Mean Value Theorem
Definition 5.2.1
Let $E\subset \mathbb{R}$ be a set and let $f:E\to\mathbb{R}$.
(a) $f$ has a local maximum at $p\in E$ if there exists a $\delta>0$ such that $f(x)\leq f(p)$ for all $x\in E\cap N_\delta (p)$. The function $f$ has a global maximum at $p\in E$ if $f(x)\leq f(p)$ for all $x\in E$.
(b) Similarly, $f$ has a local minimum at a point $q\in E$ if there exists a $\delta>0$ such that $f(x)\geq f(q)$ for all $x\in E\cap N_\delta (q)$, $f$ has a global minimum at $q\in E$ if $f(x)\geq f(q)$ for all $x\in E$.
Remark
As a consequence of Corollary 4.2.9, every continuous real-valued function defined on a compact subset $K$ of $\mathbb{R}$ has a global maximum and minimum on $K$.
Theorem 5.2.2
Let $f:I\to\mathbb{R}$ be a real-valued function. Suppose that $f$ has either a local maximum or local minimum at $p\in \text{Int}(I)$.
If $f$ is differentiable at $p$, then $f^\prime(p) = 0$.
<proof> Suppose that $f$ has a local maximum and $f$ is differentiable at $p$. Then there is a $\delta>0$ such that $f(x)\leq f(p)$ for all $x\in I$ with $|x-p| <\delta$.
If $p<x<p+\delta$, then $\frac{f(x)-f(p)}{x-p}\leq 0$. Since $f$ is differentiable at $p$, for every $\epsilon>0$, there exists $\eta>0$ such that
\[\text{for all } x\in(x,x+\eta) \Rightarrow \frac{f(x)-f(p)}{x-p}-\epsilon<f^\prime_+(p) < \frac{f(x)-f(p)}{x-p} + \epsilon\]Set $\delta^\prime = \min\{\delta, \eta\}$. Then for all $x\in (x,x+\delta^\prime)$,
\[f^\prime_+(p) < \frac{f(x)-f(p)}{x-p} + \epsilon\]Since $\epsilon$ is arbitrary, $f^\prime_+(p) \leq \frac{f(x)-f(p)}{x-p} \leq 0$.
Similarly, if $x\in (p-\delta^\prime, p)$
\[0\leq \frac{f(x)-f(p)}{x-p}-\epsilon < f^\prime_+(p)\]$f^\prime_-(p)\geq 0$. Finally, since $f^\prime_+(p)=f^\prime_-(p) = f^\prime(p)$, we have $f^\prime(p) =0$. The proof of the case where $f$ has a local minimum at $p$ is similar.
\[\tag*{$\square$}\]Corollary 5.2.3
Let $f$ be a continuous real-valued function on $[a,b]$. If $f$ has a local maximum or minimum at $p\in (a,b)$, then either the derivative of $f$ at $p$ does not exists, or $f^\prime(p)=0$.
Remark
The conclusion of Theorem 5.2. is not valid if $p\in I$ is an endpoint of the interval. For example, if $f:[a,b]\to\mathbb{R}$ has a local maximum at $a$, and if $f$ is differentiable at $a$, then we can only conclude that $f^\prime(a) =f^\prime_+(a) \leq 0$.
Theorem 5.2.5 (Rolle’s Theorem)
Let $f$ be a continuous real-valued function on $[a,b]$ with $f(a) = f(b)$, and that $f$ is differentiable on $(a,b)$. Then there exists $c\in (a,b)$ such that $f^\prime(c) = 0$.
<proof>
If $f$ is constant function on $[a,b]$, then we are done.
Suppose that $f$ is not constant. Since $[a,b]$ is compact and $f$ is continuous on $[a,b]$, $f$ has a global maximum and global minimum on $[a,b]$ by Corollary 4.2.9.
If $f(t) > f(a)$ for some $t\in (a,b)$, then $f$ has a maximum at some $c\in (a,b)$. Thus, by Theorem 5.2.2, $f^\prime(c) =0$. If $f^\prime (t) < f(a)$ for some $t$, then $f$ has a minimum at some $c\in (a,b)$, and again $f^\prime(c)=0$.
\[\tag*{$\square$}\]Rolle’s Theorem-2
Let $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$. Then if $f(b)-f(a) = g(b)-g(a)$, $\exists c \in (a,b)$ where $f^\prime(c) = g^\prime(c)$.
