3
$\begingroup$

I am trying to find a large subset of piecewise-differentiable plane curves of finite length (subsets of $\mathbb{R}^2$) with the following property:

For any pair $\gamma_1, \gamma_2$ of curves in this class, their images $\Gamma_1, \Gamma_2$ are such that $\Gamma_1\cap \Gamma_2$ has finitely many connected components.

I have attempted to prove that this is the case for this set of curves:

Piecewise-smooth curves (finite length) in which each piecewise component is either a straight line segment or a curve whose derivative is injective,

but I have failed to produce a proof or a counterexample to the claim that this satisfies the desired properties.

Could anybody suggest how to prove it, or why they believe it may be a false claim? If it is false, would some further restriction produce the desired properties?

Obviously, one can restrict to looking at just one piecewise component at a time - and this way it is easy to show that no such curve can intersect a line segment at infinitely many points (using the injectivity of the curve). And of course, any line segment can only intersect another segment at one point or the entire segment. I have failed to produce a proof for when both components are curved. I believe that having infinitely many intersection points should lead to non-injectivity of the curves at some point, but haven't been able to show that.

$\endgroup$
4
$\begingroup$

This doesn't work. See the construction below. The best restriction I can come up with is a piecewise analytic curve. Then locally around any intersection point, both curves are the graphs of analytic functions, so their difference is analytic, so if it is $0$ infinitely often, it is simply the $0$ function. Thus the intersection number is locally finite, and by compactness of the curves, it is finite.

I can construct two such curves that intersect infinitely often in an interval (each time being tangent at the intersection). If you want them to be closed curves, you can close them off in your favorite way.

They will be the graphs of functions $f_1,f_2$ on the interval $[0,1]$ such that $f_1(0)=f_2(0)=0$, $f_1'(0)=f_2'(0)=1$. As long as $f_i'(x)>0$ everywhere, your derivative condition is satisfied. We will make it so that for every natural $n$, we have $f_1(1/2^n)=f_2(1/2^n)$ and $f_1'(1/2^n)=f_2'(1/2^n)$, but that they differ in the intervals in between. To do this, we let $f_1''=0$ and $f_2''=g$ and we want to pick some $g$ such that $$ \int_{2^{-n}}^{2^{-(n-1)}}g(x)dx=0\\ \int_{2^{-n}}^{2^{-(n-1)}}xg(x)dx=0 $$ To satisfy these, we pick some nonzero smooth function $\varphi$ compactly supported on $[0,1]$. On $[0,3]$, let $\psi(x)=a\varphi(x)+b\varphi(x-1)+c\varphi(x-2)$. By linear algebra, by picking $a,b,c$ appropriately, we can make them not all 0 and have $$ \int_0^3 \psi(x)dx=\int_0^3 x\psi(x)dx=0 $$ We then make $g$ on the interval $[2^{-n},2^{-(n-1)}]$ be a horizontal and vertical rescaling of $\psi$ with the vertical rescaling factor decreasing very fast with $n$ (I think $1/n!$ is enough, but I'm not sure). The fast decrease is to guarantee that $g$ is smooth at $0$. This completes the construction. You can also get uncountably many intersections (separated by points where they don't intersect) by using a similar construction based on a Cantor set.

$\endgroup$
  • $\begingroup$ Why does having $f_i'>0$ imply the derivative condition? $\endgroup$ – MathTrain Nov 26 '19 at 21:57
  • 1
    $\begingroup$ @MathTrain It doesn't. I meant having $f_{i}''>0$. Then construction can be fixed by adding some large constant to each $f_{i}''$, that is adding $Cx^2$ to $f_i$. $\endgroup$ – Sam Zbarsky Nov 27 '19 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.