for contradiction that $\exists x \neq y \in [a, b]$ such that for any $M > 0$ :

| \frac{f (x) - f (y)}{x - y} | > M

Now notice that our chosen $M$ is > 0, so then this inequality is true. Now by the Mean Value Theorem (MVT) then $\exists c \in (x, y)$ such that:

f^{'} (c) = \frac{f (x) - f (y)}{x - y}

So then clearly:

| f^{'} (c) | > M

so we found a point $c$ that has a larger derivative (in magnitude) than the largest possible derivatives (in magnitude). This is a contradiction!

b. Yes it is! The idea is that adding the restricted slope sort of forces the function to have its range be a subset of it's domain.

To prove this, let $| f^{'} (x) | < 1$ for all $x \in [a, b]$ . We want to show that $f$ is contractive, or that there exists some $c \in (0, 1)$ such that:

| f (x) - f (y) | \leq c | x - y |

If we let $x, y \in [a, b]$ be arbitrary, then by the Mean Value Theorem (MVT) then $\exists m \in (a, b)$ such that:

| \frac{f (x) - f (y)}{x - y} | = | f^{'} (m) | < 1

This implies that:

| f (x) - f (y) | < | x - y |

Specifically, let $| f^{'} (m) | = c$ where $c \in (0, 1)$ . Then clearly:

| f (x) - f (y) | = c | x - y |

thus $f$ is contractive.

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5.3.2

Question

Let $f$ be differentiable on an interval $A$ . If $f^{'} (x) \neq 0$ on $A$ , show that $f$ is one-to-one on $A$ . Provide an example to show that the converse statement need not be true.

Proof

Let $x \neq y \in A$ be arbitrary. WLOG let $x < y$ . To show that $f (x) \neq f (y)$ , assume for contradiction that it is instead $f (x) = f (y)$ . Then using the Mean Value Theorem (MVT) then $\exists c \in (x, y)$ such that:

\frac{f (x) - f (y)}{x - y} = f^{'} (c) = \frac{0}{x - y} = 0

But that contradicts $f^{'} (x) \neq 0$ for all $x$ ! Thus then $f (x) \neq f (y)$ , so then $f$ must be injective (one-to-one).

Now for the converse, notice that $f (x) = x^{\frac{1}{3}}$ is one-to-one on $A = R$ but is not differentiable at $x = 0$ for example.
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5.3.3

Question

Let $h$ be differentiable on an interval $[0, 3]$ , and assume that $h (0) = 1$ , $h (1) = 2$ , and $h (3) = 2$ .
a. Argue that $\exists d \in [0, 3]$ where $h (d) = d$ .
b. Argue that at some point $c$ we have $h^{'} (c) = \frac{1}{3}$ .
c. Argue that $h^{'} (x) = \frac{1}{4}$ at some point in the domain.

Proof

a. Consider the function $f (x) = h (x) - x$ . Then notice that:

f (1) = h (1) - 1 = 2 - 1 = 1

f (3) = h (3) - 3 = 2 - 3 = - 1

Now using the Intermediate Value Theorem with $y = 0$ , then $0 \in f ([1, 3])$ . That means that $\exists d \in [1, 3] \subseteq [0, 3]$ such that:

f (d) = 0 \Rightarrow h (d) - d = 0 \Rightarrow h (d) = d

b. Using the Mean Value Theorem (MVT), then $\exists c \in (0, 3)$ such that:

h^{'} (c) = \frac{h (3) - h (0)}{3 - 0} = \frac{1}{3}

c. The argument comes from the fact that the derivative must be positive on $[0, 1]$ , and at most 0 over $[1, 3]$ . Using the Mean Value Theorem (MVT) over $[0, 1]$ then $\exists c_{1} \in (0, 1)$ such that:

h^{'} (c_{1}) = \frac{h (1) - h (0)}{1 - 0} = 1

and similarly via the Mean Value Theorem (MVT) then $\exists c_{2} \in (1, 3)$ such that:

h^{'} (c_{2}) = \frac{h (3) - h (1)}{3 - 1} = 0

Now, by Darboux's Theorem, then $\exists c \in [c_{1}, c_{2}]$ such that $h^{'} (c) = \frac{1}{4}$ .

