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\begin{document}
\title[Power-bounded operators]{Power-bounded
operators and related norm estimates}
\author[Kalton]{Nigel Kalton}
\makeatletter
\address{Department of Mathematics,
University of Missouri,
Columbia, MO 65211}
\email{nigel@math.missouri.edu}
\author[Montgomery-Smith]{Stephen Montgomery-Smith}
\makeatletter
\address{Department of Mathematics,
University of Missouri,
Columbia, MO 65211}
\email{stephen@math.missouri.edu}
\urladdr{http://www.math.missouri.edu/\~{}stephen}
\author[Oleszkiewicz]{Krzysztof Oleszkiewicz}
\address{Institute of Mathematics,
Warsaw University,
Banacha 2,
02-097 Warsaw,
Poland}
\email{koles@mimuw.edu.pl}
\author[Tomilov]{Yuri Tomilov}
\address{Department of Mathematics and Informatics,
Nicholas Copernicus University,
Chopin Str. 12/18,
87-100 Torun,
Poland}
\email{tomilov@mat.uni.torun.pl}
\thanks{The first and second named authors were
partially supported
by NSF grants. The third named author was visiting the
University of Missouri-Columbia
while conducting this research,
and was partially supported by
the Polish KBN Grant 2 P03A 027 22.
The fourth named author was partially supported by the Polish KBN Grant
5 P03A 027 21, and the NASA-NSF Twinning Program.}
\keywords{
Fractional Volterra operator,
fundamental bi-orthogonal system,
Lagrange's inversion formula,
Lambert $W$ function,
multiplier,
power bounded operator,
projection,
spectrum}
\subjclass{Primary 47A30, 47A10; Secondary 33E20, 42A45, 46B15}
\begin{abstract}
\noindent
We consider whether
$ L = \limsup_{n\to\infty} n \snormo{T^{n+1}-T^n} < \infty$ implies
that the operator $T$ is power bounded.
We show that this is so if $L<1/e$, but it
does not necessarily hold if $L=1/e$.
As part of our methods,
we improve a result of Esterle, showing that
if $\sigma(T) = \{1\}$ and $T \ne I$, then
$\liminf_{n\to\infty} n \snormo{T^{n+1}-T^n} \ge 1/e$.
The constant $1/e$ is sharp.
Finally we describe a way to create many generalizations of Esterle's result,
and also give many conditions on an operator which imply that its norm is
equal to its spectral radius.
\end{abstract}
\maketitle
\section{Introduction}
Let $T$ be a bounded linear operator on a complex Banach space $X$.
One of the classical problems in operator theory is to determine
the relation between the size of the resolvent $(T-\lambda I)^{-1}$
when $\lambda$ is
near the spectrum $\sigma(T)$, and the asymptotic properties of
orbits $\{T^n x: n \ge 0\}$ for each $x \in X$.
The inequality
\[
\snormo{(T-\lambda I)^{-1}} \le \frac{C}{\text{dist}(\lambda, \sigma (T))} ,
\quad \lambda \in \mathbb C \, \backslash \, \sigma (T),
\]
has been extensively studied by,
for example, Benamara and Nikolski \cite{benamara-nikolski}
and also, very recently, by Borovykh, Drissi and Spijker
\cite{borovykh et al}, and
El-Fallah and Ransford \cite{el-fallah-ransford};
see also \cite{lyubich2}, \cite{nagy-zemanek}, \cite{nevanlinna2},
\cite{tomilov-zemanek}.
Such an inequality is extreme in the sense that the converse
inequality (with $C=1$) is always satisfied.
In most cases the relationship to such an inequality and the properties
of the orbits are very difficult to determine.
Thus it is interesting
that one has a very clean equivalence for the resolvent
condition introduced by Ritt \cite{ritt}, which says
there is a constant $C>0$ such that
$$ \snormo{(T-\lambda I)^{-1}} \le \frac C{\modo{\lambda-1}}
\qquad (\modo\lambda > 1) .$$
Nagy and Zem\'anek \cite{nagy-zemanek}, and independently
Lyubich \cite{lyubich1}, proved the following result
(see also \cite[Theorem 4.5.4]{navanlinna1}).
\begin{thm}
\label{ritt} Let $T$ be an operator on a complex Banach space.
Then $T$ satisfies the Ritt resolvent condition if and only if
\begin{enumerate}
\item $T$ is power bounded, and
\item \label{c T^n+1-T^n}
$\sup_n n\snormo{T^{n+1}-T^n} < \infty$.
\end{enumerate}
\end{thm}
We recall a result of Esterle \cite{esterle} saying that if
$\sigma(T)=\{1\}$ and $T$ is not the identity operator, then
$\liminf_{n\to\infty} n\snormo{T^{n+1}-T^n} \ge 1/12$. (The
citation given only has $1/96$; this was improved by Berkani
\cite{berkani} to $1/12$.) Moreover it was noted in \cite[Theorem
4.5.1]{navanlinna1} that if $1$ is a limit point of $\sigma(T)$,
then $\limsup_{n\to\infty} n \snormo{T^{n+1} - T^n} \ge 1/e$. Thus both the
Ritt resolvent condition and condition~(\ref{c T^n+1-T^n}) are
extremal, and it is natural to ask whether these two conditions
are equivalent, at least in the case when $\sigma(T)=1$. Note it
was only recently that Lyubich \cite{lyubich2} constructed
operators satisfying the Ritt condition and $\sigma(T) = \{1\}$.
