Geodesic distance for right invariant Sobolev metrics of fractional order on the diffeomorphism group. II

Abstract

The geodesic distance vanishes on the group $\text{ Diff }_c(M)$ of compactly supported diffeomorphisms of a Riemannian manifold $M$ of bounded geometry, for the right invariant weak Riemannian metric which is induced by the Sobolev metric $H^s$ of order $0\le s<\tfrac{1}{2}$ on the Lie algebra $\mathfrak{X }_c(M)$ of vector fields with compact support.

1 Introduction

In the article [1] we studied right invariant metrics on the group $\text{ Diff }_c(M)$ of compactly supported diffeomorphisms of a manifold $M$, which are induced by the Sobolev metric $H^s$ of order $s$ on the Lie algebra $\mathfrak{X }_c(M)$ of vector fields with compact support. We showed that for $M=S^1$ the geodesic distance on $\text{ Diff }(S^1)$ vanishes if and only if $s\le \frac{1}{2}$. For other manifolds, we showed that the geodesic distance on $\text{ Diff }_c(M)$ vanishes for $M=\mathbb{R }\times N, s<\frac{1}{2}$ and for $M=S^1\times N, s\le \frac{1}{2}$, with $N$ being a compact Riemannian manifold.

Now we are able to complement this result by: The geodesic distance vanishes on $\text{ Diff }_c(M)$ for any Riemannian manifold $M$ of bounded geometry, if $0\le s<\frac{1}{2}$.

We believe that this result holds also for $s=\frac{1}{2}$, but we were able to overcome the technical difficulties only for the manifold $M=S^1$, in [1]. We also believe that it is true for the regular groups $\text{ Diff }_\mathcal{H ^\infty }(\mathbb R ^n)$ and $\text{ Diff }_\mathcal{S }(\mathbb R ^n)$ as treated in [8], and for all Virasoro groups, where we could prove it only for $s=0$ in [2].

In Sect. 2, we review the definitions for Sobolev norms of fractional orders on diffeomorphism groups as presented in [1] and extend them to diffeomorphism groups of manifolds of bounded geometry. Section 3 is devoted to the main result.

2 Sobolev metrics $H^s$ with $s\in \mathbb{R }$

2.1 Sobolev metrics $H^s$ on $\mathbb{R ^n}$

For $s\ge 0$ the Sobolev $H^s$-norm of an $\mathbb{R }^n$-valued function $f$ on $\mathbb{R }^n$ is defined as

$$\begin{aligned} \Vert f \Vert _{H^s(\mathbb{R }^n)}^2 = \Vert \mathcal{F }^{-1}(1+|\xi |^2)^{\frac{s}{2}} \mathcal{F }f \Vert _{L^2(\mathbb{R }^n)}^2, \end{aligned}$$

(1)

where $\mathcal{F }$ is the Fourier transform

$$\begin{aligned} \mathcal{F }f(\xi ) = (2\pi )^{-\frac{n}{2}} \int _{\mathbb{R }^n} {\text{ e }}^{-i \langle x,\xi \rangle } f(x)\,\,\mathrm{d}x, \end{aligned}$$

and $\xi$ is the independent variable in the frequency domain. An equivalent norm is given by

$$\begin{aligned} \Vert f \Vert ^2_{\overline{H}^s(\mathbb{R }^n)} = \Vert f \Vert ^2_{L^2(\mathbb{R }^n)} + \Vert |\xi |^s \mathcal{F }f \Vert ^2_{L^2(\mathbb{R }^n)}. \end{aligned}$$

(2)

The fact that both norms are equivalent is based on the inequality

$$\begin{aligned} \frac{1}{C} \left( 1 + \sum _j |\xi _j|^s \right) \le \left( 1 + \sum _j |\xi _j|^2 \right) ^{\frac{s}{2}} \le C \left( 1 + \sum _j |\xi _j|^s \right) , \end{aligned}$$

holding for some constant $C$. For $s>1$ this says that all $\ell ^s$-norms on $\mathbb{R }^{n+1}$ are equivalent. But the inequality is true also for $0 < s < 1$, even though the expression does not define a norm on $\mathbb{R }^{n+1}$. Using any of these norms we obtain the Sobolev spaces with non-integral $s$

$$\begin{aligned} H^s(\mathbb{R }^n) = \{ f \in L^2(\mathbb{R }^n): \Vert f \Vert _{H^s(\mathbb{R }^n)} < \infty \}. \end{aligned}$$

We will use the second version of the norm in the proof of the theorem, since it will make calculations easier.

