I'm looking for full proves for the following theorems:
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(Wold-decomposition theorem 1954.) Every isometry is a direct sum of unitary and unilateral shifts.
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Every isometry is unitary equivalent to a direct sum of unitary and unilateral shifts.
Here is a sketched proof from Wikipedia for (1).
https://en.wikipedia.org/wiki/Wold%27s_decomposition#A_sequence_of_isometries
Can I get some suggestions or references for the full proofs?
$\bf{EDIT}$: From the comments, (2) implies (1), because of the orthogonal decomposition of the space.. I still don't really see why. Could someone help me on this part?
Thank you!
Best Answer
Suppose $T : H\rightarrow H$ is an isometric linear map on a Hilbert space $H$, meaning that $\|Tx\|=\|x\|$ for all $x\in H$. Then $T$ is bounded, with an adjoint $T^*$ such that $T^*T=I$ because the polarization identity gives \begin{align} \langle T^*Tx,y\rangle &= \langle Tx,Ty\rangle \\ &=\frac{1}{4}\sum_{n=0}^{3}i^n\langle T(x+i^ny),T(x+i^ny)\rangle= \\ &= \frac{1}{4}\sum_{n=0}^{3}i^n\langle x+i^ny,x+i^ny\rangle =\langle x,y\rangle, \;\;\; x,y\in H. \end{align} $TH=H$ iff $T$ is unitary. If $TH\ne H$, there exists a unit vector $x\in H$ such that $x\perp TH$. Then $\{ x,Tx,T^2x,\cdots \}$ is an orthonormal subset of $H$ on which $T$ is a simple shift operator. If $x\perp y$ and $x,y\perp TH$, then $[\{ x,Tx,T^2x,\cdots\}]$ and $[\{y,Ty,T^2y,\cdots\}]$ are mutually orthogonal sets on which $T$ is a unilateral shift. There are as many copies of such shifts as there are in an orthonormal basis of $(TH)^{\perp}$. What's left over is $Y=\bigcap_{n=1}^{\infty}T^n H$. This is a closed invariant subspace of $T$, and $T$ is unitary on this subspace because $TY=Y$, though $Y=\{0\}$ may occur. Together this gives a decomposition of the space into a subspace where $T$ is unitary, and a collection of mutually orthogonal invariant subspaces of $T$ on which $T$ is a unilateral shift.