On the Siegel-Sternberg Linearization Theorem

Abstract

We establish a general version of the Siegel-Sternberg linearization theorem for ultradiffentiable maps which includes the analytic case, the smooth case and the Gevrey case. It may be regarded as a small divisior theorem without small divisor conditions. Along the way we give an exact characterization of those classes of ultradifferentiable maps which are closed under composition, and reprove regularity results for solutions of ode’s and pde’s.

Appendices

Appendix A: Basic Facts

We collect some basic and well known properties of the spaces $E^m$ which are determined by the asymptotic behaviour of the weight sequence m.

Lemma 9

Let $\alpha =\liminf m_n^{1/n}$.

• (i) If $\alpha =0$, then $C^\omega \smallsetminus E^m\ne \varnothing$.

• (ii) If $\alpha >0$, then $C^\omega \subset E^m$, and vice versa.

• (iii) If $\alpha =\infty$, then $C^\omega \subsetneq E^m$.

Proof

For simplicity we consider functions on some interval around 0.

• (i) If $\alpha =0$, then $m_{n_i}^{1/n_i}\rightarrow 0$ for some subsequence. Then the function $f = \sum _{i\geqslant 1} x^{n_i}$ is analytic around 0, but

  $$\begin{aligned} \mathring{M}_0f = \sum _{i\geqslant 1} \frac{x^{n_i}}{m_{n_i}} \end{aligned}$$

  is not. Hence f is an element of $C^\omega \smallsetminus E^m$.

• (ii) If $f\in C^\omega$, then its power series expansion at any point has a radius of convergence which is locally bounded away from zero, whence

  $$\begin{aligned} \limsup _{n\geqslant 1} |{f_n}|^{1/n}\leqslant R < \infty \end{aligned}$$

  locally uniformly. Hence also

  $$\begin{aligned} \limsup _{n\geqslant 1} \left( {\frac{|{f_n}|}{m_n}}\right) ^{1/n}\leqslant \frac{\limsup |{f_n}|^{1/n}}{\liminf \, m_n^{1/n}} \leqslant \frac{R}{\alpha } < \infty . \end{aligned}$$

  So the radius of convergence of $\mathring{M}^m_af$ is also locally bounded away from zero, whence $f\in E^m$. — Conversely, if $C^\omega \subset E^m$, then in particular

  $$\begin{aligned} f = \frac{1}{x+\mathrm{i}} = \sum _{n\geqslant 0} \mathrm{i}^{n-1}x^n \end{aligned}$$

  is in $E^m$. We conclude that

  $$\begin{aligned} \mathring{M}^m_0f = \sum _{n>0} \frac{x^n}{m_n} \end{aligned}$$

  has a positive radius of convergence. This implies that $\alpha >0$.

• (iii) We have $C^\omega \subset E^m$ by (ii). On the other hand, by Lemma 1 there exists for any given point a characteristic function f in $E^m$ such that at this point,

  $$\begin{aligned} |{f_n}| \geqslant m_n, \quad \quad n\geqslant 1. \end{aligned}$$

  As

  $$\begin{aligned} \liminf _{n\geqslant 1} |{f_n}|^{1/n}\geqslant \liminf _{n\geqslant 1} m_n^{1/n}= \infty , \end{aligned}$$

  its Taylor series has no positive radius of convergence, hence f is not analytic.

$\square$

Some further properties relate to the asymptotic behaviour of the associated derivative weights $M_n = n! m_n$.

Lemma 10

Let $A = \liminf M_n^{1/n}$.

• (i) If $A>0$, then $E^m \supset E^\omega$, the space of entire functions.

• (ii) If $A<\infty$, then $E^m \subset E^\omega$.

• (iii) If $A = \infty$, then $E^m = E^{\breve{m}}$, where $\breve{m}$ is the largest weakly log-convex minorant below m.

Proof

• (i) If $A>0$, then $M_n \geqslant a^n$ for all $n\geqslant 1$ with some $a>0$, hence

  $$\begin{aligned} m_n \geqslant \frac{a^n}{n!}, \quad \quad n\geqslant 1. \end{aligned}$$

  It follows that $E^m \supset E^{(1/n!)} = E^\omega$.

• (ii) For $f\in E^m$ we locally have

  $$\begin{aligned} \Vert {f^{(n)}}\Vert _U \leqslant M_nr^n, \quad \quad n\geqslant 1, \end{aligned}$$

  with some $r>0$. If $A<\infty$, then $M_n\leqslant b^n$ for infinitely many n, hence

  $$\begin{aligned} \Vert {f^{(n)}}\Vert _U \leqslant c^n \end{aligned}$$

  for infinitely many n with some $c>0$. By the Landau-Kolmogorov inequalities [9] this then also holds for all n with some larger c. Thus,

  $$\begin{aligned} \Vert {f_n}\Vert _U \leqslant \frac{c^n}{n!}, \quad \quad n\geqslant 1. \end{aligned}$$

  It follows that $f\in E^{(1/n!)}=E^\omega$.

