Let $\gamma : [0,1] \rightarrow \mathbb{R}^{2}$ be continuous path such that $\gamma |_{[0,1/2]}$ and $\gamma |_{[1/2,1]}$ are smooth. Does there exists a continuous bijection $\varphi :[0,1] \rightarrow [0,1]$ such that $\gamma \circ \varphi$ is smooth on the whole $[0,1]$ and $\varphi(0)=0,\varphi(1)=1$?
Another question that I have is: Can a smooth path $\gamma :[0,1] \rightarrow \mathbb{R}^{2}$ have infinitely many self-intersections?
Greetings, Phil
Only a partial answer for the second question. Let be $$ \begin{gathered} \gamma :\left[ {0,1} \right] \to \mathbb{R}^2 \hfill \\ t \to \left( {x(t),y(t)} \right) \hfill \\ \end{gathered} $$ such that $$ x(t) = \left\{ \begin{gathered} 0\,\,\,\,if\,\,\,t = 0 \hfill \\ t^3 \cos \left( {\frac{1} {t}} \right)\,\,\,if\,\,\,0 < t \leqslant 1 \hfill \\ \end{gathered} \right. $$ $$ y\left( t \right) = \left\{ \begin{gathered} 0\,\,\,\,if\,\,\,t = 0 \hfill \\ 2t^3 + t^3 \sin \left( {\frac{1} {t}} \right)\,\,\,if\,\,\,0 < t \leqslant 1 \hfill \\ \end{gathered} \right. $$ Then the curve is smooth. Moreover it seems to have the required property as shown in the pic. However, so far,I I'm not able to prove it analytically.