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 $L<1$?



Lipschitz map of the ellipse
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 $\|f^{-1}(C(t))-f^{-1}(C'(t))\|\leq2$.


