Doesn't this follow directly from the Calderon-Zygmund theory? If $f\in L^\infty$, then also $f\in L^p$ for all $p <\infty$. But then $u \in W^{2,p}$, which embeds into $C^{1,1-\varepsilon}$ for $p$ large enough.

Regarding your wrong fact, if both are upper triangular, then $P^{-1}e_1$ would be an eigenvector for both. So it clearly does not hold in general.

There is a discrete version of the Gauss-Bonnet theorem: For any polyhedron, consider the defect angles of the corners (i.e. the failure of angles in the corner to add up to $2\pi$). Then the sum of those defects is $2\pi$ times the Euler-characteristic. So in the case of a dodecahedron, the average angle has to be $2/3\pi-4/60\pi$ or 108°, which is ever so slightly less than the approx. 109.5° which would correspond to four edges meeting at equal angles. In other words, on average the corners of the dodecahedron are slightly acute. Not sure how to turn that into a full contradiction though.

@NateRiver In non-linear elasticity one wants to study deformations which are (almost everywhere) bijections. In a variational context if $\det \nabla f >0$ uniformly, then any small enough smooth variation gives me again a (local) bijection. Isolated points where this is not true, one could cut out and treat seperately. But the example mentioned in my last comment would have $\det \nabla f = 0$ on a dense set.

@NateRiver In general the set $\{f =0\}$ for a smooth function can be quite ugly. This is kind of the same, only for the derivative. I guess if you do the same in each component and play around a bit, you can even find a bijection $f: \mathbb{R}^n \to \mathbb{R}^n$ that has $\nabla f = 0$ on a set of dimension $n$, which also indirectly links this to half of all the unsolved open problems in nonlinear elasticity.

@NateRiver The construction should work with any Cantor set, even with the fat ones. Only with the latter the set of zero derivative will obviously have non-zero measure. As there is a Cantor-set for any dimension below $1$, that proves the supremum to be $n$. I guess there should also be a "skimmed" fat-Cantor set of full dimension but zero measure, but I leave that as an exercise for the reader.

Maybe as an additional afterthought, if $\phi$ is chosen in such a way that all derivatives are zero at the ends and the prefactor $l^2$ is replaced with something exponential, then I think that $u$ should even be smooth by the same argument.

Morgan's "Geometric Measure Theory: A Beginner's Guide" certainly has a lot of intuition about currents, but I fear that sometimes it is more eccentric than pedagogical. I would definitely not recommend reading as more than a companion to something more down to earth, so I will not put it as an answer.

Regarding the first moment, the solution is translation invariant. Fixing it simply removes that degree of freedom. The second moment in turn needs to be fixed, otherwise the solution wants to "spread out" infinitely, i.e. there is no maximum. If you do not ask about the real line, but about a finite interval, neither is necessary. But to be honest, the way your question is written now, I have no idea what precisely you are asking about. Also it feels more like an MSE than an MO-question.