Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous.

Question

The problem asks to show that $$f(x,y) = \left\{ \begin{align} \frac{x^3y^2}{x^4+y^4}, & (x,y) \neq (0,0), \\ 0, & (x,y) = (0,0), \end{align} \right.$$ is continuous at the origin, however it has resisted my bravest efforts. I have attempted using $x^4+y^4 \geq y^4$ and therefore $$\left\vert \frac{x^3y^2}{x^4+y^4} \right\vert \leq \left\vert \frac{x^3}{y^2} \right\vert$$ and similar strategies but they have failed. Trying to disprove it and see if it's discontinuous has only strengthened the belief that it's continuous.

Also: you'll want the fact that $\sin\theta$ and $\cos\theta$ are never simultaneously zero to be complete. — Adam Hughes, Jul 04 '14 at 07:13
This proves for lines. I've tried that. An $\delta-\varepsilon$ proof is probably more adequate or squeeze theorem. See my answer here. — Mark Fantini, Jul 04 '14 at 07:18
Huh? that's not true at all, that proves it period. If you look at $\delta$-$\epsilon$ you'll see that you need for $\sqrt{x^2+y^2}<\delta$, i.e $r<\delta$, so it's EXACTLY for $r\to 0$. You're assuming I'm assuming $\theta$ fixed, which I'm not. $\theta$ can follow ANY continuous path. — Adam Hughes, Jul 04 '14 at 07:24
@SomabhaMukherjee if you replace the denominator by something smaller it makes the fraction bigger, so if you're trying to dominate the function by something divergent, it would prove nothing. — Adam Hughes, Jul 04 '14 at 07:40
If $x^4+y^4 \geq 2x^2y^2$ then $$\frac{1}{x^4+y^4} \leq \frac{1}{2x^2y^2}.$$ — Mark Fantini, Jul 04 '14 at 07:43
@Fantini Ah, he was trying to do convergence. Excellent, solid answer. This is a good special case where that works. Again, I'll refer you to polar coordinates for the future though, it works better in general and is perfect for these situations. I over-explained in my answer below to make it painfully obvious that the method does work and is not just for lines (there is no assumption that $\theta=c$ anywhere in it). You can get away with much less writing than I did as well, my initial observation was enough, but I wanted an explicit minimum. — Adam Hughes, Jul 04 '14 at 07:47
I was under the impression that you had to necessarily assume $\theta$ fixed. What about when you have $$f(x,y) = \frac{x^2y}{x^4+y^2}?$$ When you apply polar coordinates you have $$g(r,\theta) = \frac{r^3\cos^2 \theta \sin\theta}{r^4\cos^4\theta +r^2\sin^2\theta} = \frac{r\cos^2\theta \sin\theta}{r^2\cos^4 \theta + \sin^2\theta}.$$ Applying $r\to0$ leads to $\lim_{r \to 0} g(r,\theta) = 0$ but the limit doesn't exist, taking $y=x^2$ shows it. — Mark Fantini, Jul 04 '14 at 08:01
@Fantini Ah, I see. You want to use polar particularly when you think it is continuous. — Adam Hughes, Jul 04 '14 at 08:06
See also: http://math.stackexchange.com/questions/1822811/continuity-of-fracx3y2x4y4-at-0-0 — Martin Sleziak, Jun 21 '16 at 18:47

Adam Hughes · Accepted Answer · 2014-07-04T07:49:32.997

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To expand on my comments, let $\epsilon > 0 $ be given, then our limit, in polar coordinates is of the form

$$f(r,\theta)={r^5\cos^3\theta\sin^2\theta\over r^4(\cos^4\theta+\sin^4\theta)}$$

this satisfies

$$|f(r,\theta)-0|\le |r|\left|{\cos^3\theta\sin^2\theta\over \cos^4\theta+\sin^4\theta}\right|\le {r\over m_\theta}={\delta\over m_\theta}$$

with $m_\theta$ the minimum value of $\cos^4\theta+\sin^4\theta$ for $0\le\theta<2\pi$. What is this minimum?

Taking derivatives, we see that it is possible when $-4\cos^3\theta\sin\theta+4\sin^3\theta\cos\theta=0\quad (*)$

three cases

(i) $\theta=0,\pi$ then the function's value is $1$.

(ii) $\theta=\pi/2, 3\pi/2$ again, value $1$.

(iii) $\cos\theta\sin\theta\ne 0$ then we have $\sin^2\theta=\cos^2\theta$ by manipulating $(*)$. This only happens when $\theta$ is on the line $y=\pm x$, whence both $\sin^4\theta$ and $\cos^4\theta$ are positive, so the minimum is a positive number, in fact it's easily seen to be $m_\theta={1\over 2}$

Hence we choose $\delta={\epsilon\over 2}$ and we are guaranteed convergence to $0$, as claimed.

edited Jul 04 '14 at 07:49

answered Jul 04 '14 at 07:36

Adam Hughes

36,777

1

By the way, I have upvoted your answer. – Mark Fantini Jul 04 '14 at 08:03
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That's cool, I just wanted it to be a useful approach for you in future things, since it can greatly simplify a lot of limits in more than one dimension since you get back a measure of distance from the point by using a radial version instead of the standard coordinates: a good tool for the toolbox, and I know it can be helpful to see all steps written out to believe an approach works, especially on the internet. – Adam Hughes Jul 04 '14 at 08:08
You don't need to determine the $\min$ of $g(\theta):=\cos^4\theta+\sin^4\theta$. It is sufficient to note that $g$ is continuous, periodic, and $>0$ for all $\theta$. – Christian Blatter Jul 04 '14 at 18:33
@ChristianBlatter Yes, I know. If you see my comments above, I wanted to make things painfully obvious. My original hint was exactly that this is positive, but out of deference to the op who didn't seem to understand that exactly, I did it a different way in my answer. – Adam Hughes Jul 05 '14 at 06:27

Pedro · Answer 2 · 2014-07-04T18:21:41.260

To apply the squeeze theorem, notice that if $x\neq 0$ and $y\neq 0$ then $$\begin{align*} \left|\frac{x^3y^2}{x^4+y^4}\right|&=\left|x\frac{x^2}{\sqrt{x^4+y^4}}\frac{y^2}{\sqrt{x^4+y^4}}\right|\\ &=\left|x\frac{x^{-2}}{x^{-2}}\frac{x^2}{\sqrt{x^4+y^4}}\frac{y^{-2}}{y^{-2}}\frac{y^2}{\sqrt{x^4+y^4}}\right|\\ & =\left|x\frac{1}{\sqrt{1+y^4/x^4}}\frac{1}{\sqrt{x^4/y^4+1}}\right|\\ &=|x|\left|\frac{1}{\sqrt{1+y^4/x^4}}\right|\left|\frac{1}{\sqrt{x^4/y^4+1}}\right|\\ &\leq |x|\cdot 1\cdot 1. \end{align*}$$

So, for all $(x,y)\in\mathbb{R}^2$ we have $0\leq |f(x,y)|\leq|x|$. It follows that $$\lim_{(x,y)\to(0,0)}f(x,y)=0=f(0,0)$$ and thus $f$ is continuous at origin.

Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous.

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