Regarding your very last question, yes, you will get faster. If you are following a topic for a while, then of that 60+ references you will have already read most of the important ones, the path of the authors took is one that you would have taken on your own after some thinking and the technicalities are precisely the ones you would have expected or have seen in a similar result. What once took you a month to understand now takes you an hour. Of course while that might give you solace, it will not help you now. The only way to get to that point is to continue reading papers.

This directed angle idea was what I meant with signed angles in the comment, so yeah, I believe this could save a bit on case distinction. On the other hand, I missed the possibility of an arc consisting of more than half of the circle. It's quite possible that this works out mathematically with angles larger than $\pi/2$, but it might need another case distinction.

@WillSawin There are some cases where an equation is used to define something, but neither side of the equation actually is the quantity defined. E.g. "Let $x_0$ be the solution to $A(x_0) =b$", where A is some complicated expression (proven to have a solution by some other statement) might be preferable to explicitly defining $x_0 := B$, when the term only ever occurs in the form $A(x_0)$ and getting there from B is non-trivial. So while I also prefer $:=$, you'd probably have to introduce variants of it, as well as $=:$ depending on context.

From the second picture it seems to me that the answer is in the base angles of the isosceles triangles given by each side and the corresponding circle center. If you take them as signed, each adjacent pair of them always add up to the given angle of the original polygon. For n points that gives you n equations in n unknowns, though the solution might not be unique, since everything is only defined up to multiples of $2\pi$.

I guess, if you care to decipher their notation, then I guess Giaquinta, Modica & Soucek "Cartesian currents in the calculus of variations" should have the state of the art when the book was written (see in particular Section I.3.4, but possibly also elsewhere). On the other hand, the book was written more than 20 years ago, so possibly there has been some newer development, so I won't post this as an answer. I believe though that in particular for $A^+_{p,q}$ the problem should still mostly be open.

@StevenStadnicki There may be another good point in there. Having only a trustable answer may not good enough. To show that P=NP you need a polynomial time algorithm that always works, not one that works 99,99999% of the time. The latter would only show NP $\subset$ BPP (which would be a big result in itself though).

Is there something missing from the problem? First of all, there is no coupling between the two integrals, so you could equally first take the supremum of one and then add it to the other. Secondly there are no additional bounds and the problem is linear, so the solution will always be either 0 or $\infty$.

Why use a screwdriver when a hammer is a much more intuitive tool? There are simply too many problems that cannot be written as a contraction and are thus inaccessible to Banach. On the other hand, there are enough problems where you are interested in uniqueness, so Leray-Schauder is no help. In some sense PDE-theory is all about having the right tool for the right job and knowing when to use it. Also, judging the importance of a tool by the amount of content in a book is never a good idea. You could equally argue that if Banach is easier to use, then it should take less pages to explain.

Toying around with thing a little, I have an explicit example where $D$ is smaller than $X$, but I fear that won't help me to avoid density. However it got me thinking: Have you tried restricting yourself to currents of the form $\delta_a-\delta_b$ first? This subspace already shows a lot of the features of the flat norm, but you can write down the topology in terms of a and b more directly.

@YaHe Ultimatively you need to be able to smooth the corners without violating this condition, so it needs to point "outwards" in some sense. But this you get automatically because it is already pointing outwards for all the flat pieces of the boundary near the corner.

@LoïcTeyssier I think quarter-singularity might be a misleading term here. Corners need not be right angle, so any number of them could join up to form a full singularity. But I guess the definition of degree as signed relative area of $S^{n-1}$ covered by the map would work safely for any angle.

@jgall no problem. Just keep in mind that I was half asleep when writing the answer and messed up the solution of the differential equation a bit. There are some squareroots missing and depending on the sign of $\lambda+\mu$ in particular, the middle part may consist of $\sinh$ and $\cosh$ instead.

Related question: mathoverflow.net/questions/358606/…

I simply assumed that velocity meant the same thing as speed in the question, because I also could find no other interpretation. And the people who always make a strict distinction between the two are usually physicists or engineers and those would never miss the harmonic oscillator as the very first example of such an ODE.