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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 :)

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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 $(1,1)$.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ How did you figure out the correct $a_k$ and $b_k$??? $\endgroup$ – EternusVia May 10 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 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 at 15:52

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