Can we run to infinity? The diameter of the diffeomorphism group with respect to right-invariant Sobolev metrics

Abstract

The group ${\text {Diff}}({\mathcal {M}})$ of diffeomorphisms of a closed manifold ${\mathcal {M}}$ is naturally equipped with various right-invariant Sobolev norms $W^{s,p}$. Recent work showed that for sufficiently weak norms, the geodesic distance collapses completely (namely, when $sp\le \dim {\mathcal {M}}$ and $s<1$). But when there is no collapse, what kind of metric space is obtained? In particular, does it have a finite or infinite diameter? This is the question we study in this paper. We show that the diameter is infinite for strong enough norms, when $(s-1)p\ge \dim {\mathcal {M}}$, and that for spheres the diameter is finite when $(s-1)p<1$. In particular, this gives a full characterization of the diameter of ${\text {Diff}}(S^1)$. In addition, we show that for ${\text {Diff}}_c({\mathbb {R}}^n)$, if the diameter is not zero, it is infinite.

Appendices

Appendix A: Proof of Lemma 3.9

We consider the family of piecewise-linear maps $\psi _{\lambda +1,\delta }$

$$\begin{aligned} \psi _{\lambda +1,\delta } = {\left\{ \begin{array}{ll} (\lambda +1)x &{} x\in \left[ 0,\frac{1-\delta }{\lambda +1}\right] \\ \frac{\delta (\lambda +1)}{\lambda +\delta }\left( x- \frac{1-\delta }{\lambda +1}\right) +1-\delta &{} x\in \left[ \frac{1-\delta }{\lambda +1},1\right] \end{array}\right. } = {\left\{ \begin{array}{ll} (\lambda +1)x &{} x\in \left[ 0,\frac{1-\delta }{\lambda +1}\right] \\ \frac{\delta (\lambda +1)x + (1-\delta )\lambda }{\lambda +\delta } &{} x\in \left[ \frac{1-\delta }{\lambda +1},1\right] \end{array}\right. }. \end{aligned}$$

Since piecewise-linear maps are not elements of the group of diffeomorphisms ${\text {Diff}}(S^1)$ we have to smoothen the maps around the break points $\frac{1-\delta }{\lambda +1}$ and $0\sim 1$. However, since the $W^{s,p}$-metric can be extended to the space of Lipschitz-maps (for $s<1+1/p$) and since the smoothening can be done in such a way that the change in the distance to identity is arbitrarily small, we ignore this in the following.

In the following we will bound the length of the linear homotopy $\varphi _t(x)$ between ${\text {Id}}$ and $\psi _{\lambda +1,\delta }$ which albeit being straightforward turns out to be a somewhat tedious calculation. We bound the length below with respect to the ${\dot{W}}^{s,p}$ norm, under the assumption that $s>1$. Boundedness with respect to the lower order parts of $W^{s,p}$ norm, as well as for $W^{s,p}$ norm for $s\le 1$, is similar, but simpler. We have

$$\begin{aligned} \varphi _t(x) = (1-t)x + t \psi _{\lambda +1,\delta } (x) = {\left\{ \begin{array}{ll} (1+\lambda t)x&{} x\in \left[ 0,\frac{1-\delta }{\lambda +1}\right] \\ x + t\frac{(1-\delta )\lambda }{\lambda +\delta }(1-x) &{} x\in \left[ \frac{1-\delta }{\lambda +1},1\right] \end{array}\right. }. \end{aligned}$$

Its inverse is then given by

$$\begin{aligned} \varphi _t^{-1}(y) = {\left\{ \begin{array}{ll} \frac{y}{1+\lambda t} &{} y\in \left[ 0,\frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right] \\ \frac{y-1}{1-t \frac{(1-\delta )\lambda }{\lambda +\delta }} + 1 &{} y\in \left[ \frac{(1-\delta )(1+\lambda t)}{\lambda +1},1\right] \end{array}\right. }, \end{aligned}$$

and its time derivative is

$$\begin{aligned} \partial _t\varphi _t(x) = \psi _{\lambda +1,\delta } (x) - x= {\left\{ \begin{array}{ll} \lambda x &{} x\in \left[ 0,\frac{1-\delta }{\lambda +1}\right] \\ \frac{(1-\delta )\lambda }{\lambda +\delta }(1-x) &{} x\in \left[ \frac{1-\delta }{\lambda +1},1\right] \end{array}\right. }. \end{aligned}$$

