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Let $T_1$ and $T_2$ be any two euclidean triangles with labeled sides. The sides are labeled respectively $e_1^1,e_2^1,e_3^1$ and $e_1^2,e_2^2,e_3^2$. Call $A:T_1\rightarrow T_2$ the affine map which sends sides to corresponding sides, i.e. $e_i^1\mapsto e_i^2$, $i=1,2,3$.

Call $\mathcal{F}$ the set of differentiable maps which send $T_1$ to $T_2$ and sides to corresponding sides, clearly $A\in \mathcal{F}$. For every $f\in \mathcal{F}$ define $L(f):=\max\{Lip(f),Lip(f^{-1})\}$ where $Lip(f)$ is the Lipschitz constant of $f$.

Question(s): Is it true $L(A)=\inf\limits_{f\in\mathcal{F}}L(f)$? If not, is there a map which realizes the infimum? Which one?

Suppose $L(A)=Lip(A)$, then the answer to the first question is yes if $Lip(A)$ is obtained along a side $e$ of $T_1$: $A(e)/e=Lip(A)$, but for every other $f\in\mathcal{F}$ it's true $L(f)\ge f(e)/e\ge A(e)/e=L(A)$.

Unfortunately in case $L(A)$ is not obtained along a side of a triangle the previous inequality can not be used and I don't know how to proceed.

Additional question: what are the answers to the previous questions in case one or two sides of $T_1$ are mapped linearly?

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  • $\begingroup$ It's not totally clear to me, how you define $A$. There are 6 possibilities to map the vertices of the triangles onto each other and not all of them lead to the same affine map. Or do you think of the triangles having numbered vertices (or edges) and the maps to preserve these numbers…? That would make sense… $\endgroup$
    – Dirk
    Commented Dec 21, 2016 at 17:10
  • $\begingroup$ yes, exactly, the sides are "labeled" (or numbered as you prefer) and the map preserves this labeling. Now it should be clear in the question, thank you for the comment. $\endgroup$
    – user99087
    Commented Dec 21, 2016 at 17:16
  • $\begingroup$ The Lipschitz constant of any such f at a vertex is equal to the Lipschitz constant of A. So your conjecture is true. $\endgroup$
    – user35593
    Commented Dec 21, 2016 at 17:23
  • $\begingroup$ Sorry I was wrong I assumed that the sides are maped affinely. $\endgroup$
    – user35593
    Commented Dec 21, 2016 at 17:25
  • $\begingroup$ yes, with the additional condition that sides are mapped affinely my conjecture is surely true. but unfortunately I want to now if it's true $without$ this additional confition $\endgroup$
    – user99087
    Commented Dec 21, 2016 at 17:27

2 Answers 2

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EDIT: My first answer falsely claimed a positive proof (the mistake was pointed out by user372511). Here is instead an explicit counterexample.

Consider the following triangles with $T_1 = ABC$ and $T_2=ABC'$

enter image description here

The following map $$ f : \begin{pmatrix} x \\ y \end{pmatrix} \rightarrow \begin{pmatrix} 1+2^y -2^{1-x} \\ 1+2^y-2^{1-y}\end{pmatrix} $$ maps $T_1$ to $T_2$ and has a smaller Lipschitz constant than the affine map.

Indeed, the affine map is given by the matrix $A=\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} $. The maximal singular value gives the Lipschitz constant $$ \mathrm{Lip}(A) = \sqrt{3+\sqrt{5}} \sim 2.288 $$

Let $J_f$ be the Jacobian of $f$. With the help of Mathematica, one can find the maximal singular value of $J_f(x,y)$ in the triangle. It is in fact reached at $(0,0)$ and gives the Lipschitz constant of $f$ in the triangle $$ \mathrm{Lip}(f) = \sqrt{7+\sqrt{13}} \ln(2) \sim 2.257$$

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  • $\begingroup$ Are you sure $Lip(A^{-1})=Lip(A)^{-1}$? To me $Lip(A)=\sigma_1$ the greatest singular value of $A$, while $Lip(A^{-1})=1/\sigma_2$ where $\sigma_2$ is the smallest singular value of $A$ and clearly $\sigma_1\neq 1/\sigma_2$. $\endgroup$
    – user99087
    Commented Jan 6, 2017 at 10:56
  • $\begingroup$ You are right. I found a counterexample while trying to fix this. I edited the previous answer. $\endgroup$
    – Vigod
    Commented Jan 6, 2017 at 23:30
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Consider the example in the comment above, i.e. $A=(-1,0)$, $B=(1,0)$, $C=(0,1)$, $A'=A=(-1,0)$, $B'=B=(1,0)$ and $C'=(3,0)$. Then the affine map mapping $A,B$ and $C$ to $A'$, $B'$ and $C'$ respectively is given by $$ x \mapsto Ax \text{ with } A=\begin{pmatrix} 1 & 2\\0 & 0\end{pmatrix}. $$ The Lipschitz constant is given by the maximal singular value of $A$ which is $\sqrt{5}$. Introduce now the points $D=(0,0)$ and $D'=(\frac{1}{2},0)$. Then the matrices of the affine maps from $ADC$ to $A'D'C'$ and from $CDB$ to $C'D'B'$ are $$ \begin{pmatrix} \frac{3}{2} & \frac{3}{2}\\0 & 0\end{pmatrix},\quad \begin{pmatrix} \frac{1}{2} & \frac{3}{2}\\0 & 0\end{pmatrix}. $$ The corresponding Lipschitz constants are $\sqrt{2}\frac{3}{2}=\frac{3}{\sqrt{2}}<\sqrt{5}$ and $\sqrt{\frac{5}{2}}<\sqrt{5}$. Hence we see that the new map has a smaller Lipschitz constant. The new Lipschitz constant is equal to the maximal ratio of the sides of the triangles and can therefore not be improved.

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  • $\begingroup$ Checking one of my last comments I noted that I've written "triangle $T_2$ with vertices $(1,0),(1,0)$ and $(2,0)$. This doesn't make sense since they are not the vertices of a triangle. Of course I meant $(0,2)$, my bad. For this reason I don't think your answer makes sense. Also, I'm not asking for which couple of triangles the lipschitz constant is equal to the ratio of sides (in which case I already know the lipschitz constant of the affine map is optimal). I'm asking which lipschitz constant is optimal for any given couple of triangles $\endgroup$
    – user99087
    Commented Dec 26, 2016 at 10:08
  • $\begingroup$ Iif you replace (2,0) by $(2, \epsilon)$ you get a valid triangle and this shows that in the affine map has not always the smallest possible Lipschitz constant. Of course this construction also works for other triangles. I doubt that one can derive the Lipschitz constant for arbitrary triangles. $\endgroup$
    – user35593
    Commented Dec 26, 2016 at 15:57
  • $\begingroup$ this still doesn't make sense to me $\endgroup$
    – user99087
    Commented Dec 29, 2016 at 23:15

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