A question on the curvature of smooth and embedded curves in a plane

I am from China. This is my first question in this website.

Given a family of embedded, smooth and closed curves $\{X_i(s)\}_{i=1}^\infty$ in a plane, let us denote by $L_i$ the perimeter and $\kappa_i(s)$ the relative curvature of $X_i$, respectively.

Suppose $\{X_i(s)\}_{i=1}^\infty$ satisfies the following:

(1) the area of the domain encloed by $X_i$ is $\pi$, $i=1, 2, \ldots$,

(2) $L_i$ tends to $+\infty$ as $i$ tends to infinity.

Can we conclude that $K_i\triangleq\max\{|\kappa_i(s)|\ |s\in[0, L_i]\}$ tends to $+\infty$ as $i$ tends to infinity? How to prove it if it is correct?

Thanks a lot!

• No, check the picture in this question: mathoverflow.net/questions/126128/… – Anton Petrunin Apr 1 '14 at 3:46
• Thank you for your counter-example. Now I know that if all of the curves are strictly convex then Ki tends to infinity. – Semon Petrus Apr 1 '14 at 4:49