The vector field $u_t$ defined by $\partial _t\varphi _t = u_t \circ \varphi _t$ is therefore

$$\begin{aligned} u_t(y)= & {} \partial _t \varphi _t (\varphi _t^{-1}(y)) = {\left\{ \begin{array}{ll} \frac{\lambda y}{1+\lambda t} &{} y\in \left[ 0,\frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right] \\ \frac{(1-\delta )\lambda }{\lambda +\delta }\frac{1-y}{1-t \frac{(1-\delta )\lambda }{\lambda +\delta }} &{} y\in \left[ \frac{(1-\delta )(1+\lambda t)}{\lambda +1},1\right] \end{array}\right. } \\= & {} {\left\{ \begin{array}{ll} \frac{y}{t+\frac{1}{\lambda }} &{} y\in \left[ 0,\frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right] \\ \frac{(1-\delta )(1-y)}{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })} &{} y\in \left[ \frac{(1-\delta )(1+\lambda t)}{\lambda +1},1\right] \end{array}\right. }, \end{aligned}$$

and therefore

$$\begin{aligned} u_t'(y) = {\left\{ \begin{array}{ll} \frac{1}{t+\frac{1}{\lambda }} &{} y\in \left[ 0,\frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right] \\ \frac{-(1-\delta )}{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })} &{} y\in \left[ \frac{(1-\delta )(1+\lambda t)}{\lambda +1},1\right] \end{array}\right. }. \end{aligned}$$

We now evaluate the ${\dot{W}}^{1+\sigma ,p}$-norm of $u_t$, for $\sigma p < 1$. That is, we evaluate the $(\sigma ,p)$-Gagliardo seminorm of $u_t'$, whose pth power is

$$\begin{aligned} \iint _{{\mathbb {R}}^2} \frac{|u_t'(x) - u_t'(y)|^p}{|x-y|^{1+\sigma p}}\,dx\,dy= & {} 2\iint _{y>x} \frac{|u_t'(x) - u_t'(y)|^p}{|x-y|^{1+\sigma p}}\,dx\,dy \\= & {} 2\int _{-\infty }^\infty \int _0^\infty \frac{|u_t'(x) - u_t'(x+s)|^p}{s^{1+\sigma p}}\,ds\,dx. \end{aligned}$$

We split this double integral into different regions:

$$\begin{aligned} \begin{aligned}&\int _{-\infty }^\infty \int _0^\infty \frac{|u_t'(x) - u_t'(x+s)|^p}{s^{1+\sigma p}}\,ds\,dx \\&\quad = \int _{-\infty }^0 \int _{-x}^{-x+\frac{(1-\delta )(1+\lambda t)}{\lambda +1}} \frac{\left( t+\frac{1}{\lambda }\right) ^{-p}}{s^{1+\sigma p}}\,ds\,dx \\&\qquad + \int _{-\infty }^0 \int _{-x+\frac{(1-\delta )(1+\lambda t)}{\lambda +1}}^{-x+1} \frac{\left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p}{s^{1+\sigma p}}\,ds\,dx \\&\qquad +\int _0^{\frac{(1-\delta )(1+\lambda t)}{\lambda +1}} \int _{-x+\frac{(1-\delta )(1+\lambda t)}{\lambda +1}}^{-x+1} \frac{\left( \frac{1}{t+\frac{1}{\lambda }}+\frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p}{s^{1+\sigma p}}\,ds\,dx \\&\qquad + \int _{\frac{(1-\delta )(1+\lambda t)}{\lambda +1}}^1 \int _{-x+1}^\infty \frac{\left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p}{s^{1+\sigma p}}\,ds\,dx . \end{aligned} \end{aligned}$$

We now evaluate each of the four integrals in the right-hand side separately. We will use repeatedly the following: for $\alpha \in (0,1)$ and $a>0$,

$$\begin{aligned} \lim _{x\rightarrow \infty } (x+a)^\alpha - x^\alpha = 0, \end{aligned}$$

and

$$\begin{aligned} (1-x)^\alpha \ge 1 - x^\alpha \qquad x\in [0,1]. \end{aligned}$$

All the constants C below are $C=C(p,\sigma )>0$, independent of $\lambda$, $\delta$ and t.

