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Schr"odinger operator, finite propagation speed, kernel estimates
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\begin{document}
\title[Finite propagation speed]{Finite propagation speed and
kernel estimates for Schr\"odinger operators}
\author{Christian Remling}
\address{Mathematics Department\\
University of Oklahoma\\
Norman, OK 73019-0315}
\email{cremling@math.ou.edu}
\urladdr{www.math.ou.edu/$\sim$cremling}
\date{\today}
\thanks{2000 {\it Mathematics Subject Classification.} Primary 81Q10 35J10 47B39 47F05}
\keywords{Schr\"odinger operator, finite propagation speed, kernel estimates}
\begin{abstract}
I point out finite propagation speed phenomena for discrete and continuous
Schr\"odinger operators and discuss various types of kernel estimates
from this point of view.
\end{abstract}
\maketitle
\section{Introduction}
In this note, I will point out finite propagation speed phenomena
associated with continuous Schr\"odinger operators
\begin{equation}
\label{soc}
(Hu)(x) = -\Delta u(x) + V(x)u(x)
\end{equation}
on $L_2(\R^d)$ and their discrete analogs
\begin{equation}
\label{sod}
(Hu)(n) = \sum_{|m-n|_1=1} u(m) + V(n)u(n) ,
\end{equation}
acting on $\ell_2(\Z^d)$. I believe that these observations lead to a very
transparent, non-technical and elegant treatment of various topics.
It is instructive to look at the one-dimensional half line problems for a moment.
So, consider the operators from \eqref{soc} and \eqref{sod}, acting on
$L_2(0,\infty)$ and $\ell_2(\N)$, respectively. Let $\rho$ be the standard
spectral measure; it can be obtained from the Weyl circle construction (see,
for example, \cite{CodLev}). Then, in the discrete case, the \textit{moments}
$\int \lambda^n \, d\rho(\lambda)$ are a very important object, and experience
has shown that a similar role is played by the function
\[
\phi(t) = \int \cos t\sqrt{\lambda}\, d\rho(\lambda) - 2\delta (t)
\]
in the continuous case.
(It can be shown that this formula, suitably interpreted, indeed defines
an absolutely continuous function. See, for example, \cite{RemdB}.)
In other words, there seems to be something special about the functions
$\lambda^n$ and $\cos t\sqrt{\lambda}$, respectively. As we will see in
Sect.\ 2, the unifying theme is \textit{finite propagation speed,} and this
in fact works in any dimension.
We will illustrate our basic observations
(see Lemmas \ref{Lbasicd}, \ref{Lbasicc} below) further by using them
to discuss the following two problems:
1.\ Let $f$ be a bounded smooth function
on the spectrum of $H$. Try to estimate $\langle \varphi_1, f(H) \varphi_2\rangle$
in terms of the separation of the supports of $\varphi_1$ and $\varphi_2$.
2.\ Suppose that $V$ is known on $B_R(0)=\{ x\in\R^d : |x|< R \}$ and $\varphi$ is supported
near zero. What can one then say about $\langle \varphi, f(H) \varphi \rangle$?
Both problems have been studied before by other methods, the first one
in fact quite extensively (see \cite{BGK} and the references cited therein).
The sample results we will prove in Sections 3, 4 below are quite similar
to what had been known before.
Actually, Corollary \ref{C5.2} below does seem to
improve results from \cite{BGK},
but that is definitely not my main point in this context.
Rather, what I'm trying to emphasize here is the realization that
finite propagation speed phenomena are at the heart of the matter,
and things become very transparent if this point of view is adopted.
This basic idea might be useful in other situations too.
A piece of closely related work is \cite{CGT}, where
finite propagation speed methods are used
to estimate kernels of functions of the Laplace-Beltrami operator on Riemannian
manifolds. See \cite{GKT,Remuniv} for work on problem 2, and it may also be
interesting to take a look at \cite{LP} or \cite[Sect.\ XI.11]{RS3} for the use of
wave phenomena in scattering theory.
