Here is Prob. 7, Sec. 6.1, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Suppose that $f \colon \mathbb{R} \to \mathbb{R}$ is differentiable at $c$ and that $f(c) = 0$. Show that $g(x) \colon= \lvert f(x) \rvert$ is differentialbe at $c$ if and only if $f^\prime(c)=0$.
My Attempt:
Suppose that $f^\prime(c)=0$.
Then, given a real number $\varepsilon > 0$, we can find a real number $\delta > 0$ such that
$$ \left\lvert \frac{ f(x) – f(c) }{ x -c } \right\rvert = \left\lvert \frac{ f(x) }{ x -c } \right\rvert < \varepsilon $$
for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta. $$Therefore for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta, $$ we find that
$$ \left\lvert \frac{g(x) – g(c) }{x-c} – 0 \right\rvert = \left\lvert \frac{ \lvert f(x) \rvert – \lvert f(c) \rvert }{ x-c} \right\rvert = \left\lvert \frac{ \lvert f(x) \rvert }{ x-c} \right\rvert = \frac{ \lvert f(x) \rvert }{ \lvert x-c \rvert } = \left\lvert \frac{ f(x) }{ x -c } \right\rvert < \varepsilon. $$
Since $\varepsilon > 0$ was arbitrary, it follows that $g$ is differentiable at $c$ and that
$$ g^\prime(c) = 0. $$Conversely, suppose that $f^\prime(c) \neq 0$. Then either $f^\prime(c) < 0$ or $f^\prime(c) > 0$.
Case 1. If $f^\prime(c) > 0$, then for $\varepsilon \colon= f^\prime(c)/2$, we can find a real number $\delta > 0$ such that
$$ \left\lvert \frac{ f(x) – f(c) }{ x -c } – f^\prime(c) \right\rvert < f^\prime(c)/2 $$
or
$$ 0 < \frac{f^\prime(c)}{2} = f^\prime(c) – f^\prime(c)/2 < \frac{ f(x) – f(c) }{ x -c } < f^\prime(c) + f^\prime(c)/2 $$
for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta. $$But $f(c) = 0$. Thus for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta, $$
we have
$$ \frac{ f(x) }{x-c} > \frac{f^\prime(c)}{2} > 0, \tag{1} $$
which implies that, for all $x \in \mathbb{R}$,
$$ f(x) \ \begin{cases} > 0 \ \mbox{ if } & c < x < c + \delta, \\ < 0 \ \mbox{ if } & c-\delta < x < c. \end{cases} $$
So from (1) it follows that
$$ \frac{g(x) – g(c) }{x-c} = \frac{ \lvert f(x) \rvert }{ x-c} = \begin{cases} \frac{f(x)}{x-c} & \mbox{ if } \ c < x < c+\delta, \\ -\frac{ f(x)}{x-c} & \mbox{ if } \ c-\delta < x < c. \end{cases} \tag{2}$$
Moreover, from (2) we can also conclude that
$$ \lim_{x \to c+} \frac{g(x) – g(c)}{x-c} = \lim_{x \to c+} \frac{ f(x) }{x-c} = \lim_{x \to c+} \frac{ f(x) – f(c) }{x-c} = f^\prime(c), $$
and
$$ \lim_{x \to c-} \frac{g(x) – g(c)}{x-c} = \lim_{x \to c-}\left(- \frac{ f(x) }{x-c} \right) = \lim_{x \to c-} \left( – \frac{ f(x) – f(c) }{x-c}\right) = – \lim_{x \to c-} \frac{ f(x) – f(c) }{x-c}=- f^\prime(c). $$
Thus if $f^\prime(c) > 0$, then
$$ \lim_{x \to c+} \frac{g(x) – g(c)}{x-c} \neq \lim_{x \to c-} \frac{g(x) – g(c)}{x-c}, $$
and so $g^\prime(c)$ does not exist. Therefore if $g$ is differentiable at $c$, then we must have $f^\prime(c) \not> 0$.Case 2. If $f^\prime(c) < 0$, then for $\varepsilon \colon= -f^\prime(c)/2 > 0$, we can find a real number $\delta > 0$ such that
$$ \left\lvert \frac{ f(x) – f(c) }{ x -c } – f^\prime(c) \right\rvert < – f^\prime(c)/2 $$
or
$$ \frac{3f^\prime(c)}{2} = f^\prime(c) – \frac{-f^\prime(c)}{2} < \frac{ f(x) – f(c) }{ x -c } < f^\prime(c) + \frac{-f^\prime(c)}{2} = \frac{ f^\prime(c)}{2} < 0 $$
for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta. $$But $f(c) = 0$. Thus for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta, $$
we have
$$ \frac{ f(x) }{x-c} < \frac{f^\prime(c)}{2} < 0, \tag{3} $$
which implies that, for all $x \in \mathbb{R}$,
$$ f(x) \ \begin{cases} < 0 \ \mbox{ if } & c < x < c + \delta, \\ > 0 \ \mbox{ if } & c-\delta < x < c. \end{cases} $$
So from (3) it follows that
$$ \frac{g(x) – g(c) }{x-c} = \frac{ \lvert f(x) \rvert }{ x-c} = \begin{cases} -\frac{f(x)}{x-c} & \mbox{ if } \ c < x < c+\delta, \\ \frac{ f(x)}{x-c} & \mbox{ if } \ c-\delta < x < c. \end{cases} \tag{4}$$
Moreover, from (4) we can also conclude that
$$ \lim_{x \to c-} \frac{g(x) – g(c)}{x-c} = \lim_{x \to c-} \frac{ f(x) }{x-c} = \lim_{x \to c-} \frac{ f(x) – f(c) }{x-c} = f^\prime(c), $$
and
$$ \lim_{x \to c+} \frac{g(x) – g(c)}{x-c} = \lim_{x \to c+}\left(- \frac{ f(x) }{x-c} \right) = \lim_{x \to c+} \left( – \frac{ f(x) – f(c) }{x-c}\right) = – \lim_{x \to c+} \frac{ f(x) – f(c) }{x-c}=- f^\prime(c). $$
Thus if $f^\prime(c) < 0$, then
$$ \lim_{x \to c+} \frac{g(x) – g(c)}{x-c} \neq \lim_{x \to c-} \frac{g(x) – g(c)}{x-c}, $$
and so $g^\prime(c)$ does not exist. Therefore if $g$ is differentiable at $c$, then we must have $f^\prime(c) \not< 0$.From the above two cases, we can conclude that if $g$ is differentiable at $c$, then we must have $f^\prime(c)=0$.
Is this proof correct? If so, then is the presentation clear and rigorous enough too? If not, then where are the issues as far as accuracy, rigor, or clarity of the argument go?
Best Answer
Of course everything you did is correct. But such a simple statement deserves a shorter proof, which then also shows what's going on here.
Assume that $c=0$, $f(0)=0$, and $f'(0)=a$. Then $$m(x):={f(x)\over x}\to a \quad (x\to0)\ ,$$ and therefore $${\bigl|f(x)\bigr|\over x}={\rm sgn}(x)\,\bigl|m(x)\bigr|\to\left\{\eqalign{|a|\quad&(x\to0+) \cr -|a|\quad&(x\to0-)\cr}\right.\quad.$$ Here the two one-sided limits coincide iff $a=0$.