For the first integral we have:

$$\begin{aligned} \begin{aligned}&\left( t+\frac{1}{\lambda }\right) ^{-p}\int _{-\infty }^0 \int _{-x}^{-x+\frac{(1-\delta )(1+\lambda t)}{\lambda +1}} \frac{1}{s^{1+\sigma p}}\,ds\,dx \\&\quad = \left( t+\frac{1}{\lambda }\right) ^{-p}\frac{1}{\sigma p} \int _{-\infty }^0 \left( (-x)^{-\sigma p} - \left( -x+\frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{-\sigma p}\right) \,dx \\&\quad = \left( t+\frac{1}{\lambda }\right) ^{-p}\frac{1}{\sigma p} \int _{0}^\infty \left( x^{-\sigma p} - \left( x+\frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{-\sigma p}\right) \,dx \\&\quad = \left( t+\frac{1}{\lambda }\right) ^{-p}\frac{1}{(1-\sigma p)\sigma p} \left( x^{1-\sigma p} - \left( x+\frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{1-\sigma p}\right) _0^\infty \\&\quad = \left( t+\frac{1}{\lambda }\right) ^{-p}\frac{1}{(1-\sigma p)\sigma p} \left( \frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{1-\sigma p} \\&\quad< C \left( t+\frac{1}{\lambda }\right) ^{-p+(1-\sigma p)} < C t^{-p+(1-\sigma p)}. \end{aligned} \end{aligned}$$

The second integral can be bounded via:

$$\begin{aligned}&\left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p\int _{-\infty }^0 \int _{-x+\frac{(1-\delta )(1+\lambda t)}{\lambda +1}}^{-x+1} \frac{1}{s^{1+\sigma p}}\,ds\,dx \\&\quad = \frac{1}{\sigma p} \left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p \\&\qquad \int _{-\infty }^0 \left( \left( \frac{(1-\delta )(1+\lambda t)}{\lambda +1}-x\right) ^{-\sigma p} - (1-x)^{-\sigma p}\right) \,dx \\&\quad = \frac{1}{\sigma p} \left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p \\&\qquad \int ^{\infty }_0 \left( \left( \frac{(1-\delta )(1+\lambda t)}{\lambda +1}+x\right) ^{-\sigma p} - (1+x)^{-\sigma p}\right) \,dx \\&\quad = \frac{1}{(1-\sigma p)\sigma p} \left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p \\&\qquad \left( \left( \frac{(1-\delta )(1+\lambda t)}{\lambda +1}+x\right) ^{1-\sigma p} - (1+x)^{1-\sigma p}\right) _0^\infty \\&\quad = \frac{1}{(1-\sigma p)\sigma p} \left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p \left( 1 - \left( \frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{1-\sigma p}\right) \\&\quad \le \frac{1}{(1-\sigma p)\sigma p} \left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p \left( 1 - \frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{1-\sigma p}\\&\quad = \frac{1}{(1-\sigma p)\sigma p} \left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p\left( \delta + (1-\delta )\frac{\lambda }{\lambda +1}(1-t)\right) ^{1-\sigma p}\\&\quad< \frac{1}{(1-\sigma p)\sigma p}\left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta }\right) ^p \left( \delta + (1-\delta )(1-t)\right) ^{1-\sigma p}\\&\quad = \frac{(1-\delta )^p}{(1-\sigma p)\sigma p} \left( \delta + (1-\delta )(1-t)\right) ^{-p+(1-\sigma p)} < C(1-t)^{-p+(1-\sigma p)}. \end{aligned}$$

Simirlarly we calcualte for the third integral:

