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Let $f:\mathbb R^2\to \mathbb R$ be defined by $$f(x,y)=\begin{cases}\frac{x|y|}{\sqrt{x^2+y^2}},& (x,y)\ne(0,0)\\ 0,& (x,y)=(0,0).\end{cases}$$

  1. For which non-zero vectors $u$ does the directional derivative exist at the point $(0,0)$?
  2. Do the partial derivatives exist?
  3. Is $f$ differentiable at $0$?
  4. Is $f$ continuous at $0$?

According to my calculations, the directional derivative is zero for all non-zero vectors $u$ at $(0,0)$ which implies the partial derivatives are zero at $(0,0)$. I am not sure how to proceed with 3. and 4. I am not allowed to use the existence of partial derivatives and their continuity to check for differentiability since the author introduces this theorem in the next section.

Thanks.

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$f$ is continuous at $(0,0)$ as you can prove the inequality $$ \vert f(x,y) \vert \le \frac{\vert xy \vert}{\sqrt{x^2+y^2}} \le \frac{\sqrt{x^2+y^2}}{2}$$

$f$ is not differentiable at $(0,0)$. If it would be the case its derivative would be equal to the zero matrix as the partial derivatives are equal to $0$. Hence all the directional derivatives would vanish which is not the case. A contradiction proving that $f$ is not differentiable at the origin.

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  • $\begingroup$ Aren't the directional derivatives zero for all non-zero vectors u?If not, why? $\endgroup$ – RagingBull Aug 12 '15 at 14:28
  • $\begingroup$ You should be more precise and say the derivative would be equal to the zero matrix rather than $0$ as the two are not the same. $\endgroup$ – user21820 Aug 12 '15 at 14:38
  • $\begingroup$ @BasantSharma: Yes in other words your part (1) is wrong. You should show how you calculated the directional derivatives, as clearly something went wrong. In particular in the direction $(1,1)$ the gradient is not $0$. $\endgroup$ – user21820 Aug 12 '15 at 14:40
  • $\begingroup$ Could you please explain what is this inequality about? $\endgroup$ – RagingBull Aug 12 '15 at 15:31
  • $\begingroup$ @user21820 The derivative is a linear map. Can't you say that a linear map is equal to $0$? Moreover you can speak of matrices for finite dimensional spaces and linear maps can exist in infinite dimensional vector spaces. $\endgroup$ – mathcounterexamples.net Aug 12 '15 at 17:07
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For 3. you should compute directly $$ \lim_{\|(x,y) \| \to 0} \frac{f(x,y)}{\sqrt{x^2+y^2}}=\lim_{\substack{x \to 0 \\ y \to 0}} \frac{x|y|}{x^2+y^2}. $$ Of course this can be done once you are sure that the partial derivatives at $(0,0)$ are both zero. By choosing the special path $x=y$, it is immediate to see that the limit is not zero, and therefore $f$ is not differentiable at $(0,0)$.

The continuity of $f$ at zero follows immediately from the inequality $$ |y| \leq \sqrt{x^2+y^2}, $$ since $$ |f(x,y)| \leq |x| \frac{|y|}{\sqrt{x^2+y^2}} \leq |x| \to 0 $$ as $x\to 0$, $y \to 0$.

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  • $\begingroup$ Why a downvote? $\endgroup$ – Siminore Aug 12 '15 at 14:24
  • $\begingroup$ The derivative is the Jacobian if it exists, your limit $\lim_{||(x,y)||\to 0} \frac{f(x,y)}{\sqrt{x^2+y^2}}$ is meaningless, since it only makes sense in one dimension. Please read en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/…. $\endgroup$ – user21820 Aug 12 '15 at 14:36
  • $\begingroup$ @user21820 Sorry to say this, but I guess you need to study a lot... The derivative is the gradient vector $(\partial_1 f(0,0),\partial_2 f(0,0))^t$, and the function is differentiable at $(0,0)$ if and only if $$\lim_{\|(x,y)\| \to 0} \frac{f(x,y)-f(0,0)-\partial_1 f(0,0)x - \partial_2 f(0,0)y}{\sqrt{x^2+y^2}}=0.$$ Please check your books before downvoting: en.wikipedia.org/wiki/… $\endgroup$ – Siminore Aug 12 '15 at 15:01
  • $\begingroup$ @user21820 Siminore is correct. That is the definition of the derivative at $(0,0)$. $\endgroup$ – Mark Viola Aug 12 '15 at 15:09
  • $\begingroup$ I know all that. But you didn't say any of that in your answer, so I thought you were computing the derivative, which is of course not the limit your wrote. Note that the other answerer made exactly the mistake I was referring to so it didn't occur to me that you were not making the same mistake. So you'll have to forgive the downvote because your answer was not sufficiently clear. I'll gladly remove the downvote if you make your answer clearer a bit. $\endgroup$ – user21820 Aug 12 '15 at 15:10

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