$X$ is an invertible bounded linear operator mapping $\ell^2\rightarrow\ell^2$. The geometric sum $Y$ is given by:
$$Y=\sum_{j=0}^{\infty}X^j$$
which is bounded, i.e. $||Y||<\infty$. My question: Is it necessarily the case that $||X||<1$?
The answer is yes for finite matrices, and is easy to prove by using the singular value decomposition of $X$. However, is this also true in the infinite dimensional case? And if not, when might it fail?
Thanks for your help! My functional analysis is not very strong yet.
Edit: a commenter correctly pointed out that this is not true when the matrix is nilpotent; I added the condition that the operator be invertible
I believe the answer is negative, even for finite matrices.
To see this let $ T= \pmatrix{\varepsilon & 0 \cr \alpha & \varepsilon }, $ where $\varepsilon $ and $\alpha $ are positive constants, with $\varepsilon <1$. One may then prove by induction that $$ T^n= \pmatrix{\varepsilon ^n & 0 \cr n\alpha \varepsilon ^{n-1} & \varepsilon ^n}, $$ for every integer $n$. From this it follows that $\Vert T^n\Vert \leq C(2\varepsilon ^n + n\alpha \varepsilon ^{n-1})$, for a suitable constant $C$, so that $ \sum_{n=0}^\infty T^n $ converges. However one can make $\Vert T\Vert $ as large as desired by choosing a sufficiently large $\alpha $.