# Almost Orthogonal Vectors given a Unitary Operator

Let $\mathit{H}$ be a (real or complex) Hilbert space and $U:\mathit{H}\rightarrow\mathit{H}$ be a unitary operator. What conditions can be placed on $U$ to guarantee a sequence $v_n$ such that $|v_n|=1$ and ($Uv_n$,$v_n$)$\rightarrow$0 as $n\rightarrow\infty$?

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I'll suppose this is a complex Hilbert space and you're using the convention that the inner product is linear in the first argument and conjugate-linear in the second. The set of all possible $\langle Uv, v\rangle$ for unit vectors $v$ is the numerical range of $U$. Since unitary operators are normal, the closure of the numerical range is the convex hull of the spectrum of $U$. Thus a necessary and sufficient condition is that 0 is in the convex hull of the spectrum.

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Well, this is probably too strong for you, but if your unitary is a bilateral shift, then you have plenty of unit vectors with equality: $\langle Uv,v\rangle = 0$. (See Sz.-Nagy-Foias: Harmonic analysis of Operators on Hilbert Space for the terminology.)

A wandering subspace is a subspace $L\subset H$ for which $UL\bot L$. See the shift in $l^2(\mathbb{Z})$.

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$1$ is an "approximate eigenvalue" of $U$ ... so $1$ is in the spectrum of $U$. If $1$ is a genuine eigenvalue, then you have $(Uv,v)=0$ for some $v$, but if not then $1$ is an approximate eigenvalue. Then $1$ should be a limit point of the spectrum.

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Isn't this true for $< Uv_n,v_n> \to 1$???? –  András Bátkai Mar 28 '11 at 5:51
Actually, having an eigenvector is the worst possible scenario for a unitary: if $Uv= \lambda v$, then $\langle Uv,v\rangle = \lambda$. You must have misunderstood something here. –  András Bátkai Mar 28 '11 at 6:18
I don't follow you. If $1$ is a genuine eigenvalue, you just get vectors such that $(Uv,v)=1$. Don't you ? And this has nothing to do with the question. –  Denis Serre Mar 28 '11 at 7:52