3
$\begingroup$

Suppose that $f(a)=0$. Prove that $g(x):=|f(x)|$ is differentiable iff $f'(a)=0$


Not sure how to go about this at all. The limit definition that I am working with is

$$ g'(a)=\lim_{x \rightarrow a} \frac{g(x)-g(a)}{x-a}= \lim_{x \rightarrow a} \frac{|f(x)|-|f(a)|}{x-a}= \lim_{x \rightarrow a} \frac{|f(x)|}{x-a}=\text{something} $$

On the other hands, if $f'(a)=0$

$$ f'(a)=\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}= \lim_{x \rightarrow a} \frac{f(x)}{x-a}=0 \quad\text{iff}\quad \lim_{x\rightarrow a} f(x)=0 $$

Then, $g'(a)=\lim\limits_{x \rightarrow a} \dfrac{|f(x)|}{x-a}=0$.

My only question is how to do the other direction.

$\endgroup$
1
  • $\begingroup$ Yes, keep going. Remember that $f(a) = 0$. Hence your last expression is equal to ... $\endgroup$ – Simon S Apr 21 '15 at 21:50
2
$\begingroup$

In what follows I'll assume that $f$ is differentiable at $a$.

Suppose $f'(a)\ne0$. Then $$ \lim_{x\to a^+}\frac{|f(x)|-|f(a)|}{x-a}= \lim_{x\to a^+}\left|\frac{f(x)}{x-a}\right|=|f'(a)| $$ while $$ \lim_{x\to a^-}\frac{|f(x)|-|f(a)|}{x-a}= \lim_{x\to a^-}-\left|\frac{f(x)}{x-a}\right|=-|f'(a)| $$ so the derivative of $g$ at $a$ doesn't exist.

So, if $g$ is differentiable at $a$, then $f'(a)=0$.

For the converse, recall that, if $f$ is differentiable at $a$, then $$ f(x)=f(a)+(x-a)f'(a)+(x-a)\varphi(x) $$ where $\lim_{x\to a}\varphi(x)=0$. If $f'(a)=0$, then $f(x)=(x-a)\varphi(x)$, so $g(x)=|x-a|\,|\varphi(x)|$ and $$ \lim_{x\to a}\frac{g(x)-g(a)}{x-a}= \lim_{x\to a}\frac{|x-a|}{x-a}|\varphi(x)|=0 $$ because $|x-a|/(x-a)$ is bounded in a punctured neighborhood of $a$.


If we don't assume $f$ is differentiable at $a$, then the result is false, in the sense that from the differentiability of $g$ at $a$ we can't conclude $f$ is differentiable at $a$.

A counterexample: the function $g(x)=|f(x)|$ can be differentiable without $f$ being even continuous; indeed, consider $$ f(x)=\begin{cases} -1 & \text{if $x$ is rational}\\ 1 & \text{if $x$ is irrational} \end{cases} $$ Then $f$ is nowhere continuous, but of course $g$ is everywhere differentiable, being constant.

Things could be arranged so that $f$ is continuous at $a$, but not differentiable: $$ f(x)=\begin{cases} -x^2 & \text{if $x$ is rational}\\ x^2 & \text{if $x$ is irrational} \end{cases} $$ The function $f$ is continuous at $0$, but not differentiable. However $g(x)=|f(x)|=x^2$ is differentiable at $0$.

$\endgroup$
3
  • $\begingroup$ Is it possible for $g'(a)$ to exist but not $f'(a)$ such that $|f'(a)|$ won't make sense? What if $f(x) = x^2+1$ when x is rational and $x^2 -1$ when x is irrational? $\endgroup$ – Ilham Apr 21 '15 at 22:45
  • 1
    $\begingroup$ @Ilham Of course, I'll add the condition and the example. $\endgroup$ – egreg Apr 21 '15 at 22:46
  • $\begingroup$ Thank you, just thought that that case was the slightly trickier bit so I was looking for a direct proof that won't have to consider the alternative cases, but I think nayrb has one below!Edit: maybe that's for the other direction though $\endgroup$ – Ilham Apr 21 '15 at 22:47
0
$\begingroup$

$$\lim_{x \rightarrow a} \frac{g(x)- g(a)}{x - a} = \lim_{x \rightarrow a} \frac{|f(x)|- |f(a)|}{x - a} = \lim_{x \rightarrow a} \frac{|f(x)|}{x - a} = \lim_{x \rightarrow a} \frac{|f(x) - f(a)|}{x - a}$$

Suppose that $\lim_{x \rightarrow a} \frac{|f(x) - f(a)|}{x - a} = g'(a) \in \mathbb R$ and $|f'(a)| > 0$ (equavalently $f'(a) \neq 0$), then

$$0 < |\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}| = \lim_{x \rightarrow a} |\frac{f(x) - f(a)}{x - a}| = \lim_{x \rightarrow a} \frac{|f(x) - f(a)|}{|x - a|} = \lim_{x \rightarrow a} \frac{|f(x) - f(a)|}{|x - a|} \frac{x-a}{x-a} = g'(a) \lim_{x \rightarrow a} \frac{x-a}{|x-a|}$$ Hence $g'(a) \neq 0$ and $\lim_{x \rightarrow a} \frac{x-a}{|x-a|} = \frac{|f'(a)|}{g'(a)} \in \mathbb{R}$, but does this limit exists ???

$\endgroup$
2
  • $\begingroup$ Can $g'(a)$ exist and $f'(a)$ not exist at all, i.e. not only is it not zero, it just doesn't exist? $\endgroup$ – Ilham Apr 21 '15 at 22:17
  • 1
    $\begingroup$ Haven't think about this @Ilham. However I guess that $f'(a)$ is supposed to exist, considering the way the question is written. $\endgroup$ – brick Apr 21 '15 at 22:23
0
$\begingroup$

By the reverse triangle inequality: $$0 \leq |g'(a)| = \lim_{x \to a} \left| \frac{|f(x)|-|f(a)|}{x-a} \right| \leq \lim_{x \to a} \frac{|f(x)-f(a)|}{|x-a|} = \left| \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \right| = |f'(a)| = 0$$

$\endgroup$
1
  • $\begingroup$ Is this the direction that supposes $f'(a) = 0$? $\endgroup$ – Ilham Apr 21 '15 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.