\medskip
\noindent\textit{Acknowledgment.} I thank Wilhelm Schlag for showing me the
proof of Lemma \ref{Lwe}.
\section{Finite propagation speed}
We make the following \textbf{basic assumptions:} In the continuous case, we assume that
$H$ is essentially self-adjoint on $C_0^{\infty}(\R^d)$ and bounded below. A popular
sufficient condition for this is $V\in L_{2,loc}$ and $V_-\in K_d$, the Kato class
($V_-$ denotes the negative part of $V$). See \cite[Sect.\ 3]{S20}. Conversely,
the fact that $C_0^{\infty}(\R^d)$ functions are in the domain of $H$ implies
that $V\in L_{2,loc}$.
In the discrete case, we assume that $V$ is bounded. It is then of course automatic that
$H$ generates a bounded, self-adjoint operator on $\ell_2(\Z^d)$.
Let us begin with the discrete case. Here the basic lemma is extremely simple.
\begin{Lemma}[finite propagation speed -- discrete case]
\label{Lbasicd}
Consider the operator $H$ from \eqref{sod}.
(a) Let $\varphi_1,\varphi_2\in \ell_2(\Z^d)$, and let $R$ be the distance,
measured in the $|\cdot |_1$ norm on $\Z^d$, between the supports of $\varphi_1$
and $\varphi_2$, respectively. Then
\[
\langle \varphi_1, H^n \varphi_2 \rangle = 0 \quad\text{for}\:\: n=0,1, \ldots , R-1.
\]
(b) Consider two (bounded) potentials $V_1$, $V_2$ and the corresponding
Schr\"odinger operators $H_1$, $H_2$, let $\varphi\in\ell_2(\Z^d)$, and,
similarly to the definition in part (a), let
\[
R = \text{\rm dist} \left( \text{\rm supp }\varphi, \text{\rm supp }(V_1-V_2) \right) .
\]
Then
\[
\langle \varphi, H_1^n \varphi \rangle = \langle \varphi, H_2^n \varphi \rangle
\quad\text{for}\:\: n=0,1,\ldots , 2R .
\]
\end{Lemma}
\begin{proof}
It is clear from the form of $H$ that the support of $H\psi$ is contained
in a $1$-neighborhood (in the $|\cdot |_1$ norm) of the support of $\psi$. Repeatedly applying
this gives (a).
It also follows that in the situation of part (b), $H_1^n\varphi = H_2^n\varphi$ for
$n\le R$ since of course $H_1\psi$ can only be different from $H_2\psi$ if $\psi$ is non-zero at
one of the points where $V_1\not= V_2$. Now it may happen that $(H_1^{R+1}\varphi)(x)\not=
(H_2^{R+1}\varphi)(x)$ for some $x\in\Z^d$, but, according to what has just been observed,
only at points $x$
from the support of $V_1-V_2$, and thus it takes at least $R$ more steps to get back to
the support of $\varphi$ (formally, we could in fact apply part (a) to see this).
\end{proof}
By the spectral theorem, part (b) also says that
\[
\int_{\R} \lambda^n \, d\rho_1(\lambda) =
\int_{\R} \lambda^n \, d\rho_2(\lambda)
\]
for $n=0,1,\ldots , 2R$. Here, $\rho_j$ is the spectral measure for $H_j$ and $\varphi$.
As already discussed in the Introduction, the appropriate continuous substitutes for the powers
$\lambda^n$ seem to be the functions $\cos t\sqrt{\lambda}$,
at least for one-dimensional problems. See also \cite{Remuniv} for further background
information.
This suggests the following continuous analog of Lemma \ref{Lbasicd}.
