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Suppose I have two points $\alpha,\beta \in \Bbb{R}^n$. Define $$ X=\{\gamma \in C^1([0,1] , \Bbb{R}^n),\ \gamma(0)=\alpha,\gamma(1)=\beta ,0 <|\gamma'|<K\}$$ parametrized curves joining $\alpha$ and $\beta$. If I have a sequence $(\gamma_n) \subset X$ such that the paths $\gamma_n$ are all contained in a compact set $K \subset \Bbb{R}^n$ then this sequence is equi-bounded and equi-continuous (in $C([0,1],\Bbb{R}^n$) and by Ascoli-Arzela theorem there is a path $\gamma :[0,1] \to \Bbb{R}^n$ such that $\gamma_n$ converges uniformly to $\gamma$. The limit $\gamma$ is continuous, but may not be of class $C^1$.

In my case I can prove that $\gamma([0,1])\subset [\alpha,\beta]$ (the line segment joining $\alpha$ and $\beta$). My question is:

Can I find a sequence of paths $\theta_n \in C^1([0,1],[\alpha,\beta])$ for which $|\theta_n'| \leq K$ and $(\theta_n)$ converges uniformly (or maybe pointwise) to $\gamma$?

My intuition says that if I can approximate $\gamma$ with a sequence of $C^1$ paths joining $\alpha$ and $\beta$ then it is sufficiently regular such that I can approximate $\gamma$ with $C^1$ curves with $\theta_n([0,1])\subset \gamma([0,1])$.

One thought was to use projections of $\gamma_n$ on the segment $[\alpha,\beta]$, but I think that in some cases the projection can destroy differentiability.

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I would imagine that since $l(\gamma) = 1$, and $|\gamma'(t)| = 1$, that you can have $\gamma([0,1])\subset [\alpha,\beta]$ only if $|\alpha- \beta| = 1$. – copper.hat May 14 '12 at 15:59
In fact, if $\gamma(t) \in [\alpha,\beta]$, then $\gamma'(t) \in \mathbb{sp}\{\beta-\alpha\}$, in which case you must have $\gamma'(t) = \pm \frac{\beta-\alpha}{|\beta-\alpha|}$. – copper.hat May 14 '12 at 16:04
@copper.hat: I understand your objection. I modified the statement of the problem. – Beni Bogosel May 14 '12 at 16:42
up vote 1 down vote accepted

You can first project onto the line through $\alpha$ and $\beta$. This preserves differentiability (since projection is a linear map) and reduces the problem to one-dimensional target. Say, you have a continuous function $F\colon [0,1]\to [0,1]$ which is a limit of smooth functions $F_n\colon [0,1]\to\mathbb R$. We want to gently truncate $F_n$ to make their values fit in $[0,1]$ without losing differentiability.

I would use $\phi_n\circ F_n$ where $\phi_n\colon \mathbb R\to[0,1]$ are smooth functions such that $|\phi'|\le 1$ and $\phi_n(t)\to t$ uniformly for $t\in [0,1]$. You can cook up $\phi_n$ in many ways. For example, $$\phi_n(t)=\begin{cases}0\quad &t\le 0\\ nt^2/2, \quad & 0\le t\le 1/n \\ t-1/(2n),\quad & 1/n\le t\le 1-1/n \\ 1-1/(2n)-n(t-1)^2/2 \quad & 1-1/n\le t\le 1 \\ 1-1/(2n) &\quad t>1 \end{cases}$$ works (unless I made any arithmetical mistakes). This is what you get by taking a trapezoidal approximation to the characteristic function of $[0,1]$ and integrating it.

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Is it true that projection preserves differentiability? – Beni Bogosel May 15 '12 at 13:13
@BeniBogosel Composition with any differentiable map preserves differentiability, by the chain rule. Projection onto a linear subspace (such as a line) is a differentiable map, because it's linear. Projection onto other convex sets (such as a line segment) is in general not differentiable. That's why I used projection onto a line followed by soft truncation. – user31373 May 15 '12 at 15:47

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