2.2 Sobolev metrics for Riemannian manifolds of bounded geometry

Following [13, Section 7.2.1] we will now introduce the spaces $H^s(M)$ on a manifold $M$. If $M$ is not compact we equip $M$ with a Riemannian metric $g$ of bounded geometry which exists by [5]. This means that

• $(I)$      The injectivity radius of $(M,g)$ is positive.

• $(B_\infty )$   Each iterated covariant derivative of the curvature is uniformly $g$-bounded: $\Vert \nabla ^i R\Vert _g<C_i$ for $i=0,1,2,\ldots$.

The following is a compilation of special cases of results collected in [3, Chapter 1], who treats Sobolev spaces only for integral order.

Proposition

[4, 6, 10] If $(M,g)$ satisfies $(I)$ and $(B_\infty )$ then the following holds:

• (1) $(M,g)$ is complete.

• (2) There exists $\varepsilon _0>0$ such that for each $\varepsilon \in (0,\varepsilon _0)$ there is a countable cover of $M$ by geodesic balls $B_\varepsilon (x_\alpha )$ such that the cover of $M$ by the balls $B_{2\varepsilon }(x_\alpha )$ is still uniformly locally finite.

• (3) Moreover, there exists a partition of unity $1= \sum _\alpha \rho _\alpha$ on $M$ such that $\rho _\alpha \ge 0$, $\rho _\alpha \in C^\infty _c(M)$, $\text{ supp }(\rho _\alpha )\subset B_{2\varepsilon }(x_\alpha )$, and $|D_u^\beta \rho _\alpha |<C_\beta$ where $u$ are normal (Riemann exponential) coordinates in $B_{2\varepsilon }(x_\alpha )$.

• (4) In each $B_{2\varepsilon }(x_\alpha )$, in normal coordinates, we have $|D_u^\beta g_{ij}|<C^{\prime }_\beta$, $|D_u^\beta g^{ij}|<C^{\prime \prime }_\beta$, and $|D_u^\beta \Gamma ^m_{ij}|<C^{\prime \prime \prime }_\beta$, where all constants are independent of $\alpha$.

We can now define the $H^s$-norm of a function $f$ on $M$:

$$\begin{aligned} \Vert f \Vert _{H^s(M,g)}^2&= \sum _{\alpha =0}^\infty \Vert (\rho _\alpha f)\circ \,\exp _{x_\alpha } \Vert ^2_{H^s(\mathbb{R }^n)} \\&= \sum _{\alpha =0}^\infty \Vert \mathcal{F }^{-1}(1+|\xi |^2)^{\frac{s}{2}} \mathcal{F }((\rho _\alpha f)\circ \,\exp _{x_\alpha }) \Vert ^2_{L^2(\mathbb{R }^n)}. \end{aligned}$$

If $M$ is compact the sum is finite. Changing the charts or the partition of unity leads to equivalent norms by the proposition above, see [13, Theorem 7.2.3]. For integer $s$ we get norms which are equivalent to the Sobolev norms treated in [3, Chapter 2]. The norms depend on the choice of the Riemann metric $g$. This dependence is worked out in detail in [3].

For vector fields we use the trivialization of the tangent bundle that is induced by the coordinate charts and define the norm in each coordinate as above. This leads to a (up to equivalence) well-defined $H^s$-norm on the Lie algebra $\mathfrak{X }_c(M)$.

2.3 Sobolev metrics on $\text{ Diff }_c(M)$

A positive definite weak inner product on $\mathfrak{X }_c(M)$ can be extended to a right-invariant weak Riemannian metric on $\text{ Diff }_c(M)$. In detail, given $\varphi \in \text{ Diff }_c(M)$ and $X, Y \in T_\varphi \text{ Diff }_c(M)$ we define

$$\begin{aligned} G^s_\varphi (X,Y) = \langle X\,\circ \,\,\varphi \,^{-1}, Y\,\circ \,\,\varphi \,^{-1}\rangle _{H^s(M)}. \end{aligned}$$

We are interested solely in questions of vanishing and non-vanishing of geodesic distance. These properties are invariant under changes to equivalent inner products, since equivalent inner products on the Lie algebra

$$\begin{aligned} \frac{1}{C} \langle X, Y \rangle _1 \le \langle X, Y \rangle _2 \le C \langle X, Y \rangle _1 \end{aligned}$$

imply that the geodesic distances will be equivalent metrics

$$\begin{aligned} \frac{1}{C} \text{ dist }_1(\varphi , \psi ) \le \text{ dist }_2(\varphi , \psi ) \le C \text{ dist }_1(\varphi , \psi ). \end{aligned}$$

Therefore the ambiguity—dependence on the charts and the partition of unity—in the definition of the $H^s$-norm is of no concern to us.

