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Suppose a function $f$ is defined as follows:

$$f(x,y)=\begin{cases} \frac{x^3y^2}{x^4+y^4}&\text{ when }(x,y)\neq(0,0),\\0 & \text{ when }(x,y)=(0,0).\end{cases}$$

Is this function continuous at $(0,0)$? How is this shown? I've tried considering limits for different $y=g(x)$ functions and I am unable to find a counterexample. But I do not see how to prove continuity in general.

Martin Sleziak
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Pierre
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  • have you tried converting whole things into polar coordinates? Usually that's the first thing you'd try when you suspect (or even when you don't have any idea) it's continuous – user340297 Jun 12 '16 at 04:37
  • See also: http://math.stackexchange.com/questions/856098/proving-fracx3y2x4y4-is-continuous – Martin Sleziak Jun 21 '16 at 18:47
  • @Did I have noticed that the other post does not have answer based on AM-GM; otherwise the answers seem to be more-or-less comparable. If this one has slightly better answers, would it perhaps make sense to close in the other direction instead? (I have asked the same thing also in chat.) – Martin Sleziak Jun 21 '16 at 19:21
  • @MartinSleziak Calling AM-GM, the inequality $2ab\leqslant a^2+b^2$? Well... (On the fascinating problem of deciding which post should be declared duplicate of which other, in this case I am rather agnostic, I am afraid. But the present situation does not seem absurd, does it?) – Did Jun 21 '16 at 19:51
  • Well, absurd is probably too strong word, especially considering that this is a totally insignificant issue. But closing a question which explicitly asks to avoid epsilon-delta proof is closed as a duplicate of a question with explicitly asks for epsilon-delta proof is definitely strange. – Martin Sleziak Jun 21 '16 at 20:36
  • @MartinSleziak Interestingly, no answer to the "explicit delta-epsilon" question even mentions any delta or epsilon... Since you asked for my opinion, let me mention that I find neither the AM-GM/no AM-GM distinction nor the delta-epsilon/no delta-epsilon distinction you mention, the least meaningful. Hence no, not strange. – Did Jun 21 '16 at 20:46

3 Answers3

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Since $x^4-2x^2y^2+y^4= (x^2-y^2)^2 \ge 0$, we have $2x^2y^2 \le x^4+y^4$.

Therefore, $\dfrac{x^2y^2}{x^4+y^4} \le \dfrac{1}{2}$ for all $(x,y) \neq (0,0)$.

Also, $\dfrac{x^2y^2}{x^4+y^4} \ge 0$ for all $(x,y) \neq (0,0)$.

From the above inequalities, we have that $\left|\dfrac{x^2y^2}{x^4+y^4}\right| \le \dfrac{1}{2}$ for all $(x,y) \neq (0,0)$

Now, multiply both sides by $|x|$ to get $\left|\dfrac{x^3y^2}{x^4+y^4}\right| \le \dfrac{1}{2}|x|$.

Can you finish the problem from here?

The Squeeze Theorem will be useful.

JimmyK4542
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    I think you mean $(x^{2} - y^{2})^{2} \ge 0$. – Matthew Cassell Jun 12 '16 at 04:29
  • I think so. Does taking $\delta = \Vert (x,y) \Vert$ in the definition of continuity suffice, keeping in mind that $|x| \leq \Vert (x,y) \Vert$? (I'm arbitrarily considering the Euclidean norm) – Pierre Jun 12 '16 at 04:30
  • @Mattos: Thanks for catching that. Pierre: Since you are only asked to show continuity at $(x,y) = (0,0)$, your formula for $\delta$ shouldn't have $x$ or $y$ in it. Try picking $\delta = \epsilon$. – JimmyK4542 Jun 12 '16 at 04:33
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A fairly efficient way to approach this problem is to transform to polar coordinates and write

$$\lim_{(x,y)\to (0,0)}\frac{x^3y^2}{x^4+y^4}=\lim_{r\to \infty}\left(r\,\,\frac{\cos^3(\phi)\sin^2(\phi)}{\cos^4(\phi)+\sin^4(\phi)}\right)$$

Noting that we can write

$$\begin{align} \left|\frac{\cos^3(\phi)\sin^2(\phi)}{\cos^4(\phi)+\sin^4(\phi)}\right|&=\left|\frac{\cos^3(\phi)\sin^2(\phi)}{2(\sin^2(\phi)-\frac12)^2+\frac12}\right|\\\\ &\le 2 \end{align}$$

then the limit of interest is $0$. Therefore, the function $f(x,y)$ is continuous at the origin.

Mark Viola
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Hint: use arithmetic-geometric mean inequality to show the function is continuous at $(0,0)$: for $a,b>0$, $$ ab \leq \frac{a^2 + b^2}{2} $$

Will Kwon
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