Example of a periodic plane curve whose image is a triangle

Question

I got the following problem as homework for my differential geometry class.

Find a $ C^\infty $ function $ \gamma: \mathbb{R} \to \mathbb{R}^2 $ satisfying

(i) $ \gamma $ is periodic with period $ 3 $;

(ii) $ \gamma $ is injective on $ [0,3) $;

(iii) The image $ \gamma(\mathbb{R}) $ is the triangle whose vertices are $ (0,0), (0,1), (1,0) $.

To me it does not seem possible, since the curve does not seem differentiable at the vertices. Can anyone comment on this problem?

Answer 1

Nice question.

Here's a comment/hint. Although you are asked to find a $C^\infty$ function with these properties, notice that you are not asked to find one whose vector valued derivative $\gamma'(t)$ is nonzero.

Your argument about nondifferentiability of the triangle at the three corner points does not imply that $\gamma$ cannot exist, but it does imply that if $\gamma(t)$ equals one of the three corner points then $\gamma'(t)$ is the zero vector. Thinking in this manner, you might be able to deduce enough necessary conditions on $\gamma$ to tell you how to actually construct the correct $\gamma$.