Piecewise smooth curve has a smooth representative 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
 A: 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.

A: Hint for first question: Find a smooth function $\varphi_1$ on $[0,1/2]$ such that $\varphi_1(0)=0, \varphi_1(1/2)=1/2,$ $\varphi_1$ is strictly increasing, and all derivatives of $\varphi_1$ at $1/2$ are $0.$ Something like $1/2-ce^{1/(t-1)}$ should do it. Hook it up with a similar $\varphi_2$ on $[1/2,1]$ in the appropriate way. Join these curves together to give $\varphi$ on $[0,1].$
For the second question, I'll build a curve in parts. Define
$$\gamma (t)= \begin{cases}(1,0)-e^{1/(t-1)}(1,\sin(1/(t-1))), \,\,t\in [0,1)\\ (1,0),\,\,t=1\\ (1-e^{1/(1-t)},0),\,\, t\in (1,2]\end{cases}$$
Note that all left and right derivatives of $\gamma$ at $1$ are $0.$ Thus $\gamma$ is smooth. Now $\gamma(0)$ is just some point in $\mathbb R^2$ we don't care too much about. But note that for $t_n=1-1/(n\pi),n=1,2,\dots,$ the $\gamma(t_n)$ are distinct points on the $x$-axis. Since the final leg of $\gamma$ just sends us to the left of $0$ along the $x$-axis, we have the desired infinite collecton of self-interesection points of $\gamma.$