<Proof>
Let $h(x):=f(x)-g(x)$. Then $h(b) = h(a)$. By Rolle’s Theorem, there is a $c\in (a,b)$ such that $h^\prime(c)=0$.
\[\tag*{$\square$}\]Theorem 5.2.6 (Mean Value Theorem)
If $f:[a,b]\to\mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in (a,b)$ such that
\[f(b)-f(a) = f^\prime(c) (b-a)\]<proof> Define $g(x) := f(x) - f(a) - \frac{f(b)-f(a)}{b-a}(x-a)$. Then $g$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $g(a)=0=g(b)$.
By Rolle’s Theorem, there is $c\in (a,b)$ such that $g^\prime (c) = 0$ for some $c\in (a,b)$.
\[g^\prime(x) = f^\prime(x) -\frac{f(b)-f(a)}{b-a}\]$\therefore f^\prime(c) = \frac{f(b)-f(a)}{b-a}$. \(\tag*{$\square$}\)
Example 5.2.7
Prove that $\frac{x}{1+x}\leq \log(1+x)\leq x$ for all $x>-1$
<proof> Define $f(x):=\log(1+x)$. Then $f(0)=0$. Suppose that $x>0$. Then $f:[0,x]\to\mathbb{R}$.
\[f^\prime(c) = \frac{1}{1+c} = \frac{ \log (1+x)}{x} \text{ for some }c\in (0,x)\]Since $c>0, \frac{1}{1+c}<1$.
$\therefore \log(1+x) < x$
$\therefore \frac{x}{1+x} < \log (1+x)$.
For $-1<x<0$,
\[f^\prime(c) = \frac{1}{1+c} = \frac{\log(1+x)}{x} \text{ for some } c\in(x,0)\]Since $-1<x<c<0, 1<\frac{1}{1+c}<\frac{1}{1+x}$, and since $x<0$
\[\frac{x}{1+x} < \log (1+x) < x\]Hence the desired inequality holds for all $x>-1$ with equality if and only if $x=0$. \(\tag*{$\square$}\)
Theorem 5.2.8 (Cauchy Mean Value Theorem)
If $f,g$ are continuous real-valued functions on $[a,b]$ that are differentiable on $(a,b)$, then there exists $c\in(a,b)$ such that
\[\left[f(b) -f(a) \right]g^\prime(c) = \left[g(b)-g(a) \right]f^\prime(c).\]<Proof>
Let
\[h(x) = [f(b) - f(a)]g(x) - [g(b)-g(a)]f(x)\]Then $h$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with
\[h(a) = f(b)g(a) - f(a)g(b) = h(b)\]Thus by Rolle’s theorem, there exists $c\in(a,b)$ such that $h^\prime(c) =0$, which gives the result. \(\tag*{$\square$}\)
<Proof-2>
Suppose that $g(b)-g(a)=0$. Then $g(b)=g(a)$. By Rolle’s Theorem, there exists $c\in (a,b)$ such taht $g^\prime(c)=0$.
$\therefore \left[f(b)-f(a) \right]g^\prime(c)=[g(b)-g(a)]f^\prime(c)$
Now assume that $g(b)\neq g(a)$. Multiply the function $g$ by
\[\begin{equation*} \alpha = \frac{f(b)-f(a)}{g(b)-g(a)} \end{equation*}\]Observe that $f(b)-f(a)=\alpha\cdot [g(b)-g(a)]$. Then by Rolle’s Theorem #2, there exists $c\in (a,b)$ such that $\alpha g^\prime(c) = f^\prime(c)$.
$\therefore [f(b)-f(a)]g^\prime(c) = [g(b)-g(a)]f^\prime(c)$
\(\tag*{$\square$}\)
Theorem 5.2.9
Suppose $f:I\to\mathbb{R}$ is differentiable on the interval $I$.
(a) If $f^\prime(x) \geq 0$ for all $x\in I$, then $f$ is monotone increasing on $I$.
