# Verification: For what $s$ is $\frac{x^4+y^4-6x^2y^2}{(x^2+y^2)^s}$ continuous and differentiable at $(0,0)$

Question

For what values of $s$ (where s is real and positive) is:
$f(x,y)=\dfrac{x^4+y^4-6x^2y^2}{(x^2+y^2)^s}$ when $(x,y) \neq (0,0)$
and
$f(x,y) = 0$ when $(x,y)=(0,0)$ ,

(a) Continuous at $(0,0)$
(b) Differentiable at $(0,0)$ .

Looking at the line $y=x$, I get:

$$f(t,t)=\frac{-4t^4}{(2t^4)^s} = \frac{-4}{(2)^s(t^{4(s-1)})}$$

However for any positive $s$, $f(t,t)$ does not tend to zero as $t$ tends to zero. Therefore $f(x,y)$ is not continuous or differentiable at $(0,0)$ for any positive real $s$.

My question is, does this prove the function is not differentiable at $(0,0)$ for any $s$?

Replace $x=rcos(\theta)$ and $y=rsin(\theta)$ in polar coordinates. Then $f(x,y)={{\cos 2\theta}\over {r^{2s}}}$. It has a limit in 0 if an only if the expression is independent from $\theta$ and bounded when $r\to 0$ which is possible only if $s<0$. It would have 1st order derivations in origin if $s\le -1$.
• How did you arrive at the expression $\cos(2\theta)/r^{2s}\>$? – Christian Blatter Jan 4 '18 at 15:07
• It's a simple substitution and using $cos^4(\theta)+sin^4(\theta)+2cos^2(\theta)sin^2(\theta)=1$ – Mostafa Ayaz Jan 4 '18 at 15:10