Lemma 10.8.244BEE08C69CE8

Proof. (1) Choose a closed ball $N$ in the $y$-coordinates, centered at $y(p)$, whose preimage lies in $V$. The map $x\circ y^{-1}$ is smooth on the convex compact set $N$, so by the mean value inequality it is Lipschitz on $N$ with some constant $L$. Moreover $N$ has some diameter $D$, and $t^\alpha \leq D^{\alpha -\alpha '}t^{\alpha '}$ for $0\leq t\leq D$. Hence, for $a,b$ in the preimage of $N$, which is a neighborhood of $p$,

(2) By (1), applied with $y=x$, every point of $M$ has a coordinate neighborhood on which $f$ is Hölder continuous with some exponent, and on which it remains so with every smaller exponent. Finitely many of these neighborhoods cover $M$, and the smallest of their exponents serves on all of them.

(3) A smooth function is Lipschitz on every closed coordinate ball, by the mean value inequality, so it is Hölder continuous with exponent one. Let $f$ and $g$ be Hölder continuous and $p\in M$. By (1) there are coordinates $x$ on a neighborhood $V$ of $p$ and $\alpha \in (0,1]$ such that $f$ and $g$ are Hölder continuous with exponent $\alpha $ on $V$ in these coordinates, say with the constant $C$; shrinking $V$, we may assume that $|f|,|g|\leq B$ on $V$. For $a,b\in V$ we have $|(f+g)(a)-(f+g)(b)|\leq 2C|x(a)-x(b)|^\alpha $,

and $|\max \{ f,g\} (a)-\max \{ f,g\} (b)|\leq \max \{ |f(a)-f(b)|,|g(a)-g(b)|\} $.

For the power, note first that $(s+t)^\gamma \leq s^\gamma +t^\gamma $ for $s,t\geq 0$: if $s+t>0$, divide by $(s+t)^\gamma $ and use $\lambda ^\gamma \geq \lambda $ for $\lambda \in [0,1]$. Hence, if $s\geq t\geq 0$, then $s^\gamma \leq (s-t)^\gamma +t^\gamma $, that is, $|s^\gamma -t^\gamma |\leq |s-t|^\gamma $ for all $s,t\geq 0$. If $f\geq 0$ has exponent $\alpha $ on $V$ with the constant $C$, it follows that $|f(a)^\gamma -f(b)^\gamma |\leq C^\gamma |x(a)-x(b)|^{\alpha \gamma }$.

(4) Let $p'\in M'$, let $V$ be a coordinate neighborhood of $\Phi (p')$, with coordinates $x$, on which $f$ is Hölder continuous with exponent $\alpha $ and constant $C$, and let $N$ be a closed coordinate ball around $p'$, in coordinates $x'$, with $\Phi (N)\subset V$. The map $x\circ \Phi $, written in the coordinates $x'$, is smooth on the convex compact set $N$, hence Lipschitz on $N$ with some constant $L$. Therefore $|f(\Phi (a))-f(\Phi (b))|\leq CL^\alpha |x'(a)-x'(b)|^\alpha $ for $a,b\in N$. $\square$