# Does an origin-centered ellipse in the plane intersect each $L^p$-circle at most 8 times?

The question is in the title: Let $E$ be an origin-centered ellipse in ${\mathbb R}^2$ and let $S$ be an "$L^p$-circle": $S = \{(x,y) : |x|^p + |y|^p = \text{const}\}$, where $1 \leq p \leq \infty$. Is it true that $E$ and $S$ have at most $8$ points of intersection?

(Of course, you should ignore the case that $p = 2$ and $E$ and $S$ coincide.)

This is easy to show if $E$ is a circle, but it's not completely obvious (to me) when $E$ is a more general ellipse, though it seems true from drawing pictures.

Motivation: I was trying to fill in (and maybe simplify) the details of "Remark 2" of the paper "On the best constant in the Khintchine-Kahane inequality" by Latala and Oleszkiewicz (http://www.mimuw.edu.pl/~rlatala/papers/chinsm.pdf). I posted a slight variant of the question at MSE (https://math.stackexchange.com/questions/408680/how-many-points-of-intersection-between-an-ellipse-and-an-l-p-circle) which would have helped with the simplification, but unfortunately the variant is false. Still, the above question seems of moderate interest, and no one at MSE answered it.

• Is it enough for your purpose to show this when $E$ has axes aligned with the coordinates, i.e. is of the form $(x/a)^2 + (y/b)^2 = 1$? In that case I can show that each quadrant contains at most two points of intersection. But maybe you knew that already. – Noam D. Elkies Jun 11 '13 at 18:21
• Yes, I think this case is not too hard (Ilya makes the same remark in his answer). – Ryan O'Donnell Jun 12 '13 at 13:50

$\def\sign{\mathop{\rm sign}}$First of all, it is enough to prove the statement when $p=u/v$ is rational, $u$ is even and $v$ is odd (such numbers are dense on the real line). We need this to simplify the last argument.

Let the equation of the ellipse be $f(x,y)=ax^2+2bxy+cy^2=1$; one may assume that $b\neq 0$ (either by a small variation argument, or by considering it directly --- this case is easy). Let us bound the number of local extrema of $f(x,y)$ on $|x|^p+|y|^p=1$; if there are at most 8 of them, then we are done: between every two, there is at most one intersection point. Since $b\neq 0$, the points where one of the coordinates vanish are not extremal (the ellipse passing through such point has a tangent not parallel to the coordinate axes).

The extremal points satisfy the system of Lagrange equations $$2(ax+by)=p\lambda|x|^{p-1}\sign x, \quad 2(bx+cy)=p\lambda|y|^{p-1}\sign y;$$ obviously, $\lambda\neq 0$, so this system yields $$\frac{at+b}{bt+c}=|t|^{p-1}\sign t,$$ where $t=x/y$. So, it suffices to show that there are at most four such values of $t$ (each corresponds to two symmetrical extremal points). Notice that $|t|^{p-1}\sign t=t^{(u-v)/v}$ by our assumptions on $u$ and $v$. Now, substituting $t=s^v$ we obtain $$as^v+b=bs^u+cs^{u-v}.$$ The last polynomial equation has at most four roots by an easy application of Descartes' rule of signs.

ADDENDUM. By the way, you have mentioned on MathSE that in your case the minor axis is in the positive quadrant, and you were interested in the intersection points lying in this quadrant as well. In this case we may assume that $a,b,c>0$, and we are interested in positive values of $t$. The rule of signs shows that there is at most one $t$ whenever $u-v>v$, that is --- if $p>2$. Otherwise it claims that there is at most one $t$ in the adjacent quadrant.

ADDENDUM2. Here is the explanation about the limiting case. Assume that you have more than 8 common pointsof an ellipse and an $\ell_p$-circle. Blowing up or away your ellipse a bit, you may reach the situation when there are more than 8 points of transversal intersection. Then you mark a point on the ellipse between every such neighboring points (or simply mark 10 of them); these marked points are alternately inside and outside the $\ell_p$-circle. Finally, if you change $p$ a bit, all these 10 points will still be alternating, guaranteeing 10 intersection points.

The same method works for $b=0$.

Finally, if it is easier for you, you may pass to rational $p$ of a desired form AFTER writing down the (almost) polynomial in $t$ --- in this case the statement is more visible...

And, even more finally, you may just repeat the proof of the Descartes' rule for an almost polynomial' --- all you (perhaps) need is that all the powers differ by at least one, which can be reached by an appropriate substitution.

ADDENDUM3. Just to mention. The same method allows to prove that the affine images of $\ell_p$- and $\ell_q$-circles also have at most 8 common points, if they have the same center.

• Very nice -- I didn't have the idea of converting to rational exponents to get actual polynomials in the Lagrange system! But actually, I have what is probably a very dumb question about that... You have established that there is a sequence of $p_i$'s with $p_i \to p$ such that $E$ has at most 8 points of intersection with each $S_{p_i}$. Is it clear why this means $E$ has at most 8 points of intersection with the limiting $S$? (Similarly, I don't see how to assume $b \neq 0$ by a limiting argument, though I agree that that $b = 0$ is easily handled directly.) – Ryan O'Donnell Jun 12 '13 at 14:03
• I've added an explanation below the main text. Since I was to type something;), I have also added some words about a generalization. – Ilya Bogdanov Jun 12 '13 at 15:14
• Thanks for the clarifications Ilya! Regarding tiny dilations of the ellipse to get transversal intersections -- I still have to think a little bit about this. Perhaps a tiny bit of care is needed -- if you have, say, two tangential intersections on "opposite sides" then when you dilate the ellipse a bit, you might lose one intersection altogether, but turn the other intersection into two. So you should be okay. OTOH, I agree with you that it's quite clear you can just wait until you've got the Lagrange solution to pass to rational p. So all is good! Thanks again! – Ryan O'Donnell Jun 12 '13 at 18:31
• Yes, I have had this dilemma in mind when I told blowing up or away'. Sorry for being not that detailed... – Ilya Bogdanov Jun 12 '13 at 19:12