Theorem 1.3 of [KKR] says that if $E\subset \mathbb{R}^n$ is closed, totally disconnected, and $\mathcal{H}^{n-1}(E)<\infty$, then $E$ is metrically removable. Can one replace $\mathcal{H}^{n-1}(E)<\infty$ with the condition that $E$ has $\sigma$-finite $\mathcal{H}^{n-1}$, i.e., is a countable union of sets of finite $\mathcal{H}^{n-1}$ measure? This was asked in follow-up paper [LPR].
Lemma 3.8 of [KKR] says that the countable union of metrically removable closed sets is metrically removable. Thus, the previous question could be settled by showing that a closed $\sigma$-finite set is a countable union of closed finite measure sets. Or, a $\sigma$-compact topological measure space with a $\sigma$-finite measure can be written as a countable union of compact sets of finite measure.
Counterexample. In general, the above is false: let $X$ be $\mathbb{R}$ with the counting measure on rationals. If $\mathbb{R} = \bigcup_{k=1}^\infty A_k$ with $A_k$ closed, then by the Baire category theorem, some $A_k$ has nonempty interior, hence infinite rational-counting measure.
In general, $X$ is the countable union of closed sets of finite measure and some null set. (Follows from interior regularity: approximate finite measure sets by compact subsets, etc.)
The union of two metrically removable sets is not always metrically removable. If at least one of them is closed, then it is (Lemma 3.7 of [KKR]). What if neither is closed, but the union is totally disconnected?
In [KKR] it is proved that a countable union of closed metrically removable sets is metrically removable. What if we drop “closed” but assume that the union is totally disconnected?
Question 9.1 of [KKR] asks: “Are metrically removable sets preserved by bi-Lipschitz homeomorphisms? Or even by quasiconformal maps?”
They are preserved by affine maps. Indeed, an affine image of a “short” curve $\gamma$ (of length $<(1+\epsilon)|a-b|$) is another short curve. To see this, consider a partition of $\gamma$ and classify its segments as “aligned” (small angle with $[a, b]$) and “non-aligned.” The total length of non-aligned segments is small, and remains small after an affine map is applied. The length of aligned segments is changed by about the same factor as the length of segment $[a, b]$.
Since a diffeomorphism is locally close to an affine map, it also preserves metrically removable sets.
Characterize harmonic polynomials $p\colon \mathbb{C}\to\mathbb{C}$ with bounded zero set $p^{-1}(0)$. This is relevant to [Wilmshurst].
Characterize harmonic polynomials $p\colon \mathbb{C}\to\mathbb{C}$ that are proper maps, i.e., $|p(z)|\to \infty$ as $|z|\to\infty$. This is also relevant to [Wilmshurst].
What can be the range of a harmonic polynomial $p\colon \mathbb{C}\to\mathbb{C}$? Trivial cases: a point, a line, or $\mathbb{C}$. What other sets may appear? For example, for $(x,y) \mapsto (x,\, x^2-y^2)$ the range consists of $(u, v)$ with $v\le u^2$. There’s a similar example in [Wilmshurst]. [Sheil-Small] observes that the range is not contained in any halfplane, and analyzes the case of quadratic harmonic polynomials completely.
The range need not be closed: $(x,y)\mapsto (x,\, xy)$ has range $\{(u, v) : u\ne 0\}\cup \{(0,0)\}$.
Can the omitted set $\mathbb{C}\setminus p(\mathbb{C})$ be a nonempty finite set? Nonempty bounded set?
In [FG] it is proved that the complement of any finite subset of $\mathbb{R}^n$ is a polynomial image of $\mathbb{R}^n$. For example, the punctured plane $\mathbb{R}^2 \setminus \{(0, 0)\}$ is the image of the polynomial map $(x, y) \mapsto (xy - 1,\, (xy - 1)x^2 - y)$. This polynomial is not harmonic (although its first component is).
