10.2 Differentiable Functions on $\mbox{{\bf R}}$

10.21   Warning. By the definition of differentiablity given in Math 111, the domain of a function was required to contain some interval $(a-\varepsilon, a+\varepsilon)$ in order for the function be differentiable at $a$. In definition 10.1 this condition has been replaced by requiring $a$ to be a limit point of the domain of the function. Now a function whose domain is a closed interval $[a,b]$ may be differentiable at $a$ and/or $b$.

10.22   Definition (Critical point.) Let $f$ be a complex function, and let $a\in\mbox{{\bf C}}$. If $f$ is differentiable at $a$ and $f'(a)=0$, we call $a$ a critical point for $f$.

10.23   Theorem (Critical Point Theorem.) Let $f\colon\mbox{{\rm dom}}(f)\to\mbox{{\bf R}}$ be a function. Suppose $f$ has a maximum at some point $c\in\mbox{{\rm dom}}(f)$, and that $\mbox{{\rm dom}}(f)$ contains an interval $(c-\varepsilon,c+\varepsilon)$ where $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$. If $f$ is differentiable at $c$, then $f'(c)=0$. The theorem also holds if we replace `` maximum" by `` minimum."

Proof: Suppose $f$ has a maximum at $c$,

$$f'(c)=\lim_{x\to c}{{f(x)-f(c)}\over {x-c}}.$$

Define two sequences $\{a_n\}$, $\{b_n\}$ in $(c-\varepsilon,c+\varepsilon)$ by

$$\begin{eqnarray*} a_n&=&c-{\varepsilon\over {(n+2)}} \mbox{ for all }n\in\mbox{{... ...&c+{\varepsilon\over {(n+2)}}\mbox{ for all }n\in\mbox{{\bf N}}. \end{eqnarray*}$$

Clearly $\{a_n\} \to c$ and $\{b_n\} \to c$, and $f(a_n)\leq f(c)$ and $f(b_n)\leq f(c)$ for all $n\in\mbox{{\bf N}}$. We have

$${{f(a_n)-f(c)}\over {a_n-c}}={{f(a_n)-f(c)}\over {-\left( {\varepsilon\over {n+2}}\right)}}\geq 0.$$

By the inequality theorem,

$$f'(c)=\lim\left\{ {{f(a_n)-f(c)}\over {a_n-c}}\right\}\geq 0.$$

Also,

$${{f(b_n)-f(c)}\over {b_n-c}}={{f(b_n)-f(c)}\over {\left( {\varepsilon\over {n+2}}\right)}}\leq 0,$$

so

$$f'(c)=\lim\left\{ {{f(b_n)-f(c)}\over {b_n-c}}\right\}\leq 0.$$

Since $0\leq f'(c)\leq 0$, we conclude that $f'(c)=0$. The proof for minimum points is left to you. $\mid\!\mid\!\mid$

10.24   Theorem (Rolle's Theorem.) Let $a,b\in\mbox{{\bf R}}$ with $a<b$ and let
$f\colon [a,b]\to\mbox{{\bf R}}$ be a function that is continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose that $f(a)=f(b)$. Then there is a number $c\in(a,b)$ such that $f'(c)=0$.

Proof: We know from the extreme value theorem that $f$ has a maximum at some point $p\in [a,b]$. If $p\in(a,b)$, then the critical point theorem says $f'(p)=0$, and we are finished. Suppose $p\in\{a,b\}$. We know there is a point $q\in[a,b]$ such that $f$ has a minimum at $q$. If $q\in(a,b)$ we get $f'(q)=0$ by the critical point theorem, so suppose $q\in\{a,b\}$. Then since $f(a)=f(b)$ and $p\in\{a,b\},q\in\{a,b\}$, we have $f(p)=f(q)$, and it follows that $f$ is a constant function on $[a,b]$, and in this case $f'(c)=0$ for all $c\in(a,b)$. $\mid\!\mid\!\mid$

10.25   Theorem (Mean Value Theorem.) Let $a,b\in\mbox{{\bf R}}$ with $a<b$, and let $f$ be a function from $[a,b]$ to $\mbox{{\bf R}}$ such that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Then there is a point $c\in(a,b)$ such that

$$f'(c)={{f(b)-f(a)}\over {b-a}}.$$

This theorem says that the tangent to the graph of $f$ at some point $\left(c,f(c)\right)$ is parallel to the chord joining $\left(a,f(a)\right)$ to $\left(b,f(b)\right)$.