Another reason that such a question is
interesting is because of the famous Esterle-Katznelson-Tzafriri Theorem
\cite{esterle}, \cite{Ka}, which states that if $T$ is power bounded,
and its spectrum meets the unit circle only at the point $1$, then
$\snormo{T^{n+1}-T^n} \to 0$ as $n \to \infty$. Thus a positive answer
to our question would provide a partial converse.
Towards this conjecture,
it is known that if $\limsup_{n\to\infty}n\snormo{T^{n+1}-T^n} < 1/12$,
then $T$ is power bounded in a rather trivial manner, that is, it is the
direct sum of an identity operator
and an operator whose spectral radius is less than $1$.
This follows
directly from the result of Esterle cited above.
In this paper, we improve these results.
We answer a conjecture of Esterle \cite{esterle} (see also \cite{berkani})
and show that in his result that $1/12$ may be replaced by $1/e$.
Furthermore an example shows that $1/e$ is sharp. As a corollary
we show that if $\limsup_{n\to\infty}n\snormo{T^{n+1}-T^n} < 1/e$, then
$T$ is power bounded. Again we provide an example to show that $1/e$ is
sharp.
In particular, the
condition $\sup_n n\snormo{T^{n+1}-T^n} < \infty$ does not necessarily imply
that $T$ is power bounded.
We leave open the question as to whether it
implies power boundedness in the case that $\sigma(T) = \{1\}$.
Finally we create a general framework which shows how to
easily create results in the same vein as Esterle's result.
For example, one can give conditions concerning $\snormo{T^n-T^m}$
that imply that an operator with $\sigma(T)=\{1\}$ is the identity.
We also give results similar to the special case of Sinclair's
Theorem \cite{sinclair} considered by Bonsall and Crabb \cite{bonsall-crabb},
giving many different
conditions on an operator that imply that its norm is equal
to its spectral radius.
Finally, we note that
the condition $\sup_n n \snormo{T^{n+1}-T^n} < \infty$ appears in
the paper by Coulhon and Saloff-Coste \cite{coulhon-saloff-coste}, and also
in the papers by
Blunck \cite{blunck1},
\cite{blunck2}, which give
many applications of
this condition
to maximal regularity problems.
Throughout this paper, we will
take the Fourier transform to be
$\hat f(\xi) = \int_{-\infty}^\infty f(x) e^{-i x \xi} \, dx$
and the inverse Fourier transform to be
$\check g(x) = \frac1{2\pi} \int_{-\infty}^\infty g(\xi) e^{i x \xi} \, d\xi$.
All Banach spaces will be complex in the remainder of the paper.
\section{Esterle's Result}
To illustrate the ideas, let us first give a
continuous time version.
The methods used are similar to those in a paper by
Bonsall and Crabb \cite{bonsall-crabb} in their proof of a special case of
Sinclair's Theorem \cite{sinclair}.
After this present article was finished, the authors learned of
the paper by Berkani, Esterle and Mokhtari \cite{berkani-esterle-mokhtari} and
the paper by Esterle and Mokhtari
\cite{esterle-mokhtari}
which use similar methods.
The function $W$ described below is often called
the Lambert function (see \cite{corless et al}).
\begin{thm}
\label{esterle continuous}
Let $A$ be a bounded operator on a Banach space such that
$\sigma(A) = \{0\}$. For each $t>0$ such that
$\snormo{A e^{tA}} \le 1/et$, we have that $\snormo{A} \le 1/t$.
In particular, if
$\liminf_{t\to\infty} t \snormo{Ae^{tA}} < 1/e$,
then $A = 0$.
\end{thm}
\begin{proof}
Let $f(z) = z e^z$.
There is analytic function $W$ such that
$W(f(z)) = z$ in some neighborhood of $0$. In particular,
by the Riesz-Dunford functional calculus,
$W(tA e^{tA}) = tA$.
Now
$$ W(z) = \sum_{m=1}^\infty p_{m} z^m $$
where, by Lagrange's inversion formula \cite[Ch.~5, Ex.~33]{asmar},
$$ p_{m} = \frac1{m!}\frac{d^{m-1}}{dz^{m-1}} \left(\frac z{f(z)}\right)^m
\Bigg |_{z=0}
= \frac{(-m)^{m-1}}{m!} .$$
The radius of convergence of $W$ is $1/e$, and
$ \sum_{m=1}^\infty \modo{p_{m}} e^{-m} = 1 $,
since $f(-1) = -1/e$.
Therefore
$\snormo{W(tA e^{tA})} \le 1$, and the result follows.
\end{proof}
\begin{thm}
\label{esterle}
Let $T$ be a bounded operator on a Banach space such that
$\sigma(T) = \{1\}$. For each positive integer $n$ such that
$\snormo{T^{n+1}-T^n} \le n^n/(n+1)^{n+1}$, we have that
$\snormo{T-I} \le 1/(n+1)$. In particular, if
$\liminf_{n\to\infty} n \snormo{T^{n+1}-T^n} < 1/e$,
then $T = I$.
\end{thm}
\begin{proof}
Let $f_n(z) = z(1+z/n)^n$.
There is analytic function $W_n$ such that
$W_n(f_n(z)) = z$ in some neighborhood of $0$. In particular,
by the Riesz-Dunford functional calculus,
$W_n(n(T^{n+1}-T^n)) = n(T-I)$.
Now
$$ W_n(z) = \sum_{m=1}^\infty p_{nm} z^m $$
where
$$ p_{nm} = \frac1{m!}\frac{d^{m-1}}{dz^{m-1}} \left(\frac z{f_n(z)}\right)^m
\Bigg |_{z=0}
= \frac{(-1)^{m-1}}{n^{m-1} (nm+m-1)}
\binom{nm+m-1}m .$$
The radius of convergence of $W_n$ is $r_n = (n/(n+1))^{n+1}$, and
$\sum_{m=1}^\infty \modo{p_{nm}} r_n^m = n/(n+1)$,
since $f_n(-n/(n+1)) = -r_n$.
