I was asked this in functional analysis class:
Let $ \mathbb{H} $ be a Hilbert space and we are given an operator T satisfying:
$ || Th || < ||h|| $ for all $ h \in H $. We are asked if the following is necessarily true: for all $ h \in \mathbb{H} $ does one have the convergence to zero $ T^nh \to 0 $ in the Hilbert space norm? How about the other direction?
Intuitively it seems to me it is untrue as the Supremum of the operator norm might be one and the inequality given here seems weaker than the conclusion given, but I have no counterexample. For the other direction I got no real idea. Could someone be so kind as to please point me to the way? Thank you.
Let $\{a_n\}$ be a sequence of positive real numbers that increases to $1$, with the property that the sequence of products $$ a_1,\ a_1a_2,\ a_1a_2a_3,\ a_1a_2a_3a_4,\ \ldots$$ converges to a positive value. It's not hard to write down a specific example.
Let $\cal H = \ell^2(\mathbf R)$ and define $T : \cal H \to \cal H$ by $$T(x_1,x_2,x_3,\ldots) = (0, a_1x_1,a_2x_2,a_3x_3,\ldots).$$ Clearly $T$ is linear and bounded, and $\|Th\| < \|h\|$ for all (nonzero) $h$. Let $h = (1,0,0,0,0,\ldots)$.
What is $\|T^nh\|$ equal to?