3 Vanishing geodesic distance

Theorem 3.1

(Vanishing geodesic distance) The Sobolev metric of order $s$ induces vanishing geodesic distance on $\text{ Diff }_c(M)$ if:

• $0\le s < \frac{1}{2}$ and $M$ is any Riemannian manifold of bounded geometry.

This means that any two diffeomorphisms in the same connected component of $\text{ Diff }_c(M)$ can be connected by a path of arbitrarily short $G^s$-length.

In the proof of the theorem we shall make use of the following lemma from [1].

Lemma 3.2

[1, Lemma 3.2] Let $\varphi \in \text{ Diff }_c(\mathbb{R })$ be a diffeomorphism satisfying $\varphi (x) \ge x$ and let $T>0$ be fixed. Then for each $0 \le s < \tfrac{1}{2}$ and $\varepsilon >0$ there exists a time-dependent vector field $u_{\mathbb{R }}^{\varepsilon }$ of the form

$$\begin{aligned} u_{\mathbb{R }}^{\varepsilon }(t,x)=1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}*G_{\varepsilon }(x), \end{aligned}$$

with $f, g \in C^\infty ([0,T])$, such that its flow $\varphi ^{\varepsilon }(t,x)$ satisfies—independently of $\varepsilon$—the properties $\varphi ^{\varepsilon }(0,x) = x$, $\varphi ^{\varepsilon }(T,x)=\varphi (x)$ and whose $H^s$-length is smaller than $\varepsilon$, i.e.,

$$\begin{aligned} \text{ Len }(\varphi {^\varepsilon })=\int _0^T \Vert u_{\mathbb{R }}^{\varepsilon }(t,\cdot )\Vert _{H^s}\,\mathrm{{d}}t\le C\Vert f^\varepsilon -g^\varepsilon \Vert _{\infty }\le \varepsilon . \end{aligned}$$

Furthermore $\{ t \,:\, f^\varepsilon (t) < g^\varepsilon (t)\} \subseteq \text{ supp }(\varphi )$ and there exists a limit function $h\in C^\infty ([0,T])$, such that $f^\varepsilon \rightarrow h$ and $g^\varepsilon \rightarrow h$ for $\varepsilon \rightarrow 0$ and the convergence is uniform in $t$.

Here, $G_\varepsilon (x) = \frac{1}{\varepsilon }G_1(\frac{x}{\varepsilon })$ is a smoothing kernel, defined via a smooth bump function $G_1$ with compact support.

Proof of Theorem 3.1

Consider the connected component $\text{ Diff }_0(M)$ of $\text{ Id }$, i.e. those diffeomorphisms of $\text{ Diff }_c(M)$, for which there exists at least one path, joining them to the identity. Denote by $\text{ Diff }_c(M)^{L=0}$ the set of all diffeomorphisms $\varphi$ that can be reached from the identity by curves of arbitrarily short length, i.e., for each $\varepsilon >0$ there exists a curve from $\text{ Id }$ to $\varphi$ with length smaller than $\varepsilon$.

• Claim A. $\text{ Diff }_c(M)^{L=0}$ is a normal subgroup of $\text{ Diff }_0(M)$.

• Claim B. $\text{ Diff }_c(M)^{L=0}$ is a non-trivial subgroup of $\text{ Diff }_0(M)$.

By [12] or [7], the group $\text{ Diff }_0(M)$ is simple. Thus claims A and B imply $\text{ Diff }_c(M)^{L=0}=\text{ Diff }_0(M)$, which proves the theorem.

The proof of claim A can be found in [1, Theorem 3.1] and works without change in the case of $M$ being an arbitrary manifold and hence we will not repeat it here. It remains to show that $\text{ Diff }_c(M)^{L=0}$ contains a diffeomorphism $\varphi \ne \text{ Id }$.

