1. 미분 (ii): 평균값 정리와 그 응용Differentiation (ii): The Mean-Value Theorems and Applications

We simply reverse the direction of the inequality in Definition 1.35. Let $f$ be a real-valued function defined on a metric space $X$ with the metric $d$. We say that $f$ has a local minimum at a point $p \in X$ iff there exists $\delta > 0$ such that

for all $q \in X$ with $d(p, q) \lt \delta$.

It is true. Define $g(x) := -f(x)$. Since $f$ has a local minimum at the point $x \in (a, b)$, obviously $g$ has a local maximum at $x$. Therefore by Theorem 1.37,

that is, $f'(x) = 0$.

Both statements are true.

(i) Fix an arbitrary $x \in (a, b)$. Put $\phi(t) := \dfrac{f(t) - f(x)}{t - x}$ for $t \in (a, b)$, $t \neq x$.

• If $t > x$, since $f$ is monotonically increasing on $(a, b)$, $f(t) \geq f(x)$, i.e. $f(t) - f(x) \geq 0$, and $t - x > 0$, so $\phi(t) \geq 0$.
• If $t \lt x$, similarly $f(x) \geq f(t)$, i.e. $f(t) - f(x) \leq 0$, and $t - x \lt 0$, so again $\phi(t) \geq 0$.

Thus $\phi(t) \geq 0$ for all $t \in (a, b) \setminus \{x\}$. Since $f$ is differentiable at $x$, $L := f'(x) = \lim_{t \to x} \phi(t)$ exists. Assume for contradiction that $L \lt 0$. Put $\varepsilon := -\dfrac{L}{2} > 0$. By the definition of the limit, there exists a $\delta > 0$ such that

for all $t \in (a, b)$ with $0 \lt |t - x| \lt \delta$. In particular,

which contradicts $\phi(t) \geq 0$. Therefore $L \geq 0$, that is,

Since $x \in (a, b)$ was arbitrary, (i) is proved.

(ii) Similarly, fix an arbitrary $x \in (a, b)$ and put $\phi(t) := \dfrac{f(t) - f(x)}{t - x}$. Since $f$ is monotonically decreasing on $(a, b)$, if $t > x$ then $f(t) \leq f(x)$ and $t - x > 0$, so $\phi(t) \leq 0$; and if $t \lt x$ then $f(t) \geq f(x)$ and $t - x \lt 0$, so again $\phi(t) \leq 0$. Thus $\phi(t) \leq 0$ for all $t \in (a, b) \setminus \{x\}$. Since $L := f'(x) = \lim_{t \to x} \phi(t)$ exists, assume that $L > 0$. Then for $\varepsilon := \dfrac{L}{2} > 0$, there exists a $\delta > 0$ such that $|\phi(t) - L| \lt \varepsilon$ for all $t$ with $0 \lt |t - x| \lt \delta$, and hence

which contradicts $\phi(t) \leq 0$. Therefore $L \leq 0$, that is, $f'(x) \leq 0$. Since $x \in (a, b)$ was arbitrary, (ii) is also proved.

Both statements are false.

(i) Define

that is,

Then $f$ is continuous on $(-1, 1)$. Moreover, one can easily check that $f(x_1) \leq f(x_2)$ whenever $x_1 \lt x_2$ in $(-1, 1)$ (consider the cases where both points are negative, both are nonnegative, and $x_1 \lt 0 \leq x_2$ separately), so $f$ is monotonically increasing on $(-1, 1)$. However, for $t \lt 0$,

so

and for $t > 0$,

so

Since the left and right limits differ, $f'(0)$ does not exist. Thus $f$ is continuous and monotonically increasing on $(-1, 1)$, yet the statement "$f'(x) \geq 0$ for all $x \in (-1, 1)$" fails, since $f'(x)$ itself does not exist at $x = 0$. This is a counterexample for (i).

(ii) Define

that is, $g = -f$ and

Then $g$ is continuous on $(-1, 1)$, and as in (i), one can check that $g(x_1) \geq g(x_2)$ whenever $x_1 \lt x_2$ in $(-1, 1)$, so $g$ is monotonically decreasing on $(-1, 1)$. As in the computation for (i),

so $g'(0)$ does not exist. Thus $g$ is continuous and monotonically decreasing on $(-1, 1)$, yet the statement "$g'(x) \leq 0$ for all $x \in (-1, 1)$" fails at $x = 0$. This is a counterexample for (ii).

Both statements are true.

(i) Let $x, y \in (a, b)$ with $x \lt y$. Since $[x, y] \subset (a, b)$, $f$ is differentiable on $[x, y]$, and by Theorem 1.21, $f$ is continuous on $[x, y]$. Thus by the mean-value theorem (Theorem 1.41), there exists a $z \in (x, y)$ such that

Since $z \in (x, y) \subset (a, b)$, the assumption gives $f'(z) > 0$, and since $y - x > 0$,

that is, $f(x) \lt f(y)$. Since $x, y \in (a, b)$ with $x \lt y$ were arbitrary, $f$ is strictly increasing on $(a, b)$.

