Problem 14 chapter 9 from PMA Rudin

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?

• if $\gamma(t_0) = (0, 0)$, you have proved it. Assume $\gamma(t_0) \ne 0$ now, it just likes chain-rule now. – runaround Feb 18 '16 at 6:13
• @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! – ZFR Feb 18 '16 at 6:27
• if $\gamma(t_0) = (0, 0)$ is zero, just apply your proof. Only left is not zero. – runaround Feb 18 '16 at 6:28
• @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 $t$? – 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

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

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