Lipschitz map of the ellipse Is there a L-Lipschitz homeomorphism of the Elipse $x^2/4+y^2=1$ onto the unit circle $x^2+y^2=1$ such that $L<1$?
 A: Assuming you mean Lipschitz with respect to the plane's Euclidean metric, as suggested by Noam D. Elkies, then no such homeomorphism exists.
The first thing to worry about is where $(0,\pm1)$ are mapped.  Note that they have distance $2$, so a mapping $f$ with $L<1$ will not send these to opposite points on the unit circle.  Put $A=f(0,1)$ and $B=f(0,-1)$.  Then $A$ and $B$ lie in a common semicircle.  
Next, we consider $A'$ and $B'$, the points on the unit circle which are opposite $A$ and $B$, respectively.  Specifically, parameterize the portion of the circle from $A$ to $B'$ as $C(t)$ and from $A'$ to $B$ as $C'(t)$ such that $C(t)$ and $C'(t)$ are opposite points on the circle for every $t$.  Then for each $t$, you need $\|f^{-1}(C(t))-f^{-1}(C'(t))\|>2$.  Note that by the homeomorphism, $f^{-1}(A')$ and $f^{-1}(B')$ lie on the same side of the $y$-axis.  Then by the intermediate value theorem, there is a $t$ such that $f^{-1}(C(t))$ and $f^{-1}(C'(t))$ have the same $x$-coordinate.  But that implies $\|f^{-1}(C(t))-f^{-1}(C'(t))\|\leq2$.
