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Let $p>1$ and $I=(0,1)$. By Rellich--Kondrachov, if $u\in W^{k,p}(I;\mathbb{R}^n)$ with $k\geq 2$, then $u\in C^{1} ([0,1];\mathbb{R}^n)$ (and in fact the embedding $W^{k,p}(I) \to C^{1}([0,1])$ is compact). In particular, we may discuss pointwise evaluation of $u$ and $u'$.

In differential geometry, we identify smooth space curves with smooth functions with non-vanishing derivative, and in particular with a function with derivative taking values in the sphere $\mathbb{S}^{n-1}$.

There are (at least) two things one might do to Sobolev-ify the set of such curves. First, let \begin{equation} H^{k,p}_{\text{curve}} =\{u \in W^{k,p}(I; \mathbb{R}^n) : |u'(s)|=1 \text{ a.e. } s \in I\,\}, \end{equation} and let \begin{equation} W^{k,p}_{\text{curve}} = \overline{ \{u \in C^{\infty}([0,1]; \mathbb{R}^n): |u'(s)|=1 \text{ for all } s\in I \,\} }. \end{equation} Here, I'm writing $\overline{A}$ for the closure of a set $A$ with respect to the strong topology on $W^{k,p}(I;\mathbb{R}^n)$.

I'd like to know: does $H_{\text{curve}}^{k,p}=W_{\text{curve}}^{k,p}$?

The nice thing about $H$ is that it is closed w.r.t. to the weak topology on $W^{k,p}(I;\mathbb{R}^n)$. The nice thing about $W$ is that smooth functions are dense with respect to the norm topology.

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I think you meant $|u'(s)|=1$, not $|u(s)|=1$.

Let's focus on $v=u'$ and forget $u$ -- the regularity of $u$ is directly inferred from the regularity of $v$. You want to define a Sobolev space of maps $v:I\to S^{n-1}$, and observe that there are two reasonably looking definitions.

In the literature on Sobolev maps between manifolds one usually proceeds by isometrically embedding the target manifold $N$ into some $\mathbb R^d$ (this is always possible by Nash's theorem, and is given in your example with $S^{n-1}$), and then defines $N$-valued Sobolev maps as $\mathbb R^d$-valued Sobolev maps $v$ such that $v(x)\in N$ for a.e. $x$. This is your $H$-definition, although I usually see the letter $W$ attached to it.

Then the natural question arises: are smooth maps dense in this Sobolev space? Sometimes yes, sometimes no. See Topology of Sobolev mappings II by Hang and Lin, which corrects some incorrect statements in an important 1991 paper by Bethuel.

In your situation, the answer is yes. Two things help a lot:

  1. There is a smooth retraction of a neighborhood of $S^{n-1}$ onto $S^{n-1}$, namely the map $\pi(x)= x/|x|$.
  2. Your Sobolev space consists of uniformly continuous maps.

In the presence of 1 and 2 every map from your $H$-space can be approximated by smooth maps via mollification-retraction. That is, given $v$ in $H^{k,p}$, we mollify it by convolution $v*\varphi_\epsilon$ and observe that, by uniform continuity of $v$, the difference $v-v*\varphi_\epsilon$ is small in the supremum norm. Therefore, $v*\varphi_\epsilon$ takes values in a small neighborhood of $S^{n-1}$ (a spherical shell). The map $\pi\circ (v*\varphi_\epsilon)$ is the desired approximation. For details see, e.g., the proof of (more general) Theorem 5.2 in Weakly Differentiable Mappings Between Manifolds by Hajlasz, Iwaniec, Malý and Onninen.

Conclusion: smooth maps are dense in your $H$-space; therefore, the two spaces you defined are the same.

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  • $\begingroup$ You're right about the typo. I've edited the question to reflect that. $\endgroup$ – AppliedSide Jun 7 '13 at 19:01

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