(b) If $f^\prime(x) >0$ for all $x\in I$, then $f$ is strictly increasing on $I$.
(c) If $f^\prime(x) \leq 0$ for all $x\in I$, then $f$ is monotone decreasing on $I$.
(d) If $f^\prime (x)<0$ for all $x\in I$, then $f$ is strictly decreasing on $I$.
(e) If $f^\prime(x) = 0$ for all $x\in I$, then $f$ is constant on $I$.
<proof> Suppose that $x_1,x_2\in I$ with $x_1 < x_2$. By the mean value theorem applied to $f$ on $[x_1,x_2]$,
\[f(x_2) - f(x_1) = f^\prime(c)(x_2-x_1)\]for some $c\in (x_1,x_2)$. If $f^\prime(c) \geq 0$, then $f(x_2) \geq f(x_1)$. Thus, if $f^\prime(x) \geq 0$ for all $x\in I$, we have $f(x_2) \geq f(x_1)$ for all $x_1,x_2\in I$ with $x_1< x_2$. Thus, $f$ is monotone increasing on $I$. The other results follow similarly.
\[\tag*{$\square$}\]Theorem 5.2.11
$f:[a,b)\to\mathbb{R}$ is continuous on $[a,b)$ and differentiable on $(a,b)$. If $\displaystyle{\lim_{x\to a^+}}f^\prime(x)$ exists, then $f^\prime_+(a)$ exists and
\[f^\prime_+(a) = \lim_{x\to a^+}f^\prime(x)\]<proof> Let $L=f^\prime(a+)$ and let $\epsilon >0$ be given. Then there exists a $\delta>0$ such that
\[0<x<a+\delta \Rightarrow \left| f^\prime(x) - L\right| <\epsilon.\]Suppose that $0<h<\delta$ such that $a+h<b$. Then $f$ is continuous on $[a,a+h]$ and $f$ is differentiable on $(a,a+h)$.
By mean value theorem, there is $\xi_h\in (a,a+h)$ such that
\[f(a+h) - f(a) = f^\prime(\xi_h)h.\]Since $\xi_h\in (a,a+h)\subset (a,a+\delta)$,
\[\left|\frac{f(a+h)-f(a)}{h} -L \right| = \left|f^\prime(\xi_h) -L\right| <\epsilon.\]$\therefore f^\prime_+(a)$ exists and $f^\prime_+(a) = L$
\[\tag*{$\square$}\]Remark
Suppose that $c\in (a,b)$, then $f$ is differentiable at $c$. If $\displaystyle{\lim_{x\to c^-}f(x) \text{ and } \lim_{x\to c^+}f(x)}$ exist, then $f^\prime_-(c), f^\prime_+(c)$ both exist and
\[\lim_{x\to c^-}f(x) = f^\prime_-(c), \lim_{x\to c^+}f(x) = f^\prime_+(x)\]Since $f$ is differentiable at $c$, $f^\prime_-(c) = f^\prime(c) = f^\prime_+(c)$. That is $f^\prime$ is continuous at $c$.
\[\lim_{x\to c^+} f^\prime(x) = \lim_{x\to c^-}f^\prime (x) = \lim_{x\to c}f^\prime(x) = f^\prime (c)\]$\because$ Let $\epsilon >0$ be given. Then there is $\delta_1, \delta_2 >0$ such that \(\begin{align} \begin{split} c<x<c+\delta_1 &\Rightarrow \left| f^\prime(x) - L \right| <\epsilon \\ c-\delta_2 < x<c &\Rightarrow \left| f^\prime(x) -L \right| <\epsilon \end{split} \end{align}\) Take $\delta :=\min {\delta_1, \delta_2}$.
\[\text{For all } x \in (c-\delta, c+\delta) \Rightarrow |f^\prime(x) - L| <\epsilon\]$\therefore \displaystyle{\lim_{x\to c}f^\prime(x) = L = \lim_{x\to c^+}f^\prime(x) = \lim_{x\to c^-}f^\prime(x)}$
So, $f^\prime$ cannot be jump-discontinuous.
Reference
- Manfred Stoll, 『Introduction to Real Analysis』, Pearson