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There is a fun reverse definition that is used for so called "currents" in geometric measure theory, objects for which then in Green's theorem always ends up trivially being true. But then using the r …

The problem is trickier than I initially thought, but with the corrected condition it can be done. I need to assume that $\sum c_n \delta_n$ is conditionally but not absolutely convergent, $0 < \delta …

If the $f_n$ are absolutely continuous, then $f'_n$ would also be the distributional derivative and $g=f'$ follows from the continuity of that. So if there is a counterexample it should probably invol …

Checking this in perfect detail might be a bit technical, but the following looks like it could work for any dimension up to n: For 1d: Take an S curve $\phi: [0,1] \to [0,1]$ such that $\phi(0) = 0$, …

If by not agree you include that the derivative may not exist, you can get any dimension and measure that does not contradict the almost everywhere. Consider the worst case $d=1$: Take the constructio …

One can use your infinite-density example, but replace the outer lines with very sparse dotted lines: $$S = (\{0\} \times \mathbb{R}) \cup \bigcup_{i=1}^\infty \{i^{-1},-i^{-1}\} \times \left[\bigcup_ …

For your interest in a minimal $f$, you might want to read a beginners textbook on Sobolev-Spaces and the calculus of variations, especially on the direct method, which is all about this. The beginner …

Building on Pietro Majer's answer to you previous question for a change, consider the following: Let $g: [0,\infty) \to \mathbb{R}$ be the unique continuous function such that $g(0)=0$, $g'(x) = 1$ fo …

It is not possible for $n>2$. Pick any point $v$ and consider the cone spanned by all tangent rays. Clearly that cone has to be a subset of the solution, as otherwise you would miss part of the ball. …

Okay, let me try a writeup of the comment chain. For any reasonable subset $A\subset \Omega_2$ and $B := f^{-1}(A)$ you get $$\int_A |f^{-1}(y)| dy = \int_B \det df dx \leq \liminf_{n\to\infty} \int_B …

There is a counterexample, however there might be ways to avoid it. Take $\mathcal{M} = \mathcal{N} =\mathbb{S}^2$, but now consider sequence of maps that cover the sphere twice, where you shrink the …

The following is not a full proof, as I skip on some calculation details, just an extension of the remark I made in a comment but I am pretty certain it is correct. The answer is no, at least not wrt. …

You can treat this as a problem with two Lagrange multipliers. Then by standard methods, a minimizer $f$ has to exist (by convexity in $f'$) and has to be a weak solution to $$-f'' + \lambda f + \mu f …

This looks like one of these examples that is more bark than bite. If we separate the first term of the sum and do a bit of reordering we get $$x_n-x_{n-1} = \frac{b^2}{n^2} x_{n-1} + \frac{b-a}{(n-1) …