$G$ is called *strongly m-displaceable* if there is $f\in G$ such that the subgroups $H$, $fHf^{-1}$$f^mHf^{-m}$ pairwise commute. Then we say that $f$ *m-displaces* $H$. Fix a conjugation-invariant norm $\nu$ on $G$ and assume that $H\s G$ is strongly $m$-displaceable. Then $e_m(H):=\inf\nu(f)$, where $f$ runs over the set of elements of $G$ that $m$-displaces $H$, is called the *order m displacement energy* of $H$.

[@BIP] Given a group $G$ equipped with a conjugation-invariant norm $\nu$ and given $H\s G$, if there exists $g\in G$ that $m$-displaces $H$ for every $m\geq 1$ then for all $h\in [H,H]$

1.  $\operatorname{cl}_G(h)\leq 2$; and

2.  $\nu(h)\leq
    14\nu(g)$.

It follows from (1) a weaker version of Lemma 3.1.

Suppose that $B$ is a ball and $\overline B\s U$, where $U$ is open. Then any homeomorphism supported in $B$ can be written as a product of two commutators of elements of $\H_U(M)$.

However, contrary to Lemma 3.1, the method based on Theorem 5.6(1) is still true in the smooth category.

Boundedness of $\hm$ and $\hdm$
===============================

The proof of the following theorem is essentially in [@BIP].

Let $B$ be a ball or a half-ball in $M$ (in the latter case we assume $n\geq 2$). Then $\H_B(M)$ 