Therefore
$\snormo{W_n(n(T^{n+1}-T^n))} \le n/(n+1)$
and the result follows.
\end{proof}
In Section~\ref{general} below, we will generalize this approach
and give many extensions of these results.
Now let us turn out attention to whether the constant $1/e$ in
Theorems~\ref{esterle continuous} and~\ref{esterle} can be
improved. By the results of Lyubich \cite{lyubich2} combined with
Theorem~\ref{ritt}, we know that there must be some upper bound on
the numbers $C>0$ such that $\sigma(T) = \{1\}$ and
$\liminf_{n\to\infty} n \snormo{T^{n+1}-T^n} < C$ imply that
$T=I$. In fact we will be able to modify the examples of Lyubich
to show that $C = 1/e$ is sharp.
We will consider the fractional Volterra operators, parameterized
by $\alpha>0$, on $L_p([0,1])$ for $1 \le p \le \infty$, given by the
formula
$$ J^\alpha f(x) = \frac1{\Gamma(\alpha)}\int_0^x (x-y)^{\alpha-1}
f(y) \, dy ,$$
and also modified fractional Volterra operators
$$ L^\alpha f(x) = \frac1{\Gamma(\alpha)}\int_0^x (x-y)^{\alpha-1}
e^{y-x} f(y) \, dy .$$
It is well known (and easy to show)
that $(J^\alpha)_{\alpha\ge0}$ is a $C_0$-semigroup
Similarly $(L^\alpha)_{\alpha\ge0}$ is also a $C_0$-semigroup.
Thus it is easily seen that $\snormo{(L^\alpha)^n} = \snormo{L^{\alpha n}}
\le 1/\Gamma(\alpha n +1)$, and hence the spectral radius of $L^\alpha$
is zero.
Let us also consider an extension of this operator $\tilde L^\alpha$
on $L_2(\R)$ given by the formula
$$ \tilde L^\alpha f(x)
= \frac1{\Gamma(\alpha)}\int_{-\infty}^x (x-y)^{\alpha-1}
e^{y-x} f(y) \, dy .$$
This is a convolution operator. Therefore,
$\widehat{\tilde L^\alpha f}(\xi) = m_\alpha(\xi) \hat f(\xi)$,
where $m_\alpha$ is the Fourier Transform of $x_+^{\alpha-1} e^{-x}
/\Gamma(\alpha)$. Direct calculation shows that
$m_\alpha(\xi) = (1+i\xi)^{-\alpha}$,
where here we are taking the principle
branch.
Next,
let $M$ denote the operator of
multiplication by the indicator function of $[0,1]$, then
it is not so hard to see that for any entire function
$f$ we have that $f(L^\alpha) = M f(\tilde L^\alpha) M$, and so
$\snormo{f(L^\alpha)} \le \snormo{f(\tilde L^\alpha)}$.
Now we see that
$\widehat{\tilde L^\alpha e^{-t\tilde L^\alpha}f}(\xi) = k(\xi) \hat f(\xi)$,
where $k(\xi) = m_\alpha(\xi) e^{-t m_\alpha(\xi)}$. If $0<\alpha<1$, then
$\text{Re}(m_\alpha(\xi)) > 0$, and
$\lim_{\xi\to\pm\infty} \text{arg}(m_\alpha(\xi)) = \alpha\pi/2$. Hence
it is easy to see that
$$ \limsup_{t\to\infty} t\snormo{L^\alpha e^{-t L^\alpha}} \le
\limsup_{t\to\infty} t\snormo{\tilde L^\alpha e^{-t\tilde L^\alpha}} \le
1/e \cos(\alpha \pi/2) .$$
This is enough to show that the constant $C = 1/e$ is sharp in
Theorem~\ref{esterle continuous}. However, we can do a little better.
\begin{thm}
\label{esterle eg}
\begin{enumerate}
\item
There exists an operator $A \ne 0$ on a Hilbert space, with $\sigma(A)=\{0\}$,
and $\limsup_{t\to\infty} t \snormo{A e^{tA}} \le 1/e$.
\item
There exists an operator $T \ne I$ on a Hilbert space, with $\sigma(T)=\{1\}$,
and $\limsup_{n\to\infty} n \snormo{T^{n+1}-T^n} \le 1/e$.
\end{enumerate}
\end{thm}
\begin{proof}
Let us consider the operator on $L_2([0,1])$
$$ A = -\int_0^{1/2} L^\alpha \, d\alpha .$$
Lyubich \cite{lyubich2} showed that the operator
$B = \int_0^\infty J^\alpha \, d\alpha$ has spectral radius equal to $0$
on $L_p([0,1])$ for all $1 \le p \le \infty$.
Now both $-A$ and $B$ are operators with positive kernels, and the kernel
of $-A$ is bounded above by the kernel of $B$. It follows that on
$L_p([0,1])$ for $p = 1$ or $p = \infty$ that
$\snormo{A^n} \le \snormo{B^n}$ for all positive integers $n$.
Thus $A$ has spectral radius equal to $0$ on $L_p([0,1])$ for $p=1$ and
$p=\infty$, and hence, by interpolation, for all $1 \le p \le \infty$.
