enter image description here

I have some problem with proving that $g$ is differentiable at $\mathbb{R}^1$. Let's prove that $g'(0)$ exists. $$\lim \limits_{t\to 0}\dfrac{g(t)-g(0)}{t}=\lim \limits_{t\to 0}\dfrac{f(\gamma(t))-f(\gamma(0))}{t}=\lim \limits_{t\to 0}\dfrac{f(\gamma(t))}{t}=$$ Let $\gamma(t)=(\gamma_1(t),\gamma_2(t))$ and since $|\gamma'(0)|>0$ then $\gamma(t)\neq \mathbf{0}$ for enough small $t$ and then $f(\gamma(t))=\dfrac{\gamma_1^3(t)}{\gamma_1^2(t)+\gamma_2^2(t)}$. Hence $$\lim \limits_{t\to 0}\dfrac{f(\gamma(t))}{t}=\lim \limits_{t\to 0}\dfrac{\gamma_1^3(t)}{t(\gamma_1^2(t)+\gamma_2^2(t))}=\lim \limits_{t\to 0}\dfrac{\left(\frac{\gamma_1(t)-\gamma_1(0)}{t}\right)^3}{\left( \frac{\gamma_1(t)-\gamma_1(0)}{t}\right)^2+\left( \frac{\gamma_2(t)-\gamma_2(0)}{t}\right)^2}=$$$$=\dfrac{\gamma_1'(0)}{|\gamma'(0)|^2}.$$ Since $|\gamma'(0)|^2>0$ then limit exists.

But how to prove that $g'(t_0)$ exists for $t_0\neq 0$? Because in this case $$\lim \limits_{t\to t_0}\dfrac{g(t)-g(t_0)}{t-t_0}=\lim \limits_{t\to t_0}\dfrac{f(\gamma(t))-f(\gamma(t_0))}{t-t_0}.$$ But in this case we have no information about $\gamma(t_0)$. I mean $\gamma(t_0)=0$ or $\gamma(t_0)\neq 0$ and this case we don't know what form has $f(\gamma(t_0))$. Also Chain rule is can not be applied here.

Can anyone please show the proof of this case?

  • 1
    $\begingroup$ if $\gamma(t_0) = (0, 0)$, you have proved it. Assume $\gamma(t_0) \ne 0$ now, it just likes chain-rule now. $\endgroup$ – runaround Feb 18 '16 at 6:13
  • $\begingroup$ @runaround, If we assume that $\gamma(t_0)\neq 0$ then OK. Why you think that $\gamma(t_0)$ is not equal to zero? We don't know that this curve is one-to-one! $\endgroup$ – ZFR Feb 18 '16 at 6:27
  • $\begingroup$ if $\gamma(t_0) = (0, 0)$ is zero, just apply your proof. Only left is not zero. $\endgroup$ – runaround Feb 18 '16 at 6:28
  • $\begingroup$ @runaround,Sorry but if $\gamma(t_0)=(0,0)$ then $f(\gamma(t_0))=0$ but what about $f(\gamma(t))$ for $t\to t_0$? What form has $f(\gamma(t))$ for such $t$? $\endgroup$ – ZFR Feb 18 '16 at 6:30

Taking a closer look to your argument, the condition $|\gamma'(t_0)|>0$ actually can be dropped. Assume that $\gamma(t_0) = 0$. You have whenever $t\neq t_0$,

$$\frac{f(\gamma(t))-f(\gamma(t_0))}{t-t_0} = \begin{cases} \dfrac{\left(\frac{\gamma_1(t)-\gamma_1(t_0)}{t-t_0}\right)^3}{\left( \frac{\gamma_1(t)-\gamma_1(t_0)}{t-t_0}\right)^2+\left( \frac{\gamma_2(t)-\gamma_2(t_0)}{t-t_0}\right)^2} & \text{if }\gamma(t)\neq 0 \\ 0 & \text{if } \gamma(t) = 0\end{cases}$$

In particular, for all $t\neq t_0$,

$$\left| \frac{f(\gamma(t)-f(\gamma(t_0))}{t-t_0}\right| \le \left| \frac{\gamma_1(t) - \gamma_1(t_0)}{t-t_0}\right|.$$

If $\gamma'(t_0) = 0$, we have

$$\lim_{t\to t_0} \frac{\gamma_1(t) -\gamma_1(t_0)}{t} = 0, $$

so squeeze theorem implies that $g'(t_0)$ exists and

$$g'(t_0) = \lim_{t\to t_0}\frac{f(\gamma(t)-f(\gamma(t_0))}{t-t_0} = 0$$

  • $\begingroup$ What about differentiability of $g(t)$ at $t\neq 0$? $\endgroup$ – ZFR Feb 18 '16 at 10:56
  • $\begingroup$ Yeah. When I was doing this, I found that the condition $|\gamma'(0)|>0$ was useless in both the two problems in part (c). $\endgroup$ – Vim Feb 18 '16 at 10:56
  • $\begingroup$ @Vim, Why it's useless? In my argument it guarantees that denominator is not equal to zero. $\endgroup$ – ZFR Feb 18 '16 at 10:58
  • $\begingroup$ @RFZ: Please see the edit. Note that one can avoid the use of $|\gamma'(0)|>0$ to show that $g$ is differentiable at $0$, this is what I did in my answer. $\endgroup$ – user99914 Feb 18 '16 at 11:00
  • $\begingroup$ @RFZ you could follow John Ma's approach. $\endgroup$ – Vim Feb 18 '16 at 11:00

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.