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I am trying to prove or to break the following statement (I assume that the statment is correct):

Assumptions: Let $H$ be a Hilbert-space (or more generally a reflexive space) and $T\in \mathcal{L}(H)$ an operator with the additional property that $\frac{1}{n}T^n$ converges to zero in the strong operator topology on $\mathcal{L}(H)$.

Assertion: The dual operator $\frac{1}{n}{T'}^n$ converges to zero in the strong operator topology on $\mathcal{L}(H')$.

I am thankful for any piece of advice, references or counterexamples.

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I hesitated for a while as to whether this was really MO material, or more math.stackexchange stuff. But no-one else commented, and I'm sort of leaning towards being lenient about closing of late (after many discussions on meta). –  Matthew Daws Dec 15 '11 at 12:29
    
Hello Matthew, thank you for your nice counterexample. I agree - this question does not really belong here. Until now i was not aware of the fact that MO has a main focus on problems at graduate or post graduate level. Sorry about that, I am new here ;) With best regards Matthias –  Matthias Dec 15 '11 at 16:36

1 Answer 1

up vote 8 down vote accepted

If we didn't have the $1/n$ term, what's the standard example here? Let $T$ be the left shift on $\ell^2$, so $T^n\rightarrow 0$ strongly, but $T^*$ is the right shift, an isometry.

To deal with the $1/n$ term, instead use a weighted shift. So something like $$ T\xi = T(\xi_1,\xi_2,\xi_3,\cdots) = (2\xi_2,\frac{3}{2}\xi_3,\frac{4}{3}\xi_4,\cdots). $$ Then $$ T^2\xi = (3\xi_3, \frac{4}{2}\xi_4, \frac{5}{3}\xi_5, \cdots), $$ and so forth. So $\frac{1}{n}T^n$ will be SOT null. But $$ T^*\xi = (0,2\xi_1,\frac{3}{2}\xi_2,\frac{4}{3}\xi_3,\cdots). $$ So I haven't quite got the numbers right so that $\frac{1}{n}(T^*)^n$ is an isometry, but "asymptotically" it will be; it certainly isn't SOT null.

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