We shall first prove claim B for $M=\mathbb{R }^n$ and then show how to extend the arguments to arbitrary manifolds. Choose a diffeomorphism $\varphi _\mathbb{R }\in \text{ Diff }_c(\mathbb{R })$ with $\varphi _{\mathbb{R }}(x)\ge x$ and $\text{ supp }(\varphi _\mathbb{R }) \subseteq [1, \infty )$. Then let

$$\begin{aligned} u^{\varepsilon }_{\mathbb{R }}(t,x):=1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}*G_{\varepsilon }(x) \end{aligned}$$

be the family of vector fields constructed in Lemma 3.2, whose flows at time $T$ equal $\varphi _\mathbb{R }$. We extend the vector field $u^\varepsilon _\mathbb{R }$ to a vector field $u^\varepsilon _{\mathbb{R }^n}$ on $\mathbb{R }^n$ via

$$\begin{aligned} u^\varepsilon _{\mathbb{R }^n}(t,x_1,\ldots ,x_n):= \left( u^\varepsilon _{\mathbb{R }}(t,|x|),0,\ldots ,0\right) . \end{aligned}$$

The flow of this vector field is given by

$$\begin{aligned} \varphi _{\mathbb{R }^n}^\varepsilon (t, x_1,\ldots ,x_n) = \left( \varphi _\mathbb{R }^\varepsilon (t, |x|), x_2, \ldots , x_n \right) , \end{aligned}$$

where $\varphi _\mathbb{R }^\varepsilon$ is the flow of $u_\mathbb{R }^\varepsilon$. In particular we see that at time $t=T$

$$\begin{aligned} \varphi _{\mathbb{R }^n}^\varepsilon (t, x_1,\ldots ,x_n) = \left( \varphi _\mathbb{R }(|x|), x_2, \ldots , x_n \right) , \end{aligned}$$

the flow is independent of $\varepsilon$. So it remains to show that for the length of the path $\varphi _{\mathbb{R }^n}^\varepsilon (t,\cdot )$ we have

$$\begin{aligned} \text{ Len }(\varphi _{\mathbb{R }^n}^\varepsilon ) \rightarrow 0 \quad \text{ as } \quad \varepsilon \rightarrow 0. \end{aligned}$$

We can estimate the length of this path via

$$\begin{aligned} \text{ Len }(\varphi _{{\mathbb{R }}^n}^{\varepsilon })^2&= \left( \int _0^T \Vert u_{{\mathbb{R }}^n}^{\varepsilon }(t,.)\Vert _{H^s({\mathbb{R }}^n)}\, \,\mathrm{d}t\right) ^2 \le T \int _0^T \Vert u_{{\mathbb{R }}^n}^{\varepsilon }(t,.)\Vert _{H^s({{\mathbb{R }}^n})}^2 \,\,\mathrm{d}t\\&= T \int _0^T \Vert u_{\mathbb{R }}^{\varepsilon }(t,|\cdot |)\Vert _{H^s({{\mathbb{R }}^n})}^2 \,\,\mathrm{d}t=T \int _0^T \Vert 1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}*G_{\varepsilon }(|x|)\Vert _{H^s({{\mathbb{R }}^n})}^2 \,\,\mathrm{d}t\\&\le C(G_1, T) \int _0^T \Vert 1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}(|\cdot |)\Vert _{H^s({{\mathbb{R }}^n})}^2 \, \,\mathrm{d}t, \end{aligned}$$

where the last estimate follows from

$$\begin{aligned}&\Vert 1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}*G_{\varepsilon }(|x|)\Vert ^2_{H^s({\mathbb{R }^n})}\\&\quad =\int _{\mathbb{R }^n} (1+|\xi |^{2s}) \left[ \mathcal{F }\left( 1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}(|\cdot |)\right) (\xi )\right] ^2\, \left[ \mathcal{F }\left( G_\varepsilon (|\cdot |)\right) (\xi )\right] ^2 \, \,\mathrm{d}\xi \\&\quad =\int _{\mathbb{R }^n} (1+|\xi |^{2s}) \left[ \mathcal{F }\left( 1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}(|\cdot |)\right) (\xi )\right] ^2\, \left[ \mathcal{F }\left( G_1(|\cdot |)\right) (\varepsilon \xi )\right] ^2 \, \,\mathrm{d}\xi \\&\quad \le \left\| \mathcal{F }G_1(|\cdot |) \right\| _{L^\infty }^2\cdot \Vert 1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}(|\cdot |)\Vert ^2_{H^s({\mathbb{R }^n})}. \end{aligned}$$

