Let $g$ be a continuous $\mathbb{R}$-valued function on the unit circle {x $\in$ $\mathbb{R^2}$ : |x| = 1}, such that $g(0,1) = g(1,0) = 0$ and $g(-x) = -g(x)$. Define $f$ by:

$f(x)$ = $|x|$$.$$g(\frac{x}{|x|}$), when $x$ $\neq$ 0 and $f(x) = 0$, when $x = 0$

(a) Show that, if x $\in$ $\mathbb{R^2}$ and $h: \mathbb{R} \rightarrow \mathbb{R}$ denotes the function $h(t) = f(tx)$, then $h$ is differentiable.

(b) Show that $f$ is not differentiable at $(0,0)$

I have been playing around with this question for a few days. For part (a), would it be better to prove that $h$ is differentiable by showing that all of its directional derivatives exist at any $h(\vec{a})$ and are given by ${\partial_u f(\vec{a})}$ = ${\triangledown}$$f(\vec{a}).\hat{u}$? Or would it be better to show that $$\lim_{h\to 0} \frac{f(ta + th) - f(ta) - c.h}{|h|} = 0$$

In either case my algebra seems to break down and I can't make an effective argument to prove this. Any assistance or tips would be greatly appreciated.

Thank you

  • $\begingroup$ For (a) show that for each $x$ there is some $k$ such that $f(t) = kt$. Differentiability follows from that. $\endgroup$
    – copper.hat
    Oct 6, 2017 at 23:10
  • $\begingroup$ Note, presumably for (b) you are supposed to show that there is some $g$ that makes $f$ not differentiable. If you choose $g=0$ then the corresponding $f$ is differentiable. $\endgroup$
    – copper.hat
    Oct 6, 2017 at 23:16

2 Answers 2


These are some rather detailed hints:

  1. On the circle, $g=g(\theta)$.

  2. On $\mathbb{R}^2$, represent $x$ as $(r,\theta)$. It's clear that $f(x)=rg(\theta)$

  3. Now $h(t)$ can be shown to be $trg(\theta)$ which is clearly differentiable in $t$.

  4. As mentioned in the comments, the function is differentiable if $g \equiv 0$

  5. Otherwise, pick some $\theta$ where $g(\theta) = a \neq 0$. Then evaluate the derivative as you approach the origin from the direction of $(1,\theta)$ and compare with the derivative as you approach the origin from the direction of $(1,-\theta)$. Do you see the discontinuity at the origin?

  • $\begingroup$ For part (b), should I try to differentiate f with respect to $\theta$? $\endgroup$ Oct 8, 2017 at 20:59
  • $\begingroup$ Nevermind, it makes sense if I look at $DF(0,0)$ while considering $(h,k)$ with $k=0$ or $h=0$. $\endgroup$ Oct 9, 2017 at 2:11
  • $\begingroup$ Yes, construct the derivative and see what value it has when approaching from diametrically opposite directions (which aren't zero). The limit as you approach the origin is different depending on the direction. $\endgroup$ Oct 10, 2017 at 2:30
  • $\begingroup$ @DPuz If you found the answer helpful, would you accept it, please? You can do that by clicking the check mark next to the answer. $\endgroup$ Oct 10, 2017 at 11:29
  • $\begingroup$ I did find the answer very helpful. I accepted as you requested. Thanks for your help. $\endgroup$ Oct 13, 2017 at 0:33

Oh I get it. For part (a). On the unit circle let $g(\theta) = g$ since $r=1$ on the unit circle, g can be defined as $g(r,\theta)=(cos(\theta), sin(\theta))$ = $g(\theta)$.

If $h$ denotes the function $h(t) = f(tx)$ then, $h(t)=f(t(r,\theta)=f(tr,t\theta)$ = $trg(\theta)$

Then $h'(a)= \lim_{b\to 0} \frac{h(a + b - h(a)}{b}$ =$\lim_{b\to 0} \frac{(a + b)rg(\theta)- arg(\theta)}{b}$ =$rg(\theta)$

Is this right?

  • $\begingroup$ Thanks for the help everyone! $\endgroup$ Oct 9, 2017 at 2:12
  • $\begingroup$ Right when not at the origin. $\endgroup$ Oct 10, 2017 at 2:28
  • 1
    $\begingroup$ . You accepted your own answer, not the one I wrote :) $\endgroup$ Oct 13, 2017 at 1:55

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