We also define the operator on $L_2(\R)$
$$ \tilde A = -\int_0^{1/2} \tilde L^\alpha \, d\alpha .$$
Following the above argument,
we see that $\snormo{A e^{tA}} \le \snormo{\tilde A e^{t \tilde A}}$, and
that $\widehat{\tilde A e^{t\tilde A}f}(\xi) = k(\xi) \hat f(\xi)$, where
$$ \modo{k(\xi)} =
\modo{h(\xi)}
\exp(- t \text{Re}(h(\xi))) ,$$
and
$$ h(\xi) = \int_0^{1/2} m_\alpha(\xi) \, d\alpha .$$
One sees that $\text{arg}(h(\xi)) \to 0$ as $\xi\to\infty$,
and hence it is an easy matter to see that
$\limsup_{t\to\infty} t\snormo{A e^{tA}} \le 1/e$.
The second example is given by $T = e^A$.
Note that $T \ne I$, because otherwise $A = \log(T) = 0$.
The estimate is easily obtained
since $T^{n+1}-T^n = \int_n^{n+1} A e^{tA} \, dt$.
\end{proof}
\section{Power Boundedness}
\begin{thm}
\label{T^n}
Let $T$ be a bounded operator on a Banach space $X$ such that
$\limsup_{n\to\infty} n \snormo{T^{n+1}-T^n} < 1/e $.
Then $X$ decomposes as the direct sum of two closed $T$-invariant
subspaces such that $T$ is the identity on one of these subspaces, and
the spectral radius of $T$ on the other subspace is strictly less
than $1$. In particular, $T^n$ converges to a projection.
\end{thm}
\begin{proof}
First note that $\sigma(T)$
must be contained in $\{1\} \cup \{z:\modo z < \alpha\}$
for some $\alpha<1$, otherwise it is easy to see that
limit superior of the spectral radius
of $T^{n+1}-T^n$ is at least $1/e$ (see, for example
\cite[Theorem 4.5.1]{navanlinna1}).
Thus there is a projection $P$ that commutes with $T$ such that
$\sigma(T|_{\text{image}(P)}) = \{1\}$, and
the spectral radius of $T|_{\text{ker}(P)}$ is strictly less than $1$.
The result now follows by applying Theorem~\ref{esterle} to
$T|_{\text{image}(P)}$.
\end{proof}
A very similar proof works also for the following continuous time version.
However, we were also able to produce a different proof of this same result.
\begin{thm}
\label{exp tA}
Let $A$ be a bounded operator on a Banach
space $X$ such that
$ L = \limsup_{t\to\infty} t \snormo{A e^{tA}} < 1/e $.
Then $X$ decomposes as the direct sum of two closed $A$-invariant
subspaces such that $A$ is the zero operator on one of these subspaces, and
on the other subspace the supremum of the real part of the spectrum is
strictly negative.
In particular, $e^{tA}$ converges to a projection.
\end{thm}
\begin{proof}%[Proof of Theorem~\ref{exp tA}]
To illustrate the ideas, let us first prove that $e^{tA}$ converges
in the case that
$L < 1/4$, that is,
there are constants $c<1/4$ and $t_0>0$ such
that $\snormo{A e^{tA}} \le c/t$ for $t\ge t_0$.
It follows that $\snormo{A^2 e^{2tA}} \le c^2/t^2$ for $t \ge t_0$, or
$\snormo{A^2 e^{tA}} \le 4c^2/t^2$ for $t \ge 2t_0$. Then for $t \ge 2t_0$
we have
$$ \snormo{A e^{tA}}
=
\normo{\lim_{\tau\to\infty} \int_t^\tau A^2 e^{sA} \, ds}
\le \frac{4c^2}t, $$
since ${A e^{\tau A}} \to 0$ as $\tau \to \infty$.
Iterating this process,
we get that
$\snormo{A e^{tA}} \le (4c)^{2^k}/4t$ for $t\ge 2^k t_0$. To put this
another way, $\snormo{A e^{tA}} \le (4c)^{t/2t_0}/4t$ for $t \ge t_0$.
It follows that
$$ e^{t_1A} - e^{t_2A} = \int_{t_2}^{t_1} A e^{sA} \, ds $$
converges to zero as $t_1,t_2 \to \infty$,
that is, $e^{tA}$ is a Cauchy sequence. Hence it
converges.
The case when $L<1/e$ is only marginally more complicated.
Again,
there are constants $c<1/e$ and $t_0>0$ such that
$\snormo{A e^{tA}} \le c/t$ for $t\ge t_0$.
For any integer $M \ge 2$ we have
that $\snormo{A^M e^{tA}} \le (cM)^M/t^M$ for $t \ge Mt_0$.
Integrating $(M-1)$ times we obtain that
$$ \snormo{A e^{tA}} \le \frac{(cM)^M}{t(M-1)!} \quad \text{for $t\ge Mt_0$}.$$
A simple computation shows that
$$ \frac{(cM)^M}{(M-1)!} \le \frac Me (ce)^M ,$$
and hence iterating we obtain that if $t>M^k t_0$ then
$$ \snormo{A e^{tA}} \le
\left(\frac Me\right)^{-1/(M-1)}
\left( ce \left(\frac Me\right)^{1/(M-1)} \right)^{M^k} \frac 1t.$$
By choosing $M$ is sufficiently large, we see that
there exist constants $c_1,c_2>1$ such that
$\snormo{A e^{tA}} \le c_1 c_2^{-t}/t$ for $t \ge t_0$,
and hence
$\snormo{e^{tA}}$ converges.
Now it is clear that $S=\lim_{t \to \infty} e^{tA}$ is a
bounded projection (because $S^{2}=S$) such that $Se^{tA}=e^{tA}S=S$.
Let $X_1 = \text{Im}(S)$, and $X_2 = \text{Ker}(S)$, so
$X = X_1 \oplus X_2$. These spaces are
clearly invariant under $e^{tA}$, and hence
invariant under $A = \lim_{t\to0}(e^{tA}-I)/t$.
