Cesaro bounded Operator which is not power bounded - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:48:34Z http://mathoverflow.net/feeds/question/83438 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83438/cesaro-bounded-operator-which-is-not-power-bounded Cesaro bounded Operator which is not power bounded Matthias 2011-12-14T17:03:10Z 2011-12-14T18:54:36Z <p>Good evening!</p> <p>Let X be a banachspace and T a bounded linear operator on X. The cesaro avearges of T are defined as:</p> <p>$A_n:=\frac{1}{n} \sum\limits_{j=0}^{n-1}T^j$</p> <p>We call T cesaro bounded if: $\sup_{n \geq 0}\Vert A_n \Vert&lt;\infty$.</p> <p>We call T power bounded if: $\sup_{n \geq 0}\Vert T^n \Vert&lt;\infty$.</p> <p>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.</p> <p><strong>Here is my question</strong>: can this be achieved in a finite dimesional setting? </p> <p>With best regards, </p> <p>Matthias</p> http://mathoverflow.net/questions/83438/cesaro-bounded-operator-which-is-not-power-bounded/83444#83444 Answer by Bill Johnson for Cesaro bounded Operator which is not power bounded Bill Johnson 2011-12-14T17:57:35Z 2011-12-14T18:54:36Z <p>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.</p> <p>EDIT: As Robert Israel points out in his answer, my answer is wrong.</p> http://mathoverflow.net/questions/83438/cesaro-bounded-operator-which-is-not-power-bounded/83451#83451 Answer by Robert Israel for Cesaro bounded Operator which is not power bounded Robert Israel 2011-12-14T18:37:29Z 2011-12-14T18:44:24Z <p>Consider $T = \pmatrix{-1 &amp; 1\cr 0 &amp; -1\cr}$. Then $T^n = \pmatrix{(-1)^n &amp; (-1)^{n+1} n\cr 0 &amp; (-1)^n\cr}$ so $T$ is not power-bounded. But $A_n = \pmatrix{\frac{1-(-1)^n}{2n} &amp; \frac{(-1)^n}{2} + \frac{1-(-1)^n}{4n}\cr 0 &amp; \frac{1-(-1)^n}{2n}\cr}$ so it is cesaro-bounded.</p> <p>You could replace $-1$ by any $\lambda \ne 1$ with $|\lambda|=1$.</p>