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I'm trying to solve the following problem:

Determine which $n\in\mathbb{N}$ make the following function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ continuous at the origin:

$ f(x,y,z)=\left\{ \begin{array}{cr} \frac{(\cos^2\left(|x|+|y|\right)-1)\sin(y^2+z^2)}{(x^2+y^2+z^2)^{n/2}} & \text{if } (x,y,z) \neq 0\\ 0 & \text{if } (x,y,z) = 0\\ \end{array} \right. $

It's obvious that $f$ is continuous at the origin if $n\leq 2$, since:

$\left|\frac{(\cos^2\left(|x|+|y|\right)-1)\sin(y^2+z^2)}{(x^2+y^2+z^2)^{n/2}}\right|\leq\frac{|\cos^2\left(|x|+|y|\right)-1|\,|y^2+z^2|}{|x^2+y^2+z^2|}\leq\frac{|\cos^2\left(|x|+|y|\right)-1|\,|y^2+z^2|}{|y^2+z^2|}=|\cos^2\left(|x|+|y|\right)-1|\rightarrow 0$

I'm almost sure the function isn't continuous if $n> 2$ but I can't seem to find a way to prove it.

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up vote 2 down vote accepted

I think it is continuous for $n=3$ too. If we let $r$ be the length of the vector $(x,y,z)$, then the expression for nonzero $r$ simplifies to $$\frac{(\cos^2(pr)-1)\sin(qr^2)}{r^n}$$ where $p\in[0,\sqrt 2]$ and $q\in[0,1]$ depend on the direction of $(x,y,z)$, but not on its magnitude. If we develop the cosine and sine to the second degree each, we get the approximation $$\frac{-(pr)^2\cdot qr^2}{r^n}=2p^2qr^{4-n}$$ so as long as $n<4$ this ought to converge towards $0$.

On the other hand for $n\ge4$ we can consider the ray $(x,y,z)=(0,r,0)$. Along this line $p=q=1$, so $f$ fails to go towards $0$ for $n=4$ and actually blows up for $n>4$.

Making this rigorous using L'Hôpital's rule is left as an exercise for the reader :-)

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Thanks for your answer! If I understood correctly, you approximated the cosine and sine using its Maclaurin polynomial, right? (my English is a bit rusty lately and I want to make sure I got your idea). I'm thinking how this could be turned into a formal argument; maybe showing the error term is small enough? – F M Oct 7 '11 at 22:30
Yes. Working out the error term explicitly should certainly work, but will probably be much more involved than just using L'Hôpital's rule. Which is easier than it may look -- note that $\frac{d^4}{dr^4}f(r)g(r)=\binom42 f''(r)g''(r)$ plus terms that include one of $f(r)$, $g(r)$, $f'(r)$, or $g'(r)$ and so vanish for $r=0$. – Henning Makholm Oct 7 '11 at 22:41
Thanks a lot! [some more text here] – F M Oct 7 '11 at 22:45

The question is for $n$ integer, but we can take $a\in\mathbb R$ instead of $n$. We denote $f_a$ the corresponding function. We have for $(x,y,z)\neq (0,0,0)$ $$f_a(x,y),z)=-\frac{\sin^2(|x|+|y|)\sin (y^2+z^2)}{(x^2+y^2+z^2)^{a/2}}.$$ Now, fix a real number $m$. We have $$f_a(x,mx,mx)=-\frac{\sin(2|mx|)\sin(2m^2x^2)}{(1+2m^2)^{a/2}|x|^a}\overset{0}{\sim}-\frac{4|m^3||x|^{3-a}}{(1+2m^2)^{a/2}},$$ hence $f_a$ cannot be continuous if $3-a\leq 0$ hence $a\geq 3$. Conversely, if $a<3$ we have thanks to then inequality $|\sin t|\leq |t|$: \begin{align*} |f_a(x,y,z)|&\leq\frac{(|x|+|y|)(y^2+z^2)}{(x^2+y^2+z^2)^{a/2}}\\ &\leq \sqrt 2\frac{\sqrt{x^2+y^2}(y^2+z^2)}{(x^2+y^2+z^2)^{a/2}}\\ &\leq \sqrt 2(x^2+y^2+z^2)^{\frac{3-a}2}, \end{align*} and we can conclude since the exponent is positive.

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Thanks a lot for your help! – F M Oct 7 '11 at 22:45
@FernandoMartin You're welcome. – Davide Giraudo Oct 7 '11 at 22:46

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