How to show that $f(x,y)$ is continuous.

$$f(x,y)=\frac{4y^3(x^2+y^2)-(x^4+y^4)2x\alpha}{(x^2+y^2)^{\alpha +1}}$$ for $\alpha <3/2$.

Please show me Thanks :)

  • 2
    $\begingroup$ It screams polar coordinates $\endgroup$ – davin Jul 16 '13 at 10:34
  • $\begingroup$ How? Please I am glad to show me step by step on the below answer part. $\endgroup$ – Mathlover4 Jul 16 '13 at 10:34
  • $\begingroup$ Hint: try $\alpha=0$ separately. Using polar coordinates you arrive at the limit of $g(\rho)h(\theta)$, with $g(\rho)\rightarrow 0$ and $h(\theta)$ bounded. Then the product $f(\rho,\theta)=g(\rho)h(\theta)\rightarrow 0$ for $\rho\rightarrow 0$. $\endgroup$ – Avitus Jul 16 '13 at 10:43

Notice that $|x|,|y|\leq (x^2+y^2)^{1/2}=||(x,y)||$ so we have

\begin{align}|f(x,y)|&=\frac{|4y^3(x^2+y^2)-(x^4+y^4)2x\alpha|}{(x^2+y^2)^{\alpha +1}}\leq \frac{4|y|^3(x^2+y^2)+(x^4+y^4)2|x||\alpha|}{(x^2+y^2)^{\alpha +1}}\\ &\leq \frac{4||(x,y)||^3||(x,y)||^2+(||(x,y)||^4+||(x,y)||^4)2||(x,y)|||\alpha|}{||(x,y)||^{2\alpha +2}}\\ &= \frac{4(|\alpha|+1)||(x,y)||^5}{||(x,y)||^{2\alpha +2}}=4(|\alpha|+1)||(x,y)||^{3-2\alpha}\rightarrow 0\end{align} So if $f(0,0)=0$ then $f$ is continuous at $(0,0)$ and hence everywhere.

  • $\begingroup$ Thank you so much:) got it :) $\endgroup$ – Mathlover4 Jul 16 '13 at 11:44
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Jul 16 '13 at 12:00
  • $\begingroup$ Please can you look at the question? math.stackexchange.com/questions/444961/… $\endgroup$ – Mathlover4 Jul 16 '13 at 12:03
  • $\begingroup$ I like $3$ more than $2$ $\endgroup$ – Namaste May 14 '14 at 19:44

The only point where we can have trouble is $(x,y)=0$. To figure out what happens there, use the polar coordinates, as suggested in the comment of davin. We have $x=r\cos\theta$, $y=r\sin\theta$ and therefore $$x^2+y^2=r^2,\qquad x^4+y^4=r^4\left(\cos^4\theta+\sin^4\theta\right)$$ so that $$f=r^{3-2\alpha}\left(4\sin^3\theta-2\alpha\cos^5\theta-2\alpha\cos\theta\sin^4\theta\right).\tag{1}$$ Obviously, as $\alpha<3/2$ and $(x,y)\rightarrow(0,0)$, the limit of $f$ in (1) exists and is equal to the value of $f$ at $(0,0)$ (equal to zero).

  • $\begingroup$ Got it well thank you:) $\endgroup$ – Mathlover4 Jul 16 '13 at 11:43

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