• (iii) This follows from the Landau-Kolmogorov interpolation inequalities

  $$\begin{aligned} \Vert {D^{\lambda p+(1-\lambda )q}f}\Vert \leqslant c_{p,q,\lambda } \Vert {D^pf }\Vert ^\lambda \Vert {D^qf }\Vert ^{1-\lambda }, \end{aligned}$$

  where $\lambda p+(1-\lambda )q$ is any integer between p and q, and the fact that M must coincide with $\breve{M}$ at infinitely many points [9].

$\square$

Lemma 11

If $A=\liminf M_n^{1/n}<\infty$, then $E^m$ is not stable under composition.

Proof

If $A > 0$, then $\exp \in E^m$. But if $A<\infty$, then $\exp \circ \exp \notin E^m$, because its $n^2$-Taylor coefficients are larger than $(1/n!)^{n+1}$, which is not of the order of $1/(n^2)!$ [1].

If $A=0$, on the other hand, then there are such that $M_n\leqslant \varepsilon _n^n$ for all n with equality holding for infinitely many n. Then

$$\begin{aligned} e = \sum _{n\geqslant 0} \frac{\varepsilon _n^n}{n!} x^n \in E^m, \end{aligned}$$

but $e\circ e\notin E^m$ by a similar calculation. $\square$

Appendix B: Examples of Weights

Example 1

There are weights, which are asm, but not fdb.

Hence there are $E^m$-spaces, which are $C^{\omega }$-closed, but not $E^m$-closed. These are obviously supersets of $C^\omega$. This seems to be a new observation.

Proof

We first construct an almost increasing sequence m depending on parameters $\lambda _1\leqslant \lambda _2\leqslant .\,.$. We subsequently choose them so that m is not fdb.

Beginnig with $\bar{n}_1=1$ and $\mu _1=1$, we proceed by induction and assume that we already determined $\bar{n}=\bar{n}_{n-1}$ and $\mu _1\leqslant \mu _2\leqslant .\,. \leqslant \mu _{\bar{n}}$ such that

$$\begin{aligned} m_{k}^{1/k} \leqslant 8m_{l}^{1/l}, \quad \quad 1\leqslant k\leqslant l\leqslant \bar{n}. \end{aligned}$$

(5)

We then set

$$\begin{aligned} \mu _k = \lambda _nk, \quad \quad \bar{n}<k\leqslant n, \end{aligned}$$

with some $\lambda _n\geqslant 8m_{\bar{n}}^{1/\bar{n}}$ and n so large that $M_{n}^{1/n} \geqslant \mu _n/4$. Subsequently we set

$$\begin{aligned} \mu _k = 2^6\lceil k/n\rceil M_{n}^{1/n}, \quad \quad n<k\leqslant n^2. \end{aligned}$$

Obviously, $\mu _k$ is increasing for $\bar{n}\leqslant k\leqslant n^2$. Similarly, $m_{k}^{1/k}$ is increasing for $\bar{n}\leqslant k\leqslant n$ by choice of $\lambda _n$. Otherwise we observe that

$$\begin{aligned} M_{in} = 2^{6(i-1)n}i!^nM_n^i, \quad \quad 2\leqslant i\leqslant n. \end{aligned}$$

As

$$\begin{aligned} \frac{i!^n n!^i}{(in)!} = \left( {\frac{\smash {i!^{1/i}\,n!^{1/n}}}{(in)!^{1/in}}}\right) ^{in} \geqslant \left( {\frac{1}{2\mathrm{e}}}\right) ^{in} \end{aligned}$$

by Stirling’s inequality, we conclude that

$$\begin{aligned} m_{in} = \frac{M_{in}}{(in)!} \geqslant 8^{in} \frac{i!^n M_n^i}{(in)!} = 8^{in} \frac{i!^n n!^i}{(in)!} m_n^i \geqslant m_n^i, \quad \quad 2\leqslant i\leqslant n, \end{aligned}$$

and therefore

$$\begin{aligned} m_{n}^{1/n} \leqslant m_{in}^{1/in}, \quad \quad 1\leqslant i\leqslant n. \end{aligned}$$

With (1) this implies that (5) now holds for $1\leqslant k\leqslant l\leqslant n^2$. Setting $\bar{n}_n=n^2$ this completes the inductive construction of the weight m.

Now consider the fdb-property. For $k_1=.\,. =k_n=n$, we have

$$\begin{aligned} \frac{m_nm_{k_1}.\,.\, m_{k_n}}{m_k} = \frac{m_n^{n+1}}{m_{n^2}} = \frac{(n^2)!}{(n!)^{n+1}} \frac{M_n^{n+1}}{M_{n^2}} \geqslant 2^{-6n^2}\frac{(n^2)!}{(n!)^{2n}} \frac{M_n}{n!}. \end{aligned}$$

As

$$\begin{aligned} \frac{n^2!}{n!^{2n}} = \left( {\frac{\smash {n^2!^{1/n^2}}}{n!^{2/n}}}\right) ^{n^2} \geqslant \left( {\frac{\mathrm{e}}{4}}\right) ^{n^2}, \end{aligned}$$

we conclude that

$$\begin{aligned} \frac{m_nm_{k_1}.\,.\, m_{k_n}}{m_k} \geqslant 2^{-7n^2} m_n. \end{aligned}$$

Choosing the $\lambda _n$ and hence the $m_n$ to increase suffciently fast, the right hand side increases faster than any power of $n^2$. Hence, the sequence m is not fdb. $\square$

Example 2

There are weights, which are strictly fdb, but which are not equivalent to any log-convex weight.

