Differentiation & the Mean Value Theorem
Differentiation & the Mean Value Theorem
The derivative $f'(x_0)=\lim_{h\to 0}(f(x_0+h)-f(x_0))/h$ measures instantaneous rate of change. The Mean Value Theorem (MVT) — $f'(c)=(f(b)-f(a))/(b-a)$ for some $c\in(a,b)$ — is the central theorem of differential calculus, connecting the average rate of change to an instantaneous one. The MVT implies monotonicity, Lipschitz conditions, and L'Hôpital's rule.
Differentiability & Lipschitz Condition
$f$ is differentiable at $x_0$ if $\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}=f'(x_0)\in\mathbb{R}$. Differentiability implies continuity (but not vice versa: $|x|$ is continuous, not differentiable at 0). Lipschitz condition: $f$ satisfies Lip($L$) on $D$ if $|f(x)-f(y)|\leq L|x-y|$ for all $x,y\in D$. If $|f'|\leq L$ on $(a,b)$, then $f$ is Lip($L$) on $[a,b]$ (by MVT). Lipschitz $\Rightarrow$ uniformly continuous; uniform continuity $\not\Rightarrow$ Lipschitz ($\sqrt{x}$ at 0).
Mean Value Theorem
Let $f:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then $\exists c\in(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$. Proof uses Rolle's theorem: $g(x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)$ satisfies $g(a)=g(b)=0$ and is differentiable; by Rolle's, $g'(c)=0$ for some $c$, giving the MVT. Corollary: if $f'=0$ on $(a,b)$, then $f$ is constant; if $f'>0$, $f$ is strictly increasing.
Example 1
Prove $|\sin x-\sin y|\leq|x-y|$ for all $x,y\in\mathbb{R}$.
Solution: By MVT: $\sin x-\sin y=\cos(c)(x-y)$ for some $c$ between $x$ and $y$. Since $|\cos c|\leq 1$, $|\sin x-\sin y|=|\cos c||x-y|\leq|x-y|$. This shows $\sin$ is Lipschitz-1 (non-expansive).
Example 2
Use L'Hôpital's rule to evaluate $\lim_{x\to 0}(e^x-1-x)/x^2$.
Solution: Numerator $\to 0$, denominator $\to 0$ (0/0 form). L'Hôpital: $\lim\frac{e^x-1}{2x}$. Still 0/0. Apply again: $\lim\frac{e^x}{2}=\frac{1}{2}$. Verify directly: $e^x=1+x+x^2/2+O(x^3)$, so $(e^x-1-x)/x^2=(x^2/2+O(x^3))/x^2\to 1/2$.
Practice
- Prove Rolle's theorem from the extreme value theorem.
- Show that if $f'(x)>0$ for all $x\in(a,b)$ then $f$ is strictly increasing on $[a,b]$.
- Prove the Cauchy mean value theorem: if $f,g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ with $g'\neq 0$, then $\exists c$ with $\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$.
- Suppose $|f'(x)|\leq M|f(x)|$ for all $x$ with $f(0)=0$. Prove $f\equiv 0$ on $[0,1/(M+1)]$.
Show Answer Key
1. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then by EVT $f$ attains max and min on $[a,b]$. If both occur at endpoints, $f$ is constant and $f'\equiv0$. Otherwise an extremum is interior at $c$, and $f'(c)=0$.
2. For $a
3. Apply Rolle's theorem to $h(x)=f(x)[g(b)-g(a)]-g(x)[f(b)-f(a)]$. Then $h(a)=h(b)$, so $\exists c$ with $h'(c)=0$, giving $f'(c)[g(b)-g(a)]=g'(c)[f(b)-f(a)]$. Divide by $g'(c)[g(b)-g(a)]\neq0$.
4. On $[0,\delta]$ with $\delta=1/(M+1)$: let $g(x)=\max_{[0,x]}|f|$. Then $|f(x)|\le\int_0^x|f'|\le M\int_0^x|f|\le Mx\cdot g(x)\le M\delta\cdot g(x)