$$\begin{aligned} \begin{aligned}&\left( \frac{1}{t+\frac{1}{\lambda }}+\frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p \int _0^{\frac{(1-\delta )(1+\lambda t)}{\lambda +1}} \int _{-x+\frac{(1-\delta )(1+\lambda t)}{\lambda +1}}^{-x+1} \frac{1}{s^{1+\sigma p}}\,ds\,dx \\&\quad = p\left( \left( t+\frac{1}{\lambda }\right) ^{-p}+\left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p\right) \int _0^{\frac{(1-\delta )(1+\lambda t)}{\lambda +1}}\int _{-x+\frac{(1-\delta )(1+\lambda t)}{\lambda +1}}^{-x+1} \frac{1}{s^{1+\sigma p}}\,ds\,dx \\&\quad = \frac{1}{\sigma }\left( \left( t+\frac{1}{\lambda }\right) ^{-p}+\left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p\right) \\&\qquad \int _0^{\frac{(1-\delta )(1+\lambda t)}{\lambda +1}} \left( \left( \frac{(1-\delta )(1+\lambda t)}{\lambda +1}-x\right) ^{-\sigma p} - \left( 1-x\right) ^{-\sigma p}\right) \,dx \\&\quad = \frac{1}{(1-\sigma p)\sigma }\left( \left( t+\frac{1}{\lambda }\right) ^{-p}+\left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p\right) \\&\qquad \left( \left( 1-x\right) ^{1-\sigma p} - \left( \frac{(1-\delta )(1+\lambda t)}{\lambda +1}-x\right) ^{1-\sigma p}\right) _0^{\frac{(1-\delta )(1+\lambda t)}{\lambda +1}} \\&\quad = \frac{1}{(1-\sigma p)\sigma }\left( \left( t+\frac{1}{\lambda }\right) ^{-p}+\left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p\right) \\&\qquad \left( \left( 1-\frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{1-\sigma p} - 1 + \left( \frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{1-\sigma p}\right) \\&\quad \le \frac{1}{(1-\sigma p)\sigma }\\&\qquad \left[ \left( t+\frac{1}{\lambda }\right) ^{-p}\left( \frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{1-\sigma p} +\left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p\left( 1-\frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{1-\sigma p}\right] \\&\quad \le \frac{1}{(1-\sigma p)\sigma }\\&\qquad \left[ \left( t+\frac{1}{\lambda }\right) ^{-p}\left( \frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{1-\sigma p} +\left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p\left( \delta + (1-\delta )\frac{\lambda }{\lambda +1}(1-t)\right) ^{1-\sigma p}\right] \\&\quad< \frac{1}{(1-\sigma p)\sigma }\\&\qquad \left[ \left( t+\frac{1}{\lambda }\right) ^{-p}\left( \frac{1+\lambda t}{\lambda +1}\right) ^{1-\sigma p} +\left( \frac{1}{(1-t)(1-\delta ) + \delta }\right) ^p\left( \delta + (1-\delta )(1-t)\right) ^{1-\sigma p}\right] \\&\quad< \frac{1}{(1-\sigma p)\sigma }\left[ \left( t+\frac{1}{\lambda }\right) ^{-p+(1-\sigma p)} +\left( \delta + (1-\delta )(1-t)\right) ^{-p+(1-\sigma p)}\right] \\&\quad < C\left( t^{-p+(1-\sigma p)} +(1-t)^{-p+(1-\sigma p)}\right) \end{aligned} \end{aligned}$$

Finally the last integral can be bounded by:

$$\begin{aligned} \begin{aligned}&\left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p \int _{\frac{(1-\delta )(1+\lambda t)}{\lambda +1}}^1 \int _{-x+1}^\infty \frac{1}{s^{1+\sigma p}}\,ds\,dx \\&\quad = \frac{1}{\sigma p} \left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p \int _{\frac{(1-\delta )(1+\lambda t)}{\lambda +1}}^1 (1-x)^{-\sigma p} \, dx \\&\quad =\frac{1}{(1-\sigma p)\sigma p} \left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p \left. (1-x)^{1-\sigma p}\right| _1^{\frac{(1-\delta )(1+\lambda t)}{\lambda +1}} \\&\quad =\frac{1}{(1-\sigma p)\sigma p} \left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p \left( 1-\frac{(1-\delta )(1+\lambda t)}{\lambda +1}\right) ^{1-\sigma p} \\&\quad =\frac{1}{(1-\sigma p)\sigma p} \left( \frac{1-\delta }{(1-t)(1-\delta ) + \delta (1+\frac{1}{\lambda })}\right) ^p \left( \delta + (1-\delta )\frac{\lambda }{\lambda +1}(1-t)\right) ^{1-\sigma p} \\&\quad<\frac{(1-\delta )^p}{(1-\sigma p)\sigma p} \left( \frac{1}{(1-t)(1-\delta ) + \delta }\right) ^p \left( \delta + (1-\delta )(1-t)\right) ^{1-\sigma p} \\&\quad =\frac{(1-\delta )^p}{(1-\sigma p)\sigma p} \left( \delta + (1-\delta )(1-t)\right) ^{-p+(1-\sigma p)} < C(1-t)^{-p+(1-\sigma p)}. \end{aligned} \end{aligned}$$