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5.3.4

Question

Let $f$ be differentiable on an interval $A$ containing zero, and assume $(x_{n}) \subseteq A$ is a Sequence with $(x_{n}) \to 0$ and $x_{n} \neq 0$ .
a. If $f (x_{n}) = 0$ for all $n \in N$ show that $f (0) = 0$ and $f^{'} (0) = 0$ .
b. Add the assumption that $f$ is twice-differentiable at zero. Show that $f^{″} (0) = 0$ as well.

Proof

a. Notice that since $f$ is differentiable then $f$ is continuous on $A$ containing zero. As such then:

lim_{x \to 0} f (x) = L

exists. By the Existence of Functional Limits then because $0$ is a limit point of $A$ , then this sequence must have $f (x_{n}) \equiv 0 \to L$ , thus $L = 0$ . Similarly, since $f$ is differentiable then:

lim_{x \to 0} \frac{f (x) - f (0)}{x - 0} = lim_{x \to 0} \frac{f (x)}{x} = m

exists. Notice that by the Existence of Functional Limits then because $0$ is a limit point of $A$ then the sequence:

m \leftarrow lim_{n \to \infty} \frac{f (x_{n})}{x_{n}} = lim_{n \to \infty} \frac{0}{x_{n}} = 0

Thus $m = 0$ similarly.

b. By the Mean Value Theorem (MVT) over $[0, x_{n}]$ then $\exists c_{n} \in (0, x_{n})$ such that:

f^{'} (c_{n}) = \frac{f (x_{n})}{x_{n}}

Then like in (a):

f^{″} (0) = lim_{n \to \infty} \frac{f^{'} (c_{n})}{c_{n}} = lim_{n \to \infty} \frac{f (x_{n})}{x_{n} c_{n}} = 0 lim_{n \to \infty} \frac{1}{c_{n}} = 0

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5.3.6

Question

a. Let $g : [0, a] \to R$ be differentiable, $g (0) = 0$ and $| g^{'} (x) | \leq M$ for all $x \in [0, a]$ . Show $| g (x) | \leq M x$ for all $x \in [0, a]$ .
b. Let $h : [0, a] \to R$ be twice differentiable, $h^{'} (0) = h (0) = 0$ and $| h^{″} (x) | \leq M$ for all $x \in [0, a]$ . Show $| h (x) | \leq \frac{M x^{2}}{2}$ for all $x \in [0, a]$ .
c. Conjecture and prove an analogous result for a function that is differentiable three times on $[0, a]$ .

Proof

a. Let $x \in [0, a]$ be arbitrary. Apply Mean Value Theorem (MVT) to find a $c \in [0, x]$ such that:

g^{'} (c) = \frac{g (x) - g (0)}{x - 0} = \frac{g (x)}{x}

Thus:

| g^{'} (c) | = \frac{| g (x) |}{| x |} \leq M = \frac{M x}{| x |}

(for the $x = 0$ case the theorem holds, so the above only considers $x \neq 0$ ). Thus $| g (x) | \leq M | x | \equiv M x$ (since $x \geq 0$ ).

b. Again let $x \in [0, a]$ be arbitrary. Notice that $h^{'}$ is differentiable and $| (h^{'})^{'} (x) | \leq M$ so then by (a) then $| h^{'} (x) | \leq M x$ . Using the Generalized Mean Value Theorem with the functions $h (x)$ and $x^{2}$ then we get that $\exists c \in (0, x)$ such that:

\begin{aligned} [h (0) - h (x)] 2 c & = [x^{2} - 0^{2}] h^{'} (c) \\ h (x) & = \frac{- x^{2} h^{'} (c)}{2 c} \\ | h (x) | & \leq | \frac{- x^{2} h^{'} (c)}{2 c} | \\ = \frac{x^{2}}{2 c} | h^{'} (c) | \\ \leq \frac{x^{2}}{2 c} M c \\ = \frac{x^{2} M}{2} \end{aligned}

c. Consider the thrice-differentiable function $f (x) = x^{4}$ . I claim and expect that:

| f (x) | \leq \frac{M x^{3}}{6}

for all $x \in [0, a]$ , where this $M$ is determined by $| f^{‴} (x) | \leq M$ . In this specific case $M = 4 \cdot 3 \cdot 2 \cdot a$ via direct calculation.