Hence log-convexity is not necessary to get stability under composition. See also [28] for an entirely different example of this kind.

Proof

If two weights m and $\tilde{m}$ are equivalent, then also their successive quotients $\alpha _n=m_{n}/m_{n-1}$ and $\tilde{\alpha }_n=\tilde{m}_{n}/\tilde{m}_{n-1}$ form equivalent sequences. If $\tilde{m}$ is log-convex, these $\tilde{\alpha }_n$ are increasing, hence for an equivalent weight $\alpha _{n+1}/\alpha _n$ can not approach zero faster than exponentially. But it is easy to construct block-convex weights where this is the case. So these weights are strictly fdb by Lemma 3, but not log-convex. $\square$

Example 3

There are weights, which are strictly fdb, but not asm.

The corresponding space $E^m$ is thus a proper subset of $C^\omega$, which is stable under composition, but not holomorphically stable.

Proof

An explicit example is

$$\begin{aligned} m_n = \log ^{-n}(1+n), \quad \quad n\geqslant 1. \end{aligned}$$

It is an elementary task to check that m is weakly log-convex and log-anticonvex. Hence,

$$\begin{aligned} m_km_l \leqslant m_{k+1}m_{l-1}, \quad \quad 1\leqslant k<l-1. \end{aligned}$$

So condition (iii) of Lemma 3 needs only be checked for $1\leqslant k\leqslant l\leqslant k+1$ – which is another elementary calculation – to show that m is strictly fdb. But m is not asm, since obviously $\lim m_n^{1/n}= 0$. $\square$

Example 4

There are weights, which are fdb and asm, but not diff-stable.

Proof

Any log-convex weight m is fdb and almost increasing, hence asm. But if the $\mu _n$ increase fast enough so that $\mu _n^{1/n}\rightarrow \infty$, then m is not closed under differentiation. $\square$

Appendix C: The Cauchy-Kowalewskaya Theorem

Reduced to normal form the problem is to find a solution to

$$\begin{aligned} \partial _su = c_0(x,u) + \sum _{1\leqslant i<s} c_i(x,u)\partial _iu, \quad \quad {u}\big |_{x_s=0} = 0, \end{aligned}$$

(6)

in a neighbourhood of the origin in s-space, where $x=(x_1,.\,. ,x_{s-1})$,

$$\begin{aligned} u = u(x) = (u_1(x),.\,. ,u_{t}(x)), \end{aligned}$$

and $c_0$ and $c_1,.\,. ,c_{s-1}$ are defined in a neighborhood of the origin in $s+t$-space and take values in $t$-space and $t\times t$-space, respectively.

Theorem 8

Suppose (6) has a smooth solution u. If the coefficients $c_0,c_1,.\,. ,c_{s-1}$ are of strict fdb class $E^m$ in some neighborhood of the origin in s-space, then u is also of class $E^m$.

Proof

We need the following two extensions of the Main Lemma. First we need to consider products of smooth functions. Assuming without loss that $m_l\geqslant 1$ for all l with $|{l}|=2$, we have

$$\begin{aligned} m_{k_1}m_{k_2} \leqslant m_{l}m_{k_1}m_{k_2} \leqslant m_{k_1+k_2} \end{aligned}$$

for a strict fdb weight. This implies that $M_a(gh) \curlyeqprec M_ag M_ah$. Note that here we have to include the constant terms.

Second, we need to consider ‘partial composition’. Suppose $g=g(x,w)$ and $h=h(x)$ are such that is well defined. We then have

This follows from the Main Lemma by extending h to a map which is the identity in the x-coordinates.

We now proceed as in the proof of Theorem 5 and expand both sides of the differential Eq. (6) into their formal power series at the origin. We get

$$\begin{aligned} \partial _s(T_0u) = T_0(\partial _su) = T_0c_0(x,u) + \sum _{1\leqslant i<s} T_0(c_i(x,u)\partial _iu). \end{aligned}$$

Using the Main Lemma and the preceding remarks to pass to their weighted majorants, we get

Here, the coefficients $M_0c_0,.\,. ,M_0c_{s-1}$ are all analytic around the origin by assumption. Hence, by the Cauchy-Kowalewskaya theorem there exists an analytic solution v to

$$\begin{aligned} \partial _sv = M_0c_0(x,v) + \sum _{1\leqslant i<s} M_0c_i(x,v)\partial _iv, \quad \quad {v}\big |_{x_s=0} = 0. \end{aligned}$$

By recursive comparison of coefficients we have $\mathring{M}_0u \curlyeqprec \dot{T}_0v$. Hence, the smooth solution u is of class $E^m$ near the origin. $\square$