Hence it is sufficient to show that

$$\begin{aligned} \left\| 1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}(|\cdot |)\right\| _{H^s({\mathbb{R }^n})} \rightarrow 0 \quad \text{ as } \quad \varepsilon \rightarrow 0 \quad \text{ uniformly } \text{ in } t. \end{aligned}$$

To compute the $H^s$-norm of $1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}(|\cdot |)$ we first Fourier transform it. The Fourier transform of a radially symmetric function $v(|\cdot |) \in L^1(\mathbb{R }^n)$ is again radially symmetric and given by the following formula, see [11, Theorem 3.3],

$$\begin{aligned} (\mathcal{F }v(|\cdot |))(\xi ) = 2\pi |\xi |^{1-n/2} \int _0^{\infty } J_{n/2-1}(2\pi |\xi | s) v(s) s^{n/2} \, \,\mathrm{d}s, \end{aligned}$$

with $J_{n/2-1}$ denoting the Bessel function of order $\tfrac{n}{2} -1$. To simplify notation we will omit the dependence of the vector field $1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}(|\cdot |)$ on $t$ and $\varepsilon$. Changing coordinates, this becomes

$$\begin{aligned} (\mathcal{F }1_{[f,g]}(|\cdot |))(\xi ) = (2\pi )^{-n/2} |\xi |^{-n} \int _{2\pi f |\xi |}^{2\pi g |\xi |} J_{n/2-1}(s) s^{n/2} \,\,\mathrm{d}s. \end{aligned}$$

This integral can be evaluated explicitly using the following integral identity for Bessel functions from [9, (10.22.1)]

$$\begin{aligned} \int z^{\nu +1} J_\nu (z) \,\mathrm{d}z = z^{\nu +1} J_{\nu +1}(z),\quad \nu \ne -\tfrac{1}{2}. \end{aligned}$$

This gives us

$$\begin{aligned} (\mathcal{F }1_{[f,g]}(|\cdot |))(\xi ) = |\xi |^{-n/2} \left( J_{n/2}(2\pi g |\xi |) g^{n/2} - J_{n/2}(2\pi f |\xi |) f^{n/2}\right) . \end{aligned}$$

The $H^s$-norm of $1_{[f,g]}(|\cdot |)$ is given by

$$\begin{aligned} \left\| 1_{[f,g]}(|\cdot |)\right\| _{H^s(\mathbb{R }^n)}^2 = \int _{\mathbb{R }^n} \left( 1 + |\xi |^{2s}\right) \mathcal{F }1_{[f,g]}(|\cdot |)(\xi )^2 \,\,\mathrm{d}\xi . \end{aligned}$$

We will only consider the term involving $|\xi |^{2s}$, since the $L^2$-term can be estimated in the same way by setting $s=0$. Transforming to polar coordinates we obtain

$$\begin{aligned} \int _{\mathbb{R }^n} |\xi |^{2s} \left( \mathcal{F }1_{[f,g]}(|\cdot |)(\xi )\right) ^2\,\,\mathrm{d}\xi&\!=\! \int _{\mathbb{R }^n} |\xi |^{2s\!-\!n} \left( J_{n/2}(2\pi g |\xi |) g^{n/2} - J_{n/2}(2\pi f |\xi |) f^{n/2}\right) ^2 \,\,\mathrm{d}\xi \\&\!=\! \text{ Vol }(S^{n\!-\!1}) \int _0^\infty r^{2s\!-\!1} \left( J_{n/2}(2\pi g r) g^{n/2}\!-\!J_{n/2}(2\pi f r) f^{n/2}\right) ^2 \,\,\mathrm{d}r. \end{aligned}$$

The above integral is non-zero only for those $t$, where $f^\varepsilon (t) \ne g^\varepsilon (t)$. From Lemma 3.2 and our assumptions on $\varphi _\mathbb{R }$ we know that

$$\begin{aligned} \{t \,:\, f^\varepsilon (t) < g^\varepsilon (t)\} \subseteq \text{ supp }(\varphi _\mathbb{R }) \subseteq [1,\infty ). \end{aligned}$$

