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Let $S$ be the bilateral shift on $\ell^2(\mathbb{Z})$ and let $T = S + S^*$. I want to show that there is no cyclic vector for the representation of $T$ on $\ell^2(\mathbb{Z})$ i.e. $\forall x\in \ell^2(\mathbb{Z})$ the set $P_x = \{f(x) : f\mbox{ is a polynomial in }T\}$ is not dense in $\ell^2(\mathbb{Z})$.

I tried to start simple, showing that if $x = \delta_n$ for $n \in \mathbb{Z}$ then $P_x$ isn't dense. This worked out, but I was exploiting a symmetry that doesn't exist when considering finite linear combinations of such elements. I'm pretty sure this idea isn't going to work out.

Then I tried considering translating this operator to an operator on $L^2(S^1)$ via the Fourier transform. The corresponding operator $\hat{T}$ is simply a multiplication operator, the multiplying function being $g(t) = 2\cos(t)$. This seemed more promising initially, but I still don't see how to approach the problem outside of explicitly constructing an element $y_x$ s.t. $y_x \notin \bar{P}_x$ for arbitrary $x$.

Any hints or pointers in the right direction would be much appreciated.

Edit: Working in the $L^2(S^1)$ setting, the problem amounts to showing that $\forall f\in L^2(S^1)$ $\exists h \in L^2(S^1)$ such that $h$ cannot be approximated in $L^2$ norm by functions of the form $(\sum_{n=0}^m c_n cos^n(t))f(t).$ This seems like it should be clear, but I have been unable to prove it for general $f$.

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Nevermind, I've figured it out. I figured I'd mention it in the comments instead of editing since that would bring it back to the front page. – Grasshopper Nov 17 '11 at 7:01
The system has bounced the question back to the front page anyway (Jan 12 2012) so why not type up your solution here? That way the question isn't left "hanging". – user16299 Jan 13 '12 at 5:20
Bounced again (Feb 11 2012) – user16299 Feb 12 '12 at 4:12
To stop the cycle of bouncing, I upvoted the incorrect answer: user19255 gets +1 for effort. – user31373 Aug 10 '12 at 19:55
That's probably not wise. We don't want to give credence to incorrect answers. – Potato Feb 17 '13 at 17:28

The idea of using $H:=L^2(\mathbb T)$ is good. Let $f\in H$ and $V:=\{P(\cos\cdot)f,P\in\Bbb C[X]\}$ is dense in $H$. One can show that the closure of $V$ for the $L^2$ norm is the set $\{gf,g\mbox{ is even and in }L^2\}$.

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I think that I have a partial answer.

Let $a = (a_i)_{i \in \mathbb{Z}}$ be cyclic vector for $T$

Claim: this forces $a$ to vanish either on all odd or all even integers.

Define two new vectors $b_1, b_2$ by the rule $$ (b_k)_i = (-1)^{i+k}a_i.$$

Notice that we must have at least one $b_k \neq a$. Otherwise $a = 0$.

Sub-Claim: We have $<b_k,T^n a> = 0$ for all integers $n > 0$ and for $k=1,2$.

Proof: Just going through by hand and checking that a bunch of stuff cancels.

Now $b_k$ is orthogonal to all of $\{f(T)a: f \in \mathbb{C}[x], f(0)=0\}$ by linearity. If $Q = \{f(T)a: f \in \mathbb{C}[x]\}$ is dense in $\ell^2(\mathbb{Z})$, then we must be able to approximate $b_k$ with elements of $Q$. Thus we must be able to approximate $b_k$ with scalar multiples of $a$.

It's not too hard to see that this forces $a$ to vanish either on all even or all odd integers. For if $a$ has a nonzero entry $a_{2k+1}$ and a nonzero entry $a_{2j}$, then if $(\lambda_m a)_{m=1}^{\infty}$ is a sequence of multiples of $a$ converging to $b_1$, then we must have $\lambda_m \to -1$ and $\lambda_m \to 1$ by division by the entry $a_{2k+1}$ or $a_{2j}$ respectively.

Thus we have shown that if $a$ is a cyclic vector, then it vanishes either on all even or all odd integers.

I think that you can keep on replicating this procedure to force all the entries of $a$ to be $0$, but I don't feel like churning out the rest of the details.

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Hm... I see how you can replicate the procedure, but I don't think it forces all of the entries of $a$ to be 0. The procedure never actually ends and for an arbitrary $n$ you don't know if repeating the procedure ever actually shows that $a_n$ = 0 for that particular $n$. – Grasshopper Nov 13 '11 at 9:21
Also, the sub-claim isn't true. Consider $a_0 = i, a_1 = 1, a_n = 0 $ else. Then $<Ta, b_2>$ = 2i. – Grasshopper Nov 15 '11 at 8:30

By Grasshopper's own work and Davide Giraudo's answer, it is sufficient to show that, for any $f\in L^2(S^1)$, $\left\{ gf:g\in L^\infty (S^1)\text{ is even.}\right\} \neq L^2(S^1)$. (Of course, when we say that $g$ is even, we just mean that $g$ has an even representative.) To show this, it suffices to exhibit a nonzero element $h\in L^2(S^1)$ that is orthogonal to every in $Ef$. It $f=0$, then $Ef=0\neq L^(S^1)$, so we may assume that $f\neq 0$.

Let $E$ and $O$ be the subspaces of $L^2(S^1)$ consisting of the even and odd functions respectively. It is easy to show that $L^2(S^1)\cong E\oplus O$, so for $h\in L^2(S^1)$, let us write $h=h_++h_-$ for $h_+\in E$ and $h_-\in O$. For $h$ to be orthogonal to every $gf$, we require $$ 0=\int _{S^1}h^*(gf)=\int _{S^1}g(f_+h_+^*+f_-h_-^*). $$ From this, we see that if we take $h_+=\sin (x)f_-^*$ and $h_-=-\sin (x)f_+^*$, then $h\neq 0$ because $f\neq 0$ and $h$ is orthogonal to every $gf$. It in turn follows that $\left\{ gf:g\in L^\infty (S^1)\text{ is even.}\right\} \neq L^2(S^1)$, competing the proof, as noted before, by Grasshopper and Davide's work.

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