I am looking for an example.
1)Find Two planar, smooth (C^2) and strictly convex curves such that their intersection counts an infinite number of points in the plane. By strictly convex I mean a curve whose curvature is strictly positive or strictly negative. In addition the curves need to be simple and not loops.
2) (Relaxed version of the problem) If you can't find the example as above, try to relax the hypothesis slightly more. Then you can take the following hypothesis. The two curves are convex. But when the curvature is 0, then you can't have a curvature locally 0 around that point. For sure this is a relaxing condition because you allow for the curvatures being zero. The rest of hypotheses are the same. Also in this case you should have an infinite number of intersections.
I am very happy if anyone finds this counterexample especially for the case 1). I am not sure if this example exists. Even you have a partial answer or you know why such curves cannot exist, please let me know..
Thanks. Francesco
Why not 1) a regular $n$-gon with smoothed corners (of vanishing radii of curvature) and 2) a circle of radius equal to the average radius of the $n$-gon, and let $n \to \infty$? You can avoid the "loop" restriction by cutting one leg from the $n$-gon and a vanishingly small segment from the circle.
Given imgur will not let me load a picture, here is the Mathematica code that generates it:
If the OP demands strictly convex (i.e., no straight line segments), then each straight line segment in the above candidate solution can be replaced by a section of a circle of arbitrarily large radius of curvature.
A "degenerate" solution would be two equal segments of any strictly convex curve that overlap along a finite segment (and hence contain infinitely many points in common).