Let $f:\mathbb{R}^2\to\mathbb{R} $ defined by $$f(x,y)= \begin{cases} \frac{x^3}{x^2+y^2} &\text{if $(x,y)\neq(0,0)$}\\ 0 &\text{if $(x,y)=(0,0)$} \end{cases}$$ It's easy to see that f is not differentiable at $(0,0)$, becasuse the derivative function at $(0,0)$ is not linear. However, I have been told that if $g:[0,1]\to\mathbb{R}^2$ is differentiable, then $f\circ g$ is differentiable. How can I prove this? (I think we are not assuming $g$ is infinitely differentiable, only differentiable, so L'Hôpital doesn't seem to help).
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1@Mehran $f$ is a homogeneous function, but not linear hence it cannot be differentiable. (It is a standard fact that a homogeneous function which is differentiable at the origin must be linear) All your argument potentially shows is continuity at the origin. – peek-a-boo Aug 07 '19 at 07:15
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@peek-a-boo thank you for pointing this out – Mehran Aug 07 '19 at 07:20
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For other counterexamples see also https://math.stackexchange.com/questions/1111319/if-every-composition-of-a-differentiable-path-and-a-function-is-differentiable-a as well as https://calculus7.org/2012/07/24/how-much-multivariable-calculus-can-be-done-along-curves/ – symplectomorphic Aug 07 '19 at 07:50
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Let $g=(g_1,g_2)$. Let $t_0 \in [0,1]$. Differentiabilty of $f\circ g$ at $t_0$ is obvious if $g(t_0) \neq (0,0)$ so assume that $g_1(t_0)=g_2(t_0)=0$. we have to show that $\frac {g_1(t-t_0)^{3}} {(t-t_0) [g_1(t-t_0)^{2}+g_2(t-t_0)^{2}]}$ has a limit as $t \to t_0$. Since $\frac {g_1(t-t_0)^{2}} {g_1(t-t_0)^{2}+g_2(t-t_0)^{2}}=\frac {g_1(t-t_0)^{2}/(t-t_0)^{2}} {g_1(t-t_0)^{2}/(t-t_0)^{2}+g_2(t-t_0)^{2}/(t-t_0)^{2}} $ which tends to a finite limit except when $g_1'(t_0)=g_2'(t_0)=0$ and $\frac {g_1(t-t_0)} {t-t_0}$ has a limit as $t \to t_0$ the proof is complete except when when $g_1'(t_0)=g_2'(t_0)=0$. When $g_1'(t_0)=g_2'(t_0)=0$ use the fact that $\frac {g_1(t-t_0)^{2}} {g_1(t-t_0)^{2}+g_2(t-t_0)^{2}}$ is bounded by $1$ to complete the proof.
Kavi Rama Murthy
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