Thus both $f^\varepsilon (t)$ and $g^\varepsilon (t)$ are different and away from 0 and we can evaluate the above integral using the identity [9, (10.22.57)],

$$\begin{aligned} \int _0^\infty \frac{J_\mu (at)J_\nu (at)}{t^\lambda } \,\,\mathrm{d}t = \frac{\left( \tfrac{1}{2} a\right) ^{\lambda - 1} \Gamma \left( \tfrac{\mu }{2} + \tfrac{\nu }{2} - \tfrac{\lambda }{2} + \tfrac{1}{2}\right) \Gamma \left( \lambda \right) }{2 \Gamma \left( \tfrac{\lambda }{2} + \tfrac{\nu }{2} - \tfrac{\mu }{2} + \tfrac{1}{2}\right) \Gamma \left( \tfrac{\lambda }{2} + \tfrac{\mu }{2} - \tfrac{\nu }{2} + \tfrac{1}{2}\right) \Gamma \left( \tfrac{\lambda }{2} + \tfrac{\mu }{2} + \tfrac{\nu }{2} + \tfrac{1}{2}\right) }, \end{aligned}$$

which holds for $\text{ Re }(\mu +\nu +1) > \text{ Re } \lambda > 0$ and the identity [9, (10.22.56)],

$$\begin{aligned} \int _0^\infty \frac{J_\mu (at)J_\nu (bt)}{t^\lambda } \,\,\mathrm{d}t\!=\!\frac{a^\mu \Gamma \left( \tfrac{\nu }{2}\!+\!\tfrac{\mu }{2}\!-\!\tfrac{\lambda }{2}\!+\!\tfrac{1}{2}\right) }{2^\lambda b^{\mu \!-\!\lambda \!+\!1} \Gamma \left( \tfrac{\nu }{2}\!-\!\tfrac{\mu }{2} \!+\!\tfrac{\lambda }{2}\!+\!\tfrac{1}{2}\right) } \mathbf F \left( \tfrac{\nu }{2}\!+\!\tfrac{\mu }{2}\!-\!\tfrac{\lambda }{2}\!+\!\tfrac{1}{2}, \tfrac{\mu }{2}\!-\!\tfrac{\nu }{2}\!-\!\tfrac{\lambda }{2}\!+\!\tfrac{1}{2}; \mu \!+\!1; \tfrac{a^2}{b^2} \right) , \end{aligned}$$

which holds for $0 < a < b$ and $\text{ Re }(\mu +\nu +1) > \text{ Re } \lambda > -1$. Here $\mathbf F (a, b;c;d)$ is the regularized hypergeometric function. Using these identities with $\lambda =1-2s$, $\mu =\nu =\tfrac{n}{2}$, $a = 2\pi f$ and $b = 2\pi g$ we obtain

$$\begin{aligned} \int _0^\infty r^{2s-1} J_{n/2}(2\pi f r)^2 \, \,\mathrm{d}r&= \frac{1}{2} (\pi f)^{-2s} \frac{\Gamma \left( \tfrac{n}{2} + s \right) \Gamma (1-2s)}{\Gamma (1-s)^2 \Gamma \left( \tfrac{n}{2} +1-s\right) } \end{aligned}$$

and

$$\begin{aligned} \int _0^\infty r^{2s\!-\!1} J_{n/2}(2\pi f r)J_{n/2}(2\pi g r) \,\mathrm{d}r\!=\!\frac{1}{2} (\pi g)^{\!-\!2s} \left( \frac{f}{g}\right) ^{n/2} \frac{\Gamma \left( \tfrac{n}{2}\!+\!s \right) }{\Gamma (1\!-\!s)} \mathbf F \left( \tfrac{n}{2}\!+\!s, s; \tfrac{n}{2}\!+\!1; \tfrac{f^2}{g^2}\right) . \end{aligned}$$