Let $\Omega^*$ be a doubly connected domain in the plane with $0<\operatorname{Mod}\Omega^*<\infty$. Does there exist $\epsilon>0$ such that for every domain $\Omega$ with $\operatorname{Mod}\Omega<\operatorname{Mod}\Omega^*+\epsilon$ there exists a harmonic diffeomorphism $h$ from $\Omega$ onto $\Omega^*$?
Same as the previous question but $h$ is required to minimize the Dirichlet energy among all diffeomorphisms from $\Omega$ onto $\Omega^*$.
Suppose $f\colon \mathbb{D}\to\mathbb{D}$ is harmonic and $\lim_{z\to 0} f(z)/z$ exists. What is the best form of Schwarz lemma for such maps? For example, $|f'(0)|\le 1$ is true.
Suppose $f\colon \mathbb{D}\to\mathbb{D}$ is harmonic and $f(0)=0$. Let $$\|f\|_{r, p} =\left(\frac{1}{2\pi} \int_0^{2\pi}|f(re^{i\theta})|^p\,d\theta\right)^{1/p}.$$ Is it true that $\|f\|_{r,4}\le r$?
True for $p\le 2$, by Parseval. True in the derivative-at-0 form for $p\le 4$, by [KY].
False for $p>4$ (even in the derivative-at-0 form); a counterexample is similar to one in Theorem 1 of [MS]. Namely, $f(z) = (z+\epsilon \bar{z})/|z+\epsilon \bar{z}|$ on the unit circle where $\epsilon>0$ is small.
Suppose $f\colon \mathbb{D}\to\mathbb{D}$ is harmonic. Is it true that the area of $f(r\mathbb D)$ is at most $\pi r^2$? The answer is yes when $f$ is also bijective [KK].
Let $X(n)$ be the space of subsets of $X$ with cardinality at most $n$, with the Hausdorff metric.
If $X$ is doubling, quasiconvex, has finite Nagata dimension, has the binary intersection property, or is an absolute Lipschitz retract, does $X(n)$ have the same property? True for quasiconvexity [Ak2024].
If a Lipschitz retraction $X(n+1)\to X(n)$ exists, does a Lipschitz retraction $X(n)\to X(n-1)$ exist?
If $X$ is an absolute Lipschitz retract, does it admit Lipschitz retractions $X(n)\to X(n-1)$ for all $n$? Would be enough to answer for $\ell^\infty$. See [Ko2022].
If $X$ admits continuous retractions $X(n)\to X(n-1)$ for all $n$, is it contractible?
Does every subset of $\mathbb{R}$ admit continuous retractions $X(n)\to X(n-1)$ for all $n$? Also, is this property preserved by homeomorphisms (or just by uniform homeomorphisms)?
Does every normed space $X$ admit Lipschitz retractions $X(n)\to X(n-1)$? True for Hilbert spaces [Ko2022]. True for small $n$ [Ak2019].
Let $X$ be a finite metric space, $|X|=N>n$. The existence of Lipschitz retractions $X(n)\to X(n-1)$ is clear (e.g., use an arbitrary order of the points of $X$, and pick the first $n-1$ elements in that order). But what is the smallest possible Lipschitz constant of such a retraction, in terms of the space?
Let $\Gamma$ be a curve (bounded or unbounded, planar or abstract) with Lipschitz retractions $\Gamma(n)\to \Gamma(n-1)$. Does it follow that $\Gamma$ is a quasiarc?
No for $n=2$: the cusp $y=\pm x^2$ admits a retraction that sends $\{(x_i, \epsilon_i x_i^2)\}$ where $\epsilon_i = \pm 1$, to $(x, \epsilon x^2)$ where $x = \min(x_1, x_2)$ and $\epsilon = \max(\epsilon_1, \epsilon_2)$.
Let $\Omega\subset \mathbb{R}^d$ be a domain (homeomorphic to a ball). When do Lipschitz retractions $\Omega(n)\to \Omega(n-1)$ exist? By [Ko2022] it is necessary for $\Omega$ to be (locally uniformly) quasiconvex. If we assume $\Omega$ bounded, then it is quasiconvex. In two dimensions, contractible and quasiconvex imply ALR (see Heinonen’s Lectures on Lipschitz Analysis).