Overall we obtained

$$\begin{aligned} \Vert u'_t\Vert _{{\dot{W}}^{\sigma ,p}({\mathbb {R}})}< C\left( (1-t)^{-p+(1-\sigma p)} + t^{-p+(1-\sigma p)}\right) ^{1/p} < C\left( (1-t)^{-1+\frac{1-\sigma p}{p}}+ t^{-1+\frac{1-\sigma p}{p}}\right) \end{aligned}$$

where we used the fact that $(1+x)^\alpha < 1 + x^\alpha$ for $x>0$ and $\alpha \in (0,1)$.

We therefore have, using the fact that $1-\sigma p >0$, that

$$\begin{aligned} \int _0^1 \Vert u'_t\Vert _{{\dot{W}}^{\sigma ,p}({\mathbb {R}})} \, dt \le C \int _0^1 C\left( (1-t)^{-1+\frac{1-\sigma p}{p}}+ t^{-1+\frac{1-\sigma p}{p}}\right) \,dt = 2C\frac{p}{1-\sigma p}, \end{aligned}$$

which is a bound independent of $\lambda$ and $\delta$.

Appendix B: Sobolev norms of radial functions

In this section we prove a technical lemma on Sobolev functions, which is used in Sect. 4.3.

Lemma B.1

Let $n>1$, and define the operator $T:C_c^\infty ((0,1))\rightarrow C_c^\infty (B_1({\mathbb {R}}^n))$ by

$$\begin{aligned} Tf(x) = f(|x|). \end{aligned}$$

Then for every $s\ge 0$ and $p\ge 1$, we have

$$\begin{aligned} \Vert Tf\Vert _{W^{s,p}} \le C \Vert f\Vert _{W^{s,p}}, \end{aligned}$$

for some $C=C(s,p,n)>0$ independent of f. That is, $T:W^{s,p}_0(0,1) \rightarrow W^{s,p}_0(B_1({\mathbb {R}}^n))$ is a bounded operator for every $s\ge 0$ and $p\ge 1$.

Proof

Step I: integer Sobolev spaces We first prove the theorem for $W^{k,p}$ norms, where k is an integer. For $k=0$, moving to polar coordinates, we have

$$\begin{aligned} \Vert Tf\Vert _{L^p}^p = \int _{B_1({\mathbb {R}}^n)} |Tf(x)|^p \,dx = \omega _n \int _0^1 |f(r)|^p r^{n-1}\,dr \le \omega _n\, \Vert f\Vert _{L^p}^p, \end{aligned}$$

where $\omega _n$ is the measure of the $(n-1)$-dimensional unit sphere. For $k=1$, we note that $D(Tf)(x) = f'(|x|)\frac{x}{|x|}$, hence $|D(Tf)(x)| = |f'(r)|$ and the estimate is similar. Differentiating further, we have for $k=2$

$$\begin{aligned} D^2(Tf)(x) = \left( f''(|x|) - \frac{f'(|x|)}{|x|}\right) \frac{x}{|x|}\otimes \frac{x}{|x|} + \frac{f'(|x|)}{|x|} {\text {Id}}, \end{aligned}$$

and for higher derivatives we obtain

$$\begin{aligned} D^k(Tf)(x) = \sum _{j=1}^k \frac{f^{(j)}(|x|)}{|x|^{k-j}} G_j^k\left( \frac{x}{|x|}\right) , \end{aligned}$$

where $G_j^k$ are smooth k-tensor-valued functions on $S^{n-1}$, which are independent of f.

