Hadamard division lemma in one normal variable

Statement

Let $f\in C^\infty(( -\varepsilon,\varepsilon)\times U)$ , with $U\subseteq\mathbb R^n$ open. If $m\ge1$ and

\partial_t^j f(0,x)=0\qquad(0\le j<m),

then $f(t,x)=t^m g(t,x)$ for a smooth function $g$ , given by

g(t,x)=\frac1{(m-1)!}\int_0^1(1-s)^{m-1}\partial_t^m f(st,x)\,ds.

This is smooth division by the normal coordinate.

Proof and boundary value

Repeated use of the fundamental theorem of calculus gives the integral remainder formula. Smoothness follows by differentiation under the integral on the compact interval $[0,1]$ . In particular,

g(0,x)=\frac{\partial_t^m f(0,x)}{m!}.

Limitation

This divides by a finite power of $t$ . Flatness of two functions at zero does not by itself make their quotient smooth or bounded; for example $e^{-1/t^2}/e^{-2/t^2}=e^{1/t^2}$ on $t>0$ .