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.



relation between SOT-convergence of T and T'
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.


