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Prove the existence or not existence of continuity, differentiability and the continuity of partial derivatives of the following function two variables at $(0,0)$

$$f(x,y) = \begin{cases} (x-y)^{2}sin \frac{1}{x^{2}-y}, & \mbox{if } (x,y) \neq (0,0) , \\ 0, & \mbox{if } (x,y) =(0,0)\end{cases}$$

At first I wanted to prove this function was not continuous at $(0,0)$ therefore I would prove this function is not differentiable at $(0,0)$, as well. I tried to compute the limit of the function as we aproach to $(0,0)$ with two different trajectories,lets say $y=x$ and $y=2x$ but this doesnt seems to work. So maybe this function is continuous and differentiable. But I cant see the $\epsilon- \delta$ trick here. In order to prove differentiability I need to compute partial derivatives and show this partial derivatives are continuous isnt? But its seems harder to prove continuity in the partia derivatives since I cannot even compute continuity over the original function. :(

As $\frac{\partial f}{ \partial x}(0,0)=lim_{h \to 0} \frac{f(h,0)-f(0,0)}{h}= lim_{h \to 0} \frac{f(h,0)}{h}=lim_{h \to 0} \frac{(h)^{2}sin \frac{1}{h^{2}}}{h} $

but I dont know how to reduce this into something that gives me an insight about the differentiability or continuity of this partial derivative. In the same way I got

$\frac{\partial f}{ \partial y}(0,0)=\frac{h^{2}sin \frac{1}{-h^{2}}}{h}.$

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    $\begingroup$ The function doesn't appear to be defined when $y = x^2$ (except when $x=y=0$). Is that intentional? Also, I'm assuming your $(x,y) = (0,0)$ and $(x,y) \neq (0,0)$ should be reversed. $\endgroup$
    – user169852
    May 7 at 20:26
  • $\begingroup$ You are absolutely right! Let me fix the typos @Bungo $\endgroup$
    – Sok
    May 7 at 20:32
  • $\begingroup$ For continuity of $f$, you can observe that $\left|(x-y)^2\sin\frac{1}{x^2-y}\right| \leq (x-y)^2$ since $|\sin \theta| \leq 1$ for any real $\theta$. And $(x-y)^2 = x^2 -2xy + y^2$ becomes arbitrarily close to zero as $(x,y) \to (0,0)$. $\endgroup$
    – user169852
    May 7 at 20:42
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    $\begingroup$ Note that $-2xy \leq x^2 + y^2$ (because if you add $2xy$ to both sides you get the equivalent $0 \leq x^2 + 2xy + y^2 = (x+y)^2$, which is obviously true). Therefore, $x^2 - 2xy + y^2 \leq x^2 + (x^2 + y^2) + y^2 = 2x^2 + 2y^2 = 2(x^2 + y^2)$. Taking square roots gives you the bound $\sqrt{x^2 - 2xy + y^2} \leq \sqrt{2} \sqrt{x^2 + y^2}$. $\endgroup$
    – user169852
    May 7 at 22:21
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    $\begingroup$ An alternative simpler argument: note that $(x,y) \to 0$ if and only if $x \to 0$ and $y \to 0$. And the latter two imply that $x^2 - 2xy + y^2 \to 0$ because of the rules involving sums and products of limits. $\endgroup$
    – user169852
    May 7 at 22:24
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According to Wikipedia $f$ is differentiable at the origin if there exists a linear map $J:\mathbb{R}^2 \rightarrow \mathbb{R}$ such that $$\lim_{h \rightarrow 0}\frac{|f(h_1,h_2)-f(0,0)-J(h_1,h_2)|}{\|(h_1,h_2)\|_{\mathbb{R}^2}}=0$$ Take $J\equiv 0$ on $\mathbb{R}^2$ so that $$\frac{|f(h_1,h_2)-f(0,0)-J(h_1,h_2)|}{\|(h_1,h_2)\|_{\mathbb{R}^2}}=\frac{(h_1-h_2)^2\Big|\sin\Big(\frac{1}{h_1^2-h_2}\Big)\Big|}{\sqrt{h_1^2+h_2^2}}\leq \frac{(h_1-h_2)^2}{\sqrt{h_1^2+h_2^2}}\longrightarrow 0$$ as $(h_1,h_2)\longrightarrow 0$. This proves that $f$ is differentiable at $(0,0)$.

Note: $(h_1,h_2)$ cannot tend to the origin along any arbitrary path. The manner in which we take $(h_1,h_2)$ to the origin must avoid intersecting the parabola $h_2=h_1^2$. Most textbooks I've seen allow for the consideration of such limits.

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  • $\begingroup$ @MathewPilling This was very helpful. The way I calculated the partial derivatives of this function at $(0,0)$ shows the derivatives are continuous at $(0,0)$? Can they be reducted and why the tend to $0$? $\endgroup$
    – Sok
    May 8 at 1:08
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    $\begingroup$ You can show that $f_x(0,0)$ and $f_y(0,0)$ exist and are both equal to $0$ by using the definition of the partial derivative. If you want to show $f_x$ and $f_y$ are continuous at $(0,0)$, you need to show that $f_x(x,y),f_y(x,y)\longrightarrow 0$ as $(x,y)\longrightarrow (0,0)$. It's not enough to show the partial derivatives exist at $(0,0)$ to say that they're continuous at $(0,0)$ $\endgroup$
    – Matthew Pilling
    May 8 at 1:41
  • $\begingroup$ Appreciate it! I will try to compute the limit of the partial derivatives as $(x,y) \to (0,0)$ $\endgroup$
    – Sok
    May 8 at 2:04

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