Approximation by $C^1$ path of a Lipschitz continuous path I was wondering if the following equality holds:
$$\inf\left\{\int_0^1 G(\gamma(t))|\gamma'(t)|dt, \gamma \in X \cap (\text{Lipschitz})\right\}\stackrel{??}{=}\inf\left\{ \int_0^1 G(\gamma(t))|\gamma'(t)|dt, \gamma \in X \cap C^1\right\}$$
where $X=\{ \gamma:[0,1]\to \Bbb{R}^d : \gamma(0)=a,\gamma(1)=b,\ |\gamma'|>0\}$ and $G$ is a continuous function $G :\Bbb{R}^d \to [0,\infty)$ with zeros only at $a,b$. I found a result which states that a Lipschitz continuous function can be uniformly approximated by smooth function in the $L^\infty$ norm, but the result on the derivative is not very strong. 
 A: This is not necessarily the most elegant approach, but it seems to work. (Of course, the best approach is to find a book where this is already done, but I haven't yet.) 


*

*Cover the image of $\gamma$ with open balls $B_i$ such that the oscillation of $G$ on each ball is at most $\epsilon$. 

*Partition $[0,1]$ into finitely many subintervals $[a_k,b_k]$ so that the image of each subinterval is contained in some $B_i$.

*Replace $\gamma$ on each $[a_k,b_k]$ by a line segment from $\gamma(a_k)$ to $\gamma(b_k)$. Let $\lambda$ denote this new piecewise linear curve. Note that $$\int_{a_k}^{b_k}G(\lambda(t))\,|\lambda'(t)|\,dt\le \sup_{B_i}G\int_{a_k}^{b_k}|\lambda'(t)|\,dt\le \sup_{B_i}G\int_{a_k}^{b_k}|\gamma'(t)|\,dt \\ \le \epsilon \int_{a_k}^{b_k}|\gamma'(t)|\,dt+\int_{a_k}^{b_k}G(\gamma(t))\,|\gamma'(t)|\,dt$$

*Thus, the infimum can be taken over piecewise linear curves. Those you can smoothen easily.


Note that $\gamma$ did not have to be Lipschitz. Absolute continuity is enough.
A: The details are slightly tedious, but it is a fact that given Lipschitz $\gamma \in X$ you can find $\sigma \in X \cap C^1$ with $\|\gamma - \sigma\|_\infty < \epsilon$ and $\|\gamma' - \sigma'\|_{L^1} < \epsilon$.  (Idea: Since continuous functions are dense in $L^1$, choose a continuous $\lambda$ with $\|\lambda - \gamma'\|_{L^1} < \epsilon/2$ and look at $\sigma_1(t) = \gamma(0) + \int_0^t \lambda(s)\,ds$.  Then tweak $\sigma_1$ a little bit to guarantee it has the correct endpoint and a nonzero derivative.)  Now you can check that if $\gamma$ comes close to achieving the infimum then so does $\sigma$.
Essentially this is the fact that $C^1([0,1], \mathbb{R}^d)$ is dense in the Sobolev space $W^{1,1}([0,1], \mathbb{R}^d)$.  As Leonid said, you only need $\gamma$ to be absolutely continuous, and if you wanted you could choose $\sigma$ to be $C^\infty$ or even polynomial.