In order to prove boundedness we need to prove that for $j\le k$ we have that

$$\begin{aligned} \int _0^1 \left| \frac{f^{(j)}(r)}{r^{k-j}}\right| ^p\,r^{n-1}\,dr \le \int _0^1 \left| f^{(k)}(r)\right| ^p\, dr. \end{aligned}$$

This follows from Jensen’s inequality: For $k=1$, we have

$$\begin{aligned} \begin{aligned} \int _0^1 \left| \frac{f'(r)}{r}\right| ^p r^{n-1} \,dr&= \int _0^1 \left| \frac{1}{r} \int _0^r f''(t)\,dt\right| ^p r^{n-1} \,dr \le \int _0^1 \left( \frac{1}{r} \int _0^r \left| f''(t)\right| ^p \,dt\right) r^{n-1} \,dr \\&= \int _0^1 \int _0^r \left| f''(t)\right| ^p \,dt \,r^{n-2} \,dr \le \int _0^1 \left| f''(t)\right| ^p \,dt \cdot \int _0^1 \,r^{n-2} \,dr \\&= \frac{1}{n-1} \int _0^1 \left| f''(t)\right| ^p \,dt. \end{aligned} \end{aligned}$$

For $k=2$ we have

$$\begin{aligned} \begin{aligned} \int _0^1 \left| \frac{f'(r)}{r^2}\right| ^p r^{n-1} \,dr&= \int _0^1 \left| \frac{1}{r} \int _0^r \frac{1}{r} \int _0^t f^{(3)}(s)\,ds\, dt\right| ^p r^{n-1} \,dr \\&\le \int _0^1 \frac{1}{r} \int _0^r \left| \frac{1}{r} \int _0^t f^{(3)}(s)\,ds\right| ^p \, dt \, r^{n-1} \,dr \\&= \int _0^1 \frac{1}{r} \int _0^r \frac{t^p}{r^p}\left| \frac{1}{t} \int _0^t f^{(3)}(s)\,ds\right| ^p \, dt \, r^{n-1} \,dr \\&\le \int _0^1 \frac{1}{r} \int _0^r \frac{t^p}{r^p} \frac{1}{t} \int _0^t \left| f^{(3)}(s)\right| ^p \,ds \, dt \, r^{n-1} \,dr \\&\le \int _0^1 \left| f^{(3)}(s)\right| ^p \,ds \cdot \int _0^1 \frac{1}{r} \int _0^r \frac{t^p}{r^p} \frac{1}{t} \, dt \, r^{n-1} \,dr \\&= \frac{1}{p(n-1)} \int _0^1 \left| f^{(3)}(s)\right| ^p \,ds. \end{aligned} \end{aligned}$$

The result for higher values of k follows in a similar manner.

Step II: Interpolation Assume for now that $s\in (0,1)$. Since $B_1({\mathbb {R}}^n)$ is a convex set, we have that the $W^{s,p}({\mathbb {R}}^n)$ norm on functions supported on $B_1({\mathbb {R}}^n)$ (the Gagliardo/Slobodeckij norm) is equivalent to the norm of the real interpolation space

$$ \chi _0^{s,p}(B_1({\mathbb {R}}^n)) = (L^p(B_1({\mathbb {R}}^n)),W^{1,p}(B_1({\mathbb {R}}^n)))_{s,p}, $$

defined by

$$\begin{aligned} \Vert f\Vert _{\chi _0^{s,p}(B_1({\mathbb {R}}^n))}^p= & {} \int _0^\infty \left( \frac{K(t,f)}{t^s}\right) \,\frac{dt}{t} \qquad K(t,f)\\= & {} \inf _{g\in C_0^\infty (B_1({\mathbb {R}}^n))} \left( \Vert f-g\Vert _{L^p(B_1({\mathbb {R}}^n))} + t\Vert g\Vert _{W^{1,p}(B_1({\mathbb {R}}^n))}\right) . \end{aligned}$$

See [15, Theorem 4.7].³ Since $\chi _0^{s,p}(B_1({\mathbb {R}}^n))$ is an interpolation space, the map T is bounded as a map $L^p([0,1])\rightarrow L^p(B_1({\mathbb {R}}^n))$ and as a map $W_0^{1,p}([0,1])\rightarrow W_0^{1,p}(B_1({\mathbb {R}}^n))$ and thus is also bounded as a map between the corresponding interpolation spaces $\chi _0^{s,p}([0,1])\rightarrow \chi _0^{s,p}(B_1({\mathbb {R}}^n))$ (see, e.g., [53, Section 2.3, Theorem 3]).

