Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $n\rightarrow\infty$?

share|improve this question

3 Answers 3

up vote 6 down vote accepted

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.

share|improve this answer

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 $l^2(\mathbb{Z})$.

share|improve this answer

$1$ 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.

http://en.wikipedia.org/wiki/Spectrum_(functional_analysis)#Approximate_point_spectrum

share|improve this answer
    
Isn't this true for $< Uv_n,v_n> \to 1$???? –  András Bátkai Mar 28 '11 at 5:51
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.