Putting it together results in

$$\begin{aligned} \int _{\mathbb{R }^n} |\xi |^{2s} (\mathcal{F }1_{[f,g]}(|\cdot |))(\xi )^2 \,\mathrm{d}\xi&= \text{ Vol }(S^{n-1}) \left( \frac{f^{-2s} + g^{-2s}}{2\pi ^{2s}} \frac{\Gamma \left( \tfrac{n}{2} + s \right) \Gamma (1-2s)}{\Gamma (1-s)^2 \Gamma \left( \tfrac{n}{2} +1-s\right) } \right. \\&\left. -\frac{g^{-2s}}{\pi ^{2s}} \frac{f^{n/2}}{g^{n/2}} \frac{\Gamma \left( \tfrac{n}{2} + s \right) }{\Gamma (1-s)} \mathbf F \left( \tfrac{n}{2} + s, s; \tfrac{n}{2} + 1; \tfrac{f^2}{g^2}\right) \right) . \end{aligned}$$

In the limit $\varepsilon \rightarrow 0$ we know from Lemma 3.2 that $f^\varepsilon (t) \rightarrow h(t)$ and $g^\varepsilon (t) \rightarrow h(t)$ uniformly in $t$ on $[0,T]$ and hence $\tfrac{f^\varepsilon (t)}{g^\varepsilon (t)} \rightarrow 1$. For the regularized hypergeometric function $\mathbf F (a, b; c; d)$ at $d=1$ we have the identity [9, (15.4.20)]

$$\begin{aligned} \mathbf F (a, b; c; 1) = \frac{\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}\,, \end{aligned}$$

for $\text{ Re }(c-a-b) > 0$. Applying the identity with $a=\tfrac{n}{2}+s$, $b=s$ and $c=\tfrac{n}{2}+1$ we get

$$\begin{aligned} \mathbf F \left( \tfrac{n}{2} + s, s; \tfrac{n}{2} + 1; 1\right) = \frac{\Gamma (1-2s)}{\Gamma (1-s)\Gamma \left( \tfrac{n}{2} +1-s\right) }. \end{aligned}$$

Using the continuity of the hypergeometric function it follows that

$$\begin{aligned} \int _{\mathbb{R }^n} |\xi |^{2s} \left( \mathcal{F }1_{[f,g]}(|\cdot |))(\xi )\right) ^2 \,\,\mathrm{d}\xi \rightarrow 0, \end{aligned}$$

as $\varepsilon \rightarrow 0$ and the convergence is uniform in $t$. This concludes the proof that

$$\begin{aligned} \left\| 1_{[f^{\varepsilon }(t), g^{\varepsilon }(t)]}(|\cdot |)\right\| _{H^s({\mathbb{R }^n})} \rightarrow 0 \quad \text{ as } \quad \varepsilon \rightarrow 0 \quad \text{ uniformly } \text{ in } t, \end{aligned}$$

and hence we have established claim B for $\text{ Diff }_c(\mathbb{R }^n)$.

To prove this result for an arbitrary manifold $M$ of bounded geometry we choose a partition of unity $(\tau _j)$ such that $\tau _0\equiv 1$ on some open subset $U\subset M$, where normal coordinates centred at $x_0\in M$ are defined. If $\varphi _{\mathbb{R }}$ is chosen with sufficiently small support, then the vector field $u^\varepsilon _{\mathbb{R }^n}$ has support in $\exp _{x_0}(U)$ and we can define the vector field $u^\varepsilon _M:=(\exp _{x_0}^{-1})^*u^\varepsilon _{\mathbb{R }^n}$ on $M$. This vector field generates a path $\varphi ^\varepsilon _M(t,\cdot ) \in \text{ Diff }_0(M)$ with an endpoint $\varphi ^\varepsilon _M(T,\cdot )=\varphi _M(\cdot )$ that does not depend on $\varepsilon$ with arbitrarily small $H^s$-length since

$$\begin{aligned} \text{ Len }(\varphi _M^\varepsilon )&\le C_1(\tau ) \int _0^T\Vert u_M^\varepsilon \Vert _{H^s(M,\tau )}\, \,\mathrm{d}t= C_1(\tau ) \int _0^T\Vert \exp _{x_0}^*(\tau _0. u_M^\varepsilon ) \Vert _{H^s({\mathbb{R }}^n)} \,\,\mathrm{d}t\\&= C_1(\tau ) \int _0^T\Vert u^\varepsilon _{{\mathbb{R }}^n} \Vert _{H^s({\mathbb{R }}^n)} \,\,\mathrm{d}t. \end{aligned}$$

Thus we can reduce the case of arbitrary manifolds to ${\mathbb{R }}^n$ and this concludes the proof. $\square$