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Let $\mathcal{E}$ be the set of smooth manifolds with boundary $E\subset \mathbb{R}^{3}$ which are perturbations of the unit ball whose volume $V$, diameter $d$ and area of the boundary $A$ satisfy: $$1\leq d\leq 4,$$ $$2\pi \leq A \leq 8\pi$$ and $$\frac{2\pi}{3}\leq V \leq \frac{8\pi}{3}.$$

Additional assumption: The total curvature, $\int_{\partial E} H^2 d\sigma$, of the boundary is smaller $8\pi$.

I am looking for a reference where I can find a theorem which states that there exists a universal constant $C$ such that for every $E\in\mathcal{E}$ that is diffeomorphic to the closed unit ball, we can find a representative diffeomorphism with $C^{1}$ norm bounded by $C$.

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A side comment: by isoperimetry the lower bounds on $d$ and $A$ are redundant: a volume of $2\pi /3$ guarantees that the surface area is at least $4\pi / 2^{2/3} > 2\pi$, and the diameter at least $2/\sqrt[3]{2}$. Similarly, the upper bound on volume is unnecessary (if one were willing to lose a constant factor) given the bounds on area and diameter. –  Willie Wong Jan 28 '13 at 11:55
    
Also on MathOverflow –  user53153 Feb 5 '13 at 20:15
    
I just now noticed the difference in formulation between Math.SE and MO. Here you asked about a diffeomorphism between manifolds with boundary ($E$ and closed unit ball). On MO it's about open sets, i.e., the map need not be defined on the boundary, let alone be smooth there. Which version are you more interested in? If smoothness on the boundary is not required, my answer can be simplified by dropping $C^{1,\alpha}$ requirement and deleting Step 3. (I'm not going to actually do this, but the remark is worth making). –  user53153 Feb 5 '13 at 22:54
    
For the version with open sets, the following result can be helpful: any $L$-biLipschitz homeomorphism between open sets in $\mathbb R^3$ can be approximated by a $C^{\infty}$-smooth diffeomorphism that is $\tilde L$-biLipschitz with $\tilde L$ depending only on $L$. This is Theorem 2 in M. Kiikka, Diffeomorphic approximation of quasiconformal and quasisymmetric homeomorphisms, Ann. Acad. Sci. Fenn., Ser. A I, Math. 8, 251-256 (1983). This result reduces the problem to finding an $L$-biLipschitz map between open sets with a uniform bound on $L$. –  user53153 Feb 5 '13 at 23:03

1 Answer 1

The curvature assumption looks unnatural when your goal is a $C^1$ bound. Ideally, $E$ should be merely $C^1$-smooth. But here is what I can prove:

Claim If $E$ is generated by rotating a $C^{1,\alpha}$-smooth closed curve $\Gamma$ (where $\alpha>0$), then such a diffeomorphism exists, with constant $C$ that depends only on the chord-arc constant of $\Gamma$.

A closed curve is called chord-arc (or Lavrentiev curve) if there is a constant $M$ such that for every two points $a,b$ on the curve the shorter arc between them has length at most $M|a-b|$, where $|a-b|$ is the Euclidean distance. In other words, subarcs cannot be much longer then the corresponding chords.

Sketch of Proof. We may and do assume that the axis of rotation is the $z$-axis. Then $\Gamma$ is a symmetric curve in the $xz$-plane which is symmetric about $z$-axis: formally, $\rho(\Gamma)=\Gamma$ where $\rho(x,z)=(-x,z)$.

Step 1. It suffices to find a diffeomorphism $g$ of the $xz$-plane such that $Dg$ and $D(g^{-1})$ are bounded by suitable $C$, and $g\circ \rho=\rho\circ g$.

Indeed, such $g$ naturally yields a diffeomorphism $G$ of $\mathbb R^3$ which, in terms of cylindrical coordinates $(r,\theta,z)$, leaves $\theta $ unchanged and applies $g$ to the pair $(r,z)$. A direct computation of derivatives shows that the derivative of $G$ also admits a two-sided bound. The key point here is that $g$ maps the $z$-axis to itself, which in view of the derivative bounds implies that the $x$-coordinate of $g(x,z)$ is comparable to $x$. Because of this fact, the transition from $2$ to $3$ dimensions does not introduce much extra distortion.

Step 2. The following theorem can be found in several places with different proofs: Theorem 7.10 in Pommerenke, also [Tukia] and [Jerison-Kenig].

Theorem A. Let $f:S^1\to\mathbb R^2$ be a bi-Lipschitz embedding: that is, $$M^{-1}|a-b|\le |f(a)-f(b)|\le M |a-b| \ \text{ for all } \ a,b\in S^1\tag{1}$$ where $M$ is a positive constant. Then $f$ extends to a homeomorphism $F:\mathbb R^2\to\mathbb R^2$ which is bi-Lipschitz with a constant $C=C(M)$.

The proof of Theorem A in [Pommerenke] follows the original paper by Tukia, but uses the Douady-Earle extension instead of Beurling-Ahlfors. Let $G$ be the domain bounded by $\Gamma$. Let $\phi:\mathbb D\to G$ be a conformal map (note that we can choose $\phi$ appropriately symmetric when $G$ is symmetric). The map $\phi$ extends to a homeomorphism of the boundaries. We consider $\phi^{-1}\circ f$, which is a homeomorphism of $\partial \mathbb D$ onto itself. The Douady-Earle extension provides a nice (quasiconformal, smooth) diffeomorphism $\psi:\mathbb D\to\mathbb D$ which agrees with $\phi^{-1}\circ f$ on the boundary. Then $F=\phi\circ \psi$ is our extension. (Or rather half of it; the process then repeats on the exterior of $\Gamma$, but you do not need this.)