Since we are assuming that $H$ from \eqref{soc} is bounded below, $\cos t\sqrt{\lambda}$ is
a bounded function on the spectrum of $H$. Also, this function clearly does not depend on a
choice of the square root and is in fact entire in $\lambda$.
\begin{Lemma}[finite propagation speed -- continuous case]
\label{Lbasicc}
Consider the operator $H$ from \eqref{soc}.
(a) Let $\varphi_1,\varphi_2\in L_2(\R^d)$ and define
\[
R = \text{\rm dist} ( \text{\rm supp }\varphi_1, \text{\rm supp }\varphi_2 ) .
\]
Then
\[
\langle \varphi_1 , \cos t\sqrt{H}\, \varphi_2 \rangle = 0 \quad \text{for}\:\: |t|\le R .
\]
(b) Consider two potentials $V_1$, $V_2$, let $\varphi\in L_2(\R^d)$ and define
\[
R = \text{\rm dist} \left( \text{\rm supp }\varphi, \text{\rm supp }(V_1-V_2) \right) .
\]
Then
\[
\langle \varphi, \cos t\sqrt{H_1}\, \varphi \rangle =
\langle \varphi, \cos t\sqrt{H_2}\, \varphi \rangle \quad \text{for}\:\: |t|\le 2R .
\]
\end{Lemma}
\begin{proof}
By a routine approximation argument, we may and will assume that all functions
$V$ and $\varphi$ are in $C_0^{\infty}(\R^d)$. Indeed, since the operators
$\cos t\sqrt{H}$ are bounded, we can certainly approximate the $\varphi$'s in
$L_2(\R^d)$ by smooth
functions, and we then have convergence of the scalar products we are interested in.
We then pick $V_n\in C_0^{\infty}(\R^d)$ so that $H_n=-\Delta+V_n\to H$ in strong
resolvent sense. This is possible by \cite[Theorem VIII.25(a)]{RS1}. But then also
$\cos t\sqrt{H_n} \to \cos t\sqrt{H}$ strongly, by \cite[Theorem VIII.20(b)]{RS1}.
Finally, we can do these approximations in such a way that $\liminf R_n\ge R$ for
the separations $R$ defined in the Lemma.
Now let $u=\cos t\sqrt{H}\varphi_2$. Then $u$ solves
\[
u_{tt} = -Hu, \quad\quad u(0)=\varphi_2,\quad u_t(0)=0 .
\]
Originally, this needs to be interpreted as an equation for functions of $t$ taking
values in $L_2(\R^d)$, but $V$ and $\varphi$ are smooth now, so the regularity results for weak
solutions of (generalized) wave equations actually show that
$u\in C^{\infty}(\R^d\times\R)$ and
\begin{equation}
\label{we}
u_{tt}(x,t)-\Delta u(x,t)=-V(x)u(x,t)
\end{equation}
holds pointwise. Compare \cite[Theorem 7.2.7]{Eva}.
We now need the following classical fact.
\begin{Lemma}[finite propagation speed -- wave equation]
\label{Lwe}
Suppose $u\in C^{\infty}$ solves \eqref{we} and $u(x,0)=u_t(x,0)=0$ for
$|x-x_0|0$,
but $u(x,-t)$ satisfies the same equation).
\end{proof}
Part (a) of Lemma \ref{Lbasicc} now follows easily: If $x_0$ is an arbitrary point from the support
of $\varphi_1$, then, by assumption, $\varphi_2(x)=0$ for all $|x-x_0| |t|$. This shows that for $|t|\le R$,
the difference function $u=u_2-u_1$ solves
\begin{equation}
\label{2.1}
u_{tt} - \Delta u = -V_1 u
\end{equation}
for all $x\in\R^d$. Indeed, if $\textrm{dist}(x,\text{supp }\varphi) R$, then, as we have
just seen, $u_1(x,t)=u_2(x,t)=0$ for $|t|\le R$.
Since $u(0)=u_t(0)=0$, Lemma \ref{Lwe}, applied to \eqref{2.1}, shows that $u\equiv 0$ for $|t|\le R$.