Since
$S|_{X_1} = I|_{X_1}$ we see immediately that $e^{tA}|_{X_1} = I|_{X_1}$,
and so $A|_{X_1} = \lim_{t\to0}(e^{tA}|_{X_1}-I|_{X_1})/t = 0$.
Furthermore, we have that $e^{tA}|_{X_2} \to 0$. Let $t_0$ be such that
$\snormo{e^{t_0 A}|_{X_2}} \le 1/2$. Then the spectral radius of
$e^{t_0 A}|_{X_2}$ is bounded by $1/2$, and so
$\sup\text{\rm Re}(A|_{X_2}) < -\log(2)/t_0$.
\end{proof}
We also point out that one could prove Theorem~\ref{T^n} in a similar
manner. But the details can be quite complicated. It is also possible
to deduce Theorem~\ref{T^n} from Theorem~\ref{exp tA}. Briefly,
if $\snormo{T^{n+1}-T^n} \le (1+\epsilon)L/(n+1)$ for large enough $n$,
then by writing out the power series for $(T-I)e^{tT}$ about $t=0$
one obtains that
$\snormo{(T-I)e^{tT}} \le (1+2\epsilon)Le^t/t$ for large enough $t$.
The result now follows quickly by applying Theorem~\ref{exp tA} to $A = T-I$,
remembering that $\sigma(T) \subset \{1\} \cup \{z:\modo z < 1\}$.
Now we give some counterexamples to show that in general the condition
$\sup_n n\snormo{T^{n+1}-T^n} < \infty $ does not necessarily imply
power boundedness.
\begin{thm}
There exists a bounded operator $T$ on $L_1(\R)$ such that
$ \sup_n n\snormo{T^{n+1}-T^n} < \infty $,
and $\snormo{T^n} \approx \log n$.
\end{thm}
\begin{proof}
The example is a multiplier on $L_1(\R)$ given by
$\widehat{Tf}(\xi) = m(\xi) \hat f(\xi)$.
It is well known that such an operator is bounded if the inverse
Fourier transform
$\check m$ is a measure of bounded variation, and indeed that the norm
is equal to the variation of $\check m$.
Let us consider the case
$$ m(\xi) = \left\{
\begin{array}{cl}
1 & \text{if $\modo\xi \le 1$}\\
\exp(1-\modo{\xi}) & \text{if $\modo\xi > 1$}.
\end{array} \right.
$$
An explicit computation
shows that the inverse Fourier transform of $m^n$ is
$$ \frac{n x\cos(x)+n^2\sin(x)}{\pi x(x^2+n^2)} $$
and that the inverse Fourier transform of $m^{n+1}-m^n$ is
$$ \frac
{(x^2-n(n+1))\cos(x) + (2n x+x)\sin(x)}
{\pi (x^2+n^2)(x^2+(n+1)^2)} ,$$
and it is now easy to verify the claims.
\end{proof}
We now show that for any infinite dimensional
Banach space we can find an operator
$T:X\to X$ with $\limsup_{n\to\infty}n\|T^n-T^{n+1}\|=\frac1e$ but
such that $\lim_{n\to\infty}\|T^n\|=\infty$. To do this we will
need to construct a special bi-orthogonal system in an arbitrary
Banach space. We recall that a family $(e_j,e_j^*)_{j\in J}$ where
$e_j\in X,\ e_j^*\in X^*$ for $j\in J$ is called a bi-orthogonal
system if $e_j^*(e_j)=1$ for $j\in J$ and $e_j^*(e_k)=0$ whenever
$j\neq k.$
We refer to \cite{LT}, \cite{OP} and \cite{P} for known results on the construction of
bi-orthogonal systems in a separable Banach space.
The following Proposition is the key to the construction. We will
give a short proof valid in a Hilbert space and then prove a lemma
which allows us to remove this restriction in an arbitrary Banach
space; the reader whose main interest is in construction of an
operator on a Hilbert space may simply omit this lemma.
\begin{prop}\label{refine} Let $X$ be an infinite dimensional
Banach space and suppose $(c_n)_{n=1}^{\infty}$ is a sequence such
$\lim_{n\to\infty}c_n=\infty$ and
$\lim_{n\to\infty}c_nn^{-\frac12}=0.$ Then $X$ contains a
bi-orthogonal system $(e_n,e_n^*)_{n=1}^{\infty}$ such that:
\newline
(a) If $P_nx=\sum_{k=1}^n e_k^*(x)e_k$ then $\|P_n\|\ge c_n$ and
\newline
(b) $\lim_{n\to\infty}\|e_n^*\|\|e_n\|=1.$
\end{prop}
\begin{proof} Let us suppose $X$ is a Hilbert space. We pick an
orthonormal sequence $(f_n)_{n=0}^{\infty}$ and a decreasing
sequence of positive reals $(\tau_m)_{m=1}^{\infty}$ such that
$\lim_{m\to\infty}\tau_m=0$ and $\tau_m\ge 2c_nn^{-\frac12}$
whenever $2^{m-1}\le n<2^{m}$. Note that this implies
$\lim_{m\to\infty}2^{\frac{m}2}\tau_m=\infty$ since
$\lim_{n\to\infty}c_n=\infty.$ Denote by $(f_n^*)_{n=0}^{\infty}$
the sequence bi-orthogonal to $(f_n)$ with $\|f_n^*\|=1$ (i.e.
$f_n^*(x)=(x,f_n)$).
