v,w\rangle$ for some self-adjoint operator $\tilde J$ and $F(v,v)\ge0$ for all $v$ such that $\langle J v,
  v\rangle=0$, then we can find $a\in{{\mathbb R}}$ such that $\tilde J \ge a J$. This means precisely that $\langle
  \tilde J v,v\rangle\ge a\langle Jv, v\rangle$ for all $v$.

If, in addition, we have $F(v,v)>0$ for all $v$ such that $\langle J v, v\rangle=0$, then we obtain a strict inequality $\tilde J > a J$ for some $a\in{{\mathbb R}}$ since the infimum in the statement of Lemma \[le:kuhne\] is strictly bigger than the supremum.

The (longer) proofs of the following results can be found in [@Wojtk01] or in [@Pota79]; see also [@Wojtk09].

For a ${\EuScript{J}}$-separated operator $L:V\to V$ and a $d$-dimensional subspace $F_+\subset C_+$, the subspaces $F_+$ and $L(F_+)\subset C_+$ have an inner product given by ${\EuScript{J}}$. Thus both subspaces are endowed with volume elements. Let $\alpha_d(L;F_+)$ be the rate of expansion of volume of $L\mid_{F_+}$ and $\sigma_d(L)$ be the infimum of $\alpha_d(L;F_+)$ over all $d$-dimensional subspaces $F_+$ of $C_+$.

\[pr:product-vol-exp\] We have $\sigma_d(L)=r_+^1 \cdots r_+^d$, where $r^i_+$ are given by Proposition \[pr:J-separated-spectrum\](2).

Moreover, if $L_1,L_2$ are ${\EuScript{J}}$-separated, then $\sigma_d(L_1L_2)\ge\sigma_d(L_1)\sigma_d(L_2)$.

The following corollary is very useful.

\[cor:compos-max-exp\] For ${\EuScript{J}}$-separated operators $L_1,L_2:V\to V$ we have 