Proof: Let

$$L(x)=f(a)+{{f(b)-f(a)}\over {b-a}}(x-a)\mbox{ for all } x\in\mbox{{\bf R}},$$

so the equation of the line joining $\left(a,f(a)\right)$ to $\left(b,f(b)\right)$ is $y=L(x)$, and

$$L'(x) = {f(b)-f(a) \over b-a} \mbox{ for all }x \in\mbox{{\bf R}}.$$

Let

$$\Delta(x)=f(x)-L(x)\mbox{ for all } x\in [a,b].$$

Then

$$\begin{eqnarray*} \Delta(a)&=&f(a)-L(a)=f(a)-f(a)=0,\\ \Delta(b)&=&f(b)-L(b)=f(b)-f(b)=0, \end{eqnarray*}$$

and $\Delta$ is continuous on $[a,b]$ and differentiable on $(a,b)$. By Rolle's theorem, there is some $c\in(a,b)$ such that $\Delta '(c)=0$; i.e., $f'(c)-L'(c)=0$; i.e.,

$$f'(c)=L'(c)={{f(b)-f(a)}\over {b-a}}.\mbox{ $\mid\!\mid\!\mid$}$$

10.26   Remark. The mean value theorem does not hold for complex valued functions. Let

$$F(t) = (1+it)^4 \mbox{ for all }t \in [-1,1].$$

Then

$$F(\pm 1) = (1\pm i)^4 = (\pm 2i)^2 = -4,$$

so

$${F(1) - F(-1) \over 1 - (-1) } = 0.$$

But

$$F'(t) = 4i(1+it)^3,$$

so $F'(t) = 0 \hspace{1ex}\Longleftrightarrow\hspace{1ex}t = -i$, and there is no point in $t \in (-1,1)$ with $F'(t) = 0.$

10.27   Definition (Interior point.) Let $J$ be an interval in $\mbox{{\bf R}}$. A number $a \in J$ is an interior point of $J$ if and only if $a$ is not an end point of $J$. The set of all interior points of $J$ is called the interior of $J$ and is denoted by $\mbox{\rm int}(J)$.

10.28   Examples. If $a<b$, then

$$\begin{eqnarray*} \mbox{\rm int}( [a,b)) = \mbox{\rm int}([a,b]) = \mbox{\rm int... ...rm int}([a,\infty)) = \mbox{\rm int}((a,\infty)) &=& (a,\infty). \end{eqnarray*}$$

If $J$ is an interval, and $s,t$ are points in $J$ with $s<t$, then every point in $(s,t)$ is in the interior of $J$.

10.29   Theorem. Let $J$ be an interval in $\mbox{{\bf R}}$, and let $f\colon J\to\mbox{{\bf R}}$ be a continuous function on $J$. Then:

a)

If $f'(x)\geq 0$ for all $x\in \mbox{\rm int}(J)$, then $f$ is increasing on $J$.

b)

If $f'(x)>0$ for all $x\in \mbox{\rm int}(J)$, then $f$ is strictly increasing on $J$.

c)

If $f'(x)\leq 0$ for all $x\in \mbox{\rm int}(J)$, then $f$ is decreasing on $J$.

d)

If $f'(x)<0$ for all $x\in \mbox{\rm int}(J)$, then $f$ is strictly decreasing on $J$.

e)

If $f'(x)=0$ for all $x\in \mbox{\rm int}(J)$, then $f$ is constant on $J$.

Proof: All five statements have similar proofs. I'll prove only part a).

