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Let $U$ be a unitary operator and $H$ a Hilbert space. $I := \{ v \in H |  Uv = v\} $ and $A := \{ Uw - w | w \in H\}$.

I would like to show that $A$ is dense in the orthogonal $I^\bot$ of $I$. I think for this I need to show (i) $A \subset I^\bot$ and (ii) $\bar{A} = I^\bot$. I've done (i) and half of (ii). Now I'm stuck with $I^\bot \subset \bar{A}$. That is for $x \in I^\bot$ I want to find a sequence $a_n \in A$ such that $\lim_{n \rightarrow \infty}a_n = x$.

Now I'm not sure how to construct such a sequence. Can someone help me with this? Thanks for your help.

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Did you try to use $\bar{A} = A^{\perp\perp}$? (this holds for every subspace $A$ of a Hilbert space and if you don't know it, you should prove it) –  t.b. Oct 24 '11 at 10:38
    
@t.b.: Doh! I wasn't aware of this equality. You're truly helpful, as always. Thank you! –  Rudy the Reindeer Oct 24 '11 at 10:44
    
@t.b.: I would like to show for $x \in I^\bot$ $\forall a_\bot \in A^\bot: \langle x, a_\bot \rangle = 0$. And to do that I was thinking of decomposing $x = a_\bot^\prime + a_{\bot \bot}^\prime$-- am I on the right track? –  Rudy the Reindeer Oct 24 '11 at 11:38

1 Answer 1

up vote 4 down vote accepted

Let me do the whole argument in detail:

Put $B = \overline{\rm span}\,\{Ux - x\,:\,x \in H\} = \overline{A}$.

It suffices to prove that $B^\perp = I = \{x\,:\,Ux=x\}$ since both $B$ and $I$ are closed by definition (so $\overline {A} = B = B^{\perp\perp} = I^{\perp}$, as desired).

  1. If $y \in B^{\perp}$ then in particular $0 = \langle y, Ux - x\rangle = \langle y - U^\ast y,x\rangle$ for all $x \in H$. This means that $y = U^\ast y$, hence $Uy = y$ because $U$ is unitary, so $y \in I$, hence $B^{\perp} \subset I$.

  2. If $y \in I$ then $y = Uy$, so $U^\ast y = y$, because $U$ is unitary. Thus, as in 1., $$ 0 = \langle U^{\ast} y - y, x \rangle = \langle y, Ux - x \rangle, $$ so $y \in A^{\perp} = B^{\perp}$ and therefore $I \subset B^{\perp}$

Taking 1. and 2. together we get $I = B^{\perp}$.

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Thank you. I thought you might leave me to it because it's homework and then I started to panic because I have to hand it in by tomorrow and I still didn't really know what I was doing. –  Rudy the Reindeer Oct 25 '11 at 7:20

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