As argued above, if $\textrm{dist}(x,\text{supp }\varphi)0$, and define, for $0<\epsilon\le\sqrt{\lambda_0}$,
\[
f_{\epsilon}(\lambda) = f \left( \frac{\sqrt{\lambda}-\sqrt{\lambda_0}}{\epsilon} \right) .
\]
Let $\varphi\in L_2(\R^d)$ and
suppose that $V_1$, $V_2$ agree on an $R$-neighborhood of the support of $\varphi$. Then,
for every $n\in\N$,
\[
\left| \langle \varphi , f_{\epsilon}(H_1) \varphi \rangle
- \langle \varphi , f_{\epsilon}(H_2) \varphi \rangle \right|
\le \frac{4\|\varphi\|^2 \|f^{(n+1)}\|_1}{n\pi(2\epsilon R)^n} .
\]
\end{Theorem}
\begin{proof}
The basic ideas are familiar by now, so it will suffice to provide a sketch
of the argument. Expand $f_{\epsilon}$:
\begin{align*}
f_{\epsilon}(\lambda) & = \int_0^{\infty} \widetilde{f}_{\epsilon}(t) \cos t\sqrt{\lambda}\, dt ,\\
\widetilde{f}_{\epsilon}(t) & = \frac{1}{\pi} \int_0^{\infty} f_{\epsilon}(\lambda)
\cos t\sqrt{\lambda} \, \frac{d\lambda}{\sqrt{\lambda}}
\end{align*}
Use $s=(\sqrt{\lambda}-
\sqrt{\lambda_0})/\epsilon$ as the variable in the integral defining $\widetilde{f}_{\epsilon}$:
\[
\widetilde{f}_{\epsilon}(t) = \frac{2\epsilon}{\pi}
\int_{-\infty}^{\infty} f(s) \cos\left( \epsilon ts+t\sqrt{\lambda_0}\right) \, ds
\]
Integrating by parts, we thus see that
\begin{equation}
\label{3.5}
|\widetilde{f}_{\epsilon}(t)| \le \frac{2\epsilon\|f^{(n+1)}\|_1}{\pi(\epsilon t)^{n+1}} .
\end{equation}
We can now argue as in the last part of the proof of Theorem \ref{T4.2} to finish the proof.
More specifically, write
\[
f_{\epsilon}(\lambda) = \int_0^{2R} \widetilde{f}_{\epsilon}(t) \cos t\sqrt{\lambda}\, dt
+ \int_{2R}^{\infty} \widetilde{f}_{\epsilon}(t) \cos t\sqrt{\lambda}\, dt \equiv g + e,
\]
note that $\langle \varphi, (g(H_1)-g(H_2)) \varphi \rangle =0$ by Lemma \ref{Lbasicc}(b),
and use \eqref{3.5} to bound the error term $e$.
\end{proof}
It is of course also possible to use the original variable $\lambda$, although things
are slightly less elegant then. We continue to assume that $H_j\ge 0$.
\begin{Theorem}
\label{T3.1}
Let $f\in C_0^{\infty}(-1,1)$, $\lambda_0>0$, and define, for $0<\epsilon\le \min\{
1, \lambda_0/2 \}$,
\[
f_{\epsilon}(\lambda) = f \left( \frac{\lambda-\lambda_0}{\epsilon} \right).
\]
Let $\varphi\in L_2(\R^d)$, and let $V_1$, $V_2$ be two potentials that
agree on an $R$-neighborhood of the support of $\varphi$. Then, for all $n\in\N$,
\begin{equation}
\label{2.3}
\left| \langle \varphi, f_{\epsilon}(H_1)\varphi\rangle
- \langle \varphi, f_{\epsilon}(H_2)\varphi\rangle \right|
\le \|\varphi\|^2\, \frac{C_n}{(\epsilon R)^n} .