Suppose $f'(x)\geq 0$ for all $x\in \mbox{\rm int}(J)$. Then for all $s,t\in J$ with $s<t$ we have $f$ is continuous on $[s,t]$ and differentiable on $(s,t)$, so by the mean value theorem

$$\begin{eqnarray*} s<t&\mbox{$\Longrightarrow$}& {{f(t)-f(s)}\over {t-s}}=f'(c)\m... ...ow$}&f(t)-f(s)\geq 0\\ &\mbox{$\Longrightarrow$}&f(t)\geq f(s). \end{eqnarray*}$$

Hence, $f$ is increasing on $J$.

10.30   Exercise. Prove part e) of the previous theorem; i.e., show that if $J$ is an interval in $\mbox{{\bf R}}$ and $f\colon J\to\mbox{{\bf R}}$ is continuous and satisfies $f'(t)=0$ for all $t\in \mbox{\rm int}(J)$, then $f$ is constant on $J$. [It is sufficient to show that $f(s)=f(t)$ for all $s,t\in J$.]

10.31   Exercise. A For each assertion below, either prove that the assertion is true for all functions $f$, or give a function $f$ for which the assertion is false. (A proof may consist of quoting a theorem.)

a) If $f$ is differentiable on $(-1,1)$ and $f$ is strictly increasing on $(-1,1)$, then $f'(t) > 0$ for all $t \in (-1,1)$.

b) If $f$ is differentiable on $[-1,1]$, and $f$ has a maximum at $t_0 \in [-1,1]$, then $f'(t_0) = 0$.

c) If $f$ is continuous on $[-1,1]$ and $f$ is differentiable on $(-1,1)$, and $f'(t) > 0$ for all $t \in (-1,1)$, then $f$ is strictly increasing on $[-1,1]$.

10.32   Theorem (Restriction theorem)

Let $S$ be a subset of $\mbox{{\bf C}}$, let $f:S\to \mbox{{\bf C}}$, and let $a\in S$ be a point such that $f$ is differentiable at $a$. Let $T$ be a subset of $S$ containing $a$, and let $f\vert _T : T \to \mbox{{\bf C}}$ be the restriction of $f$ to $T$, i.e.

$$f\vert _T(z) = f(z) \mbox{ for all }z \in T.$$

If $a$ is a limit point of $T$, then $f\vert _T$ is differentiable at $a$, and

$$f\vert _T'(a) = f'(a).$$

Proof: Let $\{z_n\}$ be any sequence in $T\setminus \{a\}$ such that $\{z_n\} \to a$. Then $\{z_n\}$ is a sequence in $S\setminus \{a\}$, and hence

$$\left\{ {f(z_n) - f(a) \over z_n-a} \right\} \to f'(a).$$

It follows that

$$\left\{ {f\vert _T(z_n) - f\vert _T(a) \over z_n - a} \right\} = \left\{ {f(z_n) - f(a) \over z_n-a}\right\} \to f'(a).$$

I've shown that

$$\lim_{z\to a} {f\vert _T(z) - f\vert _T(a) \over z-a} = f'(a).\mbox{ $\mid\!\mid\!\mid$}$$

10.33   Definition (Path, line segment.) If $a,b\in\mbox{{\bf C}}$, then the path joining $a$ to $b$ is the function $\lambda_{ab}\colon[0,1]\to\mbox{{\bf C}}$

$$\lambda_{ab}\colon t\mapsto a+t(b-a)\mbox{ for all } t\in[0,1]$$

and the set

$$\Lambda_{ab} =\lambda_{ab}([0,1])=\{a+t(b-a)\colon 0\leq t\leq 1\}$$

is called the line segment joining $a$ to $b$.

10.34   Example. We showed in example 10.9 that the function $\mbox{\rm conj}: z \mapsto z^*$ is a nowhere differentiable function on $\mbox{{\bf C}}$. I will show that for all $a$, $b$ in $\mbox{{\bf C}}$ with $a \neq b$, the restriction $\mbox{\rm conj}\vert _{\Lambda_{ab}}$ of conj to the line segment $\Lambda_{ab}$ is differentiable, and

$${\mbox{\rm conj}\vert _{\Lambda_{ab}}}'(z) = {b^*-a^* \over b-a} \mbox{ for all }z \in \Lambda_{ab}.$$

