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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

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    It is easy to arrange two strictly convex to share an entire segment, no? – 2017-02-28
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    Why not $y=x^2$ and $y=x^2+\cos x$ ? – 2017-02-28
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    @Roland yes. It could be a nice example even you should give for the second curve a compact range of values to x for being a parametrization and, so the second being a curve. Let's forget this for a while. If the curvature of the second curve is not 0, it could be interesting at least for the second case 2) of my question. – 2017-02-28
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    sorry, I do not understand what you mean in your comment. But the curvature of $y=x^2+\cos x$ is nowhere 0. – 2017-02-28
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    ok. the curve defined by the function above is convex? Is The curvature everywhere strictly positive or everywhere strictly negative? – 2017-02-28
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    $y''=2-\cos x>0$ so the curvature is strictly positive. Compute, this is not a difficult function. – 2017-02-28
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    The curvature is slightly different than the second derivative. it differs from a factor equal to 1/(1+ (2x-sin(x))^2)^3/2. However I agree with you that both curves are strictly convex and they are smooth but as I told above for being considered curves you should give a compact range of values for x. This implies automatically that the two curves will intersect just a finite number of times. For example x belongs to [0,10]. This is just how a curve is defined on a plane. However your basic idea is good. – 2017-03-01
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    @FrancescoCiardiello I didn't say $y''$ was the curvature, just that they have the same sign. I don't know why you say these aren't curves, but if you want $x\in[0;10]$, just takes $y=x^2$ and $y=x^2+f(x)$ where $f(x)$ is your favorite function with an infinite number of zeroes on $[0,10]$ with bounded second derivative. For example $f(x)=Cx^5\cos\frac{1}{x}$ with $C$ small enough so that $f''(x)<2$.. – 2017-03-01
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    @Roland I say that there are not curves just by definition of curves on a plane. However, I think that your new example might work. For sure y=x^2 +f(x) is smooth and has infinitely many intersection points with x^2. I will check the sign of the curvature or of the second derivative. I will let you know. – 2017-03-01
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    You are right @Roland. the curves x in [0,1]--> (x,x^2) and x in [0,1]--> (x, x^2 + C x^5 cos 1/x) with C small enough are simple smooth not closed with a curvature strictly positive and they intersect in infinitely many points. Thanks. – 2017-03-01

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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:

Graphics[{Line[CirclePoints[10]],
  Red, Circle[{0, 0}, .98, {-π/2 + .05, 3 π/2 - .05}]}]

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).

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    (+1) I am having troubles in updating images on imggur, but my idea was pretty the same. By using smoothed polygons we may have an arbitrary number of intersections with a circle, and by overlapping portions of smoothed polygons with an increasing number of sides we may construct and example with infinite points of intersections. – 2017-02-28
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    Yes. I programmed such a figure but imgur will not let me load it. – 2017-02-28
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    If the OP is using the standard definitions, then a strictly convex should be one with no straight line segments embedded in the boundary. So smoothing the corners of $n$-gons isn't good enough to do part (1). – 2017-02-28
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    yes @RobArthan is right. I thought about the examples that Jack was providing. actually, you can think about two straight lines whose curvature is 0, then they are convex. Two straight lines can completely overlap. For sure your example is a bit more refined in the sense it shows some geometrical continuity of what I am saying. However the hypothesis that curves are strictly convex or, in the best case, have curvatures equal to 0 at isolated points. – 2017-02-28
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    @FrancescoCiardiello: So you'll accept my solution? Run my code to see the figures. – 2017-02-28
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    no, I can't @DavidG.Stork . I really do not know if these two curves exist. Maybe such curves do not exist. – 2017-02-28
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    @DavidG.Stork: your proposed solution does not provide infinitely many points of intersection between two distinct strictly convex curves. – 2017-02-28
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    @Bob Arthan: Why not? I let $n \to \infty$ so the number of intersection points becomes countably infinite. If you seek an uncountable number of intersection points, merely put two equal convex arc segments atop one another. Done! – 2017-02-28