Given a $\Bbb R^2$-valued path $\gamma=(x(t),y(t)), \gamma :[0,1]\rightarrow\Bbb R^2$ with natural parametrization, i.e. $\mid \dot{x}(t) \mid + \mid\dot{y}(t)\mid=L $, where $L$ is the length of the path under appropriate norm. Is it possible that $\gamma$ could have infinite number of loops? If yes, I would like to see an example. (Loops are for $t_1>t_2$, $\gamma(t_1)=\gamma(t_2)$. Since the loop is at natural parametrization, there cannot be a stationary point as a loop)
My thought so far: A loop can only be defined on a non-singleton interval $[t_1,t_2]$ and there can only be countably many of them in the interval $[0,1]$. Thus I thought that the number of loops in a continuous path with finite length is at most countable. Even we have nested loops, i.e. in $[t_1,t_2]$ there is another loop, the number of loops in $[t_1,t_2]$ is at most countable. I could not think of a loop with an uncountable infinite number of loops. What would be the highest possible number of loops?
You can relatively easily find a parametrisation of an infinite earring where the perimeters are $\frac{1}{2^n}$ (so that the total length is finite).
Edit: why do you have uncountably many loops in this case? Represent a real number in $(0,1)$ by its binary expansion. That is "$0.$" followed by a sequence of $0$ and $1$. To each such number there corresponds a loop where you go around the $i$-th circle if and only if the $i$-th digit is $1$. Since the total length of the earring is finite, this always gives you a well-defined finite-length loop.
This will be smooth on $[0,1)$ only though (in the sense that this cannot be extended into a smooth loop from $S^1$ to the plane). If you want to achieve smoothness on the whole interval, there is the example of $x^n \sin(\frac1x)$ that gives you a curve like this, then you can smoothly join it to a curve that goes straight along the $x$-axis.