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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.

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    $\begingroup$ What is the motivation for this question? $\endgroup$
    – Max Horn
    Commented Feb 23, 2020 at 20:11
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    $\begingroup$ I am reading the article "Geometry of higher order realtive spectrum of linear operator in Hilbert spaces" by Eugene Shargorodsky. In Theorem 5.2 he shows that if $T$ is a bounded operator then $$\left(\sin\left(\frac{π}{2n}\right)\right)^{-1}\|T\|$$ is a bound of the $n$-th order spectrum $\mathrm{Spec}_n(T,\mathcal{L})$ for any closed linear subespace $\mathcal{L}$ of $H$. In his proof, he divides by the quantity $\|Pu_k\|$ but I do not know why this quantity is $\neq 0$. $\endgroup$
    – Andrés Felipe
    Commented Jun 13, 2020 at 17:17

1 Answer 1

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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 $Pu_k\neq 0$.

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  • $\begingroup$ Nice solution, I wonder whether there is a proof without using properties of angles. By the way, the triangle inequality for angles is not straighforward and I found the original paper by Rao in the web. Any other source? $\endgroup$
    – Giorgio Metafune
    Commented Feb 23, 2020 at 17:05
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    $\begingroup$ Triangle property for angles is easy. Since we are dealing with finitely many vectors, it is enough to consider their span, which is a finite dimensional space. Now the angle between vectors is the spherical distance on the unit sphere between the endpoints of the corresponding unit vectors. $\endgroup$
    – Alexandre Eremenko
    Commented Feb 23, 2020 at 17:20
  • $\begingroup$ You are right, I see. Then I wonder why there are papers on the subject! $\endgroup$
    – Giorgio Metafune
    Commented Feb 23, 2020 at 17:26
  • $\begingroup$ Thank you for the reply. I do not underestand why the angle between $u_k$ and $u_{k+1}$ has to be less than $\frac{\pi}{2n}$. Also, if $H$ is a complex Hilbert space. How do I underestand the angle between two vectors in this span? $\endgroup$
    – Andrés Felipe
    Commented Jun 13, 2020 at 17:06
  • $\begingroup$ The angle is $\arccos$ of (the absolute value of the dot product, divided by the product of the norms). $\endgroup$
    – Alexandre Eremenko
    Commented Jun 13, 2020 at 20:04

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