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This question is connected with my previous: Heisenberg group: function without vertical derivative. Here I am trying to look from another side: what is a difference between Sobolev space and horizontal Sobolev space on Carnot group?

Outline some notations.

A Carnot group $\mathbb G$ is a connected simply-connected nilpotent Lie group hose Lie algebra $\mathfrak g$ is graded: $\mathfrak{g} = V_1\oplus\cdots\oplus V_m$, и $[V_1,V_j]=V_{j+1}$ provided $j=1,\ldots, m-1$, and $[V_1,V_m] = \{0\}$.

$V_1$ is horizontal subbundle, $X_1, \dots X_n$ - basis of $V_1$ and $X_1, \dots X_n, X_{21}, \dots X_{2n_2},\cdots X_{m1}, \dots X_{mn_m}$ - basis of $\mathfrak g$.

Derivatives. A locally summable function $v$ is called the generalized derivative of $f$ along the vector field X, whenever $$ \int v\psi\, dx = -\int f X\psi\, dx, $$ for every $\psi\in C^{\infty}_0$. We denote $v = Xf$.

Sobolev spaces.

  • The horizontal Sobolev space $HW^1_p$ consists of all locally summable functions of finite norm $$ \|f \mid HW^1_p\| = \left(\int |f|^p\, dx \right)^{\frac1p} + \left(\int |\nabla_{\mathcal L}f|^p\, dx \right)^{\frac1p}, $$ where $\nabla_{\mathcal L}f = (X_1f, \dots X_nf)$ is the generalized subgradient of $f$ at x ∈ D which uses only the derivatives along the horizontal fields.

  • The Sobolev space $W^1_p$ consists of all locally summable functions of finite norm $$ \|f \mid W^1_p\| = \left(\int |f|^p\, dx \right)^{\frac1p} + \left(\int |\nabla f|^p\, dx \right)^{\frac1p}, $$ where $\nabla_{\mathcal L}f = (X_1f, \dots X_nf, X_{21}f, \dots X_{2n_2}f,\cdots X_{m1}f, \dots X_{mn_m}f)$.

It is clear that if $f\in W^1_p$ then $f\in HW^1_p$. The question is following: if there is a function $f\in HW^1_p$ such that $f\not\in W^1_p$?

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If $V_1 \ne \mathfrak{g}$, then $H W^{1, p} \ne W^{1, p}$.

The construction is based on the notion of dilation on the Carnot group $\mathbb{G}$. The dilation $\delta_r$ is defined by for $X \in V_i$ by $$ \delta_r \exp (t X) = \exp (i \log r X). $$ This dilation is an automorphism on $\mathbb{G}$.

If $\mu$ is the Haar measure on the group $\mathbb{G}$, then $$ \mu (\delta_r (A)) = r^Q\mu (A), $$ where $$ Q = \sum_{i = 1}^m i \dim V_i. $$

We now choose a function $\varphi \in C^\infty (\mathbb{G})$ such that $\varphi = 1$ on some compact set $K$ and $\varphi = 0$ in $\mathbb{G} \setminus \delta_2 K$.

One has then, if $X \in V_i$, $$ \int_{G} \vert X (\varphi \circ \delta_r^{-1})\vert^p \,\mathrm{d}\mu = r^{Q - pi} \int_{G} \vert X \varphi \vert^p \,\mathrm{d}\mu. $$

Therefore, if we let $f = \sum_{j = 0}^\infty a_j \varphi \circ \delta_{2^{-j}}^{-1}$, we have $$ \int_{G} \vert X f \vert^p = \sum_{j = 0}^\infty \vert a_j\vert^p 2^{j (pi - Q)} \int_{G} \vert X \varphi \vert^p \,\mathrm{d}\mu. $$ The required counterexample is obtained by choosing the sequence $(a_j)_{j \ge 0}$ in such a way that $$ \sum_{j = 0}^\infty \vert a_j\vert^p 2^{j (p - Q)} < \sum_{j = 0}^\infty \vert a_j\vert^p 2^{j (2p - Q)} = \infty, $$ for example $$ a_j = 2^{-j (2 - Q / p)}. $$

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  • $\begingroup$ Should the next to last equation be following $$\sum_{j = 0}^\infty \vert a_j\vert^p 2^{j (pi - Q)} > \sum_{j = 0}^\infty \vert a_j\vert^p 2^{j (2p - Q)} = \infty ?$$ $\endgroup$ – Nikita Evseev May 11 '15 at 9:18
  • $\begingroup$ No, the left-hand side corresponds to $\int_{\mathbb{G}} \vert \nabla_{\mathcal{L}} f\vert_{L^p}$ and the righ-hand side to $\int_{\mathbb{G}} \vert \nabla f\vert_{L^p}$. $\endgroup$ – Jean Van Schaftingen May 11 '15 at 12:23

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