When $s=k+\sigma$, the proof is similar: T is bounded as a map of between the interpolation spaces $({\dot{W}}^{k,p}(0,1),{\dot{W}}^{k+1,p}(0,1))_{\sigma ,p}\rightarrow ({\dot{W}}^{k,p}(B_1({\mathbb {R}}^n)),{\dot{W}}^{k+1,p}(B_1({\mathbb {R}}^n)))_{\sigma ,p}$, since by the previous step it is bounded as maps on the interpolating spaces; and the norm on these interpolation spaces is equivalent to the ${\dot{W}}^{s,p}({\mathbb {R}}^n)$-norm on $C_0^\infty (B_1({\mathbb {R}}^n))$ functions, by the same results as for the $k=0$ case. $\square$

Remark B.2

This lemma could probably be proven, at least for low values of k, by brute force evaluation of the Gagliardo seminorm, using the Funk–Hecke theorem (see, e.g., [29]).

An immediate corollary is the analogous result for vector fields, instead of functions:

Corollary B.3

Let $n>1$, and define the operator ${\tilde{T}}:C_c^\infty ((0,1))\rightarrow C_c^\infty (B_1({\mathbb {R}}^n);{\mathbb {R}}^n)$ by

$$\begin{aligned} {\tilde{T}}f(x) = f(|x|)\frac{x}{|x|} \end{aligned}$$

Then for every $s\ge 0$ and $p\ge 1$, we have

$$\begin{aligned} \Vert {\tilde{T}}f\Vert _{W^{s,p}} \le C \Vert f\Vert _{W^{s,p}}, \end{aligned}$$

for some $C=C(s,p,n)>0$ independent of f. That is, ${\tilde{T}}:W^{s,p}_0(0,1) \rightarrow W^{s,p}_0(B_1({\mathbb {R}}^n);{\mathbb {R}}^n)$ is a bounded operator for any $s\ge 0$ and $p\ge 1$.

Proof

Let $F\in C_c^\infty ((0,1))$ be an antiderivative of f. Then the corollary follows from Lemma B.1 since ${\tilde{T}}f = D(TF)$. $\square$

Appendix C: Diameter and displacement energy

In this section we prove a general result relating bounded displacement energy and bounded diameter, inspired by previous results relating zero displacement energy and vanishing geodesic distance [9, 23, 54]. However, as shown below, compared with the vanishing case we need stronger assumptions on the norms involved, assumptions which are too restrictive to the applications in this paper; therefore we used other means to prove boundedness of the diameter.

Let G be a (possibly infinite dimensional) manifold and topological group with neutral element e, Lie algebra ${\mathfrak {g}}=T_eG$, and left and right translations L and R given by

$$\begin{aligned} g_1 g_2= L_{g_1}(g_2)=R_{g_2}(g_1),\quad \forall g_1,g_2\in G. \end{aligned}$$

(C.1)

Assume for each $g\in G$ that $R_g:G\rightarrow G$ is smooth, and let $\Vert \cdot \Vert$ be a norm on the Lie algebra ${\mathfrak {g}}$. This gives rise to the following right-invariant Riemannian metric on G:

$$\begin{aligned} \Vert h \Vert _g = \Vert TR_{g^{-1}}h\Vert ,\quad \forall g\in G,\quad \forall h\in T_g G. \end{aligned}$$

(C.2)

The corresponding geodesic distance function is defined as

$$\begin{aligned} {\text {dist}}(g_1,g_2)={{\text {inf}}} \int _0^1 \Vert \partial _t g(t)\Vert _{g(t)} dt,\quad \forall g_1,g_2\in G, \end{aligned}$$

(C.3)

where the infimum is taken over all smooth paths in G with $g(0)=g_1$ and $g(1)=g_2$.

Theorem C.1

Let G be as above. Assume that

• 1. Any transformation g can be written as a product $g=g_1g_2$ where both $g_1$ and $g_2$ are supported on a proper closed subset of M.

• 2. For any proper closed subset $A\subset M$ the group $G_A\subset G$ of all transformations that have support in A is uniformly perfect, i.e., any $g\in G_A$ can be written as a product of n commutators, where n is independent of $g\in G$.

• 3. The geodesic distance to a commutator of g and h is uniformly controlled by the minimum of the distances to g and h, i.e.,

  $$\begin{aligned} {\text {dist}}(e,[g,h])={\text {dist}}(g\circ h,h\circ g) \le C {\text {min}}( {\text {dist}}(e,g),{\text {dist}}(e,h)),\quad \forall g,h \in G, \end{aligned}$$

  (C.4)

  where C is independent of both g and h.⁴

• 4. The displacement energy is globally bounded, i.e., for any proper closed subset $A\subset M$ we have

  $$\begin{aligned} E(A)=\inf \left\{ {\text {dist}}(e,g):g\in G, g(A)\cap A=\emptyset \right\} \le D \end{aligned}$$

  (C.5)

  where D is independent of the set A.