From (a) we know that since $| (f^{″})^{'} (x) | \leq M$ then $| f^{″} (x) | \leq M x$ . A similar argument shows that (b) can be applied via $| (f^{'})^{″} (x) | \leq M$ then $| f^{'} (x) | \leq \frac{M x^{2}}{2}$ . Use again the Generalized Mean Value Theorem with the functions $f (x)$ and $x^{3}$ to get that $\exists c \in (0, x)$ such that:

\begin{aligned} [f (0) - f (x)] 3 c^{2} & = [x^{3} - 0^{3}] f^{'} (c) \\ f (x) & = \frac{- x^{3} f^{'} (c)}{3 c^{2}} \\ | f (x) | & = | \frac{- x^{3} f^{'} (c)}{3 c^{2}} | \\ = \frac{x^{3}}{3 c^{2}} | f^{'} (c) | \\ \leq \frac{x^{3}}{3 c^{2}} \cdot \frac{M c^{2}}{2} \\ = \frac{M x^{3}}{6} \end{aligned}

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5.3.7

Question

A fixed point of a function $f$ is a value $x$ where $f (x) = x$ . Show that if $f$ is differentiable on an interval with $f^{'} (x) \neq 1$ then $f$ can have at most one fixed point.

Proof

Suppose $f$ is differentiable on some interval $(a, b)$ (at worst it's open) with $f^{'} (x) \neq 1$ . Assume for contradiction that there are at least two unique fixed points $x_{0}, x_{1}$ on that interval.

Since $x_{0}, x_{1}$ are fixed points, then $f (x_{0}) = x_{0}$ and $f (x_{1}) = x_{1}$ . Now because $x_{0} \neq x_{1}$ (since they are unique) then one is larger. WLOG let $x_{0} < x_{1}$ . Then by the Mean Value Theorem (MVT) ( $f$ is continuous on $(x_{0}, x_{1}) \subseteq (a, b)$ ) then $\exists c \in (x_{0}, x_{1}) \subseteq (a, b)$ such that:

f^{'} (c) = \frac{f (x_{1}) - f (x_{0})}{x_{1} - x_{0}} = \frac{x_{1} - x_{0}}{x_{1} - x_{0}} = 1

but this contradicts that $f^{'} (x) \neq 1$ for any $x$ (in this case we found the point $c$ ).

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5.3.8

Question

Suppose $f$ is continuous on an interval containing zero and differentiable for all $x \neq 0$ . If $lim_{x \to 0} f^{'} (x) = L$ , show $f^{'} (0)$ exists and equals $L$ .

Proof

Let $[a, b]$ be this interval where $a < 0 < b$ . We want a function $g (x)$ such that we can do:

lim_{x \to 0} g^{'} (x) = lim_{x \to 0} \frac{g (x) - g (0)}{x - 0} \overset{L H}{=} \dots = L

because going through this process with $f$ requires us determining $f (0) = 0$ which is impossible here. Choose $g (x) = f (x) - f (0)$ such that we know that $g (0) = 0$ as required for this to work. Notice $g$ is continuous on the interval and differentiable for all $x \neq 0$ just like $f$ . Then:

g^{'} (x) = f^{'} (x) - f^{'} (0)

Thus:

\begin{aligned} lim_{x \to 0} g^{'} (x) & = lim_{x \to 0} f^{'} (x) - lim_{x \to 0} f^{'} (0) \\ = L - f^{'} (0) \end{aligned}

but notice by direct calculation that $g^{'} (0)$ exists and is zero, so then $L = f^{'} (0)$ as requested.

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5.3.10

Question

Let $f (x) = x \sin (\frac{1}{x^{4}}) e^{- 1 / x^{2}}$ and $g (x) = e^{\frac{- 1}{x^{2}}}$ . Using the familiar properties of these functions, compute the limit as $x \to 0$ of $f, g, \frac{f}{g}, \frac{f^{'}}{g^{'}}$ . Explain why the results are surprising but not in conflict with L'Hopital's Rule(s).