Define $e_n=f_n+\tau_m f_0$ for $n\ge 1$ and $2^m\le n<2^{m+1}.$
Let $e_n^*=f_n^*.$ Then $(e_n,e_n^*)_{n=1}^{\infty}$ is a
bi-orthogonal system with $\lim _{n\to\infty}\|e_n\|\|e_n^*\|=1.$
Note that $\|P_1\|\ge \tau_1\ge c_1.$ Now suppose $2^m\le
n<2^{m+1}$ where $m\ge 1.$ Then
$$ \|\sum_{k=2^{m-1}}^{2^m-1}e_k\| \ge \tau_{m} 2^{m-1}.$$
On the other hand for any $r>m+1$
$$ \| \sum_{k=2^{m-1}}^{2^m-1}e_k -
\tau_{m}\tau_{r}^{-1}2^{m-r}\sum_{k=2^{r-1}}^{2^r-1}e_k\| \le
2^{(m-1)/2} + \tau_m\tau_r^{-1}2^{m-\frac12(r+1)}.$$
The second term on the right tends to zero as $r\to\infty.$ We
deduce that $\|P_n\| \ge \tau_m 2^{(m-1)/2}\ge \frac12\tau_{m+1}
\sqrt{n}\ge c_n.$ \end{proof}
Now let us indicate how to extend this to an arbitrary Banach
space. In fact it is clear the argument goes through with minor
modifications if we have the following Lemma:
\begin{lemma}\label{arbitrary} If $X$ is an infinite-dimensional
Banach space then $X$ contains a bi-orthogonal system
$(f_n,f_n^*)_{n=0}^{\infty}$ such that $\|f_n\|=1$ for $n\ge 0$,
$\|f_0^*\|=1$,
$$ \lim_{n\to\infty}\|f_n\|\|f_n^*\|=1$$ and for each
$m=1,2,\ldots$ and scalars $(a_n)_{n=2^{m-1}}^{2^m-1}$
$$\|\sum_{k=2^{m-1}}^{2^m-1}a_kf_k\|\le
2\left(\sum_{k=2^{m-1}}^{2^m-1}|a_k|^2\right)^{\frac12}.$$
\end{lemma}
\begin{proof}
We will need two basic facts from Banach space theory, which we
review for the convenience of the reader:
(1){\it Dvoretzky's theorem } \cite{MS}. If $\epsilon>0,
m\in\mathbb N$ there exists $N=N(m,\epsilon)$ so that if $X$ is an
$N$-dimensional (real or complex) Banach space then $X$ contains a
subspace $E$ of dimension $m$ whose Banach-Mazur distance to
$\ell_2^m$ is at most $1+\epsilon.$
(2){\it Lemma of Krein, Krasnoselskii and Milman} \cite{KKR} (see
also \cite{S}, p. 269). If $E$ and $F$ are two finite-dimensional
subspaces of a Banach space $X$ and $\dim F>\dim E$ then there
exists $f\in F$ with $d(f,E)=\min_{e\in E}\|f-e\|=\|f\|.$
Let $(\sigma_n)$ be a descending sequence with $\sigma_1<2$ and $\lim
\sigma_n=1.$ We will construct $(f_n,f_n^*)_{n=0}^{\infty}$
inductively to satisfy the conditions of the Lemma and
$\|f_n^*\|\le \sigma_m^2$ for $2^{m-1}\le n<2^m.$ We start by
picking $f_0,f_0^*$ so that $\|f_0\|=\|f_0^*\|=f_0^*(f_0).$ Now
suppose $(f_n,f_n^*)_{n=0}^{2^{m-1}-1}$ have been chosen (where
$m\ge 1$).
Let $F$ be the linear span $[f_n]_{n=0}^{2^{m-1}-1}.$ Let
$X_0=\{x\in X:\ f_n^*(x)=0, \ 1\le n\le 2^{m-1}-1\}$. By using
Dvoretzky's theorem twice we may find a subspace $V$ of $X_0$ of
dimension $2^{m}$ so that there are Hilbertian norms $|\cdot|_0$
and $|\cdot|_1$ on $V$ with the properties that
$$ \|x\| \le |x|_0\le \sigma_m\|x\| \qquad x\in V$$ and
$$ \sigma_m^{-1} d(x,F) \le |x|_1\le d(x,F) \qquad x\in V.$$
Let $(v_j)_{j=1}^{2^{m}}$ be an orthonormal basis of
$(V,|\cdot|_0)$ which is also orthogonal in $(V,|\cdot|_1).$ We
may assume that $|v_j|_1$ decreases in $j$; note that $|v_j|_1\le
1$ for all $j.$ Then for $x\in [v_j]_{j=2^{m-1}}^{2^{m}}$ we have
$|x|_1\le |v_{2^{m-1}}|_1|x|_0$ and hence $d(x,F)\le
\sigma_m^2|v_{2^m}|_1\|x\|.$ Since $2^m+1>\text{dim} F=2^m$ it
follows from the result of Krein, Krasnoselskii and Milman cited
above that $|v_{2^m}|_1\ge \sigma_m^{-2}.$ Let
$V_0=[v_j]_{j=1}^{2^m};$ then for $x\in V_0$ we have $|x|_0\le
\sigma_m^2|x|_1$ and hence $\|x\|\le \sigma_m^2d(x,F).$ We then
define $f_{2^{m-1}+k-1}=v_k/\|v_k\|$ for $1\le k\le 2^m;$ note
that $\sigma_m^{-1}\le \|v_k\|\le 1.$ Suppose
$a_1,\ldots,a_{2^{m}-1}$ are scalars and $2^m\le k\le 2^{m+1}-1.$
Then
\begin{align*} |a_k| &\le
|\sum_{j=2^{m-1}}^{2^{m}-1}a_jf_{j}|_0
\le \sigma_m^2|\sum_{j=2^{m-1}}^{2^{m}-1}a_jf_{j}|_1\\
&\le \sigma_m^2 d(\sum_{j=2^{m-1}}^{2^{m}-1}a_jf_{j},F)\le
\sigma_m^2 \|\sum_{j=1}^{2^{m}-1} a_j f_j\|.\end{align*} Hence by
the Hahn-Banach theorem we can define bi-orthogonal functionals
$f_k^*$ for $2^{m-1}\le k\le 2^{m}-1$ so that $\|f_k^*\|\le
\sigma_{m}^2.$ To complete the inductive step we need only
observe that
$$ \|\sum_{k=2^{m-1}}^{2^{m}-1}a_kf_k\|\le
|\sum_{k=2^{m-1}}^{2^{m}-1}a_k\|v_k\|^{-1}v_k|_0\le \sigma_m
(\sum_{k=2^{m-1}}^{2^{m}-1}|a_k|^2)^{\frac12}.$$
\end{proof}
\begin{thm} Suppose $00$ we have
$\|T^n\|\ge c(\log n)^{a}$ for all $n\ge 2.$
\end{thm}
\begin{proof} Suppose $a**0$. Hence $\limsup_n
n\|T^n-T^{n+1}\| \le 1/e$.
