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11.46.14 Smooth residuals from flat errors2C18A3499311

Proposition 11.46.15.33B8FF26FB4B Under the hypotheses of Theorem 11.46.8, suppose $g=q/q_\alpha -1$ satisfies

\[ (r\partial _ r)^ jg=O(r^ N) \quad \text{for every }N,j\ge 0. \]

Then $R-R_0$ extends smoothly across $o$, with every derivative zero there.

Proof. Put $x=-\log r$, $a(x)=Q_\alpha (\mathrm{e}^{-x})^ k$ and $E(x)=M(\mathrm{e}^{-x})-a(x)$. The exact central integral is

\[ E(x)=\int _ x^\infty g(\mathrm{e}^{-y})[-a'(y)]\, \mathrm{d}y, \qquad E'(x)=g(\mathrm{e}^{-x})a'(x). \]

Every derivative of $a$ has at most polynomial growth in $1+x$, and $a\asymp (1+x)^{-b_\alpha }$. The integral bounds $E$ by an exponential of arbitrarily large rate. Repeated differentiation of $E'$ expresses each derivative as a finite Leibniz sum of flat derivatives of $g$ and polynomially bounded derivatives of $a$; all derivatives of $E$ are therefore exponentially flat. Every derivative of $1/a$ has at most polynomial growth, by (11.46.3). Thus $E/a$ has the same flatness. Applying smooth composition near one to

\[ Q-Q_\alpha =Q_\alpha \bigl [(1+E/a)^{1/k}-1\bigr ] \]

proves flatness of every $x$ derivative of $Q-Q_\alpha $. Since $\mathrm{d}R/\mathrm{d}x=-(Q-Q_\alpha )$, integration from $x$ to infinity proves the same for $R-R_0$, including its zeroth derivative. An ordinary $j$th derivative in $r$ is $r^{-j}$ times a polynomial in $\partial _ x$; arbitrarily strong exponential decay absorbs that factor. All ordinary derivatives extend continuously as zero at $r=0$, and repeated one-dimensional integration verifies smoothness there. Composition with $r=|z|^2$ gives the conclusion. $\square$

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