(ii) Similarly, let $x, y \in (a, b)$ with $x \lt y$. As in (i), $f$ is continuous on $[x, y]$ and differentiable on $(x, y)$, so by the mean-value theorem, there exists a $z \in (x, y)$ such that

By assumption $f'(z) \lt 0$ and $y - x > 0$, so

that is, $f(x) > f(y)$. Since $x, y \in (a, b)$ with $x \lt y$ were arbitrary, $f$ is strictly decreasing on $(a, b)$.

Existence of $\inf A$. Since $f(b) > \lambda$, we have $b \in A$, so $A$ is nonempty. Also $A \subset [a, b]$, so $a$ is a lower bound of $A$ and $A$ is bounded below. Thus by Remark 1.49, $\inf A$ exists in $\mathbb{R}$. Put $x = \inf A$.

Proof that $x \in (a, b)$. Since $A \subset [a, b]$, $a$ is a lower bound of $A$, and since $x = \inf A$ is the greatest lower bound, $a \leq x$. Also $b \in A$ and $x$ is a lower bound of $A$, so $x \leq b$. Thus $x \in [a, b]$. First observe that for all $y \in [a, x)$,

Indeed, if $y \lt x = \inf A$, then $y$ cannot be an element of $A$ (otherwise $\inf A \leq y \lt x = \inf A$, a contradiction). Thus by the definition of $A$, $f(y) \leq \lambda$.

We now use the reduction to absurdity to show $x \neq a$ and $x \neq b$.

First assume $x = a$. Since $f(a) \lt \lambda$, put $\varepsilon := \dfrac{\lambda - f(a)}{2} > 0$. Since $f$ is continuous at $a$, there exists a $\delta > 0$ such that

for all $y \in [a, a + \delta) \cap [a, b]$. Then

so no point of $[a, a + \delta) \cap [a, b]$ belongs to $A$. However, since $x = a = \inf A$, by Remark 1.49 there exists a $y^* \in A$ such that $y^* \lt a + \delta$, and since $A \subset [a, b]$, $y^* \in [a, a + \delta) \cap [a, b]$. This contradicts what we just showed. Hence $x \neq a$.

Next assume $x = b$. Since $f(b) > \lambda$, put $\varepsilon := \dfrac{f(b) - \lambda}{2} > 0$. Since $f$ is continuous at $b$, there exists a $\delta > 0$ such that

for all $y \in (b - \delta, b] \cap [a, b]$. Choose a $y \in (b - \delta, b) \cap [a, b]$. Since $y \lt b = x$, the fact shown above gives $f(y) \leq \lambda$. On the other hand,

so $f(y) > \lambda$, which contradicts $f(y) \leq \lambda$. Hence $x \neq b$.

Therefore $x \in (a, b)$.

Proof that $f(x) = \lambda$. Fix an arbitrary $\varepsilon > 0$. Since $f$ is continuous at $x$, there exists a $\delta > 0$ such that

for all $y \in (x - \delta, x + \delta) \cap [a, b]$. Since $x = \inf A$, by Remark 1.49 there exists a $y^* \in A$ such that $y^* \lt x + \delta$. Since $x$ is a lower bound of $A$, $y^* \geq x$, so $y^* \in [x, x + \delta)$; moreover $y^* \in A \subset [a, b]$, hence $y^* \in (x - \delta, x + \delta) \cap [a, b]$. Since $y^* \in A$, $f(y^*) > \lambda$, and combining this with the continuity inequality above,

On the other hand, since $x \in (a, b)$, there exists a $y^{**} \in (x - \delta, x) \cap [a, b]$. Since $y^{**} \lt x$, the fact shown above gives $f(y^{**}) \leq \lambda$, and combining this with the continuity inequality,

Therefore

and since $\varepsilon > 0$ was arbitrary, we conclude

(1) From the definition of the derivative. Since $f$ is differentiable at $b$, by the definition of the derivative (the left limit),

Since $f'(b) > 0$, put $\varepsilon := \dfrac{f'(b)}{2} > 0$. By the definition of the limit, there exists a $\delta > 0$ (which we may take $\leq b - a$) such that

for all $t \in (b - \delta, b)$. In particular,

Choose a $t_2 \in (b - \delta, b)$; then $t_2 \in (a, b)$ and $\dfrac{f(t_2) - f(b)}{t_2 - b} > 0$. Since $t_2 \lt b$, we have $t_2 - b \lt 0$, and hence

that is, $f(t_2) \lt f(b)$.

(2) By applying the statement (i). Define

The map $t \mapsto a + b - t$ is a strictly decreasing bijection of $[a, b]$ onto itself, so $g$ is well-defined on $[a, b]$. Fix $t \in [a, b]$ and put $s := a + b - t \in [a, b]$. For $\tau \in [a, b]$ with $\tau \neq t$, let $\sigma := a + b - \tau \in [a, b]$, so $\sigma \neq s$, and

Since $\tau \mapsto \sigma = a + b - \tau$ is decreasing, as $\tau$ approaches $t$ from one side, $\sigma$ approaches $s$ from the opposite side (for instance, if $t = a$ then $s = b$, and $\sigma \to b^-$ as $\tau \to a^+$). Since $f$ is differentiable on $[a, b]$, the corresponding (one-sided or two-sided) limit $f'(s) = \lim \dfrac{f(\sigma) - f(s)}{\sigma - s}$ exists, and so does the limit of the expression above, giving

In particular, $g$ is differentiable on $[a, b]$.

