I want to show the following: 


Let $H$ be a Hilbert space and let $S:H\to H$ be a bounded operator such that 
$$\|S\|<\sin\frac{\pi}{2n}.$$
Let $\mathcal{L}$ be a closed subspace of $H$ and $$u\_k:=(I-S)^ku,\;\;\;\;\text{for}\;\;\;k=0,\ldots,n\;\;\;\text{and}\;\;\;u\in\mathcal{L}\setminus\{0\}.$$ 
Prove that
$$\|Pu\_k\|\neq 0\;\;\;\text{for}\;\;\;k=0,\ldots,n,$$
where $P$ denotes the orthogonal projection on $\mathcal{L}$.


An idea: 


For $k=0$ is clear because $Pu\_0=Pu=u\in\mathcal{L}\setminus\{0\}$. 


For $k=1$, suppose that $Pu\_1=0$ then $u\_1\in\mathcal{L}^\perp$ and


\begin{align\*}
 0&=\langle u\_0,u\_1\rangle=\langle u\_0,u\_0\rangle-\langle u\_0,Su\_0\rangle\\
&\geq \|u\_0\|^2-\|S\|\|u\_0\|^2\\
&>\left(1-\sin\frac{\pi}{2n}\right)\|u\_0\|^2>0
\end{align\*}
which is not posible. So, $\|Pu\_1\|\neq 0$.


For $k>2$, I don't know how to continue. Can someone give me an idea? Thanks.



If $\|S\|<\sin\frac{\pi}{2n}$ then $\|P(I-S)^ku\|\neq 0$ for all $k=0,\ldots,n$
Your condition that $\| S\|<\sin\frac{\pi}{2n}$ implies that the angle between
$u\_{{k+1}}$ and $u\_{k}$ is less than $\pi/2n$, for $k=0,...,n-1$. Therefore the angle
between $u=u\_0$ and $u\_{k}$ is $<\pi/2$ for $k=1,...,n$. Since $u\in L$, $u\_1,...,u\_n$ cannot be orthogonal to $L$ that is $Pu\_k\neq 0$.