Note that all points of $\Lambda_{ab}$ are limit points of $\Lambda_{ab}$. If $z \in \Lambda_{ab}$, then for some real number $t$

$$z = a + t(b-a)$$ (10.35)

and

$$z^* = a^* + t(b^* - a^*).$$ (10.36)

If we solve equation (10.35) for $t$ we get

$$t = {z-a \over b-a}.$$

By using this value for $t$ in equation (10.36) we get

$$z^* = a^* + {b^*-a^* \over b-a}(z-a) \mbox{ for all }z \in \Lambda_{ab}.$$

Let $H_{ab}:\mbox{{\bf C}}\to \mbox{{\bf C}}$ be defined by

$$H_{ab}(z) = a^* + {b^*- a^* \over b-a}(z - a) \mbox{ for all }z \in\mbox{{\bf C}}.$$

Then $H_{ab}$ is differentiable, and $${H'(z) = {b^*-a^* \over b-a}}$$ for all $z\in\mbox{{\bf C}}$. We have

$$H_{ab}\vert _{\Lambda_{ab}} = \mbox{\rm conj}\vert _{\Lambda_{ab}},$$

so by the restriction theorem

$${\mbox{\rm conj}\vert _{\Lambda_{ab}}}'(z) = {H\vert _{\Lambd... ...(z) = {b^* - a^* \over b-a} \mbox{ for all }z \in \Lambda_{ab}.$$

10.37   Exercise. A Let $C(0,1)$ denote the unit circle in $\mbox{{\bf C}}$. Show that $\mbox{\rm conj}\vert _{C(0,1)}$ is differentiable, and that

$${\mbox{\rm conj}_{C(0,1)}}'(z) = - (z^*)^2 \mbox{ for all }z \in C(0,1).$$

In general, the real and imaginary parts of a differentiable function are not differentiable.

10.38   Example. If $f(z)=z$ for all $z\in\mbox{{\bf C}}$, then $f$ is differentiable and $f'(z)=1$. However, $\mbox{\rm Re}f$ is nowhere differentiable. In fact, if $a\in\mbox{{\bf C}}$, $${ {{\mbox{\rm Re}(z)-\mbox{\rm Re}(z)}\over {z-a}}}$$ has no limit at $a$. To see this, let $${a_n=a+{1\over n},\; b_n=a+{i\over n}}$$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$. Then $\{a_n\}\to a$, $\{b_n\}\to a$, and

$${{\mbox{\rm Re}(a_n)-\mbox{\rm Re}(a)}\over {a_n-a}} ={{\mbox{\rm Re}( a+{1\over n})-\mbox{\rm Re}(a)}\over {a+{1\over n}-a}}=1$$

and

$${{\mbox{\rm Re}(b_n)-\mbox{\rm Re}(a)}\over {b_n-a}}={{\mbox{\rm Re}(a)-\mbox{\rm Re}(a)} \over {a+{i\over n}-a}}=0.$$

Hence, the sequences $${ \left\{ {{\mbox{\rm Re}(a_n)-\mbox{\rm Re}(a)}\over {a_n-a}}\right\}_{n\geq 1}}$$ and $${ \left\{ {{\mbox{\rm Re}( b_n)-\mbox{\rm Re}(a)}\over {b_n-a}}\right\}_{n\geq 1}}$$ have different limits, so $${\lim_{z\to a}{{\mbox{\rm Re}(z)-\mbox{\rm Re}(a)} \over {z-a}}}$$ does not exist.

However, we do have the following theorem.

10.39   Theorem. Let $J$ be an interval in $\mbox{{\bf R}}$ and let $f\colon J\to \mbox{{\bf C}}$ be a function differentiable at a point $a \in J$. Write $f(t)=u(t)+iv(t)$ where $u,v$ are real valued. Then $u$ and $v$ are differentiable at $a$, and $f'(a)=u'(a)+iv'(a)$.