\end{equation}
The constant $C_n$ depends on $\|f^{(j)}\|_1$ for $j=1,2,\ldots, n+1$ and $\lambda_0$,
but is independent of $\epsilon, R, V_1, V_2, \varphi$.
\end{Theorem}
\begin{proof}
We now use the variable $s=(\lambda-\lambda_0)/\epsilon$ in the expansion
\begin{align*}
f_{\epsilon}(\lambda) & = \int_0^{\infty} \widetilde{f}_{\epsilon}(t)
\cos t\sqrt{\lambda}\, dt , \\
\widetilde{f}_{\epsilon}(t) & = \frac{1}{\pi} \int_0^{\infty} f_{\epsilon}(\lambda)
\cos t\sqrt{\lambda}\, \frac{d\lambda}{\sqrt{\lambda}} .
\end{align*}
So
\[
\widetilde{f}_{\epsilon}(t) = \frac{\epsilon}{\pi} \int_{-\infty}^{\infty}
f(s) \cos t\sqrt{\lambda_0+\epsilon s} \, \frac{ds}{\sqrt{\lambda_0+\epsilon s}}
\]
The integral is still oscillatory and thus should
be small for large $t$. To make this precise, write $\sqrt{\lambda_0+x}\equiv\omega (x)$, note
that $1/\omega = 2\omega'$ and consider
\begin{equation}
\label{2.2}
\int_{-\infty}^{\infty} f(s)\omega'(\epsilon s) e^{it\omega(\epsilon s)}\, ds .
\end{equation}
We introduce the differential expression
\[
D = \frac{-i}{\epsilon t \omega'(\epsilon s)}\, \frac{d}{ds} ,
\]
so that $D e^{it\omega (\epsilon s)} = e^{it\omega (\epsilon s)}$. Thus the integral from
\eqref{2.2} equals
\[
\int_{-\infty}^{\infty} f(s)\omega'(\epsilon s) D^n e^{it\omega (\epsilon s)}\, ds =
\int_{-\infty}^{\infty} e^{it\omega (\epsilon s)}D'^n f(s)\omega'(\epsilon s)\, ds,
\]
where
\[
D' = \frac{d}{ds} \, \frac{i}{\epsilon t \omega'(\epsilon s)}
\]
is the transpose of $D$. We can evaluate $D'^n f\omega' = i(\epsilon t)^{-1} D'^{n-1}f'$, using the product rule.
We obtain a sum of many terms, each of which is of the form
\[
c (\epsilon t)^{-n}
\epsilon^{n-k} \frac{\omega^{(p_1)} \cdots \omega^{(p_m)}}{(\omega')^P}\, f^{(k)} ,
\]
with $1\le k\le n$, $m\ge 0$, $p_j\ge 2$ and $P\ge n$; the argument of the derivatives
of $\omega$ is $\epsilon s$. It follows that $|\eqref{2.2}| \lesssim (\epsilon t)^{-n}$,
with a constant that depends on $n$, $\lambda_0$ and bounds on the derivatives of $f$.
The rest of the argument proceeds as in the previous proof.
\end{proof}
\section{Some remarks on Gevrey type functions}
Usually, one defines the class of Gevrey functions on an interval $I\subset\R$ as
follows (we restrict to the one-dimensional case right away because that is all we will
need here): By definition, $f\in G_s(I)$ if $f\in C^{\infty}(I)$ and
\begin{equation}
\label{5.2}
|f^{(n)}(x)| \le C^{n+1} n^{sn} \quad\quad (x\in K, n\in\N_0 )
\end{equation}
for every compact subset $K\subset I$. Here, $s\ge 1$, and the constant $C$ may
depend on $f$ and $K$.
The functions $f\in G_1(I)$ are in fact real analytic (see \cite[Theorem 19.9]{Rud}).
On the other hand, if $s>1$, then $G_s(I)$
contains compactly supported functions (easily constructed with the help
of the functions $\exp(-x^{-b})$ for suitable $b>0$).
