Despite the catastrophe for the world and the many victims, at least the lockdown is favorable to two activities: Mathematics and Gardening. The difficulty of handling hedge trimmer wires and garden hoses calls to some mathematical questions relative to the (un)knots in $R^3$,
which are very natural and which I could not solve, nor find references.
Analogous questions actually hold for the Jordan curves in $R^2$.
 Any help is welcome.


For $n=2$ or $3$, consider the embeddings of the circle, $S^1\hookrightarrow R^n$, and their isotopies (1-parameter families of embeddings). By the "unknot" in $R^3$, one means the unit circle in the $xy$-plane. Let us define the *width* of any isotopy $h\_t:S^1\hookrightarrow R^n$ ($t\in I=[0,1]$) as the maximum over all $s\in S^1$ of the length of the path $t\mapsto h\_t(s)$ in $R^n$.


1) Is there a finite upper bound $A$ such that every *smooth* Jordan curve $f:S^1\hookrightarrow R^2$ which is contained in the unit disk, is isotopic to the unit circle through an isotopy of width $\le A$?


2) Is there a finite upper bound $B$ such that every *smooth* knot $f:S^1\hookrightarrow R^3$ which is contained in the unit sphere and isotopic to the unknot, is isotopic to the unknot through an isotopy of width $\le B$?


3) Is there a *continuous* Jordan curve $f:S^1\hookrightarrow R^2$ which is not isotopic to the unit circle through any $C^0$ isotopy of finite width?


4) Is there a *continuous* knot $f:S^1\hookrightarrow R^3$ which is $C^0$-isotopic to the unknot, but not through any $C^0$ isotopy of finite width?



The metric difficulty of unknotting unknots
The answer is yes for all I think, provided there is a finite width isotopy, as you can just scale the curve to make it arbitrary close to the origin, do the isotopy to a tiny circle there, and expand it back. So the infimum of $A$ or $B$ should be $2$.


Now for 1) if you work in an annulus instead of a disk the answer is no, as you can take a spiral living near the unit circle and consider the boundary of a small neighborhood of it. To isotope it to a circle within the annulus, you'll have to move the curve arbitrary far away depending on the number of turns of the spiral. Clearly that example also works for the other questions.


Another possible formulation would be to use Moebius invariant metrics on the space of curves (which make the scaling trick useless). Then the answer would be no to all I guess.