Next we estimate $\|T^n\|.$ If $(2m-1)!\le n\le (2m+1)!$ then
$$ (P_m+T^n)x= x+\sum_{k=1}^{m}\lambda_k^ne_k^*(x)e_k+
\sum_{k=m+1}^{\infty}(\lambda_k^n-1)e_k^*(x)e_k.$$ Hence
$$ \|P_m+T^n\|\le 1+
M\left(e^{-n/(2m)!}+\sum_{k=1}^{m-1}e^{-n/(2k)!}+\sum_{k=m+1}^{\infty}
\frac{n}{(2k)!}\right).$$ Again it is simple to see that
$$ \|P_m+T^n\|\le M_1$$ for some suitable constant $M_1$
independent of $n.$ Thus $\|T^n\|\ge \|P_m\|-M_1 \ge m^{b}-M_1.$
Since $\log n \le (2m+1)\log(2m+1)$ we have $(\log n)^a \le
C_1m^b$ for a suitable constant $C_1$ and the result follows.
\end{proof}
\begin{rem} {\rm It would be interesting to know if one can do
better than the growth rate for $\|T^n\|$ of $(\log
n)^{\frac12-\epsilon}$ in this theorem in the case of a Hilbert
space. If $X=\ell_p$, when $p>2$ one can use the canonical basis
in the construction and get an example where $\|T^n\|\ge c(\log
n)^{1-\frac1p-\epsilon}$, and by duality if $p<2$ one has an
example with $\|T^n\|\ge c(\log n)^{\frac1p-\epsilon}.$}
\end{rem}
\section{A general approach}
\label{general}
In this section we will discuss how to extend Theorems
\ref{esterle continuous} and \ref{esterle} by a more general
approach. We first isolate the argument used.
To do this, let us introduce a class of analytic functions. Let
$f$ be an analytic function defined on a disk $\{z:|z|1$.
\begin{cor}\label{chernoffgorin2} Suppose $T$ is an operator on a Banach
space such that $\liminf_{n\to\infty}\|I-T^n\|<1$. Suppose for
some $c>1$ there is a sequence of positive integers $(q_n)_{n=0}^{\infty}$ with $q_0=1$
and $q_{n+1}\le cq_n$ if $n\ge 0$ such that $\|I-T^{q_n}\|<
2\sin({\pi}/{(c+1)})$ for $n\ge 0$. Then $T=I$.\end{cor}
\begin{proof} This follows very simply from
Theorem~\ref{chernoffgorin}.
Indeed if $|1-\lambda^n|\le \|I-T^n\|$ for all $n$ then the fact
that $\liminf_{n\to\infty}\|I-T^n\|<1$ is enough to imply
$|\lambda|=1$. Now if $\lambda=e^{i\theta}$ where $|\theta|\le \pi$ we have
$|\theta|<2\pi/(c+1)$. If $\theta\neq 0$ let $N$ be the least integer such that
$q_{N+1}|\theta|\ge 2\pi/(c+1)$. Then $q_{N+1}|\theta|\le cq_N|\theta|\le 2c\pi/(c+1)$ so that
$|1-\lambda^{q_{N+1}}|\ge 2\sin(\pi/(c+1))$. This yields a contradiction and so $\lambda=1$.
\end{proof}
Our next Lemma gives us a recipe for constructing next examples of
functions in class $\mathcal P$, when explicit calculation of the
inverse function $\varphi$ may be difficult.
\begin{lemma}\label{newfunctions} Let $f,h$ be analytic functions on the disk $\{z:\
|z|0$, $h^{(n)}(0)\ge 0$ for all $n\ge 1$ and $h$ is
nonvanishing. Then if $F(z)=f(z)/h(z)$ we have $F\in\mathcal
P$.\end{lemma}
\begin{proof} The first three conditions are obvious. For the last condition,
let $\varphi$ be the local inverse of $f$ at the origin defined on
some disk centered at the origin. Let $0<\rho<\frac12$ be chosen
so that $\rho$ is smaller than the radius of convergence of the
power series expansions of $h$ and $\varphi$ around the origin and
let $M\ge 1$ be an upper bound for $|h|,|h'|,|\varphi|$ and
$|\varphi'|$ on the disk $ \{z:\ |z|\le \rho\}$. For fixed $w$
consider the map $\Phi_w(z)= \varphi(wh(z))$ for $|z|\le \rho$.