For our purposes, it is clear from the results discussed in Sect.\ 3, 4 that \textit{global}
control on the derivatives is more relevant. This was also (and previously)
recognized in \cite{BGK}. The following definition seems most appropriate:
\begin{Definition}
\label{D5.1}
Let $I\subset\R$ be an interval and let $s\ge 1$.
We say that $f\in G_s^1(I)$ if $f\in C^{\infty}(I)$ and there exists a constant $C>0$ so that
\[
\| f^{(n)}\|_{L_1(I)} \le C^{n+1} n^{sn} \quad\quad \textrm{for all }n\in\N_0 .
\]
\end{Definition}
The relation of this to $G_s$ can be clarified by making the following quick observations:
Suppose that $f\in G_s^1(I)$. Then there exists a constant $A>0$ (any positive $A$ with
$A|I|>1$ will do), so that for every $n\in\N_0$, we can find an $x_0^{(n)}\in I$ so that
$\left| f^{(n)}(x_0^{(n)}) \right| \le A C^{n+1} n^{sn}$.
But then for arbitrary $x\in I$, we have that
\begin{align*}
\left| f^{(n)}(x) \right| & \le \left| f^{(n)}(x_0^{(n)}) \right| + \| f^{(n+1)}\|_{L_1(I)}
\le AC^{n+1} n^{sn} + C^{n+2} (n+1)^{s(n+1)}\\
& = C^{n+1} n^{sn} \left( A
+ C(n+1)^s \left( 1 + \frac{1}{n} \right)^{sn} \right) .
\end{align*}
It follows that there exists a new constant $D$, independent of $n$, so that
$\left| f^{(n)}(x) \right| \le D^{n+1}n^{sn}$ on $I$. In particular, $f\in G_s(I)$,
but we have in fact obtained the stronger statement that the constants $C$ from
\eqref{5.2} can be taken to be independent of $K$. We denote the set of functions
satisfying such a uniform Gevrey condition by $G_s^{\text{\rm unif}}(I)$. Also,
it is obvious, by integrating the pointwise bounds, that $G_s(K) \subset G_s^1(K)$
for \textit{compact} intervals $K\subset\R$. We have thus proved the following.
\begin{Proposition}
\label{P5.1}
(a) $G_s^1(I) \subset G_s^{\text{\rm unif}}(I)\subset G_s(I)$;\\
(b) $G_s^1(K) = G_s(K)$ if $K\subset\R$ is a compact interval.
\end{Proposition}
If our functions are in $G_s^1$, we can obtain more explicit information from
the results of the preceding sections. We illustrate this
with the Gevrey versions of Theorems \ref{T4.1} and \ref{T4.2}, respectively.
Of course, there are similar Gevreyzations of Theorems \ref{T3.3} and \ref{T3.2},
which we won't make explicit.
\begin{Corollary}
\label{C5.1}
In the situation of Theorem \ref{T4.1}, suppose that $f\in G_s^1[a,b]$. Then there exist
constants $C,\gamma >0$ so that
\[
| \langle \delta_x, f(H) \delta_y \rangle | \le C \exp\left( -\gamma |x-y|_1^{1/s}\right) .
\]
\end{Corollary}
\begin{proof}
It suffices to take $n\approx (R/C)^{1/s}$ with a suitable constant $C$ in
Theorem \ref{T4.1}. To spell this out more explicitly, first note that
by Theorem \ref{T4.1} and the definition of $G_s^1$, we have that
\begin{equation}
\label{5.1}
| \langle \delta_x, f(H) \delta_y \rangle | \le \left( \frac{Bn^s}{R} \right)^n
\quad\quad (n\in\N,\: R\equiv |x-y|_1 > n)
\end{equation}
for a suitable constant $B>0$. We may assume that
$B\ge 2/e$, and we then pick $n$ so that
\[
\frac{R}{2eB} \le n^s \le \frac{R}{eB} ,
\]
provided there actually exists an integer satisfying these bounds.