Proof

\begin{aligned} lim_{x \to 0} f (x) & = lim_{x \to 0} x \sin (\frac{1}{x^{4}}) e^{\frac{- 1}{x^{2}}} \\ \leq lim_{x \to 0} x \sin (\frac{1}{x^{4}}) \\ = 0 \end{aligned}

Where we got the limit from Lecture 29 - Chain Rule#Important Examples of Peculiarly Derived Functions and considering the similar limit $lim_{x \to 0} x \sin (\frac{1}{x})$ . So using the Limits and Order (Order Limit Theorem) (since we bound below by $0$ ) then the limit is 0.

\begin{aligned} lim_{x \to 0} g (x) & = lim_{x \to 0} e^{- 1 / x^{2}} \\ = 0 \end{aligned}

Because it get's squeezed by $e^{- | x |}$ and $0$ in the interval $[- 1, 1]$ which both have limit 0.

\begin{aligned} lim_{x \to 0} \frac{f (x)}{g (x)} & = lim_{x \to 0} x \sin (\frac{1}{x^{4}}) \\ = 0 \end{aligned}

per our argument for $lim_{x \to 0} f (x)$ .

Now for $lim_{x \to 0} \frac{f^{'}}{g^{'}}$ let's get the derivatives:

f^{'} (x) = \frac{e^{\frac{- 1}{x^{2}}} (x^{2} (x^{2} + 2) \sin (\frac{1}{x^{4}}) - 4 \cos (\frac{1}{x^{4}}))}{x^{4}}

g^{'} (x) = \frac{2 e^{\frac{- 1}{x^{2}}}}{x^{3}}

Thus:

\begin{aligned} \frac{f^{'}}{g^{'}} (x) & = \frac{1}{2 x} \cdot (x^{2} (x^{2} + 2) \sin (\frac{1}{x^{4}}) - 4 \cos (\frac{1}{x^{4}})) \end{aligned}

Calculating the limit:

\begin{aligned} lim_{x \to 0} \frac{f^{'}}{g^{'}} & = lim_{x \to 0} \frac{1}{2 x} \cdot (x^{2} (x^{2} + 2) \sin (\frac{1}{x^{4}}) - 4 \cos (\frac{1}{x^{4}})) \\ = lim_{x \to 0} x (x^{2} + 2) \sin (\frac{1}{x^{4}}) - 2 lim_{x \to 0} \frac{1}{x} \cdot \cos (\frac{1}{x^{4}}) \end{aligned}

Where the right limit doesn't exist (due to the $\frac{1}{x}$ part) so then the limit doesn't exist. This shows that the converse for L'Hopital's Rule(s) doesn't necessarily hold.

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5.3.12

Question

If $f$ is twice differentiable on an open interval containing $a$ and $f^{″}$ is continuous at $a$ , show:

lim_{h \to 0} \frac{f (a + h) - 2 f (a) + f (a - h)}{h^{2}} = f^{″} (a)

Proof

We will first prove (as the book mentions) that since $f$ is differentiable at $a$ then:

f^{'} (a) = lim_{h \to 0} \frac{f (a + h) - f (a)}{h}

(ie: the alternative definition of a limit works). Simply set $h = x - a$ then:

= lim_{x \to c} \frac{f (x) - f (a)}{x - a}

as desired.

Now for the actual proof:

\begin{aligned} f^{″} (a) & = lim_{h \to 0} \frac{f^{'} (a) - f^{'} (a - h)}{h} \\ = lim_{h \to 0} \frac{(lim_{h_{1} \to 0} \frac{f (a + h_{1}) - f (a)}{h_{1}} - lim_{h_{2} \to 0} \frac{f (a) - f (a - h_{2})}{h_{2}})}{h} \\ = lim_{h \to 0} \frac{(lim_{h \to 0} \frac{f (a + h) - 2 f (a) + f (a - h)}{h})}{h} & (h_{1} = h_{2} = h \to 0) \\ = lim_{h \to 0} \frac{f (a + h) - 2 f (a) + f (a - h)}{h^{2}} \end{aligned}

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