Continuity of inversion and composition in certain topological groups For $k\in\Bbb{N}$ ($k\geqslant 1$) and $\alpha\in]0,1]$, let $H_{k,\alpha}([0,1])$ be the group of orientation preserving $C^{k,\alpha}$ diffeomorphisms of the closed unit interval $[0,1]$. We furnish this group with the $C^{k,\alpha}$ topology. Unless I am mistaken, neither the operations of inversion nor composition are continuous (I have checked the $C^{0,1}$ case, and imagine that the higher order case is similar). However, what can be said about the group $H^*_{k,\alpha}$ of $C^{k,\alpha,*}$ diffeomorphisms, where $C^{k,\alpha,*}$ is the closure of $C^\infty$ in $C^{k,\alpha}$. Are the operations of inversion and composition continuous over this group?
 A: Actually, it turns out to be a rather straightforward matter, but I find the result interesting enough to want to share it. First, a change of notation will prove helpful. Thus, let $X$ be a compact, finite-dimensional manifold. We recall first that if $C^{k,\alpha}(X)$ denotes the space of $k+\alpha$-times Hoelder differentiable functions over $X$, then $c^{k,\alpha}(X)$ denotes the little Hoelder space, defined to be the closure in $C^{k,\alpha}(X)$ of the set of smooth functions over $X$. We now define $\text{Diff}^{k,\alpha}(X)$ to be the space of $C^{k,\alpha}$ diffeomorphisms of $X$ and we define $\text{diff}^{k,\alpha}(X)$ to be the space of $c^{k,\alpha}$ diffeomorphisms of $X$, that is, the closure in $\text{Diff}^{k,\alpha}(X)$ of the set of smooth diffeomorphisms of $X$.
It turns out that although the operations of composition and inversion ARE NOT continuous over $\text{Diff}^{k,\alpha}(X)$, they ARE continuous over $\text{diff}^{k,\alpha}(X)$, provided that $k\geq 1$.
The proof is relatively straightforward, and, after a little work, reduces to proving for $\alpha<1$ that if $f\in c^{0,\alpha}(X)$ and if $(g_n)$ is a sequence in $C^{0,1}(X)$ converging to $g\in C^{0,1}(X)$, then the sequence $(f\circ g_n)$ converges to $(f\circ g)$ in $c^{0,\alpha}(X)$. Indeed, this is precisely the step that breaks down in $C^{k,\alpha}$. 
However, choose $\epsilon>0$. By definition of the little Hoelder space, there exists a smooth function, $f_1$, such that $\|f-f_1\|_{0,\alpha}<\epsilon$. Let $B>0$ be such that the Lipschitz constant of $g_n$ is bounded above by $B$ for all $n$. In particular, for all $n$
$$
\|f\circ g_n - f_1\circ g_n\|_{0,\alpha}<(1 + B^\alpha)\epsilon.
$$
Likewise,
$$
\|f\circ g - f_1\circ g\|_{0,\alpha}<(1+ B^\alpha)\epsilon.
$$
However, since $f_1$ is smooth, for sufficiently large $n$,
$$
\|f_1\circ g_n - f_1\circ g\|_{0,\alpha} < \epsilon.
$$
The result follows upon combining these estimates.