This, however, will certainly be the case if $R$ is large enough.
Since $s\ge 1$, $n$ is then not larger than $R-1$, and thus \eqref{5.1} shows that
\[
| \langle \delta_x, f(H) \delta_y \rangle | \le e^{-n} \le e^{-\gamma R^{1/s}} .
\]
This proves the asserted bound for large $R$, and validity for all $R$ is then achieved by
simply adjusting the constant $C$.
\end{proof}
As observed above in Proposition \ref{P5.1}(b), $G_s^1[a,b] = G_s[a,b]$, so we don't
really need Definition \ref{D5.1} here. The class $G_s^1$ does become relevant, however, in the continuous
case because then the spectra are unbounded and global bounds are needed.
We have the following analog of Corollary \ref{C5.1}.
\begin{Corollary}
\label{C5.2}
In the situation of Theorem \ref{T4.2}, suppose that $g\in G_s^1([0,\infty))$.
Then there exist
constants $C,\gamma >0$ so that
\[
| \langle \varphi_1, f(H) \varphi_2 \rangle | \le C \|\varphi_1 \|\,
\|\varphi_2\| \exp\left( -\gamma R^{1/s}\right) .
\]
\end{Corollary}
\noindent
The \textit{proof} is completely analogous to the proof of Corollary \ref{C5.1}.\hfill$\square$
\medskip
In \cite{BGK}, Bouclet, Germinet, and Klein introduce the class of
$C^{\infty}(I)$ functions $f$ that obey estimates of the form
\[
\int_I (1+|\lambda |)^{n-1} \left| f^{(n)}(\lambda)\right| \, d\lambda \le C^{n+1} n^{sn} ,
\]
and they go on to prove that if this holds on an open interval $I$ containing the
spectrum of $H$, then
\[
| \langle \varphi_1, f(H) \varphi_2 \rangle | \lesssim \exp\left( -\gamma R^{(1/s)-\epsilon}\right)
\]
for every $\epsilon>0$ \cite[Theorem 1.4]{BGK}.
It may therefore be interesting to relate our hypothesis that $g\in G_s^1$, where $g(k)=f(k^2)$,
to such a condition.
\begin{Theorem}
\label{T5.1}
Let $s\ge 1$. Suppose that
$f\in C^{\infty}[0,\infty)$ and
\[
\int_0^{\infty} (1+\lambda)^{(n-1)/2} \left| f^{(n)}(\lambda) \right| \, d\lambda \le
C^{n+1} n^{sn} \quad\quad (n\in\N_0) .
\]
Then $g(k)=f(k^2)\in G_s^1([0,\infty))$.
\end{Theorem}
\begin{proof}
This is elementary but a bit tedious.
Working out the derivatives with the help of the chain and product rules, we see that
$g^{(n)}$ is of the form
\[
g^{(n)}(k) = \sum a_{ij}(n) k^i f^{(j)}(k^2) ,
\]
where the sum ranges over all indices $n/2 \le j\le n$, $0\le i\le j$
satisfying $i=2j-n$. To confirm that this relation between $i$ and $j$ must hold,
one can argue as follows: To produce a contribution of the form $k^i f^{(j)}$,
we clearly must let precisely $j$ of the $n$ derivatives act on $f$. By the chain rule, this gives
$j$ factors of $k$, and the remaining $n-j$ derivatives must then act on these.
As a result, the exponent $j$ decreases by $n-j$; in other words, $i=2j-n$. The other
restrictions on $i, j$ follow from this.
Our task is to extract some information
on the coefficients $a_{ij}(n)$. By taking the derivative in the above representation,
it follows that the $a_{ij}(n)$ obey the recursion
\begin{equation}
\label{5.5}
a_{ij}(n) = (i+1) a_{i+1,j}(n-1) + 2 a_{i-1,j-1}(n-1) .