By construction $F$ is smooth on $\mathbb R^2\setminus S^1$ and therefore satisfies the desired derivative bounds there (the bi-Lipschitz property does not imply differentiability, but controls the size of derivatives when they do exist). Furthermore, $F$ has the desired symmetry property $F\circ \rho=\rho\circ F$.

Step 3. It remains to discuss the smoothness of $F$ up to the boundary $\partial \mathbb D$. One has to be careful here because in general, a conformal map onto a $C^1$ domain is not $C^1$ up to the boundary. However, if we assume a little extra smoothness - $C^{1,\alpha}$ with $\alpha>0$ - then $\phi:\mathbb D\to G$ is in $C^{1,\alpha}(\overline{\mathbb D})$, and $|f'|$ is bounded away from zero. This is a well-known theorem (Kellogg, Warschawski) which can be found in section 3.3 of the aforementioned book by Pommerenke, or in section II.4 of Garnett-Marshall. Therefore, $\phi^{-1}\circ f$ is a $C^1$-diffeomorphism of the circle.

Theorem 2 in [Hu-Pal] tells us that the Douady-Early extension of a $C^1$-diffeomorphism of the circle has continuous derivatives up to the boundary. (Interestingly, the webpage of S. Pal indicates that they are working on $C^{1,\alpha}$ case now.) This completes the puzzle: since the derivative of $\psi$ is continuous on the closed disk, the map $F=\phi\circ \psi$ has the same property. $\Box$

References (in addition to embedded links to the books):

  • [Hu-Pal] Jun Hu and Susovan Pal, Boundary differentiability of Douady-Earle extensions of diffeomorphisms of $S^n$, preprint available.
  • [Jerison-Kenig] Jerison, David S.; Kenig, Carlos E. Hardy spaces, $A_{\infty }$, and singular integrals on chord-arc domains. Math. Scand. 50 (1982), no. 2, 221--247. MR0672926 (84k:30037)
  • [Tukia] Tukia, Pekka. Extension of quasisymmetric and Lipschitz embeddings of the real line into the plane. Ann. Acad. Sci. Fenn. Ser. A I Math. 6 (1981), no. 1, 89--94. MR0639966 (83d:30022)

Still, it would be nice to have a reference (or proof) for the following quantitative $C^1$ Schoenflies theorem, without the $C^{1,\alpha}$ assumption.

Theorem/Conjecture B. Let $f:S^1\to\mathbb R^2$ be a $C^1$ embedding. Let $M$ be a constant such that $|f'|\le M$ and $|f(a)-f(b)|\ge M^{-1} |a-b|$ for all $a,b\in S^1$. Then $f$ extends to a diffeomorphism $F:\mathbb R^2\to\mathbb R^2$ such that $DF$ and $D(F^{-1})$ are bounded by a constant $C=C(M)$.


(Counterexample to an earlier version of the question): let $E$ be the union of the unit ball $B=\{x^2+y^2+z^2\le 1\}$ with the cylinder $C=\{x^2+y^2\le \epsilon^2, 0\le z\le 2\}$ (this is not a smooth manifold, but can be easily smoothed). The boundary of $E$ contains a cylinder-shaped region (topological disk) $R$ of area $\approx 2\pi \epsilon$ bounded by a curve $\partial R$ of length about $2\pi \epsilon$. If $f:E\to B$ is a diffeomorphism with derivative bounded by $C$, then $f(\partial R)$ has length $\le 2\pi C\epsilon$. The curve $f(\partial R)$ separates $\partial B$ into two regions, the smaller of which has area of order $\epsilon^2$ by the isoperimetric inequality. It follows that $f^{-1}$ maps a region of area $O(\epsilon^2)$ onto a region of area $\ge 2\pi \epsilon$, which means that the derivative of $f^{-1}$ cannot be bounded by a universal constant.

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Yes, it is a counter example to my problem. But in fact my domain is like yours replacing the condition $0\leq z \leq 2$ by $0\leq z \leq 1+\varepsilon$. In that case your argument doesn't work. If my expectation is right, what is the additionnal (geometric) assumption satisfies by this domains which will insure the existence of a uniformily bounded diffeo? the total curvature bounded? –  Paul Jan 29 '13 at 9:59
    
Huh, I interpreted the question to require that the derivative of $f$ to be bounded by a universal constant, and not $f^{-1}$. Which corresponds to $Df$ being allowed to be arbitrarily close to zero. –  Willie Wong Jan 30 '13 at 8:58
    
@Paul Does your domain have an axis of symmetry? –  user53153 Feb 2 '13 at 3:44
    
yes, we can assume that it is a domain of revolution with respect to the $z$ axis. –  Paul Feb 2 '13 at 7:40
    
@Paul Having a domain of revolution certainly helps. Let me know if anything in the answer is unclear. The chord-arc condition on the profile of a domain should be easy to verify (the profile you describe has this property with $M\approx 3$.) // I have not tried to derive the chord-arc property from the total curvature assumption, but it can be done: if the chord-arc constant is bad, the curve has a thin finger somewhere, which will create a lot of curvature when rotated. // Also, you should feel free to cross-post to MathOverflow with a link here. Without the symmetry the question looks hard. –  user53153 Feb 3 '13 at 21:20

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