Consider the function : $$f: \mathbb{R}^2 \rightarrow \mathbb{R} , (x,y) \mapsto \begin{cases} 0 & \text{for } (x,y)=(0,0) \\ \frac{x^3}{x^2+y^2} & \text{for } (x,y) \neq (0,0) \end{cases} $$

Show that $f$ not differentiable at $(0,0)$ but all directional derivatives exist.

I don't know how to tackle this problem. Can someone give some hints or solve the problem? Thanks :)


Take some (unitary) $u$. Then show that $$f'(\vec 0;u)=\lim_{h\to 0 }\frac{f(h\cdot u)-f(\vec 0)}h$$ always exists for any choice of $u$. You'll be dealing with $$\mathop {\lim }\limits_{h \to 0} \frac{\frac{{{h^3}u_1^3}}{{{h^2}(u_1^2 + u_2^2)}}}{h}$$

where $u=(u_1,u_2)$.

If your function were differentiable at the origin, then we would have $$f'(\vec 0)(u)=f'(\vec 0;u)$$ where the right hand side is the directional derivative at $\vec 0$ with direction $u$, and the left hand side is the total derivative at $(0,0)$ at evaluated at $u$. Now, $f'(\vec 0)$ would be linear, so $$f'(\vec 0)(1,1)=f'(\vec 0)(0,1)+f'(\vec 0)(1,0)$$

The right hand side is just the partial derivatives at the origin, which gives $0+1=1$. What does the left hand side give? Remember it is just the directional derivative at $\vec 0$ with direction $(1,1)$.

  • $\begingroup$ but how you gonna prove that $f$ is not differentiable at zero. By definition $\displaystyle \lim_{||(a,b)||\rightarrow 0}\frac{||f(a,b)||}{||(a,b)||}=\left( \frac{a}{\sqrt{a^2+b^2}} \right)\not=0$ I am stuck at this point.? $\endgroup$ – Cancan Apr 25 '13 at 2:17
  • $\begingroup$ @Cancan I haven't addressed the question of non-differentiability, just that of the existence of all directional derivatives, but looking at this plot you should help $\endgroup$ – Pedro Tamaroff Apr 25 '13 at 2:32
  • $\begingroup$ @cancan I added something. $\endgroup$ – Pedro Tamaroff Apr 25 '13 at 3:17

Since someone has already shown that all directional derivatives exist, I will only argue why $f$ is not differentiable at $0$.

The Jacobi Matrix $A:=Df(0,0)=(1,0)$. Therefore if $f$ is differentiable $$\lim_{|\epsilon| \to 0}\frac{f(0+\epsilon)-f(0)-A\epsilon}{|\epsilon|}=0 .$$ Since $f(0)=0$ and $A=(1,0)$ this is equivalent to, $$\lim_{|\epsilon| \to 0}\frac{f(\epsilon)-(1,0)\epsilon}{|\epsilon|}=0 $$ Let $(x_k)_{k \in \mathbb{N}} \subset \mathbb{R}^2$ be a series with $x_k=(a_k,b_k)$ and $|x_k| \to 0$ for $ (k \to \infty)$. Therefore if $f$ is differentiable.

$$0=\lim_{k \to \infty}\frac{f(x_k)-(1,0)x_k}{|x_k|}=\lim_{k \to \infty}\frac{\frac{a_k^3}{a_k^2+b_k^2}-a_k}{\sqrt{a_k^2+b_k^2}}= \lim_{k \to \infty}\frac{-a_k b_k^2}{\sqrt{a_k^2+b_k^2}^3}$$

If we set $x_k=(a_k,b_k)= (\frac {1}{k\sqrt{3}},\frac{\sqrt{2}}{k\sqrt{3}})$ then $|x_k|=\sqrt{a_k^2+b_k^2}=1/k$ and

$$\lim_{k \to \infty}\frac{-a_k b_k^2}{\sqrt{a_k^2+b_k^2}^3}=\lim_{k \to \infty}\frac{-\frac {1}{k\sqrt{3}} \frac{2}{3k^2}}{(1/k)^3}=\lim_{k \to \infty}-k^3\frac{2}{k^3\sqrt{3}^3}=\frac{-2}{\sqrt{3}^3}\neq 0$$ We get a contradiction therefore $f$ is not differentiable in 0.

  • $\begingroup$ How did you figure out the correct $a_k$ and $b_k$??? $\endgroup$ – EternusVia May 10 '19 at 15:45
  • $\begingroup$ Ah, it works just as well for $(a_k,b_k)=(\frac{1}{n},\frac{1}{n})$, or any power of $n$ in the denominator. $\endgroup$ – EternusVia May 10 '19 at 15:50
  • $\begingroup$ I think I chose the series like this so that |xk|=1/k. But yes most series should work $\endgroup$ – A. P May 10 '19 at 15:52

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.