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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.

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My gut feeling is that yes, the infima will be the same, mainly because you only request $|\gamma^'| >0 $ instead of, say, $=1$. (In the latter case I would not have a gut feeling and keep quite ;-). The point is that your functional is, quite obviously, invariant under reparametrizations of $\gamma$ This way you can 'slow' down the curves arbitrarily and approximate curves with corners quite easily with $C^1$ curves. This is no proof, though. Also note that you'll have difficulties to see that the infimum is attained in any of the two sets you are considering. Do you know anything about $G$? –  user20266 Jun 4 '12 at 12:42
    
I had the same gut feeling. I need this equality to be true in order to be able to find a convergent subsequence (in some way) of a sequence of $C^1$ paths to another path, but the limit path need not be $C^1$. It seems that Lipschitz is enough for me. $G$ has no other hypotheses than those mentioned. –  Beni Bogosel Jun 4 '12 at 18:32
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2 Answers

up vote 1 down vote accepted

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.

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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.)

  1. Cover the image of $\gamma$ with open balls $B_i$ such that the oscillation of $G$ on each ball is at most $\epsilon$.
  2. Partition $[0,1]$ into finitely many subintervals $[a_k,b_k]$ so that the image of each subinterval is contained in some $B_i$.
  3. 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 $$\int_{a_k}^{b_k}G(\lambda(t))\,|\lambda'(t)|\,dt\le \sup_{B_i}G\int_{a_k}^{b_k}|\lambda'(t)|\,dt\le \sup_{B_i}G\int_{a_k}^{b_k}|\gamma'(t)|\,dt \\ \le \epsilon \int_{a_k}^{b_k}|\gamma'(t)|\,dt+\int_{a_k}^{b_k}G(\gamma(t))\,|\gamma'(t)|\,dt$$
  4. 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.

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Oscillation of $G$, not $\sup$? –  user20266 Jun 4 '12 at 15:10
    
@Thomas I meant it to be $\sup$. The oscillation bound is used in the last inequality. –  user31373 Jun 4 '12 at 15:11
    
ok, I see...... –  user20266 Jun 4 '12 at 15:12
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