Then if $M|w|<\rho$, we have $|\Phi_w(z)|\le M|w||h(z)|\le
M^2|w|$. Thus if $|w|n$ we
have
$$\|T^m-T^n\|\le\left(1-\frac{n}{m}\right) \left(\frac{n}{m}\right)^{n/(m-n)}.$$ Then
$\|T-I\|\le 1-(\frac{n}{m})^{1/(m-n)}$.\end{thm}
\begin{proof} We show that $f(z)=(1-z)^n-(1-z)^m$ is admissible.
This follows from Lemma \ref{newfunctions} since $f(z)= (1-z)^n
(1-(1-z)^{m-n})$ and the function $1-(1-z)^{m-n}$ is in $\mathcal
P$ since its local inverse at the origin is given by
$1-(1-z)^{1/(m-n)}$. Now apply Theorem \ref{nilpotent} to $I-T$.
\end{proof}
It is possible to derive other formulas of the type of Theorem
\ref{esterle} from Theorem \ref{esterle2}. For example we have
the following Corollaries:
\begin{cor} Suppose $T$ is a bounded operator with
$\sigma(T)=\{1\}$. If $$ \liminf_{m/n\to\infty}\|T^m-T^n\|<1$$
then $T=I$.
More precisely if
$$ \limsup_{m/n\to\infty} \frac{m}{n\log(m/n)}(1-\|T^m-T^n\|)>1$$
then $T=I$.\end{cor}
\begin{cor}
Suppose $T$ is a bounded operator with $\sigma(T)=\{1\}$. If $$
\liminf_{p/n\to 0}\frac{n}{p}\|T^{n+p}-T^n\|<\frac1e$$ then $T=I$.
\end{cor}
\begin{cor}
Suppose $T$ is a bounded operator with $\sigma(T)=\{1\}$. Suppose
$0**~~0}|f(te^{i\theta})|> f(\xi)$,
or
\item
\label{12}
$\displaystyle|f(te^{i\theta})|0}\|f(tA)\|\le f(\xi).$$ Then
$ r(A)=\|A\|$. In particular, if $A$ is quasi-nilpotent then
$A=0$. Furthermore if
$$ \sup_{t>0}\|f(tA)\|< f(\xi)$$ then $A=0$.
\end{thm}
\begin{proof} We start by observing that if $\lambda\in\sigma(A)$
then $\sup_{t>0} |f(t\lambda)|\le f(\xi)$. Let $r=r(A)$. If
$tr<\xi$ then by (\ref{11}) and (\ref{12}) we have
$|f(t\lambda)|0}|f(te^{i\theta})|0$ implies that the norm and
spectral radius of $T$ coincide. Other functions are permissible
however, and lead to more general results of this type:
\begin{thm}\label{appl} Let $A$ be an operator on a Banach space $X$. Then
each of the following conditions implies that $r(A)=\|A\|$.
\begin{enumerate}
\item $\displaystyle\sup_{t>0} t\|Ae^{-tA}\|\le e^{-1}$.
\item $\displaystyle\sup_{t>0} t\|Ae^{-tA^m}\|\le (me)^{-1/m}$ for
$m>1$ an integer.
\item $\displaystyle\sup_{t>0}\|e^{-tA}-e^{-stA}\|\le (s-1) s^{-s/(s-1)}
$ for some $ s>1$.
\item $\displaystyle\sup_{t>0}\|e^{-(s+i)tA}-e^{-(s-i)tA)}\|\le
\frac{2e^{-s\arctan(1/s)}}{\sqrt{1+s^2}}$ for some $s\ge 0$.
\end{enumerate} In each case a strict inequality implies that
$A=0$.
\end{thm}
\begin{proof} The first two are immediate deductions from the
preceding Theorem \ref{sinclairtype}. We then must show for the
remaining cases that $e^{-z}-e^{-sz}$ for $s>1$ and
$e^{-sz}\sin{z}$ for $s>0$ satisfy the conditions of Theorem
\ref{sinclairtype} (the case $s=0$ is Sinclair's theorem).
Note first that $f(z)=e^{-z}(1-e^{-(s-1)z})$ is admissible by
Lemma \ref{newfunctions}, since $1-e^{(s-1)z}\in\mathcal P$. In
this case $\xi=(s-1)^{-1}\log s$ and $f(\xi)<1$. Let us assume
$-\pi<\theta<\pi$ and $\theta\neq 0$. If $|\theta|>\frac{\pi}2$
then $f(te^{i\theta})$ is unbounded; if $|\theta|=\frac{\pi}2$
then $\sup_{t>0}|f(te^{i\theta})| = 2>1$. If
$|\theta|<\frac{\pi}2$ then we observe that
$$ |f(te^{i\theta})| =
e^{-t\cos\theta}|1-e^{-(s-1)te^{i\theta}}|.$$ Assume that
$\sup_{t>0}|f(te^{i\theta})|\le f(\xi)$. Pick $t_0$ so that
$(s-1)t_0|\sin\theta|=\frac{\pi}2$. Then
$$ e^{-\xi} > f(\xi)\ge |f(t_0e^{i\theta})|\ge e^{-t_0\cos\theta}.$$ Hence
$ t_0\cos\theta>\xi$. Choose $t_10\}$. If $0<|\theta|<\frac{\pi}2$ we use the fact that if
$z=x+iy$ then $$ |f(z)|\ge e^{-sx}\cosh y|\sin x|.$$ Hence $
|f(te^{i\theta})|> |f(t\cos\theta)|$ and so
$\sup_{t>0}|f(te^{i\theta})|>f(\xi)$.\end{proof}
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\end{document}
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