Then the diameter of the group G is bounded.

Proof

Using Assumption 1 and the right invariance of the geodesic distance we can reduce the boundedness of the diameter to consider only transformations that are supported on a proper closed subset of M, since

$$\begin{aligned} {\text {dist}}(e,g)&= {\text {dist}}(e,g_1g_2) = {\text {dist}}(g_2^{-1},g_1)\le {\text {dist}}(g_2^{-1},e)+{\text {dist}}(e,g_1) \nonumber \\&={\text {dist}}(e,g_2)+{\text {dist}}(e,g_1), \end{aligned}$$

(C.6)

where both $g_1$ and $g_2$ are supported in a proper subset of M.

Thus it remains to proof the boundedness of the distance from the identity to any transformation g with support in a proper closed subset A. Using Assumption 2 we write any $g_1=[h_1,h_2][h_3,h_4]\cdots [h_{2n-1},h_{2n}]$ with $h_i \in G_A$. By the same argument as above we obtain

$$\begin{aligned} {\text {dist}}(e,g_1) \le \sum _{i=1}^n {\text {dist}}(e,[h_{2i-1},h_{2i-1}]). \end{aligned}$$

(C.7)

To bound the distance from the identity to a commutator of transformations with support in A we proceed as in [9, Theorem 1] and use Assumption 3 to obtain

$$\begin{aligned} {\text {dist}}(e,[h_{2i-1},h_{2i-1}]) \le (1+C)^2 E(A). \end{aligned}$$

(C.8)

Putting all of this together we have for each $g\in G$ that

$$\begin{aligned} {\text {dist}}(e,g)\le 2n(1+C)^2 E(A) \end{aligned}$$

(C.9)

and using assumption 4 and the triangle inequality this yields

$$\begin{aligned} {\text {dist}}(g,h)\le 4 n(1+C)^2 D \end{aligned}$$

(C.10)

for any $g,h\in G$. $\square$

Let now $M=S^n$ and let $G={\text {Diff}}(S^n)$. Then Assumptions 1 and 2 are satisfied [17, 61]. Assumption 4 is satisfied for $W^{s,p}$-metrics of low enough order, see Proposition 4.9. In the following we will however show that already in the case $s=1$ and $n=1$ condition 3 is to restrictive for our purposes as, e.g., the $\dot{H}^1$ metric on ${\text {Diff}}(S^1)$, which corresponds to bounded diameter, does not satisfy it:

Lemma C.2

There exist sequences $\psi _n,\varphi _n \in {\text {Diff}}(S^1)$ such that ${\text {dist}}_{\dot{H}^1}(\varphi _n \circ \psi _n,\psi _n\circ \varphi _n) \rightarrow \pi /2$ but ${\text {dist}}_{\dot{H}^1}({\text {Id}},\varphi _n) \rightarrow 0$.

Proof

By the analysis of Lenells [42] we have an explicit formula for the geodesic distance of the homogeneous $\dot{H}^1$-metric given by:

$$\begin{aligned} {\text {dist}}_{1,2}(\psi ,\varphi )= \arccos \left( \int _{0}^1 \sqrt{\psi '}\sqrt{\varphi '} d\theta \right) \end{aligned}$$

(C.11)

Now define the functions

$$\begin{aligned} \varphi _n:\; {\left\{ \begin{array}{ll} 0 &{} 0 \le \theta \le \frac{1}{n} \\ 2x &{} \frac{1}{n} \le \theta \le \frac{2}{n} \\ x &{} \frac{2}{n} \le \theta \le 1 \end{array}\right. }\quad \psi _n:\; {\left\{ \begin{array}{ll} nx &{} 0 \le \theta \le \frac{1}{n} \\ 1 &{} \frac{1}{n} \le \theta \le 1 \end{array}\right. } \end{aligned}$$

(C.12)

The functions $\varphi _n$ and $\psi _n$ are not diffeomorphisms, but we can smooth them with an arbitrarily small change to the ${\dot{H}}^1$ distances considered. The claim now follows by a straightforward calculation. $\square$