Setting $t = a$,

by the assumption $f'(b) > 0$. Thus applying the statement (i) to $g$, there exists a $t_1 \in (a, b)$ such that

By the definition of $g$, $g(a) = f(a + b - a) = f(b)$ and $g(t_1) = f(a + b - t_1)$, so

Put $t_2 := a + b - t_1$. Since $t_1 \in (a, b)$, we have $t_1 > a$ and $t_1 \lt b$, so $t_2 = a + b - t_1 \lt b$ and $t_2 > a$. That is, $t_2 \in (a, b)$ and

Define $g := -f$. Since $f$ is differentiable on $[a, b]$, so is $g$, with $g'(t) = -f'(t)$ for $t \in [a, b]$. Multiplying the assumption

by $-1$ gives

that is,

Thus applying the Darboux Theorem (Theorem 1.55) to $g$ and $-\lambda$, there exists an $x \in (a, b)$ such that

Since $g'(x) = -f'(x)$, we have $-f'(x) = -\lambda$, that is,

Put $L := \lim_{t \to x^+} f(t)$. Since $\lambda \lt L$, put $\varepsilon := L - \lambda > 0$. By the definition of the right limit, there exists a $\delta > 0$ (which we may take $\leq b - x$) such that

for all $t \in (x, x + \delta)$. In particular, for all such $t$,

Thus $\lambda \lt f(t)$ for all $t \in (x, x + \delta)$.

If $x = a$, then the left limit of $f'$ at $x$ cannot be defined and thus obviously the left limit does not exist. So we may assume $x \in (a, b]$. We use the reduction to absurdity, that is, assume that

Since $f'$ is not left continuous,

We additionally assume that

without loss of generality since we can consider the function $-f$. Choose a $\lambda$ such that

By the definition of the left limit, there exists a $\delta > 0$ (which we may take $\leq x - a$) such that

for all $t \in (x - \delta, x)$ (the first inequality $f'(x) \lt \lambda$ already holds regardless of $t$). Choose a $t' \in (x - \delta, x)$; then $t' \lt x$ and by (*), $f'(t') > \lambda$. Since $f$ is differentiable on $[t', x] \subset [a, b]$ and by (*)

applying Exercise 1.56 to the interval $[t', x]$ (that is, applying Exercise 1.56 with $a := t'$ and $b := x$), there exists an $x_1 \in (t', x)$ such that

However $x_1 \in (t', x) \subset (x - \delta, x)$, so by (*), $\lambda \lt f'(x_1)$. Since $f'(x_1) = \lambda$, we obtain the contradiction $\lambda \lt \lambda$. Therefore the left limit $\lim_{t \to x^-} f'(t)$ does not exist.

We use the reduction to absurdity: assume that $L := \lim_{t \to x} f'(t)$ exists.

(Case $x = a$) Since the domain of $f'$ is $[a, b]$, there are no points less than $a$ near $a$, so the existence of $\lim_{t \to a} f'(t)$ is the same as the existence of the right limit $\lim_{t \to a^+} f'(t)$. But $f'$ is not continuous at $a$, so in particular $f'$ is not right continuous at $a$, and by Corollary 1.57, the right limit $\lim_{t \to a^+} f'(t)$ does not exist. This is a contradiction.

(Case $x = b$) Similarly, the existence of $\lim_{t \to b} f'(t)$ is the same as the existence of the left limit $\lim_{t \to b^-} f'(t)$. Since $f'$ is not continuous at $b$, in particular $f'$ is not left continuous at $b$, and by Exercise 1.59, the left limit $\lim_{t \to b^-} f'(t)$ does not exist. This is a contradiction.

(Case $x \in (a, b)$) Since $L = \lim_{t \to x} f'(t)$ exists, both the left and right limits exist and equal $L$, that is,

Since the right limit exists, by the contrapositive of Corollary 1.57, $f'$ must be right continuous at $x$, that is,

Since the left limit exists, by the contrapositive of Exercise 1.59, $f'$ must be left continuous at $x$, that is,

Let $\varepsilon > 0$ be given. Since $f'$ is left continuous and right continuous at $x$, there exist $\delta_1, \delta_2 > 0$ such that

Put $\delta := \min\{\delta_1, \delta_2\}$. Since $|f'(t) - f'(x)| = 0 \lt \varepsilon$ trivially holds at $t = x$ as well,

for all $t \in (x - \delta, x + \delta)$. Since $\varepsilon > 0$ was arbitrary, $f'$ is continuous at $x$. This contradicts the assumption that $f'$ is not continuous at $x$.

Since all three cases lead to a contradiction, $\lim_{t \to x} f'(t)$ does not exist.