Proof: Since $f$ is differentiable at $a$ there is a function $D_af$ on $J$ such that $D_af$ is continuous at $a$ and

$$f(t)=f(a)+(t-a)D_af(t)\mbox{ for all }t\in J.$$

If $r\in\mbox{{\bf R}}$ and $c\in\mbox{{\bf C}}$, then $\mbox{\rm Re}(rc)=r\mbox{\rm Re}(c)$ and $\mbox{\rm Im}(rc)=r\mbox{\rm Im}(c)$, so

$$(\mbox{\rm Re}(f))(t)=(\mbox{\rm Re}(f))(a)+(t-a)(\mbox{\rm Re}(D_af))(t) \mbox{ for all }t\in J$$ (10.40)

and

$$(\mbox{\rm Im}(f))(t)=(\mbox{\rm Im}(f))(a)+(t-a)(\mbox{\rm Im}(D_af))(t) \mbox{ for all }t\in J.$$ (10.41)

Since $D_af$ is continuous at $a$, $\mbox{\rm Re}(D_af)$ and $\mbox{\rm Im}(D_af)$ are continuous at $a$, so equations (10.40) and (10.41) show that $\mbox{\rm Re}f$ and $\mbox{\rm Im}f$ are differentiable and

$$\begin{eqnarray*} (\mbox{\rm Re}f)'(a)&=&\left(\mbox{\rm Re}(D_af(a)\right)=\mbo... ...a)\right)\right)=\mbox{\rm Im}(f'(a)).\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$

10.42   Example. Let $a\in\mbox{{\bf R}}$, and let $f(t)=(2t+ia)^3$ for all $t\in\mbox{{\bf R}}$. Then $f$ is differentiable and (by the chain rule),

$$\begin{eqnarray*} f'(t)&=&3(2t+ia)^2\cdot 2\\ &=&6[(4t^2-a^2)+4iat] \\ &=&(24t^2-6a^2)+24iat. \end{eqnarray*}$$

We have by direct calculation,

$$\begin{eqnarray*} f(t)&=&8t^3+12iat^2-6ta^2-ia^3\\ &=&(8t^3-6ta^2)+(12at^2-a^3)i, \end{eqnarray*}$$

so

$$f'(t)=(24t^2-6a^2)+(24at)i.$$

(This example just illustrates that the theorem is true in a special case.)

10.43   Theorem. Let $f$ be a complex function and let $a,b\in\mbox{{\bf C}}$, and suppose $\mbox{{\rm dom}}(f)$ contains the line segment $\Lambda_{ab}$, and that $f'(z)=0$ for all $z \in \Lambda_{ab}$. Then $f$ is constant on $\Lambda_{ab}$; i.e., $f(z)=f(a)$ for all $z \in \Lambda_{ab}$.

Proof: Define a function $F\colon[0,1]\to\mbox{{\bf C}}$ by

$$F(t)=f\left(\lambda_{ab}(t)\right)=f\left(a+t(b-a)\right).$$

By the chain rule, $F$ is differentiable on $[0,1]$ and $F'(t)=f'\left(a+t(b-a)\right)\cdot(b-a)$. Since $f'(z)=0$ for all $z\in\Lambda_{ab}([0,1])$, we have $F'(t)=0$ for all $t\in[0,1]$. Hence

$$(\mbox{\rm Re}(F))'(t)=0\mbox{ and } (\mbox{\rm Im}(F))'(t)=0\mbox{ for all } t\in[0,1]$$

and hence

$$\mbox{\rm Re}(F)\mbox{ and }\mbox{\rm Im}(F)\mbox{ are constant on } [0,1].$$

If $\mbox{\rm Re}(F)=p$ and $\mbox{\rm Im}(F)=q$, then $F(t)=p+iq$ for all $t\in[0,1]$. $\mid\!\mid\!\mid$

10.44   Exercise. Let $D(a,\varepsilon)$ be a disc in $\mbox{{\bf C}}$.

a)

Show that if $b\in D(a,\varepsilon)$ then the segment $\Lambda_{ab}$ is a subset of $D(a,\varepsilon)$.

b)

Let $f\colon D(a,\varepsilon)\to\mbox{{\bf C}}$ be a function such that $f'(z)=0$ for all $z\in D(a,\varepsilon)$. Show that $f$ is constant on $D(a,\varepsilon)$.