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I want to show that $f(x) = \vert x\vert g(\frac{x}{\vert x\vert})$ is not differentiable at $(0,0)$, where $f:\mathbb{R}^2\to\mathbb{R}$, $f(0,0)=0$ and g is a continuous function on a unit circle. I've proved that $f' = 0$ so if for $f$ to be differentiable, $g$ must be constant zero. If I show that $$\lim_{(x,y)\to0}\Big\vert f\Big(\frac{(x,y)}{\big\vert(x,y)\big\vert}\Big)\Big\vert$$ does not exist unless $f$ is a constant on unit circle my prove is done. How should I show this?

Mina
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    It's not true that $f$ needs to be constant. Take $f(x,y) = |(x,y)|$, for example. It does need to be constant on the unit circle, though. Use polar coordinates. – Ennar Dec 23 '21 at 08:34
  • Actually, $f$ is constant on the unit circle if we assume continuity, otherwise we only know that $|f|$ is constant. For example, take $f(x,y) = 1$ for $x\geq 0$ and $f(x,y) = -1$ otherwise. – Ennar Dec 23 '21 at 09:39
  • @Ennar You're right. I forgot to say that f is on the unit circle. I've made changes on the post. – Mina Dec 23 '21 at 09:40
  • Lucy, you should also include your thoughts on the problem. As it stands, the question will be closed. You should edit it with some background, motivation, attempts or similar. You can check out this for some guidance on how to ask good questions. – Ennar Dec 23 '21 at 09:43
  • @Ennar Thanks for the help. I corrected my question. – Mina Dec 23 '21 at 09:48
  • How do you define $f$ at $(0,0)$? Is it $0$? Is $g$ continuous? – Ennar Dec 23 '21 at 09:58
  • What do you mean by "I've proved $f^\prime = 0$"? You wanted to prove that $f$ is not differentiable at $(0, 0)$, right? If you've really proved $f^\prime =0$, there is nothing to prove further. Please write your question carefully. – I H Dec 23 '21 at 12:51
  • @IH I added the actual question. I used the hint and proved that Df is zero. – Mina Dec 23 '21 at 13:12
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    Ah, I see. Please write your problem PRECISELY. So, what you proved is not $f^\prime =0$ but $f_x(0,0)=f_y(0,0)=0$. This cannot be proved without the assumptions $g(1,0)=g(0,1)=0$ and $g(-x)=-g(x)$ given. – I H Dec 23 '21 at 13:23

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Let $(x, y) = (r\cos \theta, r\sin \theta), r\geq0$. Then $f(x,y)=rg(\cos \theta, \sin \theta)$.

So, we have $$ \frac{f(x,y)-f(0,0)}{r} = g(\cos \theta, \sin \theta). $$

Taking $r \to 0$, we get $D_\theta f(0,0) = g(\cos \theta, \sin \theta)$, where $D_\theta$ is the directional derivative along a vector $(\cos \theta, \sin \theta)$.

Take $\theta =0, \pi$ and note that $g(1,0)=g(-1,0)=0$, we conclude $f_x(0,0)=0$. Likewise, by taking $\theta = \pi/2, 3\pi/2$, we conclude $f_y(0,0)=0$.

So, if $f$ is differentiable at $(0,0)$, $D_\theta f(0,0)$ must be $0$ for any $\theta$. However, this is not the case unless $g \equiv 0$ since $D_\theta f(0,0) = g(\cos \theta, \sin \theta)$.

I H
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