A bounded function on $\mathbb{R}^2$

Question

How to prove that the function $f(x,y)=\displaystyle\frac{xy^2}{x^2+y^4}$ if $(0,0)\not = (0,0)$ and $f(x,y)=0$ is bounded on $\mathbb{R}^2$?

I like some advice to this problem.

Thanks!

score 3 · Accepted Answer · 2014-11-05T10:08:56.307

3

Use $AM-GM$ inequality $ \frac{a + b }{2} \geq \sqrt{ab} $ with $a = x^2 $ and $b = y^4$ which are obviously positive :

$$ \frac{ x^2 + y^4}{2} \geq xy^2 \implies \frac{1}{x^2+y^4}\leq \frac{1}{2xy^2}$$

Hence,

$$ f(x,y) \leq \frac{1}{2} $$

edited Nov 05 '14 at 10:08

answered Nov 05 '14 at 09:50

That is a very simple solution! Thanks! – EQJ Nov 05 '14 at 09:52
You have a typo in the first inequality involving $(x,y)$ and you're missing absolute values throughout. – Git Gud Nov 05 '14 at 10:08

score 1 · Answer 2 · answered Nov 05 '14 at 09:50

1

Hint: define $r,\theta$ such as \begin{align} y^2 &= r\sin\theta\\x &= r\cos\theta \end{align}

answered Nov 05 '14 at 09:50

mookid

Can you elaborate a bit more on your "hint" ? thanks – Nov 05 '14 at 09:55
1

@PedroArbizu using trig identities this becomes $\sin\theta\cos\theta$ which is bounded. – mookid Nov 05 '14 at 09:56

2 Answers2