Continuity of the Derivative

Darboux Theorem

$f:[a,b]\to\mathbb{R}$ is differentiable on $[a,b]$. Assume that $f^\prime(a)<f^\prime(b)$. Then $\forall\lambda \in (f^\prime(a), f^\prime(b))$, there is $x\in (a,b)$ such that $f^\prime(x) = \lambda$.

<Proof>

Let $\lambda \in \left(f^\prime(a), f^\prime(b)\right)$ be given. Consider $g(t):= f(t)-\lambda t$. It suffices to show that $g^\prime(x) = 0$ for some in $x \in(a,b)$.

Observe that

\[\begin{align*} g^\prime(a)&=f^\prime(a)-\lambda < 0 \\ g^\prime(b)&=f^\prime(b)-\lambda >0 \end{align*}\]

Then $g$ cannot obtain a minimum or maximum at $a$ or $b$. Now $g$ is continuous on $[a,b]$ and $[a,b]$ is compact set. By extreme value theorem, $g$ obtains a minimum or maximum on $(a,b)$.

$\therefore g^\prime(x) = 0$ for some $x\in (a,b)$.

\[\tag*{$\square$}\]

L’Hôpital’s rule

$f$ and $g$ are differentiable on $(a,b)$ where $-\infty \leq a <b \leq +\infty$. Suppose that

\[\begin{equation*} g^\prime(x) \neq 0, \forall x \in (a,b) \text{ and } \lim_{x\to a^+}\frac{f^\prime(x)}{g^\prime(x)}= A\: (A\in [-\infty, +\infty]). \end{equation*}\]

If $\lim_{x\to a^+}f(x)=\lim_{x\to a^+} g(x)=0( \text{ or } \lim_{x\to a^+}g(x)=\pm\infty)$, then

\[\begin{equation*} \lim_{x\to a^+}\frac{f(x)}{g(x)}=A. \end{equation*}\]

<Proof>

For simplicity, let $A$ be finite. We want to show that $\lim_{x\to a^+}\frac{f(x)}{g(x)}=A$. Let $\epsilon >0$ be given. Since

\[\begin{equation*} \lim_{x\to a^+}\frac{f^\prime(x)}{g^\prime(x)}=A, \end{equation*}\]

there is a $\delta>0$ such that $\forall c \in (a, a+\delta)$,

\[\begin{align*} \left|\frac{f^\prime(c)}{g^\prime(c)} - A \right| < \frac{\epsilon}{2}. \end{align*}\]

Now consider any $(x,y) \subset (a, a+\delta)$. Since $g^\prime(t) \neq 0$ for all $t\in (a,b), g^\prime(c) \neq 0$. Moreover $g(y)\neq g(x)$. If $g(y)=g(x)$, then there is a $x_0\in (a,b)$ such that $g^\prime (x_0) =0$ by Rolle’s Theorem. But it is a contradiction.

Then by Cauchy mean value theorem, there is a $c\in (x,y)$ such that

\[\begin{equation*} \frac{f(y)-f(x)}{g(y)-g(x)} = \frac{f^\prime(c)}{g^\prime(c)} \end{equation*}\]

Take the limit of

\(\begin{equation*} A-\frac{\epsilon}{2} < \frac{f(y)-f(x)}{g(y)-g(x)} < A+\frac{\epsilon}{2} \end{equation*}\) as $x\to a^+$. Then by the assumption, $f(x)\to 0, g(x)\to 0$. Then

\[\begin{equation*} \left|\frac{f(y)}{g(y)}-A\right| \leq \frac{\epsilon}{2} \end{equation*}\]

It means that

\[\begin{equation*} y\in (a, a+\delta), \left| \frac{f(y)}{g(y} -A \right| \leq \frac{\epsilon}{2} < \epsilon \end{equation*}\] \[\begin{equation*} \therefore \lim_{y\to a^+}\frac{f(y)}{g(y)}=A \end{equation*}\] \[\tag*{$\square$}\]

Talyor’s Theorem

Let $f:[a,b]\to\mathbb{R}$. $f^{(k)}$ is continuous on $[a,b]$ and differentiable on $(a,b)$ for $k=0, \ldots, n-1$. Let $\alpha\in [a,b]$. We define $(n-1)$-th Taylor Polynomial of $f$ centered at $\alpha$ as

\[\begin{align*} P(t):=\sum_{k=0}^{n-1} \frac{f^{(k)}(\alpha)}{k!}(t-a)^k \end{align*}\]

Then, given any $\beta\in [a,b]$, there is a $x\in (\alpha, \beta)$ such that

\[\begin{align*} f(\beta) = P(\beta) + \frac{f^{(n)}(x)}{n!}(\beta-\alpha)^n \end{align*}\]

<Proof>

Let

\[\begin{align*} M:=\frac{f(\beta)-P(\beta)}{(\beta-\alpha)^n} \end{align*}\]

We want to show that there is a $x\in (\alpha, \beta)$ such that $f^{(n)}(x)/n! = M$. Consider $g(t) := f(t) - P(t) - M(t-\alpha)^n$. Note that $g^{(n)}(t) = f^{(n)}(t)- n!M$.

Observe that $P^{(k)}(\alpha)=f^{(k)}(\alpha)$ for $k=0,\ldots, n-1$. Thus, $g(\alpha)=g^\prime(\alpha)=g^{(2)}(\alpha)=\cdots=g^{(n-1)}(\alpha)=0$. Moreover, $g(\beta)=0$.

By mean value theorem, there is a $x_1\in (\alpha, \beta)$ such that $g^\prime(x_1)=0$. Continue to apply mean value theorem to $n-1$. Then, there is $x\in (\alpha, x_{n-1})$ such that $g^{(n)}(x)=0.$

\[\begin{align*} \therefore f(\beta) = P(\beta) + \frac{f^{(n)}(x)}{n!}(\beta-\alpha)^n \end{align*}\]

\(\tag*{$\square$}\)

Reference

• Walter Rudin, 『The Principles of Mathematical Analysis』