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Let $$f(x,y)=\begin{cases}\frac{x|y|}{\sqrt{x^2+y^2}} &\textrm{ if } (x,y)\neq(0,0), \\ 0 &\textrm{ if } (x,y)= (0,0).\end{cases} $$

I want to know: is $f$ is continuous? Are the partial derivatives exists at the origin? Is $f$ differentiable?

First, I tried some different paths to prove that $f$ is discontinuous, but none path lead me to it.

I found the partial derivatives:

$$\frac{\partial f}{\partial x}(x,y)=\frac{(y^2-x^2)|y|}{(x^2+y^2)^{3/2}},\quad \frac{\partial f}{\partial y}(x,y)=\frac{x|y|(x^2-y^2)}{y(x^2+y^2)^{3/2}}. $$

By doing $(x,y)=(0,t)$, the limit of the first partial derivative does not exist:

$$\lim_{t\rightarrow 0}\frac{t^2|t|}{t^3}=\lim_{t\rightarrow 0}\frac{|t|}{t} $$

Is this sufficient to conclude that $f$ is not differentiable? What can I do?

Arctic Char
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Mateus Rocha
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2 Answers2

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  1. Continuity: $(x-|y|)^2=x^2+y^2-2x|y|\geq0 \iff \frac{x^2+y^2}{2}\geq x|y|$

  2. Partial Derivatives: $\frac{\partial f}{\partial x}(0,0)=\lim_{h\to0} \frac{f((0,0)+h(1,0))-f(0,0)}{h}$ and it is analogous for $\frac{\partial f}{\partial y}(0,0)$.

  3. Derivability: Consider the linear application, $L$ , such that for any $(x,y)\in\mathbb{R^2}:L(x,y)=0$ then $$\lim_{(x,y)\to(0,0)} \frac{f(x,y)-f(0,0)-L((x,y)-(0,0))}{||(x,y)-(0,0)||}=\lim_{(x,y)\to(0,0)} \frac{x|y|}{x^2+y^2}$$

Can you take it from here?

  • I understood the points 1 and 2. Could you elaborate the point 3? – Mateus Rocha Sep 08 '20 at 22:48
  • I used the definition of derivability of a multivariable function and so for any $(a,b)\in\mathbb{R^2}$ then $L_{(a,b)}(x,y)=\frac{\partial f}{\partial x}(a,b)x+\frac{\partial f}{\partial y}(a,b)y$. –  Sep 08 '20 at 23:15
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Firstly $f$ is continuous, indeed by polar coordinates we have

$$\frac{x|y|}{\sqrt{x^2+y^2}}=r\cos \theta |\sin \theta| \to 0$$

For differentiability by the definition we have

$$\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$$

and

$$\lim_{(h,k)\to(0,0)} \frac{\frac{h|k|}{\sqrt{h^2+k^2}}}{\sqrt{h^2+k^2}}=\lim_{(h,k)\to(0,0)} \frac{h|k|}{h^2+k^2} \neq 0$$

and therefore $f$ is not differentiable.

As noticed, we cannot conclude for differentiability from the discontinuity of partial derivatives.

user
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  • You cannot conclude the limit at $(0,0)$ exists just by computing limits along the arrays. Polar coordinates are generally used for counter - examples. – Matcha Latte Oct 13 '23 at 13:42
  • @Invisible Do you mean that $\lim_{(x,y)\to(0,0)} \frac{x|y|}{\sqrt{x^2+y^2}}=0$ is wrong? I don't think so, please reconsider your evaluation. Thanks – user Oct 13 '23 at 19:00
  • I don't think the result is wrong (AG inequality indicates it is indeed $0$), the deduction is, though. – Matcha Latte Oct 13 '23 at 20:05
  • @Invisible Then what is your point? Using polar coordinates is fine to show that limit exists also, and in this case it works fine. – user Oct 13 '23 at 20:07
  • Limit $\frac{x^2y}{x^4+y^2}$ is found using polar coordinates but it is not supposed to exist. – Matcha Latte Oct 13 '23 at 20:13
  • @Invisible The claim by the OP that $\lim_{r \to 0} \frac{r (\cos^2\theta\sin\theta)}{r^2\cos^4\theta + \sin^2\theta}=0$ is indeed wrong (the issue is in the denominator) but it doesn't mean that in general polar coordinates fail. In this case is totally fine indeed. – user Oct 13 '23 at 20:28
  • Refer also to https://math.stackexchange.com/q/4727588/505767 – user Oct 13 '23 at 20:30
  • You cannot consider limit along some paths to conclude the limit exists and is equal along every path. If the limit exists and is equal along every path, it will, of course, be so along arrays from the origin. I'm trying to explain three dowvotes you had gotten years ago. – Matcha Latte Oct 13 '23 at 20:47
  • @Invisible I'm not taking any path, by polar coordinates we consider all the possible ways. I don't think downvotes were given for that reason. Please refer also the the link I've given here. – user Oct 13 '23 at 20:55