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Good evening!

Let X be a banachspace and T a bounded linear operator on X. The cesaro avearges of T are defined as:

$A_n:=\frac{1}{n} \sum\limits_{j=0}^{n-1}T^j $

We call T cesaro bounded if: $\sup_{n \geq 0}\Vert A_n \Vert<\infty$.

We call T power bounded if: $\sup_{n \geq 0}\Vert T^n \Vert<\infty$.

E. Hille showed in "Remarks on ergodic theorems, Trans. Amer. Math. Soc. 57, 1945, 246-269" that one can find a cesaro bounded Operator in $\mathcal{L}(L_1[0,1])$ which is not power bounded.

Here is my question: can this be achieved in a finite dimesional setting?

With best regards,

Matthias

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2 Answers 2

up vote 5 down vote accepted

Consider $T = \pmatrix{-1 & 1\cr 0 & -1\cr}$. Then $T^n = \pmatrix{(-1)^n & (-1)^{n+1} n\cr 0 & (-1)^n\cr}$ so $T$ is not power-bounded. But $A_n = \pmatrix{\frac{1-(-1)^n}{2n} & \frac{(-1)^n}{2} + \frac{1-(-1)^n}{4n}\cr 0 & \frac{1-(-1)^n}{2n}\cr}$ so it is cesaro-bounded.

You could replace $-1$ by any $\lambda \ne 1$ with $|\lambda|=1$.

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Thank you for this nice example! –  Matthias Dec 14 '11 at 20:37

I think you can read off from the Jordan canonical form that both conditions are equivalent to the spectral radius being at most one and every eigenvalue of modulus one having algebraic and geometric multiplict the same.

EDIT: As Robert Israel points out in his answer, my answer is wrong.

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That eigenvalues of modulus 1 have equal algebraic and geometric multiplicities is necessary for it to be power-bounded, but not to be cesaro-bounded: see my example. –  Robert Israel Dec 14 '11 at 18:40

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