\end{equation}
It will be convenient to use the difference $d=j-i$ as the parameter indexing the coefficients
$a$. Note that for fixed $n$, we have that $0\le d \le n/2$.
Moreover, $a_{ij}(n)$ can be different from zero
only if $i=n-2d$, $j=n-d$. We'll use the abbreviation
$C_d(n) = a_{n-2d,n-d}(n)$, and we then claim that
\begin{equation}
\label{ind}
0\le C_d(n) \le \frac{2^{n-2d} n^{2d}}{d!} .
\end{equation}
Of course, only the upper bound needs proof, and we can use
induction on $d$. The case $d=0$ corresponds to letting all the
derivatives act on $f$, so $C_0(n)=2^n$. Now assume that \eqref{ind} holds for $d-1$,
with $d\ge 1$. By \eqref{5.5} and the definition of $C_d(n)$,
\[
C_d(n) = (n-2d+1) C_{d-1}(n-1) + 2C_d(n-1) .
\]
We will now iterate this, using the induction hypothesis and the fact that $d\ge 1$
to estimate the first term on the right-hand side. In the first step, we obtain
\begin{align*}
C_d(n) & \le \frac{2^{n-2d+1}}{(d-1)!} (n-1)^{2d-1} + 2C_d(n-1) \\
& = \frac{2^{n-2d+1}}{(d-1)!} (n-1)^{2d-1} + 2((n-1)-2d+1)C_{d-1}(n-2) + 2^2 C_d(n-2) \\
& \le \frac{2^{n-2d+1}}{(d-1)!} \left( (n-1)^{2d-1} + (n-2)^{2d-1} \right) + 2^2 C_d(n-2) .
\end{align*}
Continuing in this way and recalling that $C_d(n)=0$ as soon as $n<2d$, we see that
\[
C_d(n) \le \frac{2^{n-2d+1}}{(d-1)!} \sum_{j=1}^{n-1} j^{2d-1}\le
\frac{2^{n-2d+1}}{(d-1)!} \int_1^n x^{2d-1}\, dx < \frac{2^{n-2d} n^{2d}}{d!} ,
\]
as required.
We are now ready to estimate the integrals $\int_0^{\infty} |g^{(n)}|\, dk$. To prove
Theorem \ref{T5.1}, it clearly suffices to show that
\[
C_d(n) \int_0^{\infty} k^{n-2d} \left| f^{(n-d)}(k^2)\right| \, dk \le C^{n+1} n^{sn}
\]
for $0\le d\le n/2$, with a constant $C$ independent of $d$ and $n$.
We will split these integrals
into two parts, corresponding to $0\le k\le 1$ and $k>1$, respectively.
Consider first $C_d(n) \int_1^{\infty} \cdots$. Using $\lambda=k^2$ as the
variable, we can write this as
\[
\frac{1}{2}\, C_d(n) \int_1^{\infty} \lambda^{(n-2d-1)/2} \left| f^{(n-d)}(\lambda)\right| \, d\lambda,
\]
and, by hypothesis, we have a bound of the form $C_d(n) C^{n+1} (n-d)^{s(n-d)}$.
So, taking \eqref{ind} into account,
we must now show that there exists a constant $C>0$ so that
\[
\frac{n^{2d}}{d^d}\, (n-d)^{s(n-d)} \le C^{n+1} n^{sn}
\]
for all $0\le d\le n/2$. This can be done quite easily, by simply further estimating
$(n-d)^{s(n-d)} \le n^{s(n-d)}$.
A similar argument lets us bound $C_d(n) \int_0^1 \cdots$. We can in fact work with
pointwise bounds. Indeed, by hypothesis and Proposition \ref{P5.1}(b), $f\in G_s[0,1]$,
so, reasoning along the above lines, we see that $g\in G_s[0,1]$, too.
\end{proof}
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