Given a smooth path in $\gamma: I \to \mathbb R^n$. such that $\gamma(0) = x$, $\gamma(1)=y$, $\gamma'(t) \neq 0, \forall t$, $x \neq y$. Let $z \neq x, z \neq y$, Is it always possible to extend $\gamma$ to a smooth path connecting $y$ to $z$, such that $\gamma'(t)\neq 0$ everywhere?
What is the right tool to deal with this question? Maybe should I take piecewise path first then try to smooth it using convolution fixing endpoints?
@koch,
It is definitely possible. You need a family of curves rich enough. For example, take $$ y(s)={\mathbf A}+{\mathbf B}s +{\mathbf C}s^k,\qquad s\in[0,1] $$ where ${\mathbf A}$, ${\mathbf B}$ and ${\mathbf C}$ are vectors in $\mathbb{R}^n$ and $k>1$ are to be determined. Imposing $y(0)=\gamma(1)$ and $y'(0)=\gamma'(1)$ (to guarantee continuity of the tangent vector) you get ${\mathbf A}=\gamma(1)$ and ${\mathbf B}=\gamma'(1)$. Next, you choose ${\mathbf C}$ in such a way that $y(1)=z$. You can choose then $k>1$ to guarantee $y'(s)\neq 0$ for all $s\in[0,1]$. Finally, you can easily paste both parametrizations together, namely define $x(t)=y(t-1)$ for $t\in[1,2]$.
It seems clear to me that a similar procedure can be used to produce an extension with any desired regularity. Ssy, if you want class $C